Class 12 Chemistry Chapter-Wise Subjective Question Bank PDF — 15 Questions Per Chapter (Reasoning & Reactions)

1

Solutions

Part A: Short Answer Questions (2–3 Marks Each)

Q1 • 3 Marks Short Answer

A solution of sucrose in water is

10\%

by mass. What is the mole fraction of each component in the solution? (Molar mass of sucrose

= 342 \text{ g/mol}

, water

= 18 \text{ g/mol}

)|||"HI"|||जल में सुक्रोज का एक विलयन द्रव्यमान से

10\%

है। विलयन में प्रत्येक घटक का मोल अंश क्या है? (सुक्रोज का मोलर द्रव्यमान

= 342 \text{ g/mol}

, जल

= 18 \text{ g/mol}

)

View Model Solution & Step Marking ▾

Model Answer:

Assume

100 \text{ g}

solution. Mass of sucrose

= 10 \text{ g}

, Mass of water

= 90 \text{ g}

.
Moles of sucrose

= \frac{10 \text{ g}}{342 \text{ g/mol}} = 0.0292 \text{ mol}

.
Moles of water

= \frac{90 \text{ g}}{18 \text{ g/mol}} = 5 \text{ mol}

.
Total moles

= 0.0292 + 5 = 5.0292 \text{ mol}

.
Mole fraction of sucrose

= \frac{0.0292}{5.0292} = 0.0058

.
Mole fraction of water

= \frac{5}{5.0292} = 0.9942

.

Q2 • 3 Marks Short Answer

Calculate the molarity of a solution containing

5 \text{ g}

of

\text{NaOH}

in

450 \text{ mL}

solution. (Molar mass of

\text{NaOH} = 40 \text{ g/mol}

)|||"HI"|||

450 \text{ mL}

विलयन में

5 \text{ g}

\text{NaOH}

युक्त विलयन की मोलरता की गणना कीजिए। (

\text{NaOH}

का मोलर द्रव्यमान

= 40 \text{ g/mol}

)

View Model Solution & Step Marking ▾

Model Answer:

Molar mass of

\text{NaOH} = 40 \text{ g/mol}

.
Number of moles of

\text{NaOH} = \frac{5 \text{ g}}{40 \text{ g/mol}} = 0.125 \text{ mol}

.
Volume of solution in litres

= \frac{450 \text{ mL}}{1000 \text{ mL/L}} = 0.450 \text{ L}

.
Molarity

= \frac{\text{Moles of solute}}{\text{Volume of solution in litres}} = \frac{0.125 \text{ mol}}{0.450 \text{ L}} = 0.278 \text{ M}

.

Q3 • 3 Marks Short Answer

State Raoult's law for a solution containing volatile components. How is it different from Henry's law?|||"HI"|||वाष्पशील घटकों वाले विलयन के लिए राउल्ट का नियम बताइए। यह हेनरी के नियम से किस प्रकार भिन्न है?

View Model Solution & Step Marking ▾

Model Answer:

Raoult's law states that for a solution of volatile liquids, the partial vapor pressure of each component in the solution is directly proportional to its mole fraction present in the solution.
Henry's law states that the partial pressure of the gas in the vapor phase is proportional to the mole fraction of the gas in the solution. Raoult's law is a special case of Henry's law where the proportionality constant

k_H

becomes equal to the vapor pressure of the pure solvent.

Q4 • 3 Marks Short Answer

What is osmotic pressure? Explain how measurement of osmotic pressure is preferred over other colligative properties for determining the molar mass of macromolecules.|||"HI"|||परासरण दाब क्या है? समझाइए कि मैक्रोमोलेक्यूल्स के मोलर द्रव्यमान को निर्धारित करने के लिए अन्य अणुसंख्य गुणों की तुलना में परासरण दाब का मापन क्यों पसंद किया जाता है।

View Model Solution & Step Marking ▾

Model Answer:

Osmotic pressure (

\pi

) is the excess pressure that must be applied to a solution to prevent the passage of solvent molecules into it through a semipermeable membrane.
Osmotic pressure measurement is preferred because:
1. It is measured at room temperature, so macromolecules are stable.
2. Its magnitude is large even for very dilute solutions.
3. It is useful for determining the molar mass of polymers and biomolecules which are not stable at higher temperatures.

Q5 • 3 Marks Short Answer

Define ideal and non-ideal solutions. Give one example of each.|||"HI"|||आदर्श और अनादर्श विलयनों को परिभाषित कीजिए। प्रत्येक का एक-एक उदाहरण दीजिए।

View Model Solution & Step Marking ▾

Model Answer:

Ideal solutions are solutions that obey Raoult's law over the entire range of concentrations. For ideal solutions,

\Delta H_{\text{mix}} = 0

and

\Delta V_{\text{mix}} = 0

. Example: Benzene and Toluene.
Non-ideal solutions are solutions that do not obey Raoult's law over the entire range of concentrations. For non-ideal solutions,

\Delta H_{\text{mix}} \neq 0

and

\Delta V_{\text{mix}} \neq 0

. Example: Ethanol and Acetone.

Q6 • 3 Marks Short Answer

Calculate the molality of a solution containing

18.25 \text{ g}

of

\text{HCl}

gas in

500 \text{ g}

of water.

500 \text{ g}

जल में

18.25 \text{ g}

\text{HCl}

गैस युक्त विलयन की मोललता की गणना कीजिए।

View Model Solution & Step Marking ▾

Model Answer:

Molality

(m) = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}

\nMoles of

\text{HCl} = \frac{18.25 \text{ g}}{36.5 \text{ g/mol}} = 0.5 \text{ mol}

\nMass of water in

\text{kg} = 500 \text{ g} = 0.5 \text{ kg}

\nMolality

= \frac{0.5 \text{ mol}}{0.5 \text{ kg}} = 1.0 \text{ m}

Q7 • 3 Marks Short Answer

State Raoult's Law for a solution containing volatile liquids. How does it differ from Henry's Law?

वाष्पशील द्रवों वाले विलयन के लिए राउल्ट का नियम लिखिए। यह हेनरी के नियम से किस प्रकार भिन्न है?

View Model Solution & Step Marking ▾

Model Answer:

Raoult's Law states that for a solution of volatile liquids, the partial vapor pressure of each component in the solution is directly proportional to its mole fraction present in the solution.

P_A = P_A^0 X_A

. \nHenry's Law states that the partial pressure of a gas in the vapor phase is proportional to the mole fraction of the gas in the solution.

P = K_H X

. Raoult's law is a special case of Henry's law where

K_H = P_A^0

.

Q8 • 3 Marks Short Answer

A

5\%

solution of sucrose (molecular mass

= 342

) is isotonic with a

0.877\%

solution of urea (molecular mass

= 60

). Calculate the molar mass of an unknown substance if its

1.5\%

solution is isotonic with the sucrose solution.

सुक्रोज (आणविक द्रव्यमान

= 342

) का

5\%

विलयन यूरिया (आणविक द्रव्यमान

= 60

) के

0.877\%

विलयन के साथ समपरासरी है। यदि किसी अज्ञात पदार्थ का

1.5\%

विलयन सुक्रोज विलयन के साथ समपरासरी है, तो उस अज्ञात पदार्थ के मोलर द्रव्यमान की गणना कीजिए।

View Model Solution & Step Marking ▾

Model Answer:

For isotonic solutions, osmotic pressure is equal, so

C_1 = C_2

. \nFor sucrose solution:

C_1 = \frac{5}{342} \text{ mol/L}

.\nFor unknown substance:

C_2 = \frac{1.5}{M_2} \text{ mol/L}

.\nSince they are isotonic,

\frac{5}{342} = \frac{1.5}{M_2}

.\n

M_2 = \frac{1.5 \times 342}{5} = 102.6 \text{ g/mol}

.

Q9 • 3 Marks Short Answer

An aqueous solution freezes at

-0.186^ ext{o} ext{C}

. Calculate the depression in freezing point if

K_f

for water is

1.86 \text{ K kg mol}^{-1}

. Also, determine the molality of the solution.

एक जलीय विलयन

-0.186^ ext{o} ext{C}

पर जमता है। यदि जल के लिए

K_f

1.86 \text{ K kg mol}^{-1}

है, तो हिमांक में अवनमन की गणना कीजिए। साथ ही, विलयन की मोललता भी निर्धारित कीजिए।

View Model Solution & Step Marking ▾

Model Answer:

Depression in freezing point,

\Delta T_f = T_f^0 - T_f = 0^ ext{o} ext{C} - (-0.186^ ext{o} ext{C}) = 0.186^ ext{o} ext{C}

or

0.186 \text{ K}

.\nUsing the formula

\Delta T_f = K_f \cdot m

, where

m

is molality.\n

0.186 = 1.86 \times m

\n

m = \frac{0.186}{1.86} = 0.1 \text{ mol/kg}

.

Q10 • 3 Marks Short Answer

An aqueous solution freezes at

-0.186 \text{ \degree C}

. Calculate the molality of the solution. (

K_f

for water

= 1.86 \text{ K kg/mol}

).

एक जलीय विलयन

-0.186 \text{ \degree C}

पर जमता है। विलयन की मोललता की गणना कीजिए। (जल के लिए

K_f = 1.86 \text{ K kg/mol}

)।

View Model Solution & Step Marking ▾

Model Answer:

Given

\Delta T_f = 0 - (-0.186 \text{ \degree C}) = 0.186 \text{ \degree C} = 0.186 \text{ K}

Using the formula

\Delta T_f = K_f \cdot m

0.186 \text{ K} = 1.86 \text{ K kg/mol} \cdot m

m = \frac{0.186 \text{ K}}{1.86 \text{ K kg/mol}} = 0.1 \text{ mol/kg}

Part B: Long Answer Questions & Derivations (5 Marks Each)

Q1 • 5 Marks Long Answer / Derivation

a) Define elevation in boiling point. Why is elevation in boiling point considered a colligative property?

क) क्वथनांक उन्नयन को परिभाषित कीजिए। क्वथनांक उन्नयन को अणुसंख्य गुणधर्म क्यों माना जाता है?

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Elevation in boiling point (

\Delta T_b

) is the increase in the boiling point of a solvent when a non-volatile solute is added to it. It is considered a colligative property because it depends only on the number of solute particles, not on their nature. (1+1=2 marks) b) Given:

W_B = 1.0\text{ g}

,

M_B = 100\text{ g/mol}

,

W_A = 50\text{ g}

,

K_b = 0.52\text{ K kg/mol}

. Calculate:

\Delta T_b = K_b \times \frac{W_B \times 1000}{M_B \times W_A}

(in g)

\Delta T_b = 0.52 \times \frac{1.0 \times 1000}{100 \times 50} = 0.52 \times \frac{1000}{5000} = 0.52 \times 0.2 = 0.104\text{ K}

. Boiling point of solution =

T_b^\circ + \Delta T_b = 373.15\text{ K} + 0.104\text{ K} = 373.254\text{ K}

. (3 marks)

Q2 • 5 Marks Long Answer / Derivation

a) State Henry's Law. Explain its application in deep-sea diving. b) The partial pressure of ethane over a solution containing

6.56 \times 10^{-3}\text{ g}

of ethane is

1\text{ bar}

. If the solution contains

5.00 \times 10^{-2}\text{ g}

of ethane, then what will be the partial pressure of the gas? (Atomic mass of C =

12\text{ u}

, H =

1\text{ u}

)

क) हेनरी का नियम बताइए। गहरे समुद्र में गोताखोरी में इसके अनुप्रयोग की व्याख्या कीजिए। ख)

6.56 \times 10^{-3}\text{ g}

इथेन वाले विलयन पर इथेन का आंशिक दाब

1\text{ बार}

है। यदि विलयन में

5.00 \times 10^{-2}\text{ g}

इथेन है, तो गैस का आंशिक दाब क्या होगा? (C का परमाणु द्रव्यमान =

12\text{ u}

, H =

1\text{ u}

)

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Henry's Law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the solution. (1 mark) Application: In deep-sea diving, the partial pressure of dissolved gases (like N

_2

) in the blood increases with depth. When divers ascend, the external pressure decreases, causing the dissolved gases to become less soluble and form bubbles in the blood, leading to 'bends'. To avoid this, air is diluted with helium. (2 marks) b) For ethane (

C_2H_6

),

M = 2 \times 12 + 6 \times 1 = 30\text{ g/mol}

. Case 1:

w_1 = 6.56 \times 10^{-3}\text{ g}

,

p_1 = 1\text{ bar}

.

n_1 = \frac{w_1}{M} = \frac{6.56 \times 10^{-3}}{30}\text{ mol}

. According to Henry's Law,

p = K_H x

. Since

x \propto n

, we can say

p \propto n

. So,

K_H = \frac{p_1}{n_1} = \frac{1\text{ bar}}{6.56 \times 10^{-3}/30\text{ mol}}

. Case 2:

w_2 = 5.00 \times 10^{-2}\text{ g}

.

n_2 = \frac{w_2}{M} = \frac{5.00 \times 10^{-2}}{30}\text{ mol}

.

p_2 = K_H \times n_2 = \left(\frac{1\text{ bar}}{6.56 \times 10^{-3}/30\text{ mol}}\right) \times \left(\frac{5.00 \times 10^{-2}}{30}\text{ mol}\right) = \frac{5.00 \times 10^{-2}}{6.56 \times 10^{-3}} \text{ bar} = \frac{500}{6.56} \times 10^{-3} \text{ bar} = 76.219 \times 10^{-3} \times 10 \text{ bar} \approx 7.62\text{ bar}

. (2 marks)

Q3 • 5 Marks Long Answer / Derivation

a) What is reverse osmosis? Mention two practical applications of reverse osmosis. b) A

5\%

solution of cane sugar (molar mass

342\text{ g/mol}

) is isotonic with

1\%

solution of substance 'X'. Calculate the molar mass of 'X'.

क) प्रतिलोम परासरण क्या है? प्रतिलोम परासरण के दो व्यावहारिक अनुप्रयोगों का उल्लेख कीजिए। ख) गन्ने की चीनी (मोलर द्रव्यमान

342\text{ g/mol}

) का

5\%

विलयन पदार्थ 'X' के

1\%

विलयन के साथ समपरासरी है। 'X' का मोलर द्रव्यमान ज्ञात कीजिए।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Reverse osmosis is the process where a solvent moves from a region of higher solute concentration to a region of lower solute concentration through a semi-permeable membrane, when a pressure greater than the osmotic pressure is applied to the concentrated side. (1.5 marks) Applications: Desalination of seawater; Purification of drinking water. (0.5+0.5=1 mark) b) For isotonic solutions, osmotic pressure (

\Pi

) is equal, so

C_1 = C_2

(at same temperature). For cane sugar solution:

C_1 = \frac{\text{mass of sugar}}{\text{molar mass of sugar}} \times \frac{1000}{\text{volume of solution (mL)}} = \frac{5\text{ g}}{342\text{ g/mol}} \times \frac{100}{100\text{ mL}} = \frac{5}{342}\text{ mol/L}

. For substance 'X' solution:

C_2 = \frac{\text{mass of X}}{\text{molar mass of X}} \times \frac{100}{100\text{ mL}} = \frac{1\text{ g}}{M_X\text{ g/mol}} \times \frac{100}{100\text{ mL}} = \frac{1}{M_X}\text{ mol/L}

. Since

C_1 = C_2

,

\frac{5}{342} = \frac{1}{M_X}

.

M_X = \frac{342}{5} = 68.4\text{ g/mol}

. (2 marks)

Q4 • 5 Marks Long Answer / Derivation

Calculate the mass of ascorbic acid (Vitamin C,

C_6H_8O_6

) that should be dissolved in

75 g

of acetic acid to lower its freezing point by

1.5^ ext{o}C

. The freezing point depression constant (

K_f

) for acetic acid is

3.9 K kg mol^{-1}

.

75 g

एसिटिक अम्ल में एस्कॉर्बिक अम्ल (विटामिन C,

C_6H_8O_6

) की कितनी मात्रा घोलनी चाहिए ताकि उसका हिमांक

1.5^ ext{o}C

कम हो जाए? एसिटिक अम्ल के लिए हिमांक अवनमन स्थिरांक (

K_f

)

3.9 K kg mol^{-1}

है।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

Given:

\Delta T_f = 1.5^ ext{o}C

,

K_f = 3.9 K kg mol^{-1}

, Mass of solvent (

W_A

)

= 75 g = 0.075 kg

. Molar mass of ascorbic acid (

C_6H_8O_6

):

6 \times 12.01 + 8 \times 1.01 + 6 \times 16.00 = 72.06 + 8.08 + 96.00 = 176.14 g/mol

. We use the formula:

\Delta T_f = K_f \times m

, where

m = \frac{W_B}{M_B \times W_A(kg)}

. So,

\Delta T_f = K_f \times \frac{W_B}{M_B \times W_A(kg)}

. Rearranging for

W_B

:

W_B = \frac{\Delta T_f \times M_B \times W_A(kg)}{K_f}

. Substituting the values:

W_B = \frac{1.5 K \times 176.14 g/mol \times 0.075 kg}{3.9 K kg mol^{-1}} = \frac{19.81575}{3.9} = 5.08 g

. Therefore,

5.08 g

of ascorbic acid should be dissolved.

Q5 • 5 Marks Long Answer / Derivation

Define Raoult's Law for volatile solutes. State two characteristics of an ideal solution. Explain why a solution of ethanol and acetone shows positive deviation from Raoult's Law.

वाष्पशील विलेय के लिए राउल्ट के नियम को परिभाषित कीजिए। एक आदर्श विलयन की दो विशेषताएँ बताइए। समझाइए कि एथेनॉल और एसीटोन का विलयन राउल्ट के नियम से धनात्मक विचलन क्यों दर्शाता है।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

Raoult's Law states that for a solution of volatile liquids, the partial vapor pressure of each component in the solution is directly proportional to its mole fraction present in the solution.

P_A = P_A^o X_A

and

P_B = P_B^o X_B

. Characteristics of an ideal solution:

(i)

It obeys Raoult's Law over the entire range of concentration.

(ii)

\Delta H_{mix} = 0

(no heat is absorbed or evolved on mixing).

(iii)

\Delta V_{mix} = 0

(no change in volume on mixing). A solution of ethanol and acetone shows positive deviation from Raoult's Law because in pure ethanol, molecules are hydrogen-bonded. When acetone is added, the acetone molecules get in between the ethanol molecules, breaking some of these hydrogen bonds. This reduces the intermolecular forces of attraction between ethanol molecules, and between ethanol and acetone molecules. As a result, the molecules escape more easily from the solution than from pure components, leading to a higher vapor pressure than expected.

2

Electrochemistry

Part A: Short Answer Questions (2–3 Marks Each)

Q1 • 3 Marks Short Answer

Calculate the standard cell potential (

E^ heta_{cell}

) and the standard Gibbs energy change (

riangle G^ heta

) for the following cell reaction:

2Cr(s) + 3Cd^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 3Cd(s)

. Given standard electrode potentials:

E^ heta_{Cr^{3+}/Cr} = -0.74V

and

E^ heta_{Cd^{2+}/Cd} = -0.40V

.

निम्नलिखित सेल अभिक्रिया के लिए मानक सेल विभव (

E^ heta_{cell}

) और मानक गिब्स ऊर्जा परिवर्तन (

riangle G^ heta

) की गणना कीजिए:

2Cr(s) + 3Cd^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 3Cd(s)

। दिए गए मानक इलेक्ट्रोड विभव हैं:

E^ heta_{Cr^{3+}/Cr} = -0.74V

और

E^ heta_{Cd^{2+}/Cd} = -0.40V

।

View Model Solution & Step Marking ▾

Model Answer:

E^ heta_{cell} = 0.34V

,

riangle G^ heta = -196.86 kJ/mol

Q2 • 3 Marks Short Answer

Explain why the conductivity of an electrolytic solution decreases with dilution, while molar conductivity increases.

समझाइए कि किसी विद्युत्-अपघट्य विलयन की चालकता तनुकरण के साथ क्यों घटती है, जबकि मोलर चालकता बढ़ती है।

View Model Solution & Step Marking ▾

Model Answer:

Conductivity decreases because the number of ions per unit volume decreases upon dilution. Molar conductivity increases because the degree of dissociation of the electrolyte increases, and the mobility of ions also increases due to reduced interionic attraction.

Q3 • 3 Marks Short Answer

A current of

1.5A

is passed through an aqueous solution of

CuSO_4

for

10

minutes. Calculate the mass of copper deposited at the cathode. (Molar mass of

Cu = 63.5 g/mol

,

F = 96485 C/mol

).

CuSO_4

के जलीय विलयन में

1.5A

की धारा

10

मिनट के लिए प्रवाहित की जाती है। कैथोड पर जमा हुए तांबे का द्रव्यमान ज्ञात कीजिए। (तांबे का मोलर द्रव्यमान

= 63.5 g/mol

,

F = 96485 C/mol

)।

View Model Solution & Step Marking ▾

Model Answer:

Mass of copper deposited =

0.295 g

Q4 • 3 Marks Short Answer

Distinguish between primary and secondary batteries, giving one example of each.

प्राथमिक और द्वितीयक बैटरियों के बीच अंतर स्पष्ट कीजिए, प्रत्येक का एक-एक उदाहरण देते हुए।

View Model Solution & Step Marking ▾

Model Answer:

Primary batteries are non-rechargeable and cannot be reused after their chemicals are consumed (e.g., Dry cell/Leclanché cell). Secondary batteries are rechargeable and can be reused multiple times by passing an electric current through them (e.g., Lead-acid battery/Nickel-cadmium cell).

Q5 • 3 Marks Short Answer

What is corrosion? Explain the electrochemical theory of rusting of iron.

संक्षारण क्या है? लोहे में जंग लगने के विद्युत्-रासायनिक सिद्धांत की व्याख्या कीजिए।

View Model Solution & Step Marking ▾

Model Answer:

Corrosion is the process of slow deterioration of metals due to reaction with air, moisture, or chemicals in the environment. Rusting of iron involves an electrochemical cell formation where iron acts as anode (

Fe \rightarrow Fe^{2+} + 2e^-

), oxygen acts as cathode (

O_2 + 4H^+ + 4e^- \rightarrow 2H_2O

), and the

Fe^{2+}

ions are further oxidized to

Fe_2O_3 \cdot xH_2O

(rust).

Q6 • 3 Marks Short Answer

Calculate the standard cell potential (

E^ heta_{cell}

) for a galvanic cell in which the following reaction occurs:

2Cr(s) + 3Cd^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 3Cd(s)

. Given standard electrode potentials:

E^ heta_{Cr^{3+}/Cr} = -0.74V

and

E^ heta_{Cd^{2+}/Cd} = -0.40V

.

एक गैल्वेनिक सेल के लिए मानक सेल विभव (

E^ heta_{cell}

) की गणना कीजिए जिसमें निम्न अभिक्रिया होती है:

2Cr(s) + 3Cd^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 3Cd(s)

। दिए गए मानक इलेक्ट्रोड विभव हैं:

E^ heta_{Cr^{3+}/Cr} = -0.74V

और

E^ heta_{Cd^{2+}/Cd} = -0.40V

।

View Model Solution & Step Marking ▾

Model Answer:

E^ heta_{cell} = E^ heta_{cathode} - E^ heta_{anode}

E^ heta_{cell} = (-0.40V) - (-0.74V) = 0.34V

.

Q7 • 3 Marks Short Answer

State Kohlrausch's Law of independent migration of ions. How can it be used to determine the molar conductivity of a weak electrolyte at infinite dilution?

कोल्हाराउश का आयनों के स्वतंत्र अभिगमन का नियम लिखिए। इसका उपयोग अनंत तनुता पर दुर्बल वैद्युतअपघट्य की मोलर चालकता निर्धारित करने के लिए कैसे किया जा सकता है?

View Model Solution & Step Marking ▾

Model Answer:

Kohlrausch's Law states that at infinite dilution, when the dissociation is complete, each ion makes a definite contribution to the molar conductivity of the electrolyte, irrespective of the nature of the other ion with which it is associated. For a weak electrolyte,

\Lambda^ heta_m(AB) = \lambda^ heta_A + \lambda^ heta_B

. It can be determined by adding the molar conductivities of the respective strong electrolytes. For example, to find

\Lambda^ heta_m(CH_3COOH)

, we can use

\Lambda^ heta_m(CH_3COONa) + \Lambda^ heta_m(HCl) - \Lambda^ heta_m(NaCl)

.

Q8 • 3 Marks Short Answer

The resistance of a conductivity cell containing

0.001M

KCl solution at

298K

is

1500 \Omega

. What is the cell constant if the conductivity of

0.001M

KCl solution at

298K

is

0.146 \times 10^{-3} S \cdot cm^{-1}

?

298K

पर

0.001M

KCl विलयन वाले एक चालकता सेल का प्रतिरोध

1500 \Omega

है। यदि

298K

पर

0.001M

KCl विलयन की चालकता

0.146 \times 10^{-3} S \cdot cm^{-1}

है, तो सेल स्थिरांक क्या है?

View Model Solution & Step Marking ▾

Model Answer:

Conductivity (

\kappa

) =

\frac{1}{R} \times \frac{l}{A}

. Cell constant (

G^*

) =

\frac{l}{A} = \kappa \times R

.

G^* = (0.146 \times 10^{-3} S \cdot cm^{-1}) \times (1500 \Omega) = 0.219 cm^{-1}

.

Q9 • 3 Marks Short Answer

What is meant by the term 'passivation of metals'? Give an example of a metal that undergoes passivation and explain the phenomenon.

'धातुओं का निष्क्रियकरण' शब्द से क्या अभिप्राय है? एक ऐसी धातु का उदाहरण दीजिए जो निष्क्रियकरण से गुजरती है और इस घटना की व्याख्या कीजिए।

View Model Solution & Step Marking ▾

Model Answer:

Passivation of metals refers to the formation of a thin, non-porous, and protective oxide layer on the surface of a metal, which prevents further corrosion. For example, aluminium undergoes passivation when exposed to air, forming a thin layer of aluminium oxide (

Al_2O_3

). This layer is very stable and adheres strongly to the metal surface, protecting the underlying aluminium from further oxidation.

Q10 • 3 Marks Short Answer

Explain why a dry cell becomes 'dead' after a long time, even if it has not been used. Give one reason.

समझाइए कि एक शुष्क सेल लंबे समय के बाद 'मृत' क्यों हो जाता है, भले ही इसका उपयोग न किया गया हो। एक कारण दीजिए।

View Model Solution & Step Marking ▾

Model Answer:

A dry cell becomes 'dead' after a long time because the acidic ammonium chloride, which is used as an electrolyte, corrodes the zinc container (anode) even when the cell is not in use. This leads to leakage and loss of electrolytic solution, rendering the cell inactive.

Part B: Long Answer Questions & Derivations (5 Marks Each)

Q1 • 5 Marks Long Answer / Derivation

a) State Kohlrausch's Law of independent migration of ions.b) The molar conductivities at infinite dilution for

NaCl

,

HCl

, and

CH_3COONa

are

126.4~S~cm^2~mol^{-1}

,

425.9~S~cm^2~mol^{-1}

, and

91.0~S~cm^2~mol^{-1}

respectively. Calculate the molar conductivity at infinite dilution for

CH_3COOH

. Explain why this law is essential for determining the molar conductivity of weak electrolytes at infinite dilution.

a) आयनों के स्वतंत्र अभिगमन के कोलराउश के नियम का उल्लेख कीजिए।b)

NaCl

,

HCl

, और

CH_3COONa

के लिए अनंत तनुता पर मोलर चालकताएँ क्रमशः

126.4~S~cm^2~mol^{-1}

,

425.9~S~cm^2~mol^{-1}

, और

91.0~S~cm^2~mol^{-1}

हैं।

CH_3COOH

के लिए अनंत तनुता पर मोलर चालकता की गणना कीजिए। समझाइए कि यह नियम दुर्बल विद्युत अपघट्यों की अनंत तनुता पर मोलर चालकता ज्ञात करने के लिए क्यों आवश्यक है।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Kohlrausch's Law of independent migration of ions states that at infinite dilution, when the dissociation is complete, each ion makes a definite contribution to the molar conductivity of the electrolyte, irrespective of the nature of the other ion with which it is associated.b) Given:

\Lambda^0_m(NaCl) = 126.4~S~cm^2~mol^{-1}<span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><msubsup><mi mathvariant="normal">Λ</mi><mi>m</mi><mn>0</mn></msubsup><mo stretchy="false">(</mo><mi>H</mi><mi>C</mi><mi>l</mi><mo stretchy="false">)</mo><mo>=</mo><mn>425.9</mn><mtext> </mtext><mi>S</mi><mtext> </mtext><mi>c</mi><msup><mi>m</mi><mn>2</mn></msup><mtext> </mtext><mi>m</mi><mi>o</mi><msup><mi>l</mi><mrow><mo>−</mo><mn>1</mn></mrow></msup></mrow><annotation encoding="application/x-tex">\Lambda^0_m(HCl) = 425.9~S~cm^2~mol^{-1}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="katex-base"><span class="katex-strut" style="height:1.1141em;vertical-align:-0.25em;"></span><span class="mord"><span class="mord">Λ</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.8641em;"><span style="top:-2.453em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="katex-sizing reset-size6 size3 mtight"><span class="mord mathnormal mtight">m</span></span></span><span style="top:-3.113em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="katex-sizing reset-size6 size3 mtight"><span class="mord mtight">0</span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.247em;"><span></span></span></span></span></span></span><span class="mopen">(</span><span class="mord mathnormal" style="margin-right:0.0813em;">H</span><span class="mord mathnormal" style="margin-right:0.0715em;">C</span><span class="mord mathnormal" style="margin-right:0.0197em;">l</span><span class="mclose">)</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="katex-base"><span class="katex-strut" style="height:0.8641em;"></span><span class="mord">425.9</span><span class="mspace nobreak"> </span><span class="mord mathnormal" style="margin-right:0.0576em;">S</span><span class="mspace nobreak"> </span><span class="mord mathnormal">c</span><span class="mord"><span class="mord mathnormal">m</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8641em;"><span style="top:-3.113em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="katex-sizing reset-size6 size3 mtight"><span class="mord mtight">2</span></span></span></span></span></span></span></span><span class="mspace nobreak"> </span><span class="mord mathnormal">m</span><span class="mord mathnormal">o</span><span class="mord"><span class="mord mathnormal" style="margin-right:0.0197em;">l</span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8641em;"><span style="top:-3.113em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="katex-sizing reset-size6 size3 mtight"><span class="mord mtight"><span class="mord mtight">−</span><span class="mord mtight">1</span></span></span></span></span></span></span></span></span></span></span></span></span>\Lambda^0_m(CH_3COONa) = 91.0~S~cm^2~mol^{-1}

To find

\Lambda^0_m(CH_3COOH)

:

\Lambda^0_m(CH_3COOH) = \Lambda^0_m(CH_3COONa) + \Lambda^0_m(HCl) - \Lambda^0_m(NaCl)<span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><msubsup><mi mathvariant="normal">Λ</mi><mi>m</mi><mn>0</mn></msubsup><mo stretchy="false">(</mo><mi>C</mi><msub><mi>H</mi><mn>3</mn></msub><mi>C</mi><mi>O</mi><mi>O</mi><mi>H</mi><mo stretchy="false">)</mo><mo>=</mo><mn>91.0</mn><mo>+</mo><mn>425.9</mn><mo>−</mo><mn>126.4</mn></mrow><annotation encoding="application/x-tex">\Lambda^0_m(CH_3COOH) = 91.0 + 425.9 - 126.4</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="katex-base"><span class="katex-strut" style="height:1.1141em;vertical-align:-0.25em;"></span><span class="mord"><span class="mord">Λ</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.8641em;"><span style="top:-2.453em;margin-left:0em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="katex-sizing reset-size6 size3 mtight"><span class="mord mathnormal mtight">m</span></span></span><span style="top:-3.113em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="katex-sizing reset-size6 size3 mtight"><span class="mord mtight">0</span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.247em;"><span></span></span></span></span></span></span><span class="mopen">(</span><span class="mord mathnormal" style="margin-right:0.0715em;">C</span><span class="mord"><span class="mord mathnormal" style="margin-right:0.0813em;">H</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.3011em;"><span style="top:-2.55em;margin-left:-0.0813em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="katex-sizing reset-size6 size3 mtight"><span class="mord mtight">3</span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span><span class="mord mathnormal" style="margin-right:0.0715em;">C</span><span class="mord mathnormal" style="margin-right:0.0278em;">O</span><span class="mord mathnormal" style="margin-right:0.0278em;">O</span><span class="mord mathnormal" style="margin-right:0.0813em;">H</span><span class="mclose">)</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="katex-base"><span class="katex-strut" style="height:0.7278em;vertical-align:-0.0833em;"></span><span class="mord">91.0</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">+</span><span class="mspace" style="margin-right:0.2222em;"></span></span><span class="katex-base"><span class="katex-strut" style="height:0.7278em;vertical-align:-0.0833em;"></span><span class="mord">425.9</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">−</span><span class="mspace" style="margin-right:0.2222em;"></span></span><span class="katex-base"><span class="katex-strut" style="height:0.6444em;"></span><span class="mord">126.4</span></span></span></span></span>\Lambda^0_m(CH_3COOH) = 516.9 - 126.4 = 390.5~S~cm^2~mol^{-1}

This law is essential for determining the molar conductivity of weak electrolytes at infinite dilution because weak electrolytes do not dissociate completely even at high dilutions. Their molar conductivity increases sharply with dilution and does not approach a limiting value easily. By using Kohlrausch's Law, we can calculate

\Lambda^0_m

for weak electrolytes from the

\Lambda^0_m

values of strong electrolytes, which can be determined by extrapolation.

Q2 • 5 Marks Long Answer / Derivation

a) State Faraday's first law of electrolysis.
b) A solution of

CuSO_4

is electrolysed for

10 \text{ minutes}

with a current of

1.5 \text{ amperes}

. Calculate the mass of copper deposited at the cathode.
(Given: Atomic mass of

Cu = 63.5 \text{ g mol}^{-1}

,

F = 96487 \text{ C mol}^{-1}

)
c) How does the conductivity of an electrolytic solution change with dilution? Explain why.

a) फैराडे के विद्युत अपघटन के पहले नियम का उल्लेख कीजिए।
b)

CuSO_4

के विलयन का

1.5 \text{ एम्पियर}

की धारा के साथ

10 \text{ मिनट}

तक विद्युत अपघटन किया जाता है। कैथोड पर जमा हुए तांबे के द्रव्यमान की गणना कीजिए।
(दिया गया है: तांबे का परमाणु द्रव्यमान

= 63.5 \text{ g mol}^{-1}

,

F = 96487 \text{ C mol}^{-1}

)
c) तनुकरण के साथ एक विद्युत अपघटनी विलयन की चालकता कैसे बदलती है? कारण सहित समझाइए।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Faraday's first law states that the mass of any substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed through the electrolyte.
Mathematically,

m \propto Q

or

m = ZIt

, where

Z

is the electrochemical equivalent.
b) Given:

I = 1.5 \text{ A}

,

t = 10 \text{ minutes} = 10 \times 60 = 600 \text{ s}

.
Quantity of electricity,

Q = I \times t = 1.5 \text{ A} \times 600 \text{ s} = 900 \text{ C}

.
Reaction at cathode:

Cu^{2+}(aq) + 2e^- \to Cu(s)

From the reaction,

2 \text{ moles}

of electrons (

2F

) deposit

1 \text{ mole}

of

Cu

.

2 \times 96487 \text{ C}

deposit

63.5 \text{ g}

of

Cu

.
So,

900 \text{ C}

will deposit

\frac{63.5 \text{ g}}{2 \times 96487 \text{ C}} \times 900 \text{ C} = 0.2958 \text{ g}

of

Cu

.
c) The conductivity of an electrolytic solution decreases with dilution. This is because conductivity depends on the number of ions per unit volume. Upon dilution, the number of ions per unit volume decreases, even though the total number of ions might increase due to increased dissociation. The decrease in ion concentration per unit volume outweighs the increase in molar conductivity due to increased mobility, thus leading to a decrease in specific conductivity.

Q3 • 5 Marks Long Answer / Derivation

a) Calculate the standard cell potential for the following electrochemical cell:

Mg(s)|Mg^{2+}(0.1M)||Ag^+(0.001M)|Ag(s)

. Given standard electrode potentials:

E^ heta_{Mg^{2+}/Mg} = -2.37V

and

E^ heta_{Ag^+/Ag} = +0.80V

. Also, write the overall cell reaction. b) Explain the effect of increasing the concentration of

Mg^{2+}

ions on the cell potential.

a) निम्नलिखित विद्युत रासायनिक सेल के लिए मानक सेल विभव की गणना कीजिए:

Mg(s)|Mg^{2+}(0.1M)||Ag^+(0.001M)|Ag(s)

। दिए गए मानक इलेक्ट्रोड विभव हैं:

E^ heta_{Mg^{2+}/Mg} = -2.37V

और

E^ heta_{Ag^+/Ag} = +0.80V

। साथ ही, समग्र सेल अभिक्रिया भी लिखिए। b)

Mg^{2+}

आयनों की सांद्रता बढ़ाने पर सेल विभव पर पड़ने वाले प्रभाव की व्याख्या कीजिए।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Calculation of

E^ heta_{cell}

:

E^ heta_{cell} = E^ heta_{cathode} - E^ heta_{anode} = E^ heta_{Ag^+/Ag} - E^ heta_{Mg^{2+}/Mg} = +0.80V - (-2.37V) = +3.17V

. Anode:

Mg(s) \to Mg^{2+}(aq) + 2e^-

. Cathode:

2Ag^+(aq) + 2e^- \to 2Ag(s)

. Overall cell reaction:

Mg(s) + 2Ag^+(aq) \to Mg^{2+}(aq) + 2Ag(s)

. b) According to Nernst equation,

E_{cell} = E^ heta_{cell} - \frac{0.0592}{n} \log \frac{[Mg^{2+}]}{[Ag^+]^2}

. If the concentration of

Mg^{2+}

ions increases, the value of the term

\frac{[Mg^{2+}]}{[Ag^+]^2}

increases. Consequently, the value of

\frac{0.0592}{n} \log \frac{[Mg^{2+}]}{[Ag^+]^2}

increases. Since this term is subtracted from

E^ heta_{cell}

, the cell potential (

E_{cell}

) will decrease.

Q4 • 5 Marks Long Answer / Derivation

a) What is an electrochemical cell? Differentiate between a galvanic cell and an electrolytic cell based on the following criteria:i) Energy conversionii) Spontaneity of reactioniii) Anode and Cathode polarityb) For the following cell:

Zn(s)|Zn^{2+}(aq)||Cu^{2+}(aq)|Cu(s)

, write the overall cell reaction and identify the anode and cathode.

a) एक विद्युत रासायनिक सेल क्या है? गैल्वेनिक सेल और विद्युत अपघटनी सेल के बीच निम्नलिखित मानदंडों के आधार पर अंतर स्पष्ट कीजिए:i) ऊर्जा रूपांतरणii) अभिक्रिया की स्वतः प्रवर्तनीयताiii) एनोड और कैथोड की ध्रुवीयताब) निम्नलिखित सेल के लिए:

Zn(s)|Zn^{2+}(aq)||Cu^{2+}(aq)|Cu(s)

, समग्र सेल अभिक्रिया लिखिए और एनोड व कैथोड की पहचान कीजिए।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) An electrochemical cell is a device that converts chemical energy into electrical energy or vice versa.i) Energy conversion: Galvanic cell converts chemical energy to electrical energy. Electrolytic cell converts electrical energy to chemical energy.ii) Spontaneity of reaction: Galvanic cell reactions are spontaneous (

\Delta G < 0

). Electrolytic cell reactions are non-spontaneous (

\Delta G > 0

).iii) Anode and Cathode polarity: In a galvanic cell, anode is negative and cathode is positive. In an electrolytic cell, anode is positive and cathode is negative.b) Overall cell reaction:

Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)

Anode: Zinc electrode (

Zn(s)

)Cathode: Copper electrode (

Cu(s)

)

Q5 • 5 Marks Long Answer / Derivation

a) What is the Nernst Equation? Write the Nernst equation for a general electrochemical reaction:

aA + bB \rightarrow cC + dD

.b) Calculate the EMF of the following cell at

298~K

:

Mg(s)|Mg^{2+}(0.1~M)||Ag^{+}(0.001~M)|Ag(s)

Given

E^0_{Mg^{2+}/Mg} = -2.37~V

and

E^0_{Ag^{+}/Ag} = +0.80~V

.(

R = 8.314~J~K^{-1}~mol^{-1}

,

F = 96487~C~mol^{-1}

)

a) नर्नस्ट समीकरण क्या है? एक सामान्य विद्युत रासायनिक अभिक्रिया:

aA + bB \rightarrow cC + dD

के लिए नर्नस्ट समीकरण लिखिए।b)

298~K

पर निम्नलिखित सेल के EMF की गणना कीजिए:

Mg(s)|Mg^{2+}(0.1~M)||Ag^{+}(0.001~M)|Ag(s)

दिया गया है

E^0_{Mg^{2+}/Mg} = -2.37~V

और

E^0_{Ag^{+}/Ag} = +0.80~V

।(

R = 8.314~J~K^{-1}~mol^{-1}

,

F = 96487~C~mol^{-1}

)

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) The Nernst equation relates the cell potential of an electrochemical cell to the standard electrode potentials, temperature, and concentrations (or partial pressures) of the reacting species.For the general reaction

aA + bB \rightarrow cC + dD

, the Nernst equation is:

E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln \frac{[C]^c[D]^d}{[A]^a[B]^b}

or

E_{cell} = E^0_{cell} - \frac{0.0592}{n} \log \frac{[C]^c[D]^d}{[A]^a[B]^b}

(at

298~K

).b) Anode (oxidation):

Mg(s) \rightarrow Mg^{2+}(aq) + 2e^-

Cathode (reduction):

2Ag^{+}(aq) + 2e^- \rightarrow 2Ag(s)

Overall cell reaction:

Mg(s) + 2Ag^{+}(aq) \rightarrow Mg^{2+}(aq) + 2Ag(s)

Number of electrons transferred,

n=2

.Standard cell potential,

E^0_{cell} = E^0_{cathode} - E^0_{anode} = E^0_{Ag^{+}/Ag} - E^0_{Mg^{2+}/Mg}

E^0_{cell} = +0.80~V - (-2.37~V) = +3.17~V

Using Nernst equation at

298~K

:

E_{cell} = E^0_{cell} - \frac{0.0592}{n} \log \frac{[Mg^{2+}]}{[Ag^{+}]^2}

E_{cell} = 3.17 - \frac{0.0592}{2} \log \frac{0.1}{(0.001)^2}

E_{cell} = 3.17 - 0.0296 \log \frac{0.1}{1 \times 10^{-6}}

E_{cell} = 3.17 - 0.0296 \log (1 \times 10^5)

E_{cell} = 3.17 - 0.0296 \times 5

E_{cell} = 3.17 - 0.148

E_{cell} = 3.022~V

3

Chemical Kinetics

Part A: Short Answer Questions (2–3 Marks Each)

Q1 • 3 Marks Short Answer

For a reaction

A + B \rightarrow P

, the rate law is given by

rate = k[A]^2[B]^1

. If the concentration of

A

is doubled and concentration of

B

is halved, how will the reaction rate change?

एक अभिक्रिया

A + B \rightarrow P

के लिए, दर नियम

rate = k[A]^2[B]^1

द्वारा दिया गया है। यदि

A

की सांद्रता दोगुनी कर दी जाए और

B

की सांद्रता आधी कर दी जाए, तो अभिक्रिया की दर कैसे बदलेगी?

View Model Solution & Step Marking ▾

Model Answer:

The new rate will be

2k[A]^2[B]^1

, which is twice the original rate.

Q2 • 3 Marks Short Answer

Derive the integrated rate equation for a first-order reaction. State the units of the rate constant for such a reaction.

प्रथम-कोटि अभिक्रिया के लिए समाकलित दर समीकरण व्युत्पन्न कीजिए। ऐसी अभिक्रिया के लिए दर स्थिरांक की इकाइयाँ बताइए।

View Model Solution & Step Marking ▾

Model Answer:

Integrated rate equation:

k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}

. Units of rate constant:

s^{-1}

or

time^{-1}

.

Q3 • 3 Marks Short Answer

Explain why the molecularity of a reaction is always a whole number and cannot be zero, fractional, or negative. Give an example of a unimolecular reaction.

समझाइए कि किसी अभिक्रिया की आणविकता हमेशा एक पूर्ण संख्या क्यों होती है और शून्य, भिन्नात्मक या ऋणात्मक नहीं हो सकती। एक एक-आणविक अभिक्रिया का उदाहरण दीजिए।

View Model Solution & Step Marking ▾

Model Answer:

Molecularity is the number of reacting species in an elementary step, which must be whole numbers. Example: Decomposition of ammonium nitrite,

NH_4NO_2(aq) \rightarrow N_2(g) + 2H_2O(l)

.

Q4 • 3 Marks Short Answer

The half-life of a first-order reaction is

60

minutes. Calculate the time required for

90\%

completion of the reaction. (Given:

\log 10 = 1

)

एक प्रथम-कोटि अभिक्रिया का अर्ध-आयु काल

60

मिनट है। अभिक्रिया के

90\%

पूर्ण होने में लगने वाले समय की गणना कीजिए। (दिया गया है:

\log 10 = 1

)

View Model Solution & Step Marking ▾

Model Answer:

First, calculate

k = \frac{0.693}{t_{1/2}} = \frac{0.693}{60} min^{-1}

. Then, for

90\%

completion,

t = \frac{2.303}{k} \log \frac{[R]_0}{[R]_0 - 0.90[R]_0} = \frac{2.303}{k} \log 10 = \frac{2.303 \times 60}{0.693} = 199.3

minutes.

Q5 • 3 Marks Short Answer

What is the effect of a catalyst on the activation energy and enthalpy change of a reaction? Explain with the help of a potential energy diagram.

अभिक्रिया की सक्रियण ऊर्जा और एन्थैल्पी परिवर्तन पर उत्प्रेरक का क्या प्रभाव होता है? विभव ऊर्जा आरेख की सहायता से समझाइए।

View Model Solution & Step Marking ▾

Model Answer:

A catalyst lowers the activation energy by providing an alternative reaction path but does not change the overall enthalpy change (

\Delta H

) of the reaction.

Q6 • 3 Marks Short Answer

For a reaction

2A + B \to C

, the rate of formation of

C

is

0.2 \text{ mol L}^{-1}\text{s}^{-1}

. Calculate the rate of disappearance of

A

and

B

.

एक अभिक्रिया

2A + B \to C

के लिए,

C

के बनने की दर

0.2 \text{ mol L}^{-1}\text{s}^{-1}

है।

A

और

B

के विलुप्त होने की दर की गणना कीजिए।

View Model Solution & Step Marking ▾

Model Answer:

Rate of disappearance of

A = 0.4 \text{ mol L}^{-1}\text{s}^{-1}

\nRate of disappearance of

B = 0.2 \text{ mol L}^{-1}\text{s}^{-1}

Q7 • 3 Marks Short Answer

The rate constant for a first order reaction is

60 \text{ s}^{-1}

. How much time will it take to reduce the initial concentration of the reactant to

\frac{1}{10}

th of its initial value?

एक प्रथम कोटि की अभिक्रिया के लिए दर स्थिरांक

60 \text{ s}^{-1}

है। अभिकारक की प्रारंभिक सांद्रता को उसके प्रारंभिक मान के

\frac{1}{10}

तक कम होने में कितना समय लगेगा?

View Model Solution & Step Marking ▾

Model Answer:

t = 0.0384 \text{ s}

Q8 • 3 Marks Short Answer

Define molecularity and order of a reaction. Explain why the molecularity of a reaction cannot be zero, fractional, or negative.

अभिक्रिया की आण्विकता और कोटि को परिभाषित कीजिए। समझाइए कि अभिक्रिया की आण्विकता शून्य, भिन्नात्मक या ऋणात्मक क्यों नहीं हो सकती है।

View Model Solution & Step Marking ▾

Model Answer:

Molecularity: Number of reacting species (atoms, ions, or molecules) taking part in an elementary reaction that must collide simultaneously in order to bring about a chemical reaction. \nOrder: Sum of the powers of the concentration of the reactants in the rate law expression. \nMolecularity cannot be zero, fractional, or negative because it represents the number of molecules involved in a collision, which must be a positive integer.

Q9 • 3 Marks Short Answer

The activation energy of a reaction is

100 \text{ kJ mol}^{-1}

. The rate constant at

300 \text{ K}

is

10 \text{ s}^{-1}

. Calculate the rate constant at

310 \text{ K}

if

R = 8.314 \text{ J K}^{-1}\text{mol}^{-1}

.

एक अभिक्रिया की सक्रियण ऊर्जा

100 \text{ kJ mol}^{-1}

है।

300 \text{ K}

पर दर स्थिरांक

10 \text{ s}^{-1}

है। यदि

R = 8.314 \text{ J K}^{-1}\text{mol}^{-1}

है, तो

310 \text{ K}

पर दर स्थिरांक की गणना कीजिए।

View Model Solution & Step Marking ▾

Model Answer:

k_2 = 28.5 \text{ s}^{-1}

(approximately)

Q10 • 3 Marks Short Answer

Explain the effect of catalyst on the rate of a chemical reaction using potential energy diagrams.

विभव ऊर्जा आरेखों का उपयोग करके रासायनिक अभिक्रिया की दर पर उत्प्रेरक के प्रभाव की व्याख्या कीजिए।

View Model Solution & Step Marking ▾

Model Answer:

A catalyst provides an alternative pathway or reaction mechanism by reducing the activation energy. It does not change the

\Delta H

of the reaction. The potential energy diagram will show a lower activation energy for the catalyzed reaction compared to the uncatalyzed one, with reactants and products at the same energy levels.

Part B: Long Answer Questions & Derivations (5 Marks Each)

Q1 • 5 Marks Long Answer / Derivation

For a reaction

A + B \rightarrow P

, the rate law is given by

Rate = k[A]^2[B]

. If the concentration of

A

is doubled and the concentration of

B

is halved, how will the rate of the reaction change? Justify your answer. Also, define the order of reaction and molecularity of a reaction. What is the difference between them?

एक अभिक्रिया

A + B \rightarrow P

के लिए, दर नियम

Rate = k[A]^2[B]

द्वारा दिया गया है। यदि

A

की सांद्रता दोगुनी कर दी जाए और

B

की सांद्रता आधी कर दी जाए, तो अभिक्रिया की दर में क्या परिवर्तन होगा? अपने उत्तर का औचित्य सिद्ध कीजिए। इसके अतिरिक्त, अभिक्रिया की कोटि और अभिक्रिया की आण्विकता को परिभाषित कीजिए। उनमें क्या अंतर है?

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

1. Initial rate:

R_1 = k[A]^2[B]

. New concentrations:

[A]' = 2[A]

,

[B]' = \frac{1}{2}[B]

. New rate:

R_2 = k(2[A])^2(\frac{1}{2}[B]) = k(4[A]^2)(\frac{1}{2}[B]) = 2k[A]^2[B] = 2R_1

. The rate of the reaction will double. Justification: Substituting the new concentrations into the rate law shows that the new rate is twice the initial rate.2. Order of reaction: It is the sum of the powers of the concentration terms of the reactants in the experimentally determined rate law. It can be a whole number, a fraction, or zero.Molecularity of a reaction: It is the number of reacting species (atoms, ions, or molecules) taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction. It is always a whole number (1, 2, or 3) and never zero or a fraction.Difference: Order is an experimentally determined value, while molecularity is a theoretical concept. Order can be zero or fractional, molecularity cannot. Order applies to overall reactions, molecularity applies only to elementary reactions.

Q2 • 5 Marks Long Answer / Derivation

The activation energy of a reaction is

200 kJ mol^{-1}

. The rate constant at

400 K

is

1.0 \times 10^{-5} s^{-1}

. Calculate the frequency factor (

A

) for the reaction. (Given:

R = 8.314 J K^{-1} mol^{-1}

). Explain the effect of a catalyst on the activation energy and rate of a reaction with the help of an energy profile diagram.

एक अभिक्रिया की सक्रियण ऊर्जा

200 kJ mol^{-1}

है।

400 K

पर दर स्थिरांक

1.0 \times 10^{-5} s^{-1}

है। अभिक्रिया के लिए आवृत्ति कारक (

A

) की गणना कीजिए। (दिया गया है:

R = 8.314 J K^{-1} mol^{-1}

)। एक ऊर्जा प्रोफ़ाइल आरेख की सहायता से सक्रियण ऊर्जा और अभिक्रिया की दर पर एक उत्प्रेरक के प्रभाव की व्याख्या कीजिए।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

1. Given:

E_a = 200 kJ mol^{-1} = 200 \times 10^3 J mol^{-1}

,

k = 1.0 \times 10^{-5} s^{-1}

,

T = 400 K

,

R = 8.314 J K^{-1} mol^{-1}

.According to the Arrhenius equation,

k = A e^{-E_a/RT}

.Taking natural logarithm on both sides:

\ln k = \ln A - \frac{E_a}{RT}

.Rearranging to find

A

:

\ln A = \ln k + \frac{E_a}{RT}

.

\ln A = \ln(1.0 \times 10^{-5}) + \frac{200 \times 10^3 J mol^{-1}}{8.314 J K^{-1} mol^{-1} \times 400 K}

.

\ln A = -11.5129 + \frac{200000}{3325.6} = -11.5129 + 60.133

.

\ln A = 48.6201

.

A = e^{48.6201} = 1.06 \times 10^{21} s^{-1}

.2. Effect of catalyst: A catalyst increases the rate of a chemical reaction without being consumed in the reaction. It does so by providing an alternative reaction pathway with a lower activation energy. By lowering the activation energy, a larger fraction of reactant molecules possess the minimum energy required to react, leading to an increased rate of reaction. The catalyst does not change the enthalpy of reaction or the equilibrium constant, but only speeds up the attainment of equilibrium. (An energy profile diagram would visually represent this, showing the activation energy of the catalyzed reaction being lower than that of the uncatalyzed reaction, with the reactants and products energy levels remaining the same.)

Q3 • 5 Marks Long Answer / Derivation

For a first order reaction, the time required for

99.9

\% completion is

10

times the half-life (

t_{1/2}

) of the reaction. Justify this statement with appropriate calculations. Also, write two characteristics of a first-order reaction.

एक प्रथम कोटि की अभिक्रिया के लिए,

99.9

\% पूर्ण होने में लगने वाला समय अभिक्रिया के अर्ध-आयु (

t_{1/2}

) का

10

गुना होता है। इस कथन को उचित गणनाओं के साथ न्यायोचित ठहराइए। साथ ही, प्रथम कोटि की अभिक्रिया की दो विशेषताएँ लिखिए।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

For a first order reaction, the integrated rate law is

k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}

. For

99.9

\% completion,

[R] = [R]_0 - 0.999[R]_0 = 0.001[R]_0

. So,

t_{99.9\%} = \frac{2.303}{k} \log \frac{[R]_0}{0.001[R]_0} = \frac{2.303}{k} \log 1000 = \frac{2.303}{k} \times 3 = \frac{6.909}{k}

. The half-life for a first order reaction is

t_{1/2} = \frac{0.693}{k}

. Comparing

t_{99.9\%}

and

t_{1/2}

:

\frac{t_{99.9\%}}{t_{1/2}} = \frac{6.909/k}{0.693/k} = \frac{6.909}{0.693} \approx 9.969 \approx 10

. Thus,

t_{99.9\%} \approx 10 \times t_{1/2}

. Two characteristics of a first-order reaction are: 1. Its half-life is independent of the initial concentration of the reactant. 2. The unit of the rate constant is

s^{-1}

.

Q4 • 5 Marks Long Answer / Derivation

The rate constant for a first order reaction is

60s^{-1}

. How much time will it take to reduce the initial concentration of the reactant to its

\frac{1}{10}

th value? Derive the integrated rate equation for a first order reaction.

एक प्रथम कोटि अभिक्रिया के लिए दर स्थिरांक

60s^{-1}

है। अभिकारक की प्रारंभिक सांद्रता को उसके

\frac{1}{10}

वें मान तक कम होने में कितना समय लगेगा? प्रथम कोटि अभिक्रिया के लिए समाकलित दर समीकरण व्युत्पन्न कीजिए।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

1. Given:

k = 60s^{-1}

. Let initial concentration be

[A]_0

. Final concentration

[A] = \frac{1}{10}[A]_0

.For a first order reaction,

t = \frac{2.303}{k} \log \frac{[A]_0}{[A]}

.Substituting the values:

t = \frac{2.303}{60s^{-1}} \log \frac{[A]_0}{\frac{1}{10}[A]_0} = \frac{2.303}{60} \log 10 = \frac{2.303}{60} \times 1

.

t = 0.03838

seconds.2. Derivation of integrated rate equation for a first order reaction:For a first order reaction,

Rate = -\frac{d[A]}{dt} = k[A]

.Rearranging,

\frac{d[A]}{[A]} = -k dt

.Integrating both sides:

\int \frac{d[A]}{[A]} = -k \int dt

.

\ln[A] = -kt + I

, where

I

is the integration constant.At

t=0

,

[A]=[A]_0

, so

\ln[A]_0 = -k(0) + I \Rightarrow I = \ln[A]_0

.Substituting

I

back:

\ln[A] = -kt + \ln[A]_0

.Rearranging:

\ln[A]_0 - \ln[A] = kt \Rightarrow \ln\frac{[A]_0}{[A]} = kt

.Converting to base 10 logarithm:

2.303 \log\frac{[A]_0}{[A]} = kt

.Hence,

t = \frac{2.303}{k} \log\frac{[A]_0}{[A]}

.

Q5 • 5 Marks Long Answer / Derivation

Distinguish between molecularity and order of a reaction. Explain with an example why the molecularity of a reaction can never be zero or fractional. What is the significance of the rate determining step in a complex reaction?

अभिक्रिया की आणविकता और कोटि के बीच अंतर स्पष्ट कीजिए। एक उदाहरण के साथ समझाइए कि अभिक्रिया की आणविकता कभी भी शून्य या भिन्नात्मक क्यों नहीं हो सकती। एक जटिल अभिक्रिया में दर निर्धारित करने वाले पद का क्या महत्व है?

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

Differences between molecularity and order of a reaction:Molecularity:1. It is the number of reacting species (atoms, ions, or molecules) taking part in an elementary reaction.2. It is a theoretical concept.3. It is always a whole number (1, 2, or 3) and cannot be zero or fractional.4. It is applicable only for elementary reactions.Order of Reaction:1. It is the sum of the powers of the concentration of the reactants in the rate law expression.2. It is an experimental concept.3. It can be a whole number, zero, or even fractional.4. It is applicable for both elementary and complex reactions.Molecularity cannot be zero or fractional because molecularity refers to the number of molecules participating in an elementary step. A reaction cannot occur if no molecules are participating (zero molecularity), and a fraction of a molecule cannot participate in a reaction. For example, in the reaction

2 ext{HI(g)} ightarrow ext{H}_2 ext{(g)} + ext{I}_2 ext{(g)}

, the molecularity is

2

as two

ext{HI}

molecules collide.The significance of the rate determining step:In a complex reaction, which occurs in several elementary steps, the slowest step is called the rate determining step. The overall rate of the reaction is governed by the rate of this slowest step. This is because the products of the slowest step limit the rate at which subsequent steps can proceed. Therefore, the rate law for the overall reaction is determined by the stoichiometry of the rate determining (slowest) step.

4

The d- and f-Block Elements

Part A: Short Answer Questions (2–3 Marks Each)

Q1 • 3 Marks Short Answer

Explain why phenols are more acidic than alcohols, even though both contain an -OH group. Illustrate with relevant resonance structures.

स्पष्ट कीजिए कि फिनोल, ऐल्कोहॉल की तुलना में अधिक अम्लीय क्यों होते हैं, यद्यपि दोनों में -OH समूह होता है। प्रासंगिक अनुनादी संरचनाओं के साथ समझाइए।

View Model Solution & Step Marking ▾

Model Answer:

Phenols are more acidic than alcohols because the phenoxide ion (conjugate base of phenol) is stabilized by resonance. The negative charge on oxygen is delocalized over the benzene ring, making it more stable than the alkoxide ion (conjugate base of alcohol) where the negative charge is localized on the oxygen. Resonance structures of phenoxide ion showing delocalization of negative charge.

Q2 • 3 Marks Short Answer

How will you distinguish between primary, secondary, and tertiary alcohols using the Lucas test? Describe the observations for each.

आप ल्यूकास परीक्षण का उपयोग करके प्राथमिक, द्वितीयक और तृतीयक ऐल्कोहॉलों के बीच कैसे अंतर करेंगे? प्रत्येक के लिए प्रेक्षणों का वर्णन कीजिए।

View Model Solution & Step Marking ▾

Model Answer:

The Lucas test uses a mixture of concentrated

\text{HCl}

and anhydrous

\text{ZnCl}_2

. Tertiary alcohols react immediately to form turbidity. Secondary alcohols react within 5-10 minutes to form turbidity. Primary alcohols do not react at room temperature and remain clear.

Q3 • 3 Marks Short Answer

Write the mechanism for the dehydration of ethanol to ethene in the presence of concentrated

\text{H}_2\text{SO}_4

at

443 \text{ K}

.

सांद्र

\text{H}_2\text{SO}_4

की उपस्थिति में

443 \text{ K}

पर एथेनॉल से एथीन के निर्जलीकरण की क्रियाविधि लिखिए।

View Model Solution & Step Marking ▾

Model Answer:

The mechanism involves three steps: 1. Protonation of alcohol to form protonated alcohol. 2. Formation of carbocation by loss of water. 3. Elimination of a proton to form ethene. (Each step should be shown with proper curved arrows and intermediates).

Q4 • 3 Marks Short Answer

Explain why the boiling points of ethers are much lower than those of alcohols of comparable molar masses.

स्पष्ट कीजिए कि ईथरों के क्वथनांक तुलनीय मोलर द्रव्यमान वाले ऐल्कोहॉलों की तुलना में बहुत कम क्यों होते हैं।

View Model Solution & Step Marking ▾

Model Answer:

Alcohols form intermolecular hydrogen bonds due to the presence of a highly polar -OH group, which requires more energy to break, thus leading to higher boiling points. Ethers, on the other hand, do not have a hydrogen atom directly bonded to the oxygen, so they cannot form intermolecular hydrogen bonds. They only exhibit weaker dipole-dipole interactions and van der Waals forces.

Q5 • 3 Marks Short Answer

Complete the following reaction sequence: \

\text{Phenol} \xrightarrow{\text{i) } \text{NaOH}} \text{A} \xrightarrow{\text{ii) } \text{CO}_2 \text{, } 400 \text{ K, } 4-7 \text{ atm}} \text{B} \xrightarrow{\text{iii) } \text{H}^+} \text{C}

\\Identify A, B, and C and name the overall reaction.

निम्नलिखित अभिक्रिया अनुक्रम को पूर्ण कीजिए: \

\text{फिनोल} \xrightarrow{\text{i) } \text{NaOH}} \text{A} \xrightarrow{\text{ii) } \text{CO}_2 \text{, } 400 \text{ K, } 4-7 \text{ atm}} \text{B} \xrightarrow{\text{iii) } \text{H}^+} \text{C}

\\A, B और C को पहचानिए तथा समग्र अभिक्रिया का नाम बताइए।

View Model Solution & Step Marking ▾

Model Answer:

A = Sodium phenoxide \\B = Sodium salicylate \\C = Salicylic acid \\Overall reaction is Kolbe's Reaction (Kolbe's synthesis).

Q6 • 3 Marks Short Answer

Explain why phenol is more acidic than ethanol, even though both contain an -OH group. Provide relevant resonance structures for phenol to support your answer.

स्पष्ट कीजिए कि फिनोल, इथेनॉल से अधिक अम्लीय क्यों होता है, जबकि दोनों में -OH समूह होता है। अपने उत्तर के समर्थन में फिनोल की अनुनादी संरचनाएँ प्रस्तुत कीजिए।

View Model Solution & Step Marking ▾

Model Answer:

Phenol is more acidic than ethanol because the phenoxide ion, formed after losing a proton, is stabilized by resonance. The negative charge is delocalized over the benzene ring, making it more stable than the ethoxide ion, which has no such resonance stabilization. Ethanol's ethoxide ion is destabilized by the electron-donating ethyl group. Resonance structures for phenoxide ion show delocalization of negative charge at ortho and para positions.

Q7 • 3 Marks Short Answer

How will you synthesize 1-phenyl ethanol from a suitable carbonyl compound using a Grignard reagent? Write the chemical equations involved.

आप ग्रिगनार्ड अभिकर्मक का उपयोग करके उपयुक्त कार्बोनिल यौगिक से 1-फेनिल इथेनॉल का संश्लेषण कैसे करेंगे? इसमें शामिल रासायनिक समीकरण लिखिए।

View Model Solution & Step Marking ▾

Model Answer:

1-Phenyl ethanol can be synthesized by reacting benzaldehyde with methylmagnesium bromide (Grignard reagent), followed by hydrolysis.

C_6H_5CHO + CH_3MgBr \rightarrow C_6H_5CH(OMgBr)CH_3

(Adduct formation)

C_6H_5CH(OMgBr)CH_3 + H_2O \rightarrow C_6H_5CH(OH)CH_3 + Mg(OH)Br

(Hydrolysis to form 1-phenyl ethanol)

Q8 • 3 Marks Short Answer

Explain the mechanism of dehydration of ethanol to ethene in the presence of concentrated sulfuric acid at

443 K

.

सांद्र सल्फ्यूरिक अम्ल की उपस्थिति में

443 K

पर इथेनॉल के एथीन में निर्जलीकरण की क्रियाविधि समझाइए।

View Model Solution & Step Marking ▾

Model Answer:

The mechanism involves three steps:
1. Protonation of alcohol: The oxygen atom of ethanol gets protonated by

H_2SO_4

to form protonated alcohol.
2. Formation of carbocation: The protonated alcohol loses a water molecule to form a carbocation.
3. Deprotonation: The carbocation loses a proton to form ethene. The proton lost is accepted by

HSO_4^-

to regenerate

H_2SO_4

.

Q9 • 3 Marks Short Answer

Give a chemical test to distinguish between methanol and phenol. Write the chemical equations for the reactions involved.

मेथनॉल और फिनोल के बीच अंतर करने के लिए एक रासायनिक परीक्षण दीजिए। इसमें शामिल अभिक्रियाओं के रासायनिक समीकरण लिखिए।

View Model Solution & Step Marking ▾

Model Answer:

Ferric chloride test: Phenol gives a characteristic violet, blue, or green coloration with neutral ferric chloride solution due to the formation of a complex. Methanol does not give this test.
Reaction:

6C_6H_5OH + FeCl_3 \rightarrow [Fe(OC_6H_5)_6]^{3-} + 3H^+ + 3HCl

(Violet/Blue/Green complex)
Methanol: No reaction with

FeCl_3

.

Q10 • 3 Marks Short Answer

How are ethers prepared by Williamson's synthesis? Explain with a suitable example and write the general reaction. What is the limitation of this method for preparing unsymmetrical ethers?

विलियमसन संश्लेषण द्वारा ईथर कैसे तैयार किए जाते हैं? एक उपयुक्त उदाहरण के साथ समझाइए और सामान्य अभिक्रिया लिखिए। असममित ईथर तैयार करने के लिए इस विधि की क्या सीमा है?

View Model Solution & Step Marking ▾

Model Answer:

Williamson's synthesis involves the reaction of an alkyl halide with sodium alkoxide or sodium phenoxide. It is an

S_N2

reaction.
General reaction:

R-X + R'-ONa \rightarrow R-O-R' + NaX

Example:

CH_3CH_2Br + CH_3ONa \rightarrow CH_3CH_2OCH_3 + NaBr

(Methoxyethane)
Limitation: For unsymmetrical ethers, if a primary alkoxide is reacted with a tertiary alkyl halide, elimination (

E2

) often competes with substitution (

S_N2

), leading to the formation of alkenes as major products. Therefore, for unsymmetrical ethers, the alkyl halide should be primary.

Part B: Long Answer Questions & Derivations (5 Marks Each)

Q1 • 5 Marks Long Answer / Derivation

a) Explain the mechanism of dehydration of ethanol to ethene using concentrated sulfuric acid at

443 K

.b) How will you distinguish between propan-1-ol and propan-2-ol using a chemical test? Write the chemical equations involved.

क)

443 K

पर सांद्र सल्फ्यूरिक अम्ल का उपयोग करके इथेनॉल से एथीन के निर्जलीकरण की क्रियाविधि समझाइए।ख) प्रोपेन-1-ऑल और प्रोपेन-2-ऑल के बीच रासायनिक परीक्षण का उपयोग करके आप कैसे अंतर करेंगे? इसमें शामिल रासायनिक समीकरण लिखिए।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Mechanism of dehydration of ethanol to ethene:1. Protonation of ethanol:

CH_3CH_2OH + H^+ \rightleftharpoons CH_3CH_2O^+H_2

2. Formation of carbocation:

CH_3CH_2O^+H_2 \rightarrow CH_3CH_2^+ + H_2O

3. Deprotonation to form ethene:

CH_3CH_2^+ \rightarrow CH_2=CH_2 + H^+

b) Distinction between propan-1-ol and propan-2-ol using Lucas Test:Propan-1-ol (primary alcohol) does not react with Lucas reagent (

HCl/ZnCl_2

) at room temperature, no turbidity appears.Propan-2-ol (secondary alcohol) reacts with Lucas reagent to form turbidity within 5-10 minutes.Equations:Propan-1-ol:

CH_3CH_2CH_2OH + HCl \xrightarrow{ZnCl_2} No\ reaction

Propan-2-ol:

CH_3CH(OH)CH_3 + HCl \xrightarrow{ZnCl_2} CH_3CHClCH_3 + H_2O

Q2 • 5 Marks Long Answer / Derivation

a) Arrange the following compounds in increasing order of their acidic strength and justify your answer: Phenol, 4-Nitrophenol, 4-Methylphenol.b) Explain why phenols are more acidic than alcohols, even though both have an -OH group.

क) निम्नलिखित यौगिकों को उनकी अम्लीय शक्ति के बढ़ते क्रम में व्यवस्थित करें और अपने उत्तर का औचित्य सिद्ध करें: फिनोल, 4-नाइट्रोफिनोल, 4-मेथिलफिनोल।ख) समझाइए कि फिनोल अल्कोहल की तुलना में अधिक अम्लीय क्यों होते हैं, भले ही दोनों में -OH समूह होता है।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Increasing order of acidic strength: 4-Methylphenol < Phenol < 4-Nitrophenol.Justification: Acidity of phenols is due to the stability of phenoxide ion formed after losing a proton. Electron-withdrawing groups (like

-NO_2

) stabilize the phenoxide ion by delocalization of negative charge, thus increasing acidity. Electron-donating groups (like

-CH_3

) destabilize the phenoxide ion, thus decreasing acidity. Hence, 4-Nitrophenol is most acidic, and 4-Methylphenol is least acidic among the three.b) Phenols are more acidic than alcohols because the phenoxide ion formed after losing a proton is resonance stabilized. The negative charge on oxygen is delocalized over the benzene ring, making the phenoxide ion more stable than the alkoxide ion. In alcohols, the alkoxide ion has the negative charge localized on the oxygen, which is not resonance stabilized, making it less stable and thus alcohols less acidic.

Q3 • 5 Marks Long Answer / Derivation

a) Explain the mechanism of the acid-catalyzed hydration of propene to form propan-2-ol.b) Write the main product(s) formed when anisole undergoes reaction with:i)

HI

(hot and concentrated)ii) Bromine in ethanoic acid medium.

क) प्रोपीन के अम्ल-उत्प्रेरित जलयोजन से प्रोपेन-2-ऑल बनने की क्रियाविधि समझाइए।ख) जब एनिसोल निम्नलिखित के साथ अभिक्रिया करता है तो बनने वाले मुख्य उत्पाद लिखिए:i)

HI

(गर्म और सांद्र)ii) एथेनोइक अम्ल माध्यम में ब्रोमीन।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Mechanism of acid-catalyzed hydration of propene to form propan-2-ol:1. Protonation of alkene:

CH_3CH=CH_2 + H^+ \rightleftharpoons CH_3CH^+CH_3

(Markovnikov's rule followed to form more stable carbocation).2. Nucleophilic attack by water:

CH_3CH^+CH_3 + H_2O \rightarrow CH_3CH(O^+H_2)CH_3

3. Deprotonation:

CH_3CH(O^+H_2)CH_3 \rightarrow CH_3CH(OH)CH_3 + H^+

b) Main product(s) formed when anisole reacts:i) With

HI

(hot and concentrated): Anisole is an ether, and

HI

cleaves it. Since one group is methyl and the other is phenyl, the cleavage occurs such that the smaller alkyl group forms alkyl iodide, and the phenyl group forms phenol. Products: Phenol and Methyl iodide.

C_6H_5OCH_3 + HI \xrightarrow{heat} C_6H_5OH + CH_3I

ii) With Bromine in ethanoic acid medium: Anisole is an activated aromatic compound due to the

-OCH_3

group (an ortho-para directing group). Bromination occurs at ortho and para positions. Products: 2-Bromoanisole and 4-Bromoanisole (major product).

Q4 • 5 Marks Long Answer / Derivation

a) Explain why phenols are more acidic than alcohols, despite both containing an -OH group. (3 marks)
b) Give a chemical test to distinguish between phenol and ethanol. (2 marks)||HI||a) समझाइए कि फिनोल अल्कोहल की तुलना में अधिक अम्लीय क्यों होते हैं, जबकि दोनों में

-\text{OH}

समूह होता है। (3 अंक)
b) फिनोल और इथेनॉल के बीच अंतर करने के लिए एक रासायनिक परीक्षण दीजिए। (2 अंक)

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Phenols are more acidic than alcohols due to the resonance stabilization of the phenoxide ion formed after the loss of a proton. In phenol, the lone pair of electrons on the oxygen atom participates in resonance with the benzene ring, delocalizing the negative charge over the ring. This stabilization makes the removal of

H^+

easier. In contrast, for alcohols, the alkoxide ion formed is not resonance stabilized; instead, the alkyl group's electron-donating inductive effect destabilizes the alkoxide ion, making alcohols weaker acids. (3 marks for clear explanation of resonance stabilization in phenoxide ion and lack thereof in alkoxide ion)
b) A common chemical test is the Ferric Chloride Test. Phenol reacts with neutral ferric chloride solution to give a characteristic violet/purple/green coloration due to the formation of a complex, while ethanol does not show any such color change. (1 mark for test name, 1 mark for observation/explanation).||HI||a) फिनोल अल्कोहल की तुलना में अधिक अम्लीय होते हैं क्योंकि प्रोटॉन के निष्कासन के बाद बनने वाले फिनोक्साइड आयन का अनुनाद स्थिरीकरण होता है। फिनोल में, ऑक्सीजन परमाणु पर एकाकी इलेक्ट्रॉन युग्म बेंजीन वलय के साथ अनुनाद में भाग लेता है, जिससे ऋणात्मक आवेश वलय पर विस्थानीकृत हो जाता है। यह स्थिरीकरण

H^+

के निष्कासन को आसान बनाता है। इसके विपरीत, अल्कोहलों के लिए, बनने वाला एल्कोक्साइड आयन अनुनाद द्वारा स्थिर नहीं होता है; इसके बजाय, एल्किल समूह का इलेक्ट्रॉन-दाता प्रेरणिक प्रभाव एल्कोक्साइड आयन को अस्थिर करता है, जिससे अल्कोहल दुर्बल अम्ल होते हैं। (फिनोक्साइड आयन में अनुनाद स्थिरीकरण की स्पष्ट व्याख्या और एल्कोक्साइड आयन में इसकी कमी के लिए 3 अंक)
b) एक सामान्य रासायनिक परीक्षण फेरिक क्लोराइड परीक्षण है। फिनोल उदासीन फेरिक क्लोराइड विलयन के साथ अभिक्रिया करके एक संकुल के निर्माण के कारण एक विशिष्ट बैंगनी/जामुनी/हरा रंग देता है, जबकि इथेनॉल ऐसा कोई रंग परिवर्तन नहीं दिखाता है। (परीक्षण के नाम के लिए 1 अंक, अवलोकन/व्याख्या के लिए 1 अंक)।

Q5 • 5 Marks Long Answer / Derivation

a) Explain the mechanism of hydration of ethene to form ethanol in the presence of an acid catalyst. (3 marks)b) Give a chemical test to distinguish between ethanol and phenol. (2 marks)||HHI||a) एथीन के अम्ल उत्प्रेरित जलयोजन द्वारा एथेनॉल के निर्माण की क्रियाविधि समझाइए। (3 अंक)b) एथेनॉल और फीनॉल में विभेद करने के लिए एक रासायनिक परीक्षण दीजिए। (2 अंक)

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Mechanism of hydration of ethene to form ethanol:1. Protonation of ethene to form carbocation:

CH_2=CH_2 + H_3O^+ \rightleftharpoons CH_3-CH_2^+ + H_2O

2. Nucleophilic attack by water on carbocation:

CH_3-CH_2^+ + H_2O \rightleftharpoons CH_3-CH_2-OH_2^+

3. Deprotonation to form ethanol:

CH_3-CH_2-OH_2^+ + H_2O \rightleftharpoons CH_3-CH_2-OH + H_3O^+

(Each step 1 mark)b) Chemical test to distinguish between ethanol and phenol:Phenol gives a characteristic violet/purple colour with neutral ferric chloride solution, while ethanol does not. (2 marks)Alternative: Phenol reacts with bromine water to give a white precipitate of 2,4,6-tribromophenol, while ethanol does not. (2 marks)||HHI||a) एथीन के एथेनॉल में जलयोजन की क्रियाविधि:1. एथीन का प्रोटोनीकरण करके कार्बधनायन बनाना:

CH_2=CH_2 + H_3O^+ \rightleftharpoons CH_3-CH_2^+ + H_2O

2. कार्बधनायन पर जल द्वारा नाभिकस्नेही आक्रमण:

CH_3-CH_2^+ + H_2O \rightleftharpoons CH_3-CH_2-OH_2^+

3. एथेनॉल बनाने के लिए विप्रोटोनीकरण:

CH_3-CH_2-OH_2^+ + H_2O \rightleftharpoons CH_3-CH_2-OH + H_3O^+

(प्रत्येक चरण के लिए 1 अंक)b) एथेनॉल और फीनॉल में विभेद करने के लिए रासायनिक परीक्षण:फीनॉल उदासीन फेरिक क्लोराइड विलयन के साथ एक विशिष्ट बैंगनी/जामुनी रंग देता है, जबकि एथेनॉल नहीं देता है। (2 अंक)वैकल्पिक: फीनॉल ब्रोमीन जल के साथ अभिक्रिया करके 2,4,6-ट्राइब्रोमोफीनॉल का सफेद अवक्षेप देता है, जबकि एथेनॉल नहीं देता है। (2 अंक)

5

Coordination Compounds

Part A: Short Answer Questions (2–3 Marks Each)

Q1 • 3 Marks Short Answer

Explain the terms 'ligand' and 'coordination number' with suitable examples.

'लिगेंड' और 'उपसहसंयोजन संख्या' पदों को उपयुक्त उदाहरणों सहित समझाइए।

View Model Solution & Step Marking ▾

Model Answer:

Ligand: A ligand is an ion or molecule capable of donating a pair of electrons to the central metal atom or ion to form a coordinate bond. Example:

NH_3

,

Cl^-

,

H_2O

. (1.5 marks)
Coordination Number: The coordination number of a central metal atom or ion in a complex is the number of ligand donor atoms to which the metal is directly bonded. Example: In

[Co(NH_3)_6]^{3+}

, the coordination number of

Co

is 6. (1.5 marks)

Q2 • 3 Marks Short Answer

Using Valence Bond Theory (VBT), explain the magnetic behavior and geometry of the complex ion

[NiCl_4]^{2-}

. (Atomic number of

Ni = 28

)

संयोजकता बंध सिद्धांत (VBT) का उपयोग करके संकुल आयन

[NiCl_4]^{2-}

के चुंबकीय व्यवहार और ज्यामिति की व्याख्या कीजिए। (

Ni

का परमाणु क्रमांक

= 28

)

View Model Solution & Step Marking ▾

Model Answer:

In

[NiCl_4]^{2-}

,

Ni

is in

+2

oxidation state. The electronic configuration of

Ni^{2+}

is

3d^8

.

Cl^-

is a weak field ligand, so it does not cause pairing of electrons in

3d

orbitals. The hybridization is

sp^3

, leading to a tetrahedral geometry. Due to the presence of two unpaired electrons, the complex is paramagnetic.

Q3 • 3 Marks Short Answer

Draw the structures of all possible isomers for the complex ion

[Co(en)_2Cl_2]^+. (en = ethane-1,2-diamine)

Identify and name the type of isomerism exhibited.

[Co(en)_2Cl_2]^+. (en = इथेन-1,2-डाईएमीन)

संकुल आयन के सभी संभावित समावयवों की संरचनाएँ बनाइए। प्रदर्शित समावयवता के प्रकार को पहचानिए और उसका नाम लिखिए।

View Model Solution & Step Marking ▾

Model Answer:

The complex ion

[Co(en)_2Cl_2]^+

exhibits geometrical isomerism. It has two geometrical isomers: cis and trans.
1. cis-

[Co(en)_2Cl_2]^+

: In this isomer, the two chlorine ligands are adjacent to each other.
2. trans-

[Co(en)_2Cl_2]^+

: In this isomer, the two chlorine ligands are opposite to each other.
(Structures should be drawn correctly for both cis and trans isomers).

Q4 • 3 Marks Short Answer

Write the IUPAC names for the following coordination compounds:
(i)

K_3[Fe(C_2O_4)_3]

(ii)

[Co(NH_3)_5Cl]Cl_2

निम्नलिखित उपसहसंयोजक यौगिकों के IUPAC नाम लिखिए:
(i)

K_3[Fe(C_2O_4)_3]

(ii)

[Co(NH_3)_5Cl]Cl_2

View Model Solution & Step Marking ▾

Model Answer:

(i) Potassium trioxalatoferrate(III)
(ii) Pentaamminechlorocobalt(III) chloride

Q5 • 3 Marks Short Answer

Describe crystal field splitting in an octahedral complex. How does the presence of strong field and weak field ligands affect the splitting energy (

\Delta_o

)?

एक अष्टफलकीय संकुल में क्रिस्टल क्षेत्र विपाटन का वर्णन कीजिए। प्रबल क्षेत्र और दुर्बल क्षेत्र लिगेंड की उपस्थिति विपाटन ऊर्जा (

\Delta_o

) को कैसे प्रभावित करती है?

View Model Solution & Step Marking ▾

Model Answer:

In an octahedral complex, the five degenerate

d

-orbitals split into two sets: three lower energy

t_{2g}

orbitals (

d_{xy}, d_{yz}, d_{zx}

) and two higher energy

e_g

orbitals (

d_{x^2-y^2}, d_{z^2}

). The energy difference between these two sets is called crystal field splitting energy (

\Delta_o

). Strong field ligands cause a large splitting (large

\Delta_o

), leading to pairing of electrons in

t_{2g}

orbitals (low spin complexes). Weak field ligands cause small splitting (small

\Delta_o

), leading to electrons occupying

e_g

orbitals before pairing in

t_{2g}

(high spin complexes).

Q6 • 3 Marks Short Answer

Explain the difference between a double salt and a coordination compound with suitable examples.

उपयुक्त उदाहरणों के साथ द्विक लवण और उपसहसंयोजन यौगिक के बीच अंतर स्पष्ट कीजिए।

View Model Solution & Step Marking ▾

Model Answer:

Double salts dissociate into their constituent ions in solution, losing their identity (e.g., Carnallite, Mohr's salt). Coordination compounds retain their identity in solution, even though they contain complex ions (e.g.,

[Fe(CN)_6]^{4-}

,

[Cu(NH_3)_4]^{2+}

).

Q7 • 3 Marks Short Answer

For the complex ion

[Co(en)_2Cl_2]^+

(where 'en' is ethane-1,2-diamine), draw all possible stereoisomers and identify the type of isomerism exhibited.

संकुल आयन

[Co(en)_2Cl_2]^+

('en' एथेन-1,2-डाइऐमीन है) के लिए, सभी संभावित त्रिविम समावयवों को आरेखित कीजिए और प्रदर्शित समावयवता के प्रकार की पहचान कीजिए।

View Model Solution & Step Marking ▾

Model Answer:

The complex exhibits geometrical (cis-trans) isomerism and optical isomerism. Cis-isomer will show optical isomerism (enantiomers), while trans-isomer will be optically inactive (meso compound if chiral centers are present, but here it's achiral). (Diagrams of cis and trans isomers, and enantiomers of cis-isomer are expected).

Q8 • 3 Marks Short Answer

Using Valence Bond Theory, explain the hybridization, geometry, and magnetic behaviour of the complex ion

[NiCl_4]^{2-}

. (Atomic number of Ni = 28).

संयोजकता बंध सिद्धांत का उपयोग करके, संकुल आयन

[NiCl_4]^{2-}

के संकरण, ज्यामिति और चुंबकीय व्यवहार की व्याख्या कीजिए। (Ni का परमाणु क्रमांक = 28)।

View Model Solution & Step Marking ▾

Model Answer:

Ni is in

+2

oxidation state. Electronic configuration of

Ni^{2+}

is

[Ar]3d^8

.

Cl^-

is a weak field ligand. Therefore, no pairing of electrons occurs. Hybridization is

sp^3

. Geometry is tetrahedral. Due to two unpaired electrons, it is paramagnetic.

Q9 • 3 Marks Short Answer

Explain why carbon monoxide (CO) is a stronger ligand than ammonia (

NH_3

) in forming coordination compounds with transition metals, based on Crystal Field Theory.

क्रिस्टल क्षेत्र सिद्धांत के आधार पर समझाइए कि संक्रमण धातुओं के साथ उपसहसंयोजन यौगिक बनाने में कार्बन मोनोऑक्साइड (CO) अमोनिया (

NH_3

) की तुलना में एक प्रबल लिगेंड क्यों है।

View Model Solution & Step Marking ▾

Model Answer:

According to Crystal Field Theory, CO is a strong field ligand because it can form

\pi

-backbonding with the metal. This

\pi

-backbonding enhances the metal-ligand bond strength and increases the crystal field splitting energy (

\Delta_o

), making it a stronger ligand than

NH_3

which only forms

\sigma

-bonds.

Q10 • 3 Marks Short Answer

Write the IUPAC names for the following coordination compounds: (i)

[Cr(NH_3)_3Cl_3]

and (ii)

K_3[Fe(CN)_6]

.

निम्नलिखित उपसहसंयोजन यौगिकों के IUPAC नाम लिखिए: (i)

[Cr(NH_3)_3Cl_3]

और (ii)

K_3[Fe(CN)_6]

।

View Model Solution & Step Marking ▾

Model Answer:

(i) Triamminetrichlorochromium(III) (ii) Potassium hexacyanoferrate(III)

Part B: Long Answer Questions & Derivations (5 Marks Each)

Q1 • 5 Marks Long Answer / Derivation

a) State the primary and secondary valencies of the central metal ion in the coordination compound

K_3[Co(C_2O_4)_3]

.b) Differentiate between double salts and coordination compounds with two suitable examples for each.c) Write the IUPAC name for the following coordination compounds:i)

[Co(NH_3)_5Cl]Cl_2

ii)

K_3[Fe(CN)_6]

a) उपसहसंयोजन यौगिक

K_3[Co(C_2O_4)_3]

में केंद्रीय धातु आयन की प्राथमिक और द्वितीयक संयोजकताएँ बताइए।b) द्विक लवणों और उपसहसंयोजन यौगिकों के बीच अंतर को प्रत्येक के दो उपयुक्त उदाहरणों सहित स्पष्ट कीजिए।c) निम्नलिखित उपसहसंयोजन यौगिकों के IUPAC नाम लिखिए:i)

[Co(NH_3)_5Cl]Cl_2

ii)

K_3[Fe(CN)_6]

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) In

K_3[Co(C_2O_4)_3]

:
Primary valency (Oxidation state of Co) =

+3

Secondary valency (Coordination number of Co) =

6

(since oxalate is a bidentate ligand,

3 \times 2 = 6

).
b) Double salts dissociate completely into their constituent ions in solution, losing their identity. For example, Mohr's salt (

FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O

) and Carnallite (

KCl \cdot MgCl_2 \cdot 6H_2O

).
Coordination compounds retain their identity in solution and do not dissociate completely into all constituent ions. They form complex ions. For example,

[Cu(NH_3)_4]SO_4

and

K_4[Fe(CN)_6]

.
c) i) Pentaamminechloridocobalt(III) chloride
ii) Potassium hexacyanoferrate(III)

Q2 • 5 Marks Long Answer / Derivation

Using Valence Bond Theory (VBT), explain the bonding, hybridization, magnetic nature, and geometry of the complex ion

[Ni(CN)_4]^{2-}

. (Given: Atomic number of Ni =

28

).

संयोजकता बंध सिद्धांत (VBT) का उपयोग करते हुए, संकुल आयन

[Ni(CN)_4]^{2-}

में बंधन, संकरण, चुंबकीय प्रकृति और ज्यामिति की व्याख्या कीजिए। (दिया गया है: Ni का परमाणु क्रमांक =

28

)।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

1. **Central metal ion and its oxidation state:** The central metal ion is Ni. Let its oxidation state be

x

.

x + 4(-1) = -2 \implies x = +2

. So, it is

Ni^{2+}

ion.
2. **Electronic configuration of

Ni^{2+}

:** Atomic number of Ni =

28

, so its configuration is

[Ar]3d^84s^2

. For

Ni^{2+}

, it is

[Ar]3d^8

.
3. **Ligand type:**

CN^-

(cyanide) is a strong field ligand. It causes pairing of electrons in the

3d

orbitals.
4. **Hybridization:** The

3d^8

electrons pair up, leaving one

3d

orbital vacant. One

3d

, one

4s

, and two

4p

orbitals hybridize to form four

dsp^2

hybrid orbitals.
5. **Geometry:** The

dsp^2

hybridization leads to square planar geometry.
6. **Magnetic nature:** Since all electrons are paired after the strong field ligand

CN^-

causes pairing, the complex is diamagnetic (zero unpaired electrons).

Q3 • 5 Marks Long Answer / Derivation

Using Werner's theory of coordination compounds, explain the postulates and their implications for the compound

CoCl_3 \cdot 6NH_3

. How would you represent its structure based on Werner's theory, and what would be the number of ions produced when it dissolves in water?

उपसहसंयोजन यौगिकों के वर्नर सिद्धांत का उपयोग करते हुए, यौगिक

CoCl_3 \cdot 6NH_3

के लिए अभिधारणाओं और उनके निहितार्थों की व्याख्या कीजिए। वर्नर सिद्धांत के आधार पर आप इसकी संरचना को कैसे निरूपित करेंगे, और जब यह पानी में घुलता है तो उत्पन्न होने वाले आयनों की संख्या क्या होगी?

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

Werner's theory postulates that metal ions exhibit two types of valencies: primary (ionizable, satisfied by anions, corresponds to oxidation state) and secondary (non-ionizable, satisfied by ligands, corresponds to coordination number). In

CoCl_3 \cdot 6NH_3

, Cobalt (Co) has a primary valency of

+3

and a secondary valency of

6

. The six ammonia molecules (

NH_3

) satisfy the secondary valency, forming the coordination sphere

[Co(NH_3)_6]^{3+}

. The three chloride ions (

Cl^-

) satisfy the primary valency and are outside the coordination sphere.
Structure representation:

[Co(NH_3)_6]Cl_3

.
When dissolved in water, it ionizes as:

[Co(NH_3)_6]Cl_3 \rightarrow [Co(NH_3)_6]^{3+} + 3Cl^-

.
Total number of ions produced =

1

complex ion

+ 3

chloride ions

= 4

ions.

Q4 • 5 Marks Long Answer / Derivation

(a) For the complex ion

[Fe(CN)_6]^{4-}

, determine the following:
(i) IUPAC name
(ii) Oxidation state of the central metal ion
(iii) Coordination number of the central metal ion
(iv) Magnetic behaviour
(v) Hybridisation of the central metal ion (3 marks)
(b) Draw the structures of geometrical isomers for

[Co(en)_2Cl_2]^+

(where 'en' is ethane-1,2-diamine). (2 marks)

(a) संकुल आयन

[Fe(CN)_6]^{4-}

के लिए निम्नलिखित का निर्धारण कीजिए:
(i) IUPAC नाम
(ii) केंद्रीय धातु आयन की ऑक्सीकरण अवस्था
(iii) केंद्रीय धातु आयन की उपसहसंयोजन संख्या
(iv) चुंबकीय व्यवहार
(v) केंद्रीय धातु आयन का संकरण (3 अंक)
(b)

[Co(en)_2Cl_2]^+

(जहाँ 'en' इथेन-1,2-डाइऐमीन है) के ज्यामितीय समावयवों की संरचनाएँ बनाइए। (2 अंक)

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

(a) For

[Fe(CN)_6]^{4-}

:
(i) IUPAC name: Hexacyanidoferrate(II) ion
(ii) Oxidation state of Fe:

+2

(iii) Coordination number of Fe:

6

(iv) Magnetic behaviour: Diamagnetic (due to strong field ligand

CN^-

, all electrons are paired)
(v) Hybridisation of Fe:

d^2sp^3

(b) Geometrical isomers of

[Co(en)_2Cl_2]^+

:
Cis-isomer: Both

Cl

ligands are adjacent to each other.
Trans-isomer: Both

Cl

ligands are opposite to each other.

Q5 • 5 Marks Long Answer / Derivation

(a) Explain Werner's Postulates for coordination compounds with suitable examples. (3 marks)
(b) Differentiate between primary and secondary valencies. (2 marks)

(a) उपयुक्त उदाहरणों के साथ उपसहसंयोजन यौगिकों के लिए वर्नर के अभिगृहीतों की व्याख्या कीजिए। (3 अंक)
(b) प्राथमिक और द्वितीयक संयोजकता के बीच अंतर स्पष्ट कीजिए। (2 अंक)

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

(a) Werner's Postulates:
1. Metal ions exhibit two types of valencies: primary (ionizable) and secondary (non-ionizable).
2. Primary valencies are satisfied by negative ions and correspond to the oxidation state of the metal.
3. Secondary valencies are satisfied by neutral molecules or negative ions and correspond to the coordination number of the metal.
4. The secondary valencies are directed towards fixed positions in space, giving a definite geometry to the coordination compound.
Example: In

[Co(NH_3)_6]Cl_3

, primary valency is

+3

(satisfied by

3 Cl^-

ions) and secondary valency is

6

(satisfied by

6 NH_3

molecules).
(b) Differentiation:
- Primary valency: Ionizable, corresponds to oxidation state, satisfied by negative ions, non-directional.
- Secondary valency: Non-ionizable, corresponds to coordination number, satisfied by ligands (neutral or negative), directional, determines geometry.

6

Haloalkanes And Haloarenes

Part A: Short Answer Questions (2–3 Marks Each)

Q1 • 3 Marks Short Answer

Explain why haloalkanes are more reactive towards nucleophilic substitution reactions than haloarenes. Illustrate with an example.

स्पष्ट कीजिए कि हेलोऐल्केन, हेलोएरीन की तुलना में नाभिकरागी प्रतिस्थापन अभिक्रियाओं के प्रति अधिक अभिक्रियाशील क्यों होते हैं। एक उदाहरण सहित समझाइए।

View Model Solution & Step Marking ▾

Model Answer:

Haloalkanes have a partially positive carbon atom bonded to a halogen, making it susceptible to nucleophilic attack. In haloarenes, the C-X bond acquires partial double bond character due to resonance, making it stronger and less reactive. Also, the carbon atom is

sp^2

hybridized in haloarenes, making it more electronegative and holding the electron pair more tightly. Example:

CH_3Cl + OH^- \rightarrow CH_3OH + Cl^-

(haloalkane) vs. Chlorobenzene +

OH^-

under harsh conditions.

Q2 • 3 Marks Short Answer

Complete the following reactions and identify the major product formed in each case:
(a)

CH_3CH_2Br + KOH(aq) \rightarrow

(b)

CH_3CH_2Br + KOH(alc) \rightarrow

निम्नलिखित अभिक्रियाओं को पूर्ण कीजिए और प्रत्येक स्थिति में बनने वाले मुख्य उत्पाद की पहचान कीजिए:
(a)

CH_3CH_2Br + KOH(aq) \rightarrow

(b)

CH_3CH_2Br + KOH(alc) \rightarrow

View Model Solution & Step Marking ▾

Model Answer:

(a)

CH_3CH_2Br + KOH(aq) \rightarrow CH_3CH_2OH + KBr

(Ethanol)
(b)

CH_3CH_2Br + KOH(alc) \rightarrow CH_2=CH_2 + KBr + H_2O

(Ethene)

Q3 • 3 Marks Short Answer

How will you convert chlorobenzene to phenol? Write the chemical equations involved. Name the reaction if any.

आप क्लोरोबेंजीन को फिनोल में कैसे परिवर्तित करेंगे? इसमें शामिल रासायनिक समीकरण लिखिए। यदि कोई अभिक्रिया हो तो उसका नाम बताइए।

View Model Solution & Step Marking ▾

Model Answer:

Chlorobenzene can be converted to phenol by Dow's process.

C_6H_5Cl + NaOH \xrightarrow{623K, 300atm} C_6H_5ONa + HCl

C_6H_5ONa + H^+ \rightarrow C_6H_5OH + Na^+

This reaction is known as Dow's process.

Q4 • 3 Marks Short Answer

Identify the chirality of the following molecules. Which one would show optical activity?
(a) 2-chloropropane
(b) 2-chlorobutane
(c) 1-bromopropane

निम्नलिखित अणुओं की काइरलता की पहचान कीजिए। इनमें से कौन प्रकाशिक सक्रियता दिखाएगा?
(a) 2-क्लोरोप्रोपेन
(b) 2-क्लोरोब्यूटेन
(c) 1-ब्रोमोप्रोपेन

View Model Solution & Step Marking ▾

Model Answer:

(a) 2-chloropropane: Achiral. Does not show optical activity.
(b) 2-chlorobutane: Chiral. Shows optical activity.
(c) 1-bromopropane: Achiral. Does not show optical activity.

Q5 • 3 Marks Short Answer

Describe the Wurtz-Fittig reaction. Write the general chemical equation for this reaction.

वुर्ट्ज़-फिटिग अभिक्रिया का वर्णन कीजिए। इस अभिक्रिया के लिए सामान्य रासायनिक समीकरण लिखिए।

View Model Solution & Step Marking ▾

Model Answer:

The Wurtz-Fittig reaction is a coupling reaction that involves treating a mixture of an alkyl halide and an aryl halide with sodium metal in dry ether to give an alkylarene.
General equation:

RX + R'X' + 2Na \xrightarrow{dry \ ether} R-R' + 2NaX

(where R is alkyl and R' is aryl).

Q6 • 3 Marks Short Answer

Explain why haloarenes are less reactive towards nucleophilic substitution reactions compared to haloalkanes. Give two reasons.

स्पष्ट कीजिए कि हैलोऐरीन, हैलोऐल्केन की तुलना में नाभिकरागी प्रतिस्थापन अभिक्रियाओं के प्रति कम अभिक्रियाशील क्यों होते हैं। दो कारण दीजिए।

View Model Solution & Step Marking ▾

Model Answer:

Reasons for lower reactivity of haloarenes: 1. Resonance effect: The C-X bond acquires partial double bond character due to resonance, making it shorter and stronger, hence difficult to cleave. 2. Difference in hybridization of carbon atom: The carbon atom attached to halogen in haloarene is

sp^2

hybridized, which is more electronegative than

sp^3

hybridized carbon in haloalkane, holding the electron pair of C-X bond more tightly.

Q7 • 3 Marks Short Answer

How will you distinguish between

CH_3CH_2Br

and

CH_2=CHBr

using a simple chemical test? Write the chemical equation for the reaction involved.

आप एक साधारण रासायनिक परीक्षण का उपयोग करके

CH_3CH_2Br

और

CH_2=CHBr

के बीच कैसे अंतर करेंगे? इसमें शामिल अभिक्रिया के लिए रासायनिक समीकरण लिखिए।

View Model Solution & Step Marking ▾

Model Answer:

Add aqueous KOH and then acidify with

HNO_3

and add

AgNO_3

solution.

CH_3CH_2Br

(bromoethane) will react with aqueous KOH to form

CH_3CH_2OH

and

KBr

. The

KBr

will then react with

AgNO_3

to form a pale yellow precipitate of

AgBr

.

CH_2=CHBr

(bromoethene) is a vinyl halide and is less reactive towards nucleophilic substitution, so it will not react under these conditions to give

Br^-

ions and thus no precipitate will be formed. Reaction:

CH_3CH_2Br + KOH(aq) \rightarrow CH_3CH_2OH + KBr

.

KBr + AgNO_3 \rightarrow AgBr(s) \downarrow + KNO_3

.

Q8 • 3 Marks Short Answer

Starting from an appropriate haloalkane, how would you prepare

1

-ethoxypropane? Write the chemical equation and name the reaction.

एक उपयुक्त हैलोऐल्केन से प्रारंभ करते हुए, आप

1

-एथॉक्सीप्रोपेन कैसे तैयार करेंगे? रासायनिक समीकरण लिखिए और अभिक्रिया का नाम बताइए।

View Model Solution & Step Marking ▾

Model Answer:

To prepare

1

-ethoxypropane, we can use Williamson synthesis. The appropriate haloalkane would be

1

-bromopropane (or

1

-chloropropane/iodopropane) and sodium ethoxide. Reaction:

CH_3CH_2CH_2Br + NaOCH_2CH_3 \rightarrow CH_3CH_2CH_2OCH_2CH_3 + NaBr

. Reaction name: Williamson Synthesis.

Q9 • 3 Marks Short Answer

Identify the major product formed when

2

-bromopentane reacts with alcoholic KOH. Write the chemical equation for the reaction. Name the type of reaction.

2

-ब्रोमोपेंटेन की अल्कोहोलिक KOH के साथ अभिक्रिया करने पर बनने वाले मुख्य उत्पाद को पहचानिए। अभिक्रिया के लिए रासायनिक समीकरण लिखिए। अभिक्रिया के प्रकार का नाम बताइए।

View Model Solution & Step Marking ▾

Model Answer:

Major product: Pent-

2

-ene. Chemical equation:

CH_3-CH_2-CH_2-CH(Br)-CH_3 + Alc. KOH \rightarrow CH_3-CH_2-CH=CH-CH_3 + KBr + H_2O

. Type of reaction: Dehydrohalogenation or

\beta

-elimination reaction.

Q10 • 3 Marks Short Answer

Explain why haloalkanes are more reactive towards nucleophilic substitution reactions than haloarenes. Provide two reasons.

स्पष्ट कीजिए कि हैलोऐल्केन, हैलोऐरीन की तुलना में नाभिकरागी प्रतिस्थापन अभिक्रियाओं के प्रति अधिक क्रियाशील क्यों होते हैं। दो कारण दीजिए।

View Model Solution & Step Marking ▾

Model Answer:

1. In haloarenes, the C-X bond acquires partial double bond character due to resonance, making it shorter and stronger than the C-X bond in haloalkanes. This makes cleavage difficult. 2. The phenyl cation formed after removal of halogen in haloarenes is less stable than the carbocation formed in haloalkanes.

Part B: Long Answer Questions & Derivations (5 Marks Each)

Q1 • 5 Marks Long Answer / Derivation

a) Explain why haloarenes are less reactive towards nucleophilic substitution reactions than haloalkanes. Give any two reasons. (3 marks)|||
HI|||a) समझाइए कि हैलोऐरीन, हैलोऐल्केनों की तुलना में नाभिकरागी प्रतिस्थापन अभिक्रियाओं के प्रति कम अभिक्रियाशील क्यों होते हैं। कोई दो कारण दीजिए। (3 अंक)
b) How will you convert chlorobenzene to: i) Toluene ii) Phenol? (Write chemical equations only). (2 marks)|||
HI|||b) आप क्लोरोबेंजीन को: i) टोलूईन ii) फीनॉल में कैसे परिवर्तित करेंगे? (केवल रासायनिक समीकरण लिखिए)। (2 अंक)

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Haloarenes are less reactive than haloalkanes towards nucleophilic substitution reactions due to:
i) Resonance effect: The lone pair of electrons on the halogen atom is in conjugation with the

\pi

-electrons of the benzene ring, leading to partial double bond character between the carbon and halogen. This makes the C-X bond stronger and shorter, hence difficult to break. (1.5 marks)
ii) Difference in hybridization of carbon atom: In haloalkanes, the carbon atom attached to the halogen is

sp^3

hybridized, while in haloarenes, it is

sp^2

hybridized. An

sp^2

hybridized carbon is more electronegative and holds the C-X bond electrons more tightly, making the bond shorter and stronger. (1.5 marks)
iii) Polarity of C-X bond: The

sp^2

hybridized carbon in haloarenes is more electronegative than

sp^3

hybridized carbon in haloalkanes. This reduces the polarity of the C-X bond in haloarenes, making nucleophilic attack more difficult.
iv) Instability of phenyl carbocation: Phenyl carbocation, if formed, is less stable due to the non-aromatic nature and high energy.
(Any two reasons, 1.5 marks each).
b) i) Chlorobenzene to Toluene:

C_6H_5Cl + CH_3Cl \xrightarrow{Na/Dry\ Ether} C_6H_5CH_3 + NaCl

(Wurtz-Fittig reaction) (1 mark)
ii) Chlorobenzene to Phenol:

C_6H_5Cl + NaOH \xrightarrow{623K/300atm} C_6H_5ONa \xrightarrow{H^+} C_6H_5OH

(Dow's process) (1 mark)|||
HI|||a) हैलोऐरीन, हैलोऐल्केनों की तुलना में नाभिकरागी प्रतिस्थापन अभिक्रियाओं के प्रति कम अभिक्रियाशील होते हैं, इसके कारण हैं:
i) अनुनाद प्रभाव: हैलोजन परमाणु पर एकाकी इलेक्ट्रॉन युगल बेंजीन वलय के

\pi

-इलेक्ट्रॉनों के साथ संयुग्मन में होता है, जिससे कार्बन और हैलोजन के बीच आंशिक द्विबंध गुण आ जाता है। यह C-X बंध को मजबूत और छोटा बनाता है, इसलिए इसे तोड़ना मुश्किल होता है। (1.5 अंक)
ii) कार्बन परमाणु के संकरण में अंतर: हैलोऐल्केनों में, हैलोजन से जुड़ा कार्बन परमाणु

sp^3

संकरित होता है, जबकि हैलोऐरीनों में, यह

sp^2

संकरित होता है। एक

sp^2

संकरित कार्बन अधिक विद्युतऋणात्मक होता है और C-X बंध इलेक्ट्रॉनों को अधिक कसकर पकड़ता है, जिससे बंध छोटा और मजबूत हो जाता है। (1.5 अंक)
iii) C-X बंध की ध्रुवीयता: हैलोऐरीनों में

sp^2

संकरित कार्बन हैलोऐल्केनों में

sp^3

संकरित कार्बन की तुलना में अधिक विद्युतऋणात्मक होता है। यह हैलोऐरीनों में C-X बंध की ध्रुवीयता को कम करता है, जिससे नाभिकरागी आक्रमण अधिक कठिन हो जाता है।
iv) फेनिल कार्बधनायन का अस्थायित्व: यदि फेनिल कार्बधनायन बनता है, तो यह गैर-ऐरोमैटिक प्रकृति और उच्च ऊर्जा के कारण कम स्थिर होता है।
(कोई भी दो कारण, प्रत्येक के लिए 1.5 अंक)।
b) i) क्लोरोबेंजीन से टोलूईन:

C_6H_5Cl + CH_3Cl \xrightarrow{Na/शुष्क\ ईथर} C_6H_5CH_3 + NaCl

(वुर्ट्ज़-फिटिग अभिक्रिया) (1 अंक)
ii) क्लोरोबेंजीन से फीनॉल:

C_6H_5Cl + NaOH \xrightarrow{623K/300atm} C_6H_5ONa \xrightarrow{H^+} C_6H_5OH

(डाऊ प्रक्रम) (1 अंक)

Q2 • 5 Marks Long Answer / Derivation

a) Explain the mechanism of nucleophilic substitution reactions (

S_N1

and

S_N2

) with suitable examples. Which factor determines the order of reactivity of haloalkanes towards these reactions? (3 marks)|||
HI|||a) नाभिकरागी प्रतिस्थापन अभिक्रियाओं (

S_N1

और

S_N2

) की क्रियाविधि को उपयुक्त उदाहरणों सहित समझाइए। इन अभिक्रियाओं के प्रति हैलोऐल्केनों की अभिक्रियाशीलता का क्रम कौन-सा कारक निर्धारित करता है? (3 अंक)
b) Give the major product(s) formed when 2-bromopentane is treated with alcoholic KOH. Write the mechanism involved. (2 marks)|||
HI|||b) जब 2-ब्रोमोपेन्टेन को अल्कोहलिक KOH के साथ अभिकृत किया जाता है, तो बनने वाले मुख्य उत्पाद लिखिए। इसमें शामिल क्रियाविधि भी लिखिए। (2 अंक)

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a)

S_N1

Mechanism: It is a two-step mechanism. In the first step, the leaving group departs to form a carbocation (slow and rate-determining step). In the second step, the nucleophile attacks the carbocation. Carbocation stability determines reactivity (

3^ ext{o} > 2^ ext{o} > 1^ ext{o}

). Example: Hydrolysis of tert-butyl bromide.

S_N2

Mechanism: It is a single-step, concerted mechanism where the nucleophile attacks from the backside, and the leaving group departs simultaneously, forming a transition state. Steric hindrance determines reactivity (

1^ ext{o} > 2^ ext{o} > 3^ ext{o}

). Example: Hydrolysis of methyl bromide. (1.5 marks for each mechanism explanation with example and reactivity order).
b) When 2-bromopentane is treated with alcoholic KOH, elimination reaction (dehydrohalogenation) occurs, following Saytzeff's rule. The major product is Pent-2-ene. (1 mark for product). Mechanism: E2 mechanism. The strong base (KOH) abstracts a

eta

-hydrogen, and simultaneously the leaving group (Br) departs, forming a double bond. (1 mark for mechanism).|||
HI|||a)

S_N1

क्रियाविधि: यह दो-चरणीय क्रियाविधि है। पहले चरण में, निष्कासित होने वाला समूह हट जाता है जिससे एक कार्बधनायन बनता है (धीमा और दर-निर्धारक चरण)। दूसरे चरण में, नाभिकरागी कार्बधनायन पर आक्रमण करता है। कार्बधनायन का स्थायित्व अभिक्रियाशीलता निर्धारित करता है (

3^ ext{o} > 2^ ext{o} > 1^ ext{o}

)। उदाहरण: तृतीयक-ब्यूटिल ब्रोमाइड का जल-अपघटन।

S_N2

क्रियाविधि: यह एक-चरणीय, संकेंद्रित क्रियाविधि है जहाँ नाभिकरागी पीछे से आक्रमण करता है, और निष्कासित होने वाला समूह एक साथ निकलता है, जिससे एक संक्रमण अवस्था बनती है। त्रिविम बाधा अभिक्रियाशीलता निर्धारित करती है (

1^ ext{o} > 2^ ext{o} > 3^ ext{o}

)। उदाहरण: मेथिल ब्रोमाइड का जल-अपघटन। (प्रत्येक क्रियाविधि की व्याख्या, उदाहरण और अभिक्रियाशीलता क्रम के लिए 1.5 अंक)।
b) जब 2-ब्रोमोपेन्टेन को अल्कोहलिक KOH के साथ अभिकृत किया जाता है, तो विलोपन अभिक्रिया (विरोहैलोजनीकरण) होती है, जो सेत्ज़ेफ के नियम का पालन करती है। मुख्य उत्पाद पेन्ट-2-ईन है। (उत्पाद के लिए 1 अंक)। क्रियाविधि: E2 क्रियाविधि। प्रबल क्षार (KOH) एक

eta

-हाइड्रोजन को निकालता है, और साथ ही निष्कासित होने वाला समूह (Br) निकल जाता है, जिससे एक द्विबंध बनता है। (क्रियाविधि के लिए 1 अंक।)

Q3 • 5 Marks Long Answer / Derivation

a) Explain the mechanism of

S_N1

reaction with an example. What factors favor

S_N1

over

S_N2

reaction? (3 marks)|||
HI|||क)

S_N1

अभिक्रिया की क्रियाविधि को उदाहरण सहित समझाइए। कौन से कारक

S_N2

अभिक्रिया की तुलना में

S_N1

को प्राथमिकता देते हैं? (3 अंक)
b) Why is it difficult to carry out a nucleophilic substitution reaction on chloroarenes under ordinary conditions? (2 marks)|||
HI|||ख) सामान्य परिस्थितियों में क्लोरोएरीन पर नाभिकरागी प्रतिस्थापन अभिक्रिया करना क्यों कठिन है? (2 अंक)

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a)

S_N1

reaction mechanism: It is a two-step process. In the first step, the leaving group departs slowly to form a carbocation (rate-determining step). In the second step, the nucleophile attacks the carbocation rapidly. Example: Hydrolysis of tert-butyl bromide. Factors favoring

S_N1

: Tertiary alkyl halides, weak nucleophiles, polar protic solvents. (3 marks)|||
HI|||क)

S_N1

अभिक्रिया क्रियाविधि: यह एक दो-चरणीय प्रक्रिया है। पहले चरण में, निष्कासित होने वाला समूह धीरे-धीरे पृथक होकर एक कार्बोकैटायन बनाता है (दर-निर्धारक चरण)। दूसरे चरण में, नाभिकरागी तेजी से कार्बोकैटायन पर आक्रमण करता है। उदाहरण: तृतीयक-ब्यूटिल ब्रोमाइड का जल-अपघटन।

S_N1

को प्राथमिकता देने वाले कारक: तृतीयक ऐल्किल हैलाइड, दुर्बल नाभिकरागी, ध्रुवीय प्रोटिक विलायक। (3 अंक)
b) Reasons for difficulty in nucleophilic substitution on chloroarenes: Resonance effect (partial double bond character between C-Cl), difference in hybridization of carbon atom (

sp^2

carbon in chloroarenes is more electronegative and holds the electron pair of C-Cl bond more tightly, making it shorter and stronger), instability of phenyl carbocation, and repulsion between nucleophile and electron-rich arene. (2 marks)|||
HI|||ख) क्लोरोएरीन पर नाभिकरागी प्रतिस्थापन में कठिनाई के कारण: अनुनाद प्रभाव (C-Cl के बीच आंशिक द्विबंध गुण), कार्बन परमाणु के संकरण में अंतर (क्लोरोएरीन में

sp^2

संकरित कार्बन अधिक विद्युत ऋणात्मक होता है और C-Cl बंध के इलेक्ट्रॉन युग्म को अधिक कसकर पकड़े रहता है, जिससे यह छोटा और मजबूत होता है), फेनिल कार्बोकैटायन की अस्थिरता, और नाभिकरागी व इलेक्ट्रॉन-समृद्ध एरीन के बीच प्रतिकर्षण। (2 अंक)

Q4 • 5 Marks Long Answer / Derivation

a) Write the chemical equations for the following reactions: (3 marks)
i) Finkelstein reaction
ii) Swarts reaction
iii) Wurtz reaction
|||
HI|||a) निम्नलिखित अभिक्रियाओं के लिए रासायनिक समीकरण लिखिए: (3 अंक)
i) फिंकेलस्टीन अभिक्रिया
ii) स्वार्ट्स अभिक्रिया
iii) वुर्ट्ज़ अभिक्रिया

b) Explain why Grignard reagents should be prepared under anhydrous conditions. Write the general preparation of a Grignard reagent and its reaction with water. (2 marks)|||
HI|||b) समझाइए कि ग्रिगनार्ड अभिकर्मकों को निर्जल परिस्थितियों में क्यों तैयार किया जाना चाहिए। ग्रिगनार्ड अभिकर्मक के सामान्य निर्माण और जल के साथ इसकी अभिक्रिया लिखिए। (2 अंक)

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) i) Finkelstein reaction:

R-X + NaI \xrightarrow{Dry\ Acetone} R-I + NaX

(where X = Cl, Br) (1 mark)
ii) Swarts reaction:

R-X + AgF \xrightarrow{Heat} R-F + AgX

(where X = Cl, Br) (1 mark)
iii) Wurtz reaction:

2R-X + 2Na \xrightarrow{Dry\ Ether} R-R + 2NaX

(1 mark)
b) Grignard reagents are highly reactive and readily react with any source of proton such as water, alcohols, amines, etc., to form alkanes. If prepared in the presence of moisture, they would react with water to give alkanes, thus destroying the reagent. (1 mark for explanation).
General preparation:

R-X + Mg \xrightarrow{Dry\ Ether} R-MgX

(1st 0.5 mark)
Reaction with water:

R-MgX + H_2O \rightarrow R-H + Mg(OH)X

(2nd 0.5 mark)|||
HI|||a) i) फिंकेलस्टीन अभिक्रिया:

R-X + NaI \xrightarrow{शुष्क\ एसीटोन} R-I + NaX

(जहाँ X = Cl, Br) (1 अंक)
ii) स्वार्ट्स अभिक्रिया:

R-X + AgF \xrightarrow{ऊष्मा} R-F + AgX

(जहाँ X = Cl, Br) (1 अंक)
iii) वुर्ट्ज़ अभिक्रिया:

2R-X + 2Na \xrightarrow{शुष्क\ ईथर} R-R + 2NaX

(1 अंक)
b) ग्रिगनार्ड अभिकर्मक अत्यधिक अभिक्रियाशील होते हैं और जल, ऐल्कोहॉल, ऐमीन आदि जैसे प्रोटॉन के किसी भी स्रोत के साथ आसानी से अभिक्रिया करके ऐल्केन बनाते हैं। यदि नमी की उपस्थिति में तैयार किया जाता है, तो वे जल के साथ अभिक्रिया करके ऐल्केन देंगे, जिससे अभिकर्मक नष्ट हो जाएगा। (स्पष्टीकरण के लिए 1 अंक)।
सामान्य निर्माण:

R-X + Mg \xrightarrow{शुष्क\ ईथर} R-MgX

(पहला 0.5 अंक)
जल के साथ अभिक्रिया:

R-MgX + H_2O \rightarrow R-H + Mg(OH)X

(दूसरा 0.5 अंक)

Q5 • 5 Marks Long Answer / Derivation

a) Explain the term 'optical isomerism' with reference to haloalkanes. What is a chiral carbon?b) Draw the enantiomers of 2-chlorobutane and assign their R/S configuration.c) Give one chemical test to distinguish between

CHCl_3

and

CCl_4

. Write the relevant chemical equation.

a) हैलोऐल्केनों के संदर्भ में 'प्रकाशिक समावयवता' शब्द की व्याख्या कीजिए। किरेल कार्बन क्या होता है?b) 2-क्लोरोब्यूटेन के प्रतिबिंब रूपी समावयवी बनाइए और उनके R/S विन्यास निर्धारित कीजिए।c)

CHCl_3

और

CCl_4

के बीच अंतर करने के लिए एक रासायनिक परीक्षण दीजिए। प्रासंगिक रासायनिक समीकरण लिखिए।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Optical isomerism: It is a type of stereoisomerism shown by compounds that are non-superimposable mirror images of each other. These isomers are called enantiomers and they rotate the plane of plane-polarized light in opposite directions.Chiral carbon: A carbon atom bonded to four different atoms or groups is called a chiral carbon or an asymmetric carbon. The presence of a chiral carbon usually leads to optical isomerism.b) The enantiomers of 2-chlorobutane are: (Structures would be drawn with wedges and dashes here, representing the 3D arrangement).R-2-chlorobutane and S-2-chlorobutane. (Detailed R/S assignment based on CIP rules would be explained if space permitted, but for a 5-mark question, drawing and naming usually suffices).c) Chemical test to distinguish

CHCl_3

and

CCl_4

:Carbylamine reaction (Isocyanide test): Chloroform (

CHCl_3

) gives a positive carbylamine test, whereas carbon tetrachloride (

CCl_4

) does not. When

CHCl_3

is heated with a primary amine and alcoholic KOH, it forms an offensive-smelling isocyanide.Chemical equation:

R-NH_2 + CHCl_3 + 3KOH(alc.) \xrightarrow{\Delta} R-NC + 3KCl + 3H_2O

.Carbon tetrachloride (

CCl_4

) does not contain an

\alpha

-hydrogen, hence it does not give this test.

7

Alcohols, Phenols and Ethers

Part A: Short Answer Questions (2–3 Marks Each)

Q1 • 3 Marks Short Answer

Explain why phenols are more acidic than alcohols, even though both contain an -OH group. Illustrate with relevant resonance structures.

स्पष्ट कीजिए कि फिनोल, ऐल्कोहॉल की तुलना में अधिक अम्लीय क्यों होते हैं, यद्यपि दोनों में -OH समूह होता है। प्रासंगिक अनुनादी संरचनाओं के साथ समझाइए।

View Model Solution & Step Marking ▾

Model Answer:

Phenols are more acidic than alcohols because the phenoxide ion (conjugate base of phenol) is stabilized by resonance. The negative charge on oxygen is delocalized over the benzene ring, making it more stable than the alkoxide ion (conjugate base of alcohol) where the negative charge is localized on the oxygen. Resonance structures of phenoxide ion showing delocalization of negative charge.

Q2 • 3 Marks Short Answer

How will you distinguish between primary, secondary, and tertiary alcohols using the Lucas test? Describe the observations for each.

आप ल्यूकास परीक्षण का उपयोग करके प्राथमिक, द्वितीयक और तृतीयक ऐल्कोहॉलों के बीच कैसे अंतर करेंगे? प्रत्येक के लिए प्रेक्षणों का वर्णन कीजिए।

View Model Solution & Step Marking ▾

Model Answer:

The Lucas test uses a mixture of concentrated

\text{HCl}

and anhydrous

\text{ZnCl}_2

. Tertiary alcohols react immediately to form turbidity. Secondary alcohols react within 5-10 minutes to form turbidity. Primary alcohols do not react at room temperature and remain clear.

Q3 • 3 Marks Short Answer

Write the mechanism for the dehydration of ethanol to ethene in the presence of concentrated

\text{H}_2\text{SO}_4

at

443 \text{ K}

.

सांद्र

\text{H}_2\text{SO}_4

की उपस्थिति में

443 \text{ K}

पर एथेनॉल से एथीन के निर्जलीकरण की क्रियाविधि लिखिए।

View Model Solution & Step Marking ▾

Model Answer:

The mechanism involves three steps: 1. Protonation of alcohol to form protonated alcohol. 2. Formation of carbocation by loss of water. 3. Elimination of a proton to form ethene. (Each step should be shown with proper curved arrows and intermediates).

Q4 • 3 Marks Short Answer

Explain why the boiling points of ethers are much lower than those of alcohols of comparable molar masses.

स्पष्ट कीजिए कि ईथरों के क्वथनांक तुलनीय मोलर द्रव्यमान वाले ऐल्कोहॉलों की तुलना में बहुत कम क्यों होते हैं।

View Model Solution & Step Marking ▾

Model Answer:

Alcohols form intermolecular hydrogen bonds due to the presence of a highly polar -OH group, which requires more energy to break, thus leading to higher boiling points. Ethers, on the other hand, do not have a hydrogen atom directly bonded to the oxygen, so they cannot form intermolecular hydrogen bonds. They only exhibit weaker dipole-dipole interactions and van der Waals forces.

Q5 • 3 Marks Short Answer

Complete the following reaction sequence: \

\text{Phenol} \xrightarrow{\text{i) } \text{NaOH}} \text{A} \xrightarrow{\text{ii) } \text{CO}_2 \text{, } 400 \text{ K, } 4-7 \text{ atm}} \text{B} \xrightarrow{\text{iii) } \text{H}^+} \text{C}

\\Identify A, B, and C and name the overall reaction.

निम्नलिखित अभिक्रिया अनुक्रम को पूर्ण कीजिए: \

\text{फिनोल} \xrightarrow{\text{i) } \text{NaOH}} \text{A} \xrightarrow{\text{ii) } \text{CO}_2 \text{, } 400 \text{ K, } 4-7 \text{ atm}} \text{B} \xrightarrow{\text{iii) } \text{H}^+} \text{C}

\\A, B और C को पहचानिए तथा समग्र अभिक्रिया का नाम बताइए।

View Model Solution & Step Marking ▾

Model Answer:

A = Sodium phenoxide \\B = Sodium salicylate \\C = Salicylic acid \\Overall reaction is Kolbe's Reaction (Kolbe's synthesis).

Q6 • 3 Marks Short Answer

Explain why phenol is more acidic than ethanol, even though both contain an -OH group. Provide relevant resonance structures for phenol to support your answer.

स्पष्ट कीजिए कि फिनोल, इथेनॉल से अधिक अम्लीय क्यों होता है, जबकि दोनों में -OH समूह होता है। अपने उत्तर के समर्थन में फिनोल की अनुनादी संरचनाएँ प्रस्तुत कीजिए।

View Model Solution & Step Marking ▾

Model Answer:

Phenol is more acidic than ethanol because the phenoxide ion, formed after losing a proton, is stabilized by resonance. The negative charge is delocalized over the benzene ring, making it more stable than the ethoxide ion, which has no such resonance stabilization. Ethanol's ethoxide ion is destabilized by the electron-donating ethyl group. Resonance structures for phenoxide ion show delocalization of negative charge at ortho and para positions.

Q7 • 3 Marks Short Answer

How will you synthesize 1-phenyl ethanol from a suitable carbonyl compound using a Grignard reagent? Write the chemical equations involved.

आप ग्रिगनार्ड अभिकर्मक का उपयोग करके उपयुक्त कार्बोनिल यौगिक से 1-फेनिल इथेनॉल का संश्लेषण कैसे करेंगे? इसमें शामिल रासायनिक समीकरण लिखिए।

View Model Solution & Step Marking ▾

Model Answer:

1-Phenyl ethanol can be synthesized by reacting benzaldehyde with methylmagnesium bromide (Grignard reagent), followed by hydrolysis.

C_6H_5CHO + CH_3MgBr \rightarrow C_6H_5CH(OMgBr)CH_3

(Adduct formation)

C_6H_5CH(OMgBr)CH_3 + H_2O \rightarrow C_6H_5CH(OH)CH_3 + Mg(OH)Br

(Hydrolysis to form 1-phenyl ethanol)

Q8 • 3 Marks Short Answer

Explain the mechanism of dehydration of ethanol to ethene in the presence of concentrated sulfuric acid at

443 K

.

सांद्र सल्फ्यूरिक अम्ल की उपस्थिति में

443 K

पर इथेनॉल के एथीन में निर्जलीकरण की क्रियाविधि समझाइए।

View Model Solution & Step Marking ▾

Model Answer:

The mechanism involves three steps:
1. Protonation of alcohol: The oxygen atom of ethanol gets protonated by

H_2SO_4

to form protonated alcohol.
2. Formation of carbocation: The protonated alcohol loses a water molecule to form a carbocation.
3. Deprotonation: The carbocation loses a proton to form ethene. The proton lost is accepted by

HSO_4^-

to regenerate

H_2SO_4

.

Q9 • 3 Marks Short Answer

Give a chemical test to distinguish between methanol and phenol. Write the chemical equations for the reactions involved.

मेथनॉल और फिनोल के बीच अंतर करने के लिए एक रासायनिक परीक्षण दीजिए। इसमें शामिल अभिक्रियाओं के रासायनिक समीकरण लिखिए।

View Model Solution & Step Marking ▾

Model Answer:

Ferric chloride test: Phenol gives a characteristic violet, blue, or green coloration with neutral ferric chloride solution due to the formation of a complex. Methanol does not give this test.
Reaction:

6C_6H_5OH + FeCl_3 \rightarrow [Fe(OC_6H_5)_6]^{3-} + 3H^+ + 3HCl

(Violet/Blue/Green complex)
Methanol: No reaction with

FeCl_3

.

Q10 • 3 Marks Short Answer

How are ethers prepared by Williamson's synthesis? Explain with a suitable example and write the general reaction. What is the limitation of this method for preparing unsymmetrical ethers?

विलियमसन संश्लेषण द्वारा ईथर कैसे तैयार किए जाते हैं? एक उपयुक्त उदाहरण के साथ समझाइए और सामान्य अभिक्रिया लिखिए। असममित ईथर तैयार करने के लिए इस विधि की क्या सीमा है?

View Model Solution & Step Marking ▾

Model Answer:

Williamson's synthesis involves the reaction of an alkyl halide with sodium alkoxide or sodium phenoxide. It is an

S_N2

reaction.
General reaction:

R-X + R'-ONa \rightarrow R-O-R' + NaX

Example:

CH_3CH_2Br + CH_3ONa \rightarrow CH_3CH_2OCH_3 + NaBr

(Methoxyethane)
Limitation: For unsymmetrical ethers, if a primary alkoxide is reacted with a tertiary alkyl halide, elimination (

E2

) often competes with substitution (

S_N2

), leading to the formation of alkenes as major products. Therefore, for unsymmetrical ethers, the alkyl halide should be primary.

Part B: Long Answer Questions & Derivations (5 Marks Each)

Q1 • 5 Marks Long Answer / Derivation

a) Explain the mechanism of dehydration of ethanol to ethene using concentrated sulfuric acid at

443 K

.b) How will you distinguish between propan-1-ol and propan-2-ol using a chemical test? Write the chemical equations involved.

क)

443 K

पर सांद्र सल्फ्यूरिक अम्ल का उपयोग करके इथेनॉल से एथीन के निर्जलीकरण की क्रियाविधि समझाइए।ख) प्रोपेन-1-ऑल और प्रोपेन-2-ऑल के बीच रासायनिक परीक्षण का उपयोग करके आप कैसे अंतर करेंगे? इसमें शामिल रासायनिक समीकरण लिखिए।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Mechanism of dehydration of ethanol to ethene:1. Protonation of ethanol:

CH_3CH_2OH + H^+ \rightleftharpoons CH_3CH_2O^+H_2

2. Formation of carbocation:

CH_3CH_2O^+H_2 \rightarrow CH_3CH_2^+ + H_2O

3. Deprotonation to form ethene:

CH_3CH_2^+ \rightarrow CH_2=CH_2 + H^+

b) Distinction between propan-1-ol and propan-2-ol using Lucas Test:Propan-1-ol (primary alcohol) does not react with Lucas reagent (

HCl/ZnCl_2

) at room temperature, no turbidity appears.Propan-2-ol (secondary alcohol) reacts with Lucas reagent to form turbidity within 5-10 minutes.Equations:Propan-1-ol:

CH_3CH_2CH_2OH + HCl \xrightarrow{ZnCl_2} No\ reaction

Propan-2-ol:

CH_3CH(OH)CH_3 + HCl \xrightarrow{ZnCl_2} CH_3CHClCH_3 + H_2O

Q2 • 5 Marks Long Answer / Derivation

a) Arrange the following compounds in increasing order of their acidic strength and justify your answer: Phenol, 4-Nitrophenol, 4-Methylphenol.b) Explain why phenols are more acidic than alcohols, even though both have an -OH group.

क) निम्नलिखित यौगिकों को उनकी अम्लीय शक्ति के बढ़ते क्रम में व्यवस्थित करें और अपने उत्तर का औचित्य सिद्ध करें: फिनोल, 4-नाइट्रोफिनोल, 4-मेथिलफिनोल।ख) समझाइए कि फिनोल अल्कोहल की तुलना में अधिक अम्लीय क्यों होते हैं, भले ही दोनों में -OH समूह होता है।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Increasing order of acidic strength: 4-Methylphenol < Phenol < 4-Nitrophenol.Justification: Acidity of phenols is due to the stability of phenoxide ion formed after losing a proton. Electron-withdrawing groups (like

-NO_2

) stabilize the phenoxide ion by delocalization of negative charge, thus increasing acidity. Electron-donating groups (like

-CH_3

) destabilize the phenoxide ion, thus decreasing acidity. Hence, 4-Nitrophenol is most acidic, and 4-Methylphenol is least acidic among the three.b) Phenols are more acidic than alcohols because the phenoxide ion formed after losing a proton is resonance stabilized. The negative charge on oxygen is delocalized over the benzene ring, making the phenoxide ion more stable than the alkoxide ion. In alcohols, the alkoxide ion has the negative charge localized on the oxygen, which is not resonance stabilized, making it less stable and thus alcohols less acidic.

Q3 • 5 Marks Long Answer / Derivation

a) Explain the mechanism of the acid-catalyzed hydration of propene to form propan-2-ol.b) Write the main product(s) formed when anisole undergoes reaction with:i)

HI

(hot and concentrated)ii) Bromine in ethanoic acid medium.

क) प्रोपीन के अम्ल-उत्प्रेरित जलयोजन से प्रोपेन-2-ऑल बनने की क्रियाविधि समझाइए।ख) जब एनिसोल निम्नलिखित के साथ अभिक्रिया करता है तो बनने वाले मुख्य उत्पाद लिखिए:i)

HI

(गर्म और सांद्र)ii) एथेनोइक अम्ल माध्यम में ब्रोमीन।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Mechanism of acid-catalyzed hydration of propene to form propan-2-ol:1. Protonation of alkene:

CH_3CH=CH_2 + H^+ \rightleftharpoons CH_3CH^+CH_3

(Markovnikov's rule followed to form more stable carbocation).2. Nucleophilic attack by water:

CH_3CH^+CH_3 + H_2O \rightarrow CH_3CH(O^+H_2)CH_3

3. Deprotonation:

CH_3CH(O^+H_2)CH_3 \rightarrow CH_3CH(OH)CH_3 + H^+

b) Main product(s) formed when anisole reacts:i) With

HI

(hot and concentrated): Anisole is an ether, and

HI

cleaves it. Since one group is methyl and the other is phenyl, the cleavage occurs such that the smaller alkyl group forms alkyl iodide, and the phenyl group forms phenol. Products: Phenol and Methyl iodide.

C_6H_5OCH_3 + HI \xrightarrow{heat} C_6H_5OH + CH_3I

ii) With Bromine in ethanoic acid medium: Anisole is an activated aromatic compound due to the

-OCH_3

group (an ortho-para directing group). Bromination occurs at ortho and para positions. Products: 2-Bromoanisole and 4-Bromoanisole (major product).

Q4 • 5 Marks Long Answer / Derivation

a) Explain why phenols are more acidic than alcohols, despite both containing an -OH group. (3 marks)
b) Give a chemical test to distinguish between phenol and ethanol. (2 marks)||HI||a) समझाइए कि फिनोल अल्कोहल की तुलना में अधिक अम्लीय क्यों होते हैं, जबकि दोनों में

-\text{OH}

समूह होता है। (3 अंक)
b) फिनोल और इथेनॉल के बीच अंतर करने के लिए एक रासायनिक परीक्षण दीजिए। (2 अंक)

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Phenols are more acidic than alcohols due to the resonance stabilization of the phenoxide ion formed after the loss of a proton. In phenol, the lone pair of electrons on the oxygen atom participates in resonance with the benzene ring, delocalizing the negative charge over the ring. This stabilization makes the removal of

H^+

easier. In contrast, for alcohols, the alkoxide ion formed is not resonance stabilized; instead, the alkyl group's electron-donating inductive effect destabilizes the alkoxide ion, making alcohols weaker acids. (3 marks for clear explanation of resonance stabilization in phenoxide ion and lack thereof in alkoxide ion)
b) A common chemical test is the Ferric Chloride Test. Phenol reacts with neutral ferric chloride solution to give a characteristic violet/purple/green coloration due to the formation of a complex, while ethanol does not show any such color change. (1 mark for test name, 1 mark for observation/explanation).||HI||a) फिनोल अल्कोहल की तुलना में अधिक अम्लीय होते हैं क्योंकि प्रोटॉन के निष्कासन के बाद बनने वाले फिनोक्साइड आयन का अनुनाद स्थिरीकरण होता है। फिनोल में, ऑक्सीजन परमाणु पर एकाकी इलेक्ट्रॉन युग्म बेंजीन वलय के साथ अनुनाद में भाग लेता है, जिससे ऋणात्मक आवेश वलय पर विस्थानीकृत हो जाता है। यह स्थिरीकरण

H^+

के निष्कासन को आसान बनाता है। इसके विपरीत, अल्कोहलों के लिए, बनने वाला एल्कोक्साइड आयन अनुनाद द्वारा स्थिर नहीं होता है; इसके बजाय, एल्किल समूह का इलेक्ट्रॉन-दाता प्रेरणिक प्रभाव एल्कोक्साइड आयन को अस्थिर करता है, जिससे अल्कोहल दुर्बल अम्ल होते हैं। (फिनोक्साइड आयन में अनुनाद स्थिरीकरण की स्पष्ट व्याख्या और एल्कोक्साइड आयन में इसकी कमी के लिए 3 अंक)
b) एक सामान्य रासायनिक परीक्षण फेरिक क्लोराइड परीक्षण है। फिनोल उदासीन फेरिक क्लोराइड विलयन के साथ अभिक्रिया करके एक संकुल के निर्माण के कारण एक विशिष्ट बैंगनी/जामुनी/हरा रंग देता है, जबकि इथेनॉल ऐसा कोई रंग परिवर्तन नहीं दिखाता है। (परीक्षण के नाम के लिए 1 अंक, अवलोकन/व्याख्या के लिए 1 अंक)।

Q5 • 5 Marks Long Answer / Derivation

a) Explain the mechanism of hydration of ethene to form ethanol in the presence of an acid catalyst. (3 marks)b) Give a chemical test to distinguish between ethanol and phenol. (2 marks)||HHI||a) एथीन के अम्ल उत्प्रेरित जलयोजन द्वारा एथेनॉल के निर्माण की क्रियाविधि समझाइए। (3 अंक)b) एथेनॉल और फीनॉल में विभेद करने के लिए एक रासायनिक परीक्षण दीजिए। (2 अंक)

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Mechanism of hydration of ethene to form ethanol:1. Protonation of ethene to form carbocation:

CH_2=CH_2 + H_3O^+ \rightleftharpoons CH_3-CH_2^+ + H_2O

2. Nucleophilic attack by water on carbocation:

CH_3-CH_2^+ + H_2O \rightleftharpoons CH_3-CH_2-OH_2^+

3. Deprotonation to form ethanol:

CH_3-CH_2-OH_2^+ + H_2O \rightleftharpoons CH_3-CH_2-OH + H_3O^+

(Each step 1 mark)b) Chemical test to distinguish between ethanol and phenol:Phenol gives a characteristic violet/purple colour with neutral ferric chloride solution, while ethanol does not. (2 marks)Alternative: Phenol reacts with bromine water to give a white precipitate of 2,4,6-tribromophenol, while ethanol does not. (2 marks)||HHI||a) एथीन के एथेनॉल में जलयोजन की क्रियाविधि:1. एथीन का प्रोटोनीकरण करके कार्बधनायन बनाना:

CH_2=CH_2 + H_3O^+ \rightleftharpoons CH_3-CH_2^+ + H_2O

2. कार्बधनायन पर जल द्वारा नाभिकस्नेही आक्रमण:

CH_3-CH_2^+ + H_2O \rightleftharpoons CH_3-CH_2-OH_2^+

3. एथेनॉल बनाने के लिए विप्रोटोनीकरण:

CH_3-CH_2-OH_2^+ + H_2O \rightleftharpoons CH_3-CH_2-OH + H_3O^+

(प्रत्येक चरण के लिए 1 अंक)b) एथेनॉल और फीनॉल में विभेद करने के लिए रासायनिक परीक्षण:फीनॉल उदासीन फेरिक क्लोराइड विलयन के साथ एक विशिष्ट बैंगनी/जामुनी रंग देता है, जबकि एथेनॉल नहीं देता है। (2 अंक)वैकल्पिक: फीनॉल ब्रोमीन जल के साथ अभिक्रिया करके 2,4,6-ट्राइब्रोमोफीनॉल का सफेद अवक्षेप देता है, जबकि एथेनॉल नहीं देता है। (2 अंक)

8

Aldehydes, Ketones and Carboxylic Acids

Part A: Short Answer Questions (2–3 Marks Each)

Q1 • 3 Marks Short Answer

Explain why aldehydes are generally more reactive than ketones towards nucleophilic addition reactions. Give one example of a nucleophilic addition reaction common to both.

स्पष्ट कीजिए कि नाभिकरागी योग अभिक्रियाओं के प्रति ऐल्डिहाइड सामान्यतः कीटोन की अपेक्षा अधिक क्रियाशील क्यों होते हैं। नाभिकरागी योग अभिक्रिया का एक उदाहरण दीजिए जो दोनों के लिए सामान्य हो।

View Model Solution & Step Marking ▾

Model Answer:

Aldehydes are more reactive than ketones due to a combination of steric and electronic factors. Sterically, the presence of two larger alkyl groups in ketones hinders the approach of a nucleophile more than the one alkyl group and one hydrogen atom in aldehydes. Electronically, the two alkyl groups in ketones are electron-donating, which reduces the partial positive charge on the carbonyl carbon, making it less electrophilic compared to aldehydes where only one alkyl group donates electrons. Example: Reaction with HCN to form cyanohydrins.

Q2 • 3 Marks Short Answer

How will you convert propanone to 2-methylpropan-2-ol? Write the chemical equation for the reaction involved.

आप प्रोपेनोन को 2-मेथिलप्रोपेन-2-ऑल में कैसे परिवर्तित करेंगे? इसमें शामिल अभिक्रिया के लिए रासायनिक समीकरण लिखिए।

View Model Solution & Step Marking ▾

Model Answer:

Propanone can be converted to 2-methylpropan-2-ol by reacting it with methylmagnesium bromide (Grignard reagent) followed by hydrolysis.

CH_3COCH_3 + CH_3MgBr \xrightarrow{dry ether} (CH_3)_3COMgBr \xrightarrow{H_3O^+} (CH_3)_3COH

Q3 • 3 Marks Short Answer

Give a chemical test to distinguish between benzaldehyde and benzoic acid. Write the chemical equations for the reactions involved.

बेंजैल्डिहाइड और बेंजोइक अम्ल के बीच अंतर करने के लिए एक रासायनिक परीक्षण दीजिए। इसमें शामिल अभिक्रियाओं के लिए रासायनिक समीकरण लिखिए।

View Model Solution & Step Marking ▾

Model Answer:

Tollens' Reagent Test: Benzaldehyde, being an aldehyde, will give a positive Tollens' test, forming a silver mirror. Benzoic acid will not react.

C_6H_5CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow C_6H_5COO^- + 2Ag(s) + 4NH_3 + 2H_2O

Benzoic acid does not react with Tollens' reagent.

Q4 • 3 Marks Short Answer

Arrange the following compounds in increasing order of their acidic strength and give a brief reason: Acetic acid, Chloroacetic acid, Formic acid.

निम्नलिखित यौगिकों को उनकी अम्लीय शक्ति के बढ़ते क्रम में व्यवस्थित कीजिए और एक संक्षिप्त कारण दीजिए: एसिटिक अम्ल, क्लोरोएसिटिक अम्ल, फॉर्मिक अम्ल।

View Model Solution & Step Marking ▾

Model Answer:

Increasing order of acidic strength: Acetic acid < Formic acid < Chloroacetic acid.
Reason: Electron-withdrawing groups increase acidic strength by stabilizing the carboxylate ion through the inductive effect. Chlorine is an electron-withdrawing group, making chloroacetic acid most acidic. The methyl group in acetic acid is electron-donating, destabilizing the carboxylate ion. Formic acid has a hydrogen atom instead of an alkyl group, making it more acidic than acetic acid but less acidic than chloroacetic acid.

Q5 • 3 Marks Short Answer

Identify A, B, and C in the following reaction sequence:

CH_3COOH \xrightarrow{PCl_5} A \xrightarrow{H_2/Pd-BaSO_4} B \xrightarrow{dil. NaOH} C

निम्नलिखित अभिक्रिया अनुक्रम में A, B और C को पहचानिए:

CH_3COOH \xrightarrow{PCl_5} A \xrightarrow{H_2/Pd-BaSO_4} B \xrightarrow{dil. NaOH} C

View Model Solution & Step Marking ▾

Model Answer:

A:

CH_3COCl

(Acetyl chloride)
B:

CH_3CHO

(Acetaldehyde)
C:

CH_3CH(OH)CH_2CHO

(3-Hydroxybutanal) or Aldol product

Q6 • 3 Marks Short Answer

Explain why carboxylic acids are stronger acids than alcohols of comparable molecular masses. Give a reason for your answer.

समझाइए कि कार्बोक्सिलिक अम्ल तुलनीय आणविक द्रव्यमान वाले ऐल्कोहॉलों की तुलना में प्रबल अम्ल क्यों होते हैं। अपने उत्तर का कारण दीजिए।

View Model Solution & Step Marking ▾

Model Answer:

Carboxylic acids are stronger acids than alcohols because the carboxylate ion (

R-COO^-

) formed after losing a proton is resonance stabilized. The negative charge is delocalized over two electronegative oxygen atoms. In contrast, the alkoxide ion (

R-O^-

) formed from an alcohol is not resonance stabilized.

Q7 • 3 Marks Short Answer

An organic compound with molecular formula

C_3H_6O

gives a positive iodoform test but does not reduce Tollens' reagent. Identify the compound and write the chemical equations for the reactions involved in the iodoform test.

एक कार्बनिक यौगिक जिसका आणविक सूत्र

C_3H_6O

है, आयोडोफॉर्म परीक्षण देता है लेकिन टोलन अभिकर्मक को अपचयित नहीं करता है। यौगिक की पहचान कीजिए और आयोडोफॉर्म परीक्षण में शामिल अभिक्रियाओं के रासायनिक समीकरण लिखिए।

View Model Solution & Step Marking ▾

Model Answer:

The compound is propanone (

CH_3COCH_3

). It gives a positive iodoform test because it contains a methyl ketone group, but it does not reduce Tollens' reagent as it is a ketone. Reactions for iodoform test:

CH_3COCH_3 + 3I_2 + 4NaOH \rightarrow CHI_3 (yellow~ppt) + CH_3COONa + 3NaI + 3H_2O

.

Q8 • 3 Marks Short Answer

Identify the products A and B formed in the following reaction sequence:

CH_3CHO \xrightarrow{PCl_5} A \xrightarrow{H_2O} B

. Explain the type of reaction involved in the formation of B from A.

निम्नलिखित अभिक्रिया अनुक्रम में बनने वाले उत्पादों A और B को पहचानिए:

CH_3CHO \xrightarrow{PCl_5} A \xrightarrow{H_2O} B

। A से B के निर्माण में शामिल अभिक्रिया के प्रकार की व्याख्या कीजिए।

View Model Solution & Step Marking ▾

Model Answer:

A is 1,1-dichloroethane (

CH_3CHCl_2

). B is ethanal (

CH_3CHO

). The reaction from A to B is hydrolysis.

Q9 • 3 Marks Short Answer

How will you distinguish between propanal and propanone using a chemical test? Write the chemical equations for the reactions involved.

आप प्रोपेनल और प्रोपेनोन के बीच एक रासायनिक परीक्षण का उपयोग करके कैसे अंतर करेंगे? इसमें शामिल अभिक्रियाओं के रासायनिक समीकरण लिखिए।

View Model Solution & Step Marking ▾

Model Answer:

Propanal (an aldehyde) will give a positive Tollens' test (silver mirror formed) or Fehlings' test (red precipitate). Propanone (a ketone) will not react. For Tollens' test:

CH_3CH_2CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow CH_3CH_2COO^- + 2Ag(s) + 4NH_3 + 2H_2O

.

Q10 • 3 Marks Short Answer

Complete the following reaction and name the product formed:

CH_3COOH + NH_3 \xrightarrow{\Delta} A

. What is the common name of product A?

निम्नलिखित अभिक्रिया को पूर्ण कीजिए और बनने वाले उत्पाद का नाम बताइए:

CH_3COOH + NH_3 \xrightarrow{\Delta} A

। उत्पाद A का सामान्य नाम क्या है?

View Model Solution & Step Marking ▾

Model Answer:

The reaction product A is acetamide. The reaction is:

CH_3COOH + NH_3 \rightarrow CH_3COONH_4 \xrightarrow{\Delta} CH_3CONH_2 + H_2O

. The common name of product A (

CH_3CONH_2

) is acetamide.

Part B: Long Answer Questions & Derivations (5 Marks Each)

Q1 • 5 Marks Long Answer / Derivation

a) An organic compound 'A' with molecular formula

C_3H_6O

is an aldehyde. It undergoes a reaction with dilute

NaOH

solution to give compound 'B'. Compound 'B' on heating gives compound 'C' with molecular formula

C_6H_{10}O

. Identify A, B, and C and write the chemical reactions involved. What is the name of this reaction?

a) एक कार्बनिक यौगिक 'A' जिसका अणुसूत्र

C_3H_6O

है, एक एल्डिहाइड है। यह तनु

NaOH

विलयन के साथ अभिक्रिया करके यौगिक 'B' देता है। यौगिक 'B' को गर्म करने पर यौगिक 'C' प्राप्त होता है, जिसका अणुसूत्र

C_6H_{10}O

है। A, B और C को पहचानिए तथा इसमें शामिल रासायनिक अभिक्रियाएँ लिखिए। इस अभिक्रिया का नाम क्या है?

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Compound 'A' is propanal (

CH_3CH_2CHO

). When propanal reacts with dilute

NaOH

, it undergoes aldol condensation. The product 'B' is 3-hydroxy-2-methylpentanal (

CH_3CH_2CH(OH)CH(CH_3)CHO

). On heating, 'B' undergoes dehydration to form 'C', which is 2-methylpent-2-enal (

CH_3CH_2CH=C(CH_3)CHO

). The reaction is Aldol Condensation.

Reactions:

2CH_3CH_2CHO \xrightarrow{Dilute\ NaOH} CH_3CH_2CH(OH)CH(CH_3)CHO

(Compound B)

CH_3CH_2CH(OH)CH(CH_3)CHO \xrightarrow{Heat} CH_3CH_2CH=C(CH_3)CHO + H_2O

(Compound C)

(1 mark for identifying A, B, C each, 1 mark for reactions, 1 mark for reaction name)

Q2 • 5 Marks Long Answer / Derivation

a) How will you convert ethanoic acid to methane? Write the chemical equations.
b) Give a chemical test to distinguish between propanal and propanone.
c) Explain why carboxylic acids are stronger acids than alcohols.

a) आप एथेनोइक अम्ल को मेथेन में कैसे परिवर्तित करेंगे? रासायनिक समीकरण लिखिए।
b) प्रोपेनल और प्रोपेनोन के बीच अंतर करने के लिए एक रासायनिक परीक्षण दीजिए।
c) समझाइए कि कार्बोक्सिलिक अम्ल, ऐल्कोहॉलों की तुलना में अधिक प्रबल अम्ल क्यों होते हैं।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Ethanoic acid to methane:

CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O

(Neutralization)

CH_3COONa + NaOH \xrightarrow{CaO, Heat} CH_4 + Na_2CO_3

(Decarboxylation/Kolbe's electrolysis is also acceptable).

b) Propanal and Propanone distinction: Tollen's reagent test.
Propanal gives a silver mirror with Tollen's reagent (

Ag(NH_3)_2OH

). Propanone does not.

CH_3CH_2CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow CH_3CH_2COO^- + 2Ag(s) + 4NH_3 + 2H_2O

c) Carboxylic acids are stronger acids than alcohols because the carboxylate ion (

RCOO^-

) formed after losing a proton is stabilized by resonance. The negative charge is delocalized over two electronegative oxygen atoms, making it more stable. In alcohols, the alkoxide ion (

RO^-

) is not resonance stabilized. The electron-donating alkyl group further destabilizes the alkoxide ion, making alcohols weaker acids.

(1.5 marks for a, 1.5 marks for b, 2 marks for c)

Q3 • 5 Marks Long Answer / Derivation

a) Give reasons for the following:
i) Carboxylic acids do not give characteristic reactions of carbonyl group.
ii) Benzaldehyde does not undergo Aldol condensation.
b) Complete the following reactions:
i)

CH_3COCH_3 \xrightarrow{i) CH_3MgBr, ii) H_2O}

ii)

C_6H_5CHO + conc. NaOH \rightarrow

iii)

CH_3COOH + PCl_5 \rightarrow

a) निम्नलिखित के कारण दीजिए:
i) कार्बोक्सिलिक अम्ल कार्बोनिल समूह की अभिलाक्षणिक अभिक्रियाएँ नहीं देते हैं।
ii) बेन्जैल्डिहाइड एल्डोल संघनन अभिक्रिया नहीं देता है।
b) निम्नलिखित अभिक्रियाओं को पूर्ण कीजिए:
i)

CH_3COCH_3 \xrightarrow{i) CH_3MgBr, ii) H_2O}

ii)

C_6H_5CHO + सांद्र. NaOH \rightarrow

iii)

CH_3COOH + PCl_5 \rightarrow

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) i) In carboxylic acids, the carbonyl carbon is attached to an -OH group. The lone pair of electrons on the oxygen of the -OH group is delocalized with the

\pi

-electrons of the carbonyl group (resonance), making the carbonyl group less electrophilic and thus it does not give characteristic reactions of aldehydes/ketones (like nucleophilic addition). The carbon-oxygen double bond acquires partial single bond character.
ii) Benzaldehyde does not contain any

\alpha

-hydrogen atoms. Aldol condensation requires the presence of at least one

\alpha

-hydrogen atom in the aldehyde or ketone to form a carbanion in the presence of a base.
b)
i)

CH_3COCH_3 \xrightarrow{i) CH_3MgBr, ii) H_2O} (CH_3)_3COH

(tert-Butyl alcohol)
ii)

C_6H_5CHO + conc. NaOH \rightarrow C_6H_5COONa + C_6H_5CH_2OH

(Cannizzaro reaction)
iii)

CH_3COOH + PCl_5 \rightarrow CH_3COCl + POCl_3 + HCl

(1 mark for each explanation, 1 mark for each reaction product)

Q4 • 5 Marks Long Answer / Derivation

a) How will you convert Ethanoic acid to Ethanal? Write the chemical equations involved.b) Explain why carboxylic acids do not give characteristic reactions of carbonyl compounds (like aldehydes and ketones).c) Write the chemical equations for the following conversions:i) Propanone to Propeneii) Benzoic acid to Benzaldehyde

क) एथेनोइक अम्ल को एथेनल में कैसे परिवर्तित करेंगे? इसमें शामिल रासायनिक समीकरण लिखिए।ख) समझाइए कि कार्बोक्सिलिक अम्ल कार्बोनिल यौगिकों (जैसे एल्डिहाइड और कीटोन) की अभिलाक्षणिक अभिक्रियाएँ क्यों नहीं देते हैं।ग) निम्नलिखित रूपांतरणों के लिए रासायनिक समीकरण लिखिए:i) प्रोपेनोन से प्रोपीनii) बेंजोइक अम्ल से बेंजलडिहाइड

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Conversion of Ethanoic acid to Ethanal:Ethanoic acid (

CH_3COOH

) is first reduced to ethanol (

CH_3CH_2OH

) using

LiAlH_4

or by passing vapours over red hot copper. Then ethanol is oxidized to ethanal (

CH_3CHO

) using

PCC

(Pyridinium chlorochromate) or by controlled oxidation with

K_2Cr_2O_7/H_2SO_4

.

CH_3COOH \xrightarrow{LiAlH_4} CH_3CH_2OH \xrightarrow{PCC} CH_3CHO

b) Carboxylic acids (

RCOOH

) have a carbonyl group, but they do not give characteristic reactions of carbonyl compounds (like nucleophilic addition reactions) because of resonance. The carbonyl carbon of carboxylic acids is less electrophilic than that of aldehydes and ketones due to the electron-donating resonance effect from the lone pair of electrons on the oxygen atom of the -OH group. This resonance stabilizes the carboxylic acid and reduces the partial positive charge on the carbonyl carbon, making it less susceptible to nucleophilic attack.c) i) Propanone to Propene:

CH_3COCH_3 \xrightarrow{LiAlH_4} CH_3CH(OH)CH_3

(Propan-2-ol)

CH_3CH(OH)CH_3 \xrightarrow{Conc. H_2SO_4, \Delta} CH_3CH=CH_2

(Propene)ii) Benzoic acid to Benzaldehyde:

C_6H_5COOH \xrightarrow{SOCl_2} C_6H_5COCl

(Benzoyl chloride)

C_6H_5COCl \xrightarrow{H_2, Pd/BaSO_4} C_6H_5CHO

(Benzaldehyde, Rosenmund reduction)|||HI|क) एथेनोइक अम्ल को एथेनल में परिवर्तित करना:एथेनोइक अम्ल (

CH_3COOH

) को पहले

LiAlH_4

का उपयोग करके या लाल गर्म तांबे पर वाष्प प्रवाहित करके एथेनॉल (

CH_3CH_2OH

) में अपचयित किया जाता है। फिर एथेनॉल को

PCC

(पिरिडिनियम क्लोरोक्रोमेट) का उपयोग करके या

K_2Cr_2O_7/H_2SO_4

के साथ नियंत्रित ऑक्सीकरण द्वारा एथेनल (

CH_3CHO

) में ऑक्सीकृत किया जाता है।

CH_3COOH \xrightarrow{LiAlH_4} CH_3CH_2OH \xrightarrow{PCC} CH_3CHO

ख) कार्बोक्सिलिक अम्लों (

RCOOH

) में कार्बोनिल समूह होता है, लेकिन वे अनुनाद के कारण कार्बोनिल यौगिकों (जैसे न्यूक्लियोफिलिक योग अभिक्रियाएँ) की अभिलाक्षणिक अभिक्रियाएँ नहीं देते हैं। कार्बोक्सिलिक अम्लों का कार्बोनिल कार्बन एल्डिहाइड और कीटोन की तुलना में कम इलेक्ट्रोफिलिक होता है क्योंकि -OH समूह के ऑक्सीजन परमाणु पर एकाकी इलेक्ट्रॉन युग्म से इलेक्ट्रॉन-दाता अनुनाद प्रभाव होता है। यह अनुनाद कार्बोक्सिलिक अम्ल को स्थायित्व प्रदान करता है और कार्बोनिल कार्बन पर आंशिक धनात्मक आवेश को कम करता है, जिससे यह न्यूक्लियोफिलिक आक्रमण के प्रति कम संवेदनशील हो जाता है।ग) i) प्रोपेनोन से प्रोपीन:

CH_3COCH_3 \xrightarrow{LiAlH_4} CH_3CH(OH)CH_3

(प्रोपेन-2-ऑल)

CH_3CH(OH)CH_3 \xrightarrow{सांद्र H_2SO_4, \Delta} CH_3CH=CH_2

(प्रोपीन)ii) बेंजोइक अम्ल से बेंजलडिहाइड:

C_6H_5COOH \xrightarrow{SOCl_2} C_6H_5COCl

(बेंजोयल क्लोराइड)

C_6H_5COCl \xrightarrow{H_2, Pd/BaSO_4} C_6H_5CHO

(बेंजलडिहाइड, रोज़ेनमुंड अपचयन)

Q5 • 5 Marks Long Answer / Derivation

a) Give a chemical test to distinguish between Benzaldehyde and Acetophenone.b) Arrange the following compounds in increasing order of their boiling points: Ethanal, Propan-1-ol, Ethanoic acid. Give reasons.c) What happens when:

CH_3CHO

is treated with

NaOH

(dilute) and then heated?

CH_3COOH

is treated with

PCl_5

?Write chemical equations for each reaction.|||HI|क) बेंजलडिहाइड और एसीटोफेनोन के बीच अंतर करने के लिए एक रासायनिक परीक्षण दीजिए।ख) निम्नलिखित यौगिकों को उनके क्वथनांकों के बढ़ते क्रम में व्यवस्थित कीजिए: एथेनल, प्रोपेन-1-ऑल, एथेनोइक अम्ल। कारण भी दीजिए।ग) क्या होता है जब:

CH_3CHO

को

NaOH

(तनु) के साथ उपचारित किया जाता है और फिर गर्म किया जाता है?

CH_3COOH

को

PCl_5

के साथ उपचारित किया जाता है?प्रत्येक अभिक्रिया के लिए रासायनिक समीकरण लिखिए।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Iodoform test:Acetophenone (

C_6H_5COCH_3

) contains a methyl ketone group (

CH_3CO-

) and will give a positive iodoform test, forming a yellow precipitate of iodoform (

CHI_3

) when treated with

I_2/NaOH

.Benzaldehyde (

C_6H_5CHO

) does not have a methyl ketone group and will not give a positive iodoform test.Equation:

C_6H_5COCH_3 + 3I_2 + 4NaOH \rightarrow C_6H_5COONa + CHI_3 \downarrow + 3NaI + 3H_2O

b) Increasing order of boiling points: Ethanal < Propan-1-ol < Ethanoic acid.Reason:Ethanal has a dipole-dipole interaction due to its polar carbonyl group.Propan-1-ol has stronger intermolecular hydrogen bonding due to the presence of a hydroxyl group.Ethanoic acid has the strongest intermolecular hydrogen bonding because it forms dimeric structures through two hydrogen bonds, making its effective molecular mass much higher than its actual molecular mass, requiring more energy to break these bonds.c) i) When

CH_3CHO

is treated with dilute

NaOH

and then heated, Aldol Condensation occurs, forming 3-Hydroxybutanal, which on heating gives But-2-enal.

2CH_3CHO \xrightarrow{Dil. NaOH} CH_3CH(OH)CH_2CHO

(3-Hydroxybutanal)

CH_3CH(OH)CH_2CHO \xrightarrow{\Delta} CH_3CH=CHCHO + H_2O

(But-2-enal)ii) When

CH_3COOH

is treated with

PCl_5

, acetyl chloride (

CH_3COCl

) is formed along with

POCl_3

and

HCl

.

CH_3COOH + PCl_5 \rightarrow CH_3COCl + POCl_3 + HCl

|||HI|क) आयोडोफॉर्म परीक्षण:एसीटोफेनोन (

C_6H_5COCH_3

) में एक मेथिल कीटोन समूह (

CH_3CO-

) होता है और यह

I_2/NaOH

के साथ उपचारित करने पर आयोडोफॉर्म (

CHI_3

) का पीला अवक्षेप बनाकर एक धनात्मक आयोडोफॉर्म परीक्षण देगा।बेंजलडिहाइड (

C_6H_5CHO

) में मेथिल कीटोन समूह नहीं होता है और यह धनात्मक आयोडोफॉर्म परीक्षण नहीं देगा।समीकरण:

C_6H_5COCH_3 + 3I_2 + 4NaOH \rightarrow C_6H_5COONa + CHI_3 \downarrow + 3NaI + 3H_2O

ख) क्वथनांकों का बढ़ता क्रम: एथेनल < प्रोपेन-1-ऑल < एथेनोइक अम्ल।कारण:एथेनल में ध्रुवीय कार्बोनिल समूह के कारण द्विध्रुव-द्विध्रुव अंतःक्रिया होती है।प्रोपेन-1-ऑल में हाइड्रॉक्सिल समूह की उपस्थिति के कारण प्रबल अंतर-आणविक हाइड्रोजन बंधन होता है।एथेनोइक अम्ल में सबसे प्रबल अंतर-आणविक हाइड्रोजन बंधन होता है क्योंकि यह दो हाइड्रोजन बंधों के माध्यम से द्विलकी संरचनाएँ बनाता है, जिससे इसका प्रभावी आणविक द्रव्यमान इसके वास्तविक आणविक द्रव्यमान से बहुत अधिक हो जाता है, और इन बंधों को तोड़ने के लिए अधिक ऊर्जा की आवश्यकता होती है।ग) i) जब

CH_3CHO

को तनु

NaOH

के साथ उपचारित किया जाता है और फिर गर्म किया जाता है, तो एल्डोल संघनन होता है, जिससे 3-हाइड्रॉक्सीब्यूटेनल बनता है, जो गर्म करने पर ब्यूट-2-एनल देता है।

2CH_3CHO \xrightarrow{तनु NaOH} CH_3CH(OH)CH_2CHO

(3-हाइड्रॉक्सीब्यूटेनल)

CH_3CH(OH)CH_2CHO \xrightarrow{\Delta} CH_3CH=CHCHO + H_2O

(ब्यूट-2-एनल)ii) जब

CH_3COOH

को

PCl_5

के साथ उपचारित किया जाता है, तो एसिटिल क्लोराइड (

CH_3COCl

) का निर्माण

POCl_3

और

HCl

के साथ होता है।

CH_3COOH + PCl_5 \rightarrow CH_3COCl + POCl_3 + HCl

9

Amines

Part A: Short Answer Questions (2–3 Marks Each)

Q1 • 3 Marks Short Answer

Explain why aniline is less basic than ethylamine. Give appropriate reasons for your answer.

स्पष्ट कीजिए कि एनिलीन एथिलएमीन से कम क्षारीय क्यों है। अपने उत्तर के लिए उचित कारण दीजिए।

View Model Solution & Step Marking ▾

Model Answer:

Aniline is less basic than ethylamine due to the delocalization of the lone pair of electrons on the nitrogen atom of aniline into the benzene ring through resonance. This makes the lone pair less available for protonation. In contrast, in ethylamine, the ethyl group is an electron-donating group, which increases the electron density on the nitrogen atom, making it more available for protonation and thus more basic.

Q2 • 3 Marks Short Answer

How will you convert methylamine to methanol? Write the chemical equations involved.

आप मेथिलएमीन को मेथनॉल में कैसे परिवर्तित करेंगे? इसमें शामिल रासायनिक समीकरण लिखिए।

View Model Solution & Step Marking ▾

Model Answer:

Methylamine can be converted to methanol by reacting it with nitrous acid (

HNO_2

), which is generated in situ from

NaNO_2

and

HCl

. This reaction forms a diazonium salt, which then decomposes to produce methanol and nitrogen gas.

CH_3NH_2 + HNO_2 \rightarrow CH_3N_2^+Cl^- + 2H_2O

CH_3N_2^+Cl^- \rightarrow CH_3OH + N_2 + HCl

Q3 • 3 Marks Short Answer

Give the mechanism of ammonolysis of alkyl halides for the preparation of amines.

ऐमीन के विरचन के लिए ऐल्किल हैलाइडों के अमोनोलिसिस की क्रियाविधि दीजिए।

View Model Solution & Step Marking ▾

Model Answer:

The mechanism of ammonolysis involves the nucleophilic attack of ammonia on the alkyl halide. The lone pair of electrons on the nitrogen atom of ammonia attacks the electrophilic carbon atom of the alkyl halide, leading to the displacement of the halide ion. This forms an alkylammonium salt, which on treatment with a strong base yields the primary amine. Further reaction can occur to form secondary, tertiary amines, and quaternary ammonium salts.

Q4 • 3 Marks Short Answer

How will you convert aniline to

p

-bromoaniline? Explain the steps involved.

आप ऐनिलीन को

p

-ब्रोमोऐनिलीन में कैसे परिवर्तित करेंगे? इसमें शामिल चरणों की व्याख्या कीजिए।

View Model Solution & Step Marking ▾

Model Answer:

Aniline reacts with bromine water to give 2,4,6-tribromoaniline due to strong activating effect of

-\text{NH}_2

group. To get

p

-bromoaniline, the amino group must be protected by acetylation with acetic anhydride to form acetanilide. Then, bromination gives

p

-bromoacetanilide. Finally, hydrolysis with

\text{H}^+/\text{OH}^-

regenerates the amino group, yielding

p

-bromoaniline.

Q5 • 3 Marks Short Answer

Explain why aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis.

समझाइए कि गैब्रियल थैलिमाइड संश्लेषण द्वारा ऐरोमैटिक प्राथमिक एमीन क्यों नहीं बनाए जा सकते हैं।

View Model Solution & Step Marking ▾

Model Answer:

Gabriel phthalimide synthesis involves the nucleophilic substitution of an alkyl halide by the phthalimide anion. Aromatic halides (e.g., bromobenzene) do not undergo nucleophilic substitution with the phthalimide anion under normal conditions because the carbon-halogen bond has partial double bond character due to resonance, making it less reactive towards nucleophiles. Therefore, aryl amines cannot be prepared by this method.

Q6 • 3 Marks Short Answer

Distinguish between primary, secondary, and tertiary amines using the Hinsberg's reagent. Describe the observations for each case.

हिन्सबर्ग अभिकर्मक का उपयोग करके प्राथमिक, द्वितीयक और तृतीयक ऐमीन के बीच अंतर स्पष्ट कीजिए। प्रत्येक स्थिति के लिए प्रेक्षणों का वर्णन कीजिए।

View Model Solution & Step Marking ▾

Model Answer:

Primary amines react with Hinsberg's reagent (benzenesulphonyl chloride) to form N-alkylbenzenesulphonamide, which is soluble in alkali. Secondary amines react to form N,N-dialkylbenzenesulphonamide, which is insoluble in alkali. Tertiary amines do not react with Hinsberg's reagent under normal conditions.

Q7 • 3 Marks Short Answer

Arrange the following in increasing order of their basic strength: Methylamine, Aniline,

N

-methylaniline. Justify your answer.

निम्नलिखित को उनकी क्षारीय शक्ति के बढ़ते क्रम में व्यवस्थित कीजिए: मेथिलैमीन, ऐनिलीन,

N

-मेथिलऐनिलीन। अपने उत्तर का औचित्य सिद्ध कीजिए।

View Model Solution & Step Marking ▾

Model Answer:

The increasing order of basic strength is Aniline <

N

-methylaniline < Methylamine. Aniline is less basic due to resonance stabilization of the lone pair on nitrogen with the benzene ring.

N

-methylaniline is more basic than aniline due to the electron-donating effect of the methyl group, but less basic than methylamine due to partial resonance and steric hindrance. Methylamine is most basic due to the positive inductive effect of the methyl group stabilizing the conjugate acid.

Q8 • 3 Marks Short Answer

Write the chemical equations for the Carbylamine reaction. What is the characteristic feature of this reaction, and what type of amines give this test?

कार्बिलैमीन अभिक्रिया के लिए रासायनिक समीकरण लिखिए। इस अभिक्रिया की विशिष्ट विशेषता क्या है, और किस प्रकार के एमीन यह परीक्षण देते हैं?

View Model Solution & Step Marking ▾

Model Answer:

The Carbylamine reaction (Isocyanide test) is:

\text{R-NH}_2 + \text{CHCl}_3 + 3\text{KOH} \xrightarrow{\text{heat}} \text{R-NC} + 3\text{KCl} + 3\text{H}_2\text{O}

. The characteristic feature is the formation of foul-smelling isocyanides (carbylamines). This reaction is given only by primary aliphatic and primary aromatic amines.

Q9 • 3 Marks Short Answer

Explain why aniline is less basic than ethylamine. Give appropriate reasons for your answer.

स्पष्ट कीजिए कि एनिलीन एथिलऐमीन की तुलना में कम क्षारीय क्यों है। अपने उत्तर के लिए उचित कारण दीजिए।

View Model Solution & Step Marking ▾

Model Answer:

Aniline is less basic than ethylamine due to the delocalization of the lone pair of electrons on the nitrogen atom of aniline into the benzene ring through resonance. This makes the lone pair less available for protonation. In ethylamine, there is no such delocalization, and the ethyl group's electron-donating inductive effect

(+I)

increases the electron density on the nitrogen, making it more basic.

Q10 • 3 Marks Short Answer

Write the balanced chemical equations for the following reactions:
(a) Acetylation of ethylamine
(b) Carbylamine reaction with aniline

निम्नलिखित अभिक्रियाओं के लिए संतुलित रासायनिक समीकरण लिखिए:
(a) एथिलएमीन का ऐसीटिलीकरण
(b) ऐनिलीन के साथ कार्बिलएमीन अभिक्रिया

View Model Solution & Step Marking ▾

Model Answer:

(a) Acetylation of ethylamine:

CH_3CH_2NH_2 + (CH_3CO)_2O \rightarrow CH_3CH_2NHCOCH_3 + CH_3COOH

(b) Carbylamine reaction with aniline:

C_6H_5NH_2 + CHCl_3 + 3KOH \xrightarrow{\text{heat}} C_6H_5NC + 3KCl + 3H_2O

Part B: Long Answer Questions & Derivations (5 Marks Each)

Q1 • 5 Marks Long Answer / Derivation

Account for the following:
(a) Arrange the following in increasing order of their basic strength in aqueous solution:

(CH_3)_2NH, (CH_3)_3N, CH_3NH_2, NH_3

.
(b) Aniline does not undergo Friedel-Crafts reaction.
(c) The

pK_b

value of aniline is much higher than that of methylamine.
(d) Gabriel phthalimide synthesis is preferred for synthesising primary amines.

निम्नलिखित के लिए कारण बताइए:
(a) निम्नलिखित को जलीय विलयन में उनकी क्षारकीय प्रबलता के बढ़ते क्रम में व्यवस्थित कीजिए:

(CH_3)_2NH, (CH_3)_3N, CH_3NH_2, NH_3

।
(b) एनिलिन फ्रीडेल-क्राफ्ट्स अभिक्रिया नहीं करती है।
(c) एनिलिन का

pK_b

मान मेथिलऐमीन की तुलना में बहुत अधिक होता है।
(d) प्राथमिक ऐमीन के संश्लेषण के लिए गैब्रिएल थैलिमाइड संश्लेषण को प्राथमिकता दी जाती है।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

(a) The increasing order of basic strength in aqueous solution is:

NH_3 < (CH_3)_3N < CH_3NH_2 < (CH_3)_2NH

. This order is due to the combined effect of inductive effect, solvation effect, and steric hindrance of the alkyl groups.

(b) Aniline is a Lewis base. It reacts with the Lewis acid catalyst,

AlCl_3

, used in Friedel-Crafts reaction to form a salt. This deactivates the benzene ring by producing a positive charge on the nitrogen atom, making it highly resistant to electrophilic substitution.

(c) In aniline, the lone pair of electrons on the nitrogen atom is delocalised over the benzene ring due to resonance. This makes the lone pair less available for protonation. In methylamine, the

+I

effect of the methyl group increases the electron density on the nitrogen atom, making it more basic. Higher basicity means a lower

pK_b

value. Therefore, aniline has a higher

pK_b

value.

(d) Gabriel phthalimide synthesis yields pure primary amines without any contamination of secondary or tertiary amines. Aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.

Q2 • 5 Marks Long Answer / Derivation

(a) Write the chemical reactions involved in the preparation of a diazonium salt from aniline. Why is the temperature maintained between

273\ K

and

278\ K

during this reaction?
(b) Illustrate the Sandmeyer reaction and the Gattermann reaction by taking benzenediazonium chloride as the starting material to prepare chlorobenzene.
(c) How can you convert benzenediazonium chloride to phenol?

(a) एनिलिन से डाइएज़ोनियम लवण बनाने में शामिल रासायनिक अभिक्रियाएँ लिखिए। इस अभिक्रिया के दौरान तापमान

273\ K

से

278\ K

के बीच क्यों बनाए रखा जाता है?
(b) क्लोरोबेंजीन तैयार करने के लिए बेंजीनडाइएज़ोनियम क्लोराइड को प्रारंभिक पदार्थ के रूप में लेते हुए सैंडमेयर अभिक्रिया और गाटरमान अभिक्रिया का वर्णन कीजिए।
(c) आप बेंजीनडाइएज़ोनियम क्लोराइड को फीनॉल में कैसे परिवर्तित कर सकते हैं?

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

(a) Preparation of benzenediazonium chloride from aniline:
Aniline is treated with nitrous acid (

NaNO_2 + HCl

) at low temperature (

273-278\ K

).

C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273-278\ K} C_6H_5N_2^+Cl^- + NaCl + 2H_2O

The temperature is kept low because benzenediazonium chloride is unstable and decomposes at higher temperatures.

(b) Sandmeyer reaction:

C_6H_5N_2^+Cl^- \xrightarrow{CuCl/HCl} C_6H_5Cl + N_2

Gattermann reaction:

C_6H_5N_2^+Cl^- \xrightarrow{Cu/HCl} C_6H_5Cl + N_2

(c) Conversion to phenol:
The diazonium salt solution is warmed with water.

C_6H_5N_2^+Cl^- + H_2O \xrightarrow{Warm} C_6H_5OH + N_2 + HCl

Q3 • 5 Marks Long Answer / Derivation

(a) Give a chemical test to distinguish between Aniline and N-methylaniline.
(b) How will you carry out the following conversions?
(i) Ethanoic acid to methanamine
(ii) Aniline to p-bromoaniline
(c) Write the reaction of ethanamine with benzoyl chloride and name the product formed.

(a) एनिलिन और N-मेथिलएनिलिन के बीच अंतर करने के लिए एक रासायनिक परीक्षण दीजिए।
(b) आप निम्नलिखित रूपांतरणों को कैसे करेंगे?
(i) एथेनोइक अम्ल से मेथेनामीन
(ii) एनिलिन से p-ब्रोमोएनिलिन
(c) एथेनामीन की बेंज़ोयल क्लोराइड के साथ अभिक्रिया लिखिए और बनने वाले उत्पाद का नाम बताइए।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

(a) Carbylamine test: Aniline (a primary amine) on heating with chloroform and alcoholic potassium hydroxide gives a foul-smelling isocyanide (phenyl isocyanide). N-methylaniline (a secondary amine) does not give this test.
Reaction:

C_6H_5NH_2 + CHCl_3 + 3KOH(alc.) \xrightarrow{Heat} C_6H_5NC + 3KCl + 3H_2O

(b) (i) Ethanoic acid to methanamine (Hofmann bromamide degradation):

CH_3COOH \xrightarrow{SOCl_2} CH_3COCl \xrightarrow{NH_3} CH_3CONH_2 \xrightarrow{Br_2/NaOH} CH_3NH_2

(ii) Aniline to p-bromoaniline (Protection of amino group):
First, aniline is acetylated with acetic anhydride to form acetanilide. This reduces the activating effect of the amino group.

C_6H_5NH_2 \xrightarrow{(CH_3CO)_2O} C_6H_5NHCOCH_3

Then, bromination is carried out, which gives p-bromoacetanilide as the major product.

C_6H_5NHCOCH_3 \xrightarrow{Br_2/CH_3COOH} p-Br-C_6H_4NHCOCH_3

Finally, hydrolysis of p-bromoacetanilide gives p-bromoaniline.

p-Br-C_6H_4NHCOCH_3 \xrightarrow{H^+ or OH^-} p-Br-C_6H_4NH_2

(c) Ethanamine reacts with benzoyl chloride in a reaction called benzoylation to form N-ethylbenzamide.

C_6H_5COCl + C_2H_5NH_2 \rightarrow C_6H_5CONHC_2H_5 + HCl

Product name: N-ethylbenzamide

Q4 • 5 Marks Long Answer / Derivation

(a) Write the chemical reactions of aniline with the following reagents:
(i) Bromine water
(ii) Acetyl chloride
(iii) Benzoyl chloride
(b) How will you convert ethanenitrile into ethanamine?
(c) How will you convert aniline to chlorobenzene?

(क) निम्नलिखित अभिकर्मकों के साथ एनिलिन की रासायनिक अभिक्रियाएँ लिखिए:
(i) ब्रोमीन जल
(ii) एसिटिल क्लोराइड
(iii) बेन्जॉयल क्लोराइड
(ख) आप एथेननाइट्राइल को एथेनऐमीन में कैसे परिवर्तित करेंगे?
(ग) आप एनिलिन को क्लोरोबेंजीन में कैसे परिवर्तित करेंगे?

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

(a) (i) Aniline reacts with bromine water at room temperature to give a white precipitate of 2,4,6-tribromoaniline.

C_6H_5NH_2 + 3Br_2(aq) \rightarrow C_6H_2Br_3NH_2(s) + 3HBr

(ii) Aniline reacts with acetyl chloride in the presence of pyridine to form N-phenylethanamide (acetanilide).

C_6H_5NH_2 + CH_3COCl \xrightarrow{Pyridine} C_6H_5NHCOCH_3 + HCl

(iii) Aniline reacts with benzoyl chloride (Schotten-Baumann reaction) to form N-phenylbenzamide.

C_6H_5NH_2 + C_6H_5COCl \rightarrow C_6H_5NHCOC_6H_5 + HCl

(b) Ethanenitrile can be converted into ethanamine by reduction with lithium aluminium hydride (

LiAlH_4

) or by catalytic hydrogenation using

H_2/Ni

.

CH_3C\equiv N \xrightarrow{LiAlH_4 / H_2O} CH_3CH_2NH_2

(c) Aniline is first converted to benzenediazonium chloride by diazotisation with

NaNO_2 + HCl

at

273-278

K. The diazonium salt is then treated with cuprous chloride (

CuCl

) dissolved in

HCl

(Sandmeyer reaction) to give chlorobenzene.

C_6H_5NH_2 \xrightarrow{NaNO_2 + HCl, 273-278 K} C_6H_5N_2^+Cl^- \xrightarrow{CuCl/HCl} C_6H_5Cl + N_2

Q5 • 5 Marks Long Answer / Derivation

An aromatic compound 'A' on treatment with aqueous ammonia and heating forms compound 'B', which on heating with

Br_2

and

KOH

forms a compound 'C' of molecular formula

C_6H_7N

. Write the structures and IUPAC names of compounds A, B, and C. Also, write the chemical equations for the reactions involved. Explain why compound 'C' is a weaker base than cyclohexylamine.

एक ऐरोमैटिक यौगिक 'A' जलीय अमोनिया के साथ उपचार और गर्म करने पर यौगिक 'B' बनाता है, जिसे

Br_2

और

KOH

के साथ गर्म करने पर आणविक सूत्र

C_6H_7N

का एक यौगिक 'C' बनता है। यौगिकों A, B, और C की संरचनाएँ और IUPAC नाम लिखिए। इसमें शामिल अभिक्रियाओं के लिए रासायनिक समीकरण भी लिखिए। समझाइए कि यौगिक 'C' साइक्लोहेक्सिलएमीन की तुलना में एक दुर्बल क्षार क्यों है।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

The compound 'C' with molecular formula

C_6H_7N

is aniline (

C_6H_5NH_2

).
Compound 'C' is formed from 'B' by Hofmann bromamide degradation reaction (

Br_2/KOH

), which means 'B' is an amide with one more carbon atom than 'C'. So, 'B' is Benzamide (

C_6H_5CONH_2

).
Compound 'B' is formed from 'A' by reaction with aqueous ammonia and heating. This indicates 'A' is a carboxylic acid. So, 'A' is Benzoic acid (

C_6H_5COOH

).

Structures and IUPAC Names:
A: Benzoic acid (

C_6H_5COOH

)
B: Benzamide (

C_6H_5CONH_2

)
C: Aniline or Benzenamine (

C_6H_5NH_2

)

Reactions:
(i)

C_6H_5COOH (A) + NH_3(aq) \rightarrow C_6H_5COONH_4^+ \xrightarrow{\Delta} C_6H_5CONH_2 (B) + H_2O

(ii)

C_6H_5CONH_2 (B) + Br_2 + 4KOH \rightarrow C_6H_5NH_2 (C) + K_2CO_3 + 2KBr + 2H_2O

Basicity Comparison: Aniline ('C') is a weaker base than cyclohexylamine. In aniline, the lone pair of electrons on the nitrogen atom is delocalized into the benzene ring due to resonance, making it less available for protonation. In cyclohexylamine, the lone pair is localized on the nitrogen atom and the cyclohexyl group exerts a +I effect, increasing the electron density on nitrogen and making it a stronger base.

10

Biomolecules

Part A: Short Answer Questions (2–3 Marks Each)

Q1 • 3 Marks Short Answer

Explain the denaturation of proteins. What are the factors that can cause denaturation and what is its effect on the biological activity of proteins?

प्रोटीन के विकृतीकरण की व्याख्या कीजिए। ऐसे कौन से कारक हैं जो विकृतीकरण का कारण बन सकते हैं और प्रोटीन की जैविक गतिविधि पर इसका क्या प्रभाव पड़ता है?

View Model Solution & Step Marking ▾

Model Answer:

Denaturation is the process where a protein loses its specific three-dimensional structure (secondary, tertiary, and quaternary structures) due to physical or chemical changes, while the primary structure remains intact. Factors causing denaturation include changes in temperature, pH, presence of salts, and heavy metal ions. Denaturation leads to the loss of biological activity of the protein because the specific shape required for its function is disrupted.

Q2 • 3 Marks Short Answer

Differentiate between essential and non-essential amino acids. Give one example of each. Why are essential amino acids crucial for human health?

आवश्यक और गैर-आवश्यक अमीनो अम्लों में अंतर स्पष्ट कीजिए। प्रत्येक का एक-एक उदाहरण दीजिए। आवश्यक अमीनो अम्ल मानव स्वास्थ्य के लिए क्यों महत्वपूर्ण हैं?

View Model Solution & Step Marking ▾

Model Answer:

Essential amino acids are those that cannot be synthesized by the human body and must be obtained from the diet (e.g., Valine). Non-essential amino acids can be synthesized by the body (e.g., Glycine). Essential amino acids are crucial because they serve as building blocks for proteins and other biomolecules, and their deficiency can lead to various health issues.

Q3 • 3 Marks Short Answer

Describe the primary and secondary structures of proteins. How do peptide bonds contribute to the formation of the primary structure?

प्रोटीन की प्राथमिक और द्वितीयक संरचनाओं का वर्णन कीजिए। पेप्टाइड बंध प्राथमिक संरचना के निर्माण में कैसे योगदान करते हैं?

View Model Solution & Step Marking ▾

Model Answer:

The primary structure of a protein refers to the specific sequence of amino acids linked together by peptide bonds. Peptide bonds (

—CONH—

) are formed between the carboxyl group of one amino acid and the amino group of another, forming a linear chain. The secondary structure refers to the regular folding patterns of the polypeptide chain, such as

\alpha

-helix and

\beta

-pleated sheet, stabilized by hydrogen bonds between the carbonyl oxygen and amide hydrogen atoms.

Q4 • 3 Marks Short Answer

Explain the biological functions of nucleic acids, specifically DNA and RNA. How do they differ in their primary roles within a cell?

न्यूक्लिक अम्लों, विशेष रूप से DNA और RNA के जैविक कार्यों की व्याख्या कीजिए। वे एक कोशिका के भीतर अपनी प्राथमिक भूमिकाओं में कैसे भिन्न होते हैं?

View Model Solution & Step Marking ▾

Model Answer:

DNA's primary biological function is to store and transmit genetic information from one generation to the next. It carries the blueprint for all proteins. RNA's primary biological functions include carrying genetic information from DNA to ribosomes (mRNA), translating genetic information into proteins (tRNA), and forming structural components of ribosomes (rRNA). Thus, DNA is the genetic repository, while RNA is involved in gene expression.

Q5 • 3 Marks Short Answer

Carbohydrates are broadly classified into monosaccharides, oligosaccharides, and polysaccharides. Give one example for each class and explain their general characteristics.

कार्बोहाइड्रेट को मोटे तौर पर मोनोसैकराइड, ओलिगोसैकराइड और पॉलीसैकराइड में वर्गीकृत किया जाता है। प्रत्येक वर्ग का एक-एक उदाहरण दीजिए और उनकी सामान्य विशेषताओं की व्याख्या कीजिए।

View Model Solution & Step Marking ▾

Model Answer:

Monosaccharides are the simplest carbohydrates that cannot be hydrolyzed further (e.g., Glucose). They are typically sweet, water-soluble, and reducing sugars. Oligosaccharides yield 2 to 10 monosaccharide units upon hydrolysis (e.g., Sucrose, which yields glucose and fructose). Polysaccharides are complex carbohydrates formed by a large number of monosaccharide units linked together (e.g., Starch). They are generally not sweet, often insoluble in water, and can be storage or structural components.

Q6 • 3 Marks Short Answer

Identify and explain the structural difference between glucose and fructose, both being monosaccharides. How does this difference affect their classification?

ग्लूकोज और फ्रक्टोज, दोनों मोनोसैकेराइड होते हुए भी, उनकी संरचनात्मक भिन्नता को पहचानिए और समझाइए। यह भिन्नता उनके वर्गीकरण को कैसे प्रभावित करती है?

View Model Solution & Step Marking ▾

Model Answer:

Glucose is an aldohexose, containing an aldehyde group (

- ext{CHO}

). Fructose is a ketohexose, containing a ketone group (

> ext{C=O}

). This structural difference classifies glucose as an aldose and fructose as a ketose.

Q7 • 3 Marks Short Answer

Explain the significance of peptide bonds in the formation of proteins. How are these bonds formed between amino acids?

प्रोटीन के निर्माण में पेप्टाइड बंधों के महत्व की व्याख्या कीजिए। ये बंध अमीनो अम्लों के बीच कैसे बनते हैं?

View Model Solution & Step Marking ▾

Model Answer:

Peptide bonds are crucial for linking amino acids together to form polypeptide chains, which are the primary structure of proteins. They are formed by the condensation reaction between the carboxyl group of one amino acid and the amino group of another, with the elimination of a water molecule.

Q8 • 3 Marks Short Answer

Differentiate between DNA and RNA based on their sugar component and nitrogenous bases. Explain the functional implications of these differences.

डीएनए और आरएनए के बीच उनके शर्करा घटक और नाइट्रोजनस क्षारों के आधार पर अंतर स्पष्ट कीजिए। इन अंतरों के कार्यात्मक निहितार्थों की व्याख्या कीजिए।

View Model Solution & Step Marking ▾

Model Answer:

DNA contains deoxyribose sugar, while RNA contains ribose sugar. DNA has adenine, guanine, cytosine, and thymine as nitrogenous bases, whereas RNA has adenine, guanine, cytosine, and uracil. These differences contribute to DNA's stability for genetic information storage and RNA's versatility in gene expression.

Q9 • 3 Marks Short Answer

Define denaturation of proteins. List two factors that can cause protein denaturation and explain how it affects the protein's biological activity.

प्रोटीन के विकृतीकरण को परिभाषित कीजिए। प्रोटीन विकृतीकरण का कारण बनने वाले किन्हीं दो कारकों को सूचीबद्ध कीजिए और समझाइए कि यह प्रोटीन की जैविक गतिविधि को कैसे प्रभावित करता है।

View Model Solution & Step Marking ▾

Model Answer:

Denaturation is the process where a protein loses its native three-dimensional structure (secondary, tertiary, and quaternary) due to disruption of non-covalent bonds, without breaking peptide bonds. Factors include heat and changes in pH. Denaturation causes the protein to lose its biological activity because the specific 3D structure is essential for its function.

Q10 • 3 Marks Short Answer

Explain why vitamin D is crucial for bone health. What happens if there is a deficiency of vitamin D in the human body?

समझाइए कि विटामिन

ext{D}

हड्डियों के स्वास्थ्य के लिए क्यों महत्वपूर्ण है। यदि मानव शरीर में विटामिन

ext{D}

की कमी हो जाए तो क्या होता है?

View Model Solution & Step Marking ▾

Model Answer:

Vitamin D is crucial for bone health as it regulates the absorption of calcium and phosphorus in the intestines, which are essential for bone mineralization. A deficiency of vitamin D can lead to rickets in children, characterized by soft and weak bones, and osteomalacia in adults, causing bone pain and muscle weakness.

Part B: Long Answer Questions & Derivations (5 Marks Each)

Q1 • 5 Marks Long Answer / Derivation

a) What are essential and non-essential amino acids? Give one example of each.

क) आवश्यक और अनावश्यक अमीनो अम्ल क्या होते हैं? प्रत्येक का एक-एक उदाहरण दीजिए।
b) Draw the Zwitterionic form of glycine and explain why amino acids show amphoteric behavior.|||HI|||ख) ग्लाइसिन का ज़्विटर आयनिक रूप बनाइए और समझाइए कि अमीनो अम्ल उभयधर्मी व्यवहार क्यों प्रदर्शित करते हैं।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Essential amino acids are those that cannot be synthesized by the human body and must be obtained from the diet. Example: Valine, Leucine, Isoleucine. Non-essential amino acids are those that can be synthesized by the body from other compounds and thus do not need to be supplied through the diet. Example: Glycine, Alanine, Serine.

Q2 • 5 Marks Long Answer / Derivation

a) Explain the term 'denaturation of proteins'. What are the factors responsible for denaturation?

क) 'प्रोटीन का विकृतीकरण' पद को समझाइए। विकृतीकरण के लिए कौन से कारक जिम्मेदार हैं?
b) How is the primary structure of a protein different from its secondary structure?|||HI|||ख) प्रोटीन की प्राथमिक संरचना उसकी द्वितीयक संरचना से किस प्रकार भिन्न है?

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Denaturation of proteins refers to the process where the three-dimensional structure of a protein (secondary, tertiary, and quaternary structures) is disrupted, leading to a loss of its biological activity. The primary structure (sequence of amino acids) remains intact. Factors responsible for denaturation include changes in temperature (e.g., heating), changes in pH (addition of acid or base), and exposure to certain chemicals (e.g., organic solvents, heavy metal salts).

Q3 • 5 Marks Long Answer / Derivation

a) Differentiate between DNA and RNA based on their structure and functions. (Any three points)

क) DNA और RNA में उनकी संरचना और कार्यों के आधार पर अंतर स्पष्ट कीजिए। (कोई तीन बिंदु)
b) What are the products formed when adenine undergoes hydrolysis in DNA?|||HI|||ख) जब DNA में एडिनीन का जल-अपघटन होता है तो कौन से उत्पाद बनते हैं?

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Differences between DNA and RNA:
1. **Sugar**: DNA contains deoxyribose sugar, while RNA contains ribose sugar.
2. **Bases**: DNA contains A, G, C, T. RNA contains A, G, C, U (uracil replaces thymine).
3. **Structure**: DNA is typically a double-stranded helix. RNA is usually a single-stranded molecule.
4. **Location**: DNA is primarily found in the nucleus (and mitochondria/chloroplasts). RNA is found in the nucleus, cytoplasm, and ribosomes.
5. **Function**: DNA is the genetic material, responsible for heredity and long-term storage of genetic information. RNA is involved in protein synthesis (mRNA, tRNA, rRNA) and gene regulation.

Q4 • 5 Marks Long Answer / Derivation

a) Define peptide linkage. Illustrate the formation of a dipeptide from two amino acids with a suitable example.b) Explain what is meant by the primary and secondary structure of proteins.

a) पेप्टाइड लिंकेज को परिभाषित कीजिए। दो अमीनो अम्लों से एक डाइपेप्टाइड के निर्माण को एक उपयुक्त उदाहरण के साथ समझाइए।b) प्रोटीन की प्राथमिक और द्वितीयक संरचना से क्या अभिप्राय है, समझाइए।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Peptide linkage: It is an amide bond formed between the carboxyl group of one amino acid and the amino group of another amino acid with the elimination of a water molecule. For example, the formation of glycylalanine from glycine and alanine:

ext{H}_2 ext{N}- ext{CH}_2- ext{COOH} + ext{H}_2 ext{N}- ext{CH}( ext{CH}_3)- ext{COOH} ightarrow ext{H}_2 ext{N}- ext{CH}_2- ext{CO}- ext{NH}- ext{CH}( ext{CH}_3)- ext{COOH} + ext{H}_2 ext{O}

(Glycine) (Alanine) (Glycylalanine)b) Primary structure: It refers to the specific sequence of amino acids in a polypeptide chain. Any change in this sequence creates a different protein.Secondary structure: It refers to the shape in which a long polypeptide chain can exist. It arises due to the regular folding of the backbone of the polypeptide chain due to hydrogen bonding between the

- ext{C}= ext{O}

and

- ext{NH}

groups of the peptide bond. The two common secondary structures are

ext{a}

-helix and

ext{b}

-pleated sheet.

Q5 • 5 Marks Long Answer / Derivation

a) Differentiate between DNA and RNA based on their chemical composition and structure.b) Write the main functions of DNA and RNA in living organisms.

a) DNA और RNA के रासायनिक संघटन और संरचना के आधार पर उनमें अंतर स्पष्ट कीजिए।b) सजीवों में DNA और RNA के मुख्य कार्य लिखिए।

View Step-by-Step Proof & Solution ▾

Complete Derivation / Solution:

a) Differences between DNA and RNA:1. Sugar: DNA contains deoxyribose sugar, while RNA contains ribose sugar.2. Bases: DNA contains A, G, C, T. RNA contains A, G, C, U (uracil replaces thymine).3. Structure: DNA is typically a double-stranded helical structure. RNA is usually a single-stranded structure, though it can fold into complex 3D shapes.4. Stability: DNA is more stable due to the presence of deoxyribose and its double-helical structure. RNA is less stable and more reactive.b) Functions:DNA: It is the genetic material in most organisms. It carries genetic information from one generation to the next. It controls the synthesis of proteins.RNA: It primarily functions in protein synthesis (mRNA carries genetic code, tRNA carries amino acids, rRNA is part of ribosomes). Some RNAs also have catalytic activity (ribozymes) or regulatory roles.

For Students

For Institutes

Company

Class 12 Chemistry Chapter-Wise Subjective Question Bank PDF — 15 Questions Per Chapter (Reasoning & Reactions)

Chapter-Wise Distribution (150 Total Questions)

Solutions

Part A: Short Answer Questions (2–3 Marks Each)

Part B: Long Answer Questions & Derivations (5 Marks Each)

Electrochemistry

Part A: Short Answer Questions (2–3 Marks Each)

Part B: Long Answer Questions & Derivations (5 Marks Each)

Chemical Kinetics

Part A: Short Answer Questions (2–3 Marks Each)

Part B: Long Answer Questions & Derivations (5 Marks Each)

The d- and f-Block Elements

Part A: Short Answer Questions (2–3 Marks Each)

Part B: Long Answer Questions & Derivations (5 Marks Each)

Coordination Compounds

Part A: Short Answer Questions (2–3 Marks Each)

Part B: Long Answer Questions & Derivations (5 Marks Each)

Haloalkanes And Haloarenes

Part A: Short Answer Questions (2–3 Marks Each)

Part B: Long Answer Questions & Derivations (5 Marks Each)

Alcohols, Phenols and Ethers

Part A: Short Answer Questions (2–3 Marks Each)

Part B: Long Answer Questions & Derivations (5 Marks Each)

Aldehydes, Ketones and Carboxylic Acids

Part A: Short Answer Questions (2–3 Marks Each)

Part B: Long Answer Questions & Derivations (5 Marks Each)

Amines

Part A: Short Answer Questions (2–3 Marks Each)

Part B: Long Answer Questions & Derivations (5 Marks Each)

Biomolecules

Part A: Short Answer Questions (2–3 Marks Each)

Part B: Long Answer Questions & Derivations (5 Marks Each)

Get Instant AI Grading on Your Handwritten Derivations

Frequently Asked Questions

Get Your Handwritten Answers Evaluated by AI

Explore Related Study Materials

CBSE & BSEB Class 12 Physics Notes PDF — Chapter-wise Revision Notes & Formula Sheets (All 14 Chapters)

CBSE & BSEB Class 12 Chemistry Notes PDF — Chapter-wise Revision Notes & Reactions (All 9 Chapters)

CBSE & BSEB Class 12 Maths Notes PDF — Chapter-wise Formula Sheets & PYQs (All 12 Chapters)