CBSE Sample Papers for Class 12 Chemistry Set 5 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Chemistry with Solutions Set 5 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Chemistry Set 5 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 70

General Instructions:

There are 35 questions in this question paper with internal choices.
Section A consists of 18 multiple-choice questions carrying 1 mark each.
Section B consists of 7 very short answer questions carrying 2 marks each.
Section C consists of 5 short answer questions carrying 3 marks each.
Section D consists of 2 case-based questions carrying 4 marks each.
Section E consists of 3 long answer questions carrying 5 marks each.
All questions are compulsory.
Use of log tables and calculators are not allowed.

SECTION – A (18 Marks)
(The following questions are multiple-choice questions with one correct answer.
Each question carries 1 mart There is no internal choice in this section.)

Question 1.
If limiting molar conductivity of Ca²⁺ and Cl^– are 119.0 and 76.3 S cm² mol^-1, then the value of limiting molar conductivity of CaCl₂ will be: [1]
(a) 195.3 S cm² mol^-1
(b) 271.6 S cm² mol^-1
(c) 43.3 S cm² mol^-1
(d) 314.3 S cm² mol^-1
Answer:
(b) 271.6 S cm² mol^-1

Explanation:
From Kohlrausch’s Law,
∧caCl₂ = ∧_Ca²⁺ + 2∧_cl^–
⇒ ∧cacl₂ = 119 + 2 × 76.3
= 119 + 152.6
= 271.6 cm² mol^-1
The limiting molar conductivity of CaCl₂ is 271.6 cm² mol^-1

Question 2.
Curdling of milk is an example of: [1]
(a) breaking of peptide linkage
(b) hydrolysis of lactose
(c) breaking of protein into amino acids
(d) denaturation of protein
Answer:
(d) denaturatior, of protein

Explanation:
Curdling of milk is an example of denaturation of proteins where the formation of lactic acid by microbial action results in denaturation.

Question 3.
Complete the following analogy:

(a) X: Catechol, Y: Hydroquinone
(b) X: Catechol, Y: Resorcinol
(c) X: Resorcinol, Y: Catechol
(d) X: Hydroquinone, Y: Catechol
Answer:
(a) X: Catechol, Y: Hydroquinone

Explanation:
When -OH group is attached at the oetho position of phenol, then the compound formed is called catechol when – OH group is attached at the para position, then the compound is called hydroquinone.

Question 4.
Diamagnetic species is weakly repelled by external magnetic field. Which of the following is not a diamagnetic ion: (Atomic numbers of Sc, Co, Mn and Cu are 21, 27, 25 and 29 respectively)?
(a) Co²⁺
(b) Sc³⁺
(c) Cu²⁺
(d) Mn²⁺
Answer:
(b) Sc³⁺

Explanation:
Co²⁺ (Z = 27): [Ar]₁₈3d⁷
(3 unpaired electrons)
Cu²⁺ (Z = 29): [Ar]₁₈3d⁹ (1 unpaired electrons)
Mn²⁺ (Z = 25): [Ar]₁₈3d⁵ (5 unpaired electrons)
Sc³⁺ (Z = 21): [Ar]₁₈3d⁰ (No unpaired electro
Sc³⁺ with no unpaired electron will be diamagnetic

Question 5.
Order of a chemical reaction can be fractional also. Which of the following, is an example of a fractional order reaction?
(a) NH₄NO₂ → N₂ + 2H₂O
(b) NO + O₃ → NO₂ + O₂
(c) 2NO + Br₂ → 2NOBr
(d) CH₃CHO → CH₄ + CO
Answer:
(d) CH₃CHO → CH₄ + CO

Explanation:
CH₃CHO → CH₄ + CO
Rate = k[CH₃CHO]^3/2

Question 6.
When CH₃CHBrCH₂CH₃ is reacted with alcoholic KOH, the major product is:
(a) CH₃ – CH = CH – CH₃
(b) CH₂ = CH – CH₂ – CH₃
(c) CH₃ – CH(OH) – CH₂ – CH₃
(d) CH₃ – CH₂ – CH₂ – CH₃
Answer:
(a) CH₃-CH = CH-CH₃

Explanation:
If the dehydrohalogenation of an alkyl halide can yield more than one alkene, then according to the Saytzeff rule, the main product is the most highly substituted alkene.

Question 7.
IUPAC name of [Pt(NH₃)₃ Br(NO₂)Cl] Cl is: [1]
(a) Triamminechloridobromidonitraplatinum (IV)chloride
(b) Triamminebromidonitrochloridoplatinum (IV)chloride
(c) Triamminebromidochloridonitroplatinum (IV)chloride
(d) Triamminenitrochloridobromiodoplatinum (IV)chloride
Answer:
(c) Triamminebromidochloridonitroplatinum (IV)chioride

Question 8.
The correct acidic strength order of the following is: [1]

(a) (I) > (II) > (III)
(b) (III) > (I) > (II)
(c) (II) > (III) > (I)
(d) (I) > (III) > (II)
Answer:
(b) (III) > (I) > (II)

Explanation:
Presence of EWG increases the acidic character.

Question 9.
Aniline is less basic than ethylamine. This is due to: [1]
(a) conjugation of lone pair of nitrogen with the ring
(b) the insoluble nature of aniline
(c) more Kb value of aniline
(d) hydrogen bonding
Answer:
(a) conjugation of lone pair of nitrogen with the ring

Explanation:
Lone pair of electrons on the nitrogen atom of aniline is involved in resonance and is not easily available for donation to protons. So, aniline is less basic than ethylamine.

Question 10.
Which of the following alcohol is most soluble in water? [1]
(a) Propanol
(b) Butanol
(c) Pentanol
(d) Hexanol
Answer:
(a) Propanol

Explanation:
Propanol is most soluble in water because the lower alcohols are highly soluble in water due to the presence of -OH group in the alcohols it forms H-bond with itself and molecular association takes place which causes the increase in the boiling point of the corresponding alcohols due to the increase in the number of carbon atoms and thus high temperature is required to break this association of bonds and thus the solubility in water increases. The extent of the hydrogen bonding in alcohols depends on the number of the carbon atoms attached in the chain and takes place as shown below:

Question 11.
Which of the ligands has largest CFSE value? [1]
(a) CO
(b) NH₃
(c) P
(d) H₂O
Answer:
(a) CO

Explanation:
CO is the strongest field ligand causes the highest CFSE, ∆₀ as a ligand.

Question 12.
What is the coordination number of Pt in [Pt(NH₃)₄Br₂]²⁺? [1]
(a) 4
(b) 6
(c) 2
(d) 3
Answer:
(a) 4

Explanation:
Let the coordination number of Pt in [Pt(NH₃)₄Br₂]²⁺ be x
x + (0 × 4) – 2 = 2
x – 2 = 2
× = 2 + 2
x = 4

Question 13.
What do you conclude about the order of the reaction? [1]
(a) It is a first order reaction.
(b) It is a second order reaction.
(c) It is a third order reaction.
(d) It is a pseudo unimolecular reaction.
Answer:
(a) It is a first order reaction.

Explanation:
The plot of in [R] vs time is a straight line with negative slope. It is characteristic of first order reaction.

Question 14.
What is the unit of the order of the reaction depicted in the graph? [1]

(a) mol^-1 sec^-1
(b) sec^-1
(c) mol^-1
(d) mol^-2 Ls^-1
Answer:
(b) sec^-1

Explanation:
For first order reaction, the unit of the rate constant is sec^-1.

In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(c) (A) is true but (R) is false
(d) (A) is false but (R) is true

Question 15.
Assertion (A): Fructose reduces Fehling’s solution and Tollens’ reagent.
Reason (R): Fructose does not contain any aldehyde group. [1]
Answer:
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).

Explanation:
Fructose on warming with dilute alkali, gives rise to an equilibrium mixture of glucose, fructose and mannose. The ability of fructose to reduce Fehling solution and Tollen’s reagent is probably due to the isomerisation of fructose to glucose and mannose (this is called Lobry de Bruyn and Elkenstein rearrangement).

Question 16.
Assertion (A): It is possible to make a cell with a pair of same type of half cells.
Reason (R): Change in concentration varies the electrode potential. [1]
Answer:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Explanation:
It is possible to make a cell with a pair of same type of half cells as change in concentration varies the electrode potential.

Question 17.
Assertion (A): All collisions of reactant molecules lead to product formation.
Reason (R): Only those collisions in which molecules have correct orientation and sufficient kinetic energy lead to the compound formation. [1]
Answer:
(d) (A) is false but (R) is true

Explanation:
Correct assertion is “only effective collision lead to formation of product.” Reason defines correct meaning of effective collision, and criterion of collision theory for completion of reaction. Only those collisions in which molecules have correct orientation and sufficient energy lead to formation of product.

Question 18.
Assertion (A): [Ti(H₂O)₆]³⁺ is coloured while [SC(H₂O)₆]³⁺ is colourless.
Reason (R): d-d transition is not possible in [SC(H₂O)₆]³⁺. [1]
Answer:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Explanation:
[Ti(H₂O)₆]³⁺ is coloured while [Sc(H₂O)₆]³⁺ is colourless because [Ti(H₂O)₆]³⁺ has unpaired election for d-d transition. So, it is coloured.

Section – B (14 Marks)
(The following questions are very short answer type with internal choice in two questions and carry 2 marks each.)

Question 19.
(A) Give one example of coordination isomerism.
(B) Calculate the number of unpaired electeon in Cr³⁺, Ti³⁺ and V³⁺ and identify the most stable ion in aqueous solution. [2]
Answer:
(A) Coordination isomerism arises when both (B) No. of unpaired e , the cation and anion are complexes and the ligands are exchanged. For example, [Co(NH₃)₆] [Cr(CN)₆] and [Cr(NH₃)₆] [Co(CN)₆]

(B) No.of unpaired e^–
Cr³⁺ : 3
Ti³⁺ : 1
V³⁺ : 2
Among them, Cr³⁺ is the most stable as it has stable configuration i.e., t³_2g.

Question 20.
Give simple tests to distinguish between the following pairs of compounds: [2]
(A) Pentan-2-one and Pentan-3-one
(B) Benzaldehyde and Acetophenone
Answer:
(A) Iodoform test: Pentan-2-one contains CH₃CO- group and as such it will give iodoform test with NaOl [NaOH + I₂] while no such group is there in pentan -3- one and so it will not give iodoform test.
CH₃COCH₂CH₂CH₃ + 2NaOl → CH₃CH₂CH₂COONa + CHI₃ + 2 NaOH Iodoform (yellow ppt)

(B) Benzaldehyde is an aromatic aldehyde while acetophenone is a methyl ketone. These may be distinguished as follows. Iodoform test: Acetophenone, due to the presence of CH₃CO^– group, will give iodoform test with NaOl (NaOH + l₂), while benzaldehyde will not give this test.

C₆H₅CHO + NaOl → No yellow ppt

Question 21.
What is meant by the following terms? [2]
(A) Cyanohydrin
(B) Acetal
OR
Arrange the following compounds in increasing order of their property as indicated:
(A) CH₃COCH₃, C₆H₅ – CO – C₆H₅, CH₃CHO (reactivity towards nucleophilic addition reactions)
(B)

Answer:
(A) -OH group and cyano group are present on same carbon atom. Addition of HCN to carbonyl group in weakly acidic medium forms cyanohydrin.

(B) Acetal:
Terminal C atom has two alkoxy groups. Two equivalents of monohydric alcohol add to 1 equivalent of aldehyde in presence of dry HCl gas.

(A) C₆H₅-CO-C₆H₆ < CH₃COCH₃ < CH₃CHO
Explanation:
Ketones are less reactive towards nucleophilic addition reactions due to more steric hinderance.

(B)

Explanation:
More the number of electron with drawing groups (Cl^–), more the acidic character.

Question 22.
(A) Differentiate between hormones and vitamins.
(B) Give the chemical name of the following:
(i) Vitamin E
(ii) Vitamin C [2]
Answer:
(A) Hormones:

Hormones are produced in the endo-crine or ductless glands.
Hormones are not stored in the body. These are produced as and when re-quired.
Example: Insulin

Vitamins:

Vitamins are stored in the body upto certain extent. 3. They are needed in small quantity.
Excess vitamins are excreted.
Example: Vitamin A, B, C, D, E and K

(B) (i) Tocopherol
(ii) Ascorbic acid

Question 23.
On the basis of electrochemical series, elaborate the following:
(A) Compound AU₂O₃ decomposes on hearing.
(B) Zn can displace hydrogen from dil HCl but Cu cannot. [2]
Answer:
(A) Oxide gold (III) on thermal breakdown to create gold and oxygen. The temperature range for this reaction is 160 – 290°C.

(B) In electrochemical series, Zn is placed above hydrogen but Cu is placed below hydrogen. Therefore, Zn can displace hydrogen from dil. HCl but Cu cannot.

Question 24.
(A) Draw the plot of In [R] vs t for a first order reaction. What is the slope?
(B) A first order reaction takes 30 minutes for 90% decomposition. Calculate k.
Given: log₁₀ = 1 [2]
Answer:
(A)

(B) For first order reaction,
k = \(\frac{2.303}{t}\) log \(\frac{\left[R_0\right]}{[R]}\)
Given, t = 30 min
[R₁] = 100
[R] = 10
So, k = \(\frac{2.303}{30}\) log \(\frac{100}{10}\)
= 0.076 × log 10
= 0.076 min^-1

Question 25.
(A) Give any two differences between double salts and complexes?
(B) Explain how [FeF₆]^3- is paramagnetic and outer orbital complex.
OR
(A) On the basis of crystal field theory, write the electronic configuration for d⁴ ion if ∆₀ < P.
(B) Write the hybridization and shape of [Ni(CN)₄]^2-. (Atomic number of Ni = 28) [2]
Answer:
(A) The difference between Double salt and Complex salt are:

Double Salt	Complex Salt
(1) These exist only in solid state and dissociate into constituent species in their solution.	They retain their identity in solid as well as in solution state.
(2) They lose their identity in dissolved state.	They do not lose their identity in dissolved state.

(B) In [FeF₆]^3- complex
Oxidation number of Fe = + 3
Coordination Number Fe = 6
Coordination orbitals = 6
Fe → Fe³⁺ + 3e^–
Fe³⁺ → (26 – 3) – [Ar]₁₈
[FeF₆]^3- (Strongly Paramagnetic)

Here F^– is a weak field ligand and it obeys. Hund’s rule of leaving maximum number of unpaired electron (n = 5) and it uses one 40, three 4p and two 4d orbitals which are sp³d² hybridised. So it is an outer orbital complex.

(A) t³_2g e_g ∆_o < P, the configuration will be t_2g, e_g and it is in the case of weak field ligands and high spin complex will be formed.

(B) dsp², square planar.

Section – C (15 Marks)
(The following questions are short answer type with internal choice in two questions and carry 3 marks each.)

Question 26.
Write the mechanism for the formation of ethoxy ethane from ethanol. [3]
Answer:

Mechanism: Formation of ether follows S_N² mechanism.

Question 27.
How would you account for the following? [3]
(A) Many of the transition elements are known to form interstitial compounds.
(B) With the same d-orbital configuration (d⁴) Cr²⁺ is a reducing agent while Mn³⁺ is an oxidizing agent.
(C) The enthalpies of atomization of transition elements are quite high.
Answer:
(A) Many of the transition elements are known to form interstitial compounds because of unpaired electrons in the d-orbital. Transition elements have vacant interstitial sites and are able to trap small atoms like H, C or N to form such compound.

(B) As Cr³⁺ has stable t³_2g configuration so Cr²⁺ has a tendency to change to Cr³⁺ hence it is a reducing agent. Mn³⁺ changes to Mn²⁺ (3d)⁵ which is half filled state configuration. Hence, Mn³⁺ acts as oxidising agent.

(C) The enthalpies of atomization of a transition metal are high because they have a large number of unpaired electrons and hence have strong metallic bonding.

Question 28.
Answer the following questions: [3]
(A) Write the formula of an oxo-anion of chromium (Cr) in which it shows the oxidation state equal to its group number.
(B) Write one similarity and one difference between the chemistry of lanthanoids and that of actinoids.
(C) Name the element showing the maximum number of oxidation states among the first series of transition metals from Sc (Z = 21) to Zn (Z = 30).
OR
The standard electrode potential for the M³⁺/M²⁺ half-cell gives the relative stability between M³⁺ and M²⁺. The reduction potential values are tabulated as below.

Reaction	Standard reduction potential (V)
Ti³⁺ + e^– → Ti²⁺	– 0.37
V³⁺ + e^–→ V²⁺	– 0.26
Cr³⁺ + e^– → Cr²⁺	– 0.41
Mn³⁺ + e^–→ Mn²⁺	+ 1.51
Fe³⁺ + e^–→ Fe²⁺	+ 0.77
Co³⁺ + e^–→ Co²⁺	+ 1.81

(A) What does the negative values for titanium, vanadium and chromium indicate?
(B) What does reduction potential of Mn³⁺/Mn²⁺ indicate about the stability of these oxidation states?
(C) The E° value of the Mn³⁺/Mn²⁺ couple is much more positive than that for Cr³⁺/Cr²⁺ couple or Fe³⁺/Fe²⁺ couple.
Answer:
(A) An oxo-anion of chromium (Cr) in which it shows the oxidation state equal to its group number is Cr2 O^2-₇. Here, oxidation number of Cr is + 6 which is equal to its group number.

(B) Similarity: Both lanthanoids and actinoids show contraction in size and irregularity in their electronic configuration.
Difference: Actinoids show wide range of oxidation states but lanthanoids do not.

(C) Mn (Z = 25) has the maximum number of unpaired electrons present in the d-subshell so it shows maximum oxidation states (+ 2 to + 7)

(A) The negative values for titanium, vanadium and chromium indicate that the higher oxidation state is preferred.

(B) The high reduction potential of Mn³⁺/Mn²⁺ indicates Mn²⁺ is more stable than Mn³⁺. Mn³⁺ has a 3d⁴ configuration while that of Mn²⁺ is 3d⁵. The extra stability associated with a half filled d-sub shell makes the reduction of Mn³⁺ very feasible
(E° = + 1.51V).

(C) Because Mn³⁺ has the outer electronic configuration of 3d⁴ and Mn²⁺ has the outer electronic configuration of 3d⁵. Thus, the conversion of Mn³⁺ to Mn²⁺ will be a favourable reaction since 3d⁵ is a very stable configuration as it is half filled configuration. Hence, E° value for Mn³⁺/Mn²⁺ couple is positive.
Fe³⁺ to Fe²⁺ undergoes a change in outer electronic configuration from 3d⁵ to 3d⁶. The configuration of the resultant are not stable and hence have a lower E° value.

Question 29.
Answer any three questions: [3]
(A) What type of linkage is present in disaccharides?
(B) Write one source and deficiency disease of vitamin B₁₂.
(C) Write the difference between DNA and RNA.
(D) What are the expected products of hydrolysis of lactose?
Answer:
(A) Glycosidic linkage, an oxide linkage, is present in disaccharides and polysaccharides to connect monosaccharides units.

(B) Vitamin B₁₂ is present in animal tissues, eggs, curd and almonds.

(D) On hydrolysis, lactose gives D-galactose and β-D-glucose

Question 30.
A solution containing 15 g urea (molar mass = 60 g mol^-1) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (motar mass = 180 g mol^-1) in water. Calculate the mass of glucose present in one litre of its solution. [3]
Answer:
For isotonic solutions,
π(urea) = π(glucose)

Section – D (8 Marks)
(The following questions are case-based questions. Each question has an internal choice and carries 4 (1+1+2) marks each. Read the passage carefully and answer the questions that follow.)

Question 31.
The various classes of organic compounds containing carbonyl groups are aldehydes, ketones, carboxylic acids and their derivatives. The carbonyl carbon of the simplest aldehyde, formaldehyde is bonded to two hydrogen atoms. All other aldehydes contain the carbonyl carbon bonded to a hydrogen atom and to an alkyl group. The carbonyl carbon of a ketone is bonded to two alkyl groups. Carboxylic acids are obtained by the oxidation of primary alcohols or aldehydes. They are also obtained by the hydrolysis of nitriles, acid chlorides, esters, anhydrides and amides.
(A) Why carboxylic acids contain a carbonyl group but do not give characteristic reactions of the carbonyl group? [1]
(B) State the geometry of formaldehyde. [1]
(C) How will you differentiate the following compound.
(i) Benzaldehyde and acetaldehyde
(ii) Formic acid and acetic acid.
OR
Explain the feature of following:
(i) Iodoform
(ii) Schiff s test [2]
Answer:
(A) Carboxylic acids contain a carbonyl group but do not give characteristic reaction of the carbonyl group because of resonance, the electrophilic nature of the carboxyl carbon is greatly reduced as compared to the carbonyl carbon is aldehyde and Vetones.

(B) Formaldehyde is planar molecule due to sp2 hybridized carbon atom.

(C) (i) Fehling’s solution can be used to distinguish benzaldehyde from acetaldehyde as benzaldehyde (aromatic aldehyde) does not reduce Fehling’s solution, but acetldehyde (aliphatic aldehyde) reduces Fehling’s solution.
(ii) Formic acid gives silver mirror with Tollen’s reagent wile acetic acid does not.
OR
(i) Iodoform test is used to check the presence of carbonyl compounds with the structure R-CO-CH₃ or alcohols with the structure R-CH(OH)-CH₃ in a given unknown substance.
(ii) Schiffs reagent is the aqueous solution of Rosaniline hydrochloride decolorised by SO₂.

Question 32.
An ideal solution may be defined as the solution which obeys Raoult’s law exactly over the entire range of concentration. The solutions for which vapour pressure is either higher or lower than that predicted by Raoult’s law are called non-ideal solutions.

Non-ideal solutions can show either positive or negative deviations from Raoult’s law depending on whether the A-B interactions in solution are stronger or weaker than A – A and B – B interactions.
(A) Name any two examples of ideal solution(s)? [1]
(B) If liquids A and B form an ideal solution, then what is the Gibbs free energy of mixing? [1]
(C) (i) Draw a graph for water and nitric acid mixture.
(ii) Water-HCl mixture shows maximum or minimum boiling azeotropes?
OR
A 2 litre flask contains 4g of hydrogen and 128 g of hydrogen iodide. Find out their active masses. [2]
Answer:
(A) (1) Bromoethane and iodoethane
(2) n-Heptane and n-hexane

Explanation:
A solution where the interaction of component molecules does not vary from the interactions of each component’s molecules. The ideal solutions are those at all concentrations and temperatures that obey Raoult’s law.

(B) For ideal solutions, the value of the Gibbs free energy is always negative as mixing of ideal solutions is a spontaneous process.

Explanation:
Two liquids HNO₃ (A) and water (B) form a maximum boiling azeotrope when mixed in the ratio of 68% and 32% respectively because the forces between the particles in the mixture are stronger than the mean of the forces between the particles in the pure liquids.

(ii) This mixture shows maximum boiling azeoteopes.
Explanation: Negative deviations from Raoult’s Law: New stronger forces must exist in the mixture than in the original liquids. These are cases where the molecules break away from the mixture less easily than they do from the pure liquids.

Mass of hydrogen = 4 g
Mol. mass of hydrogen = 2
Volume of the flask = 2 litre
Active mass of hydrogen = 422 × = 1 mol L^-1
Mass of HI = 128 g
Mol. mass of HI = 128
Volume of the flask = 2 litre
Active mass of hydrogen iodide
= \(\frac{128}{128 \times 2}\) = 0.5 mol L^-1

Section – E (15 Marks)
(The following questions are long answer type and carry 5 marks each. Two questions have an internal choice.)

Question 33.
(A) Arrange the following in the decreasing order of their basic strength in aqueous solutions:
CH₃NH₂, (CH₃)₂NH, (CH₃)₃N and NH₃ (B) Give the chemical tests to distinguish sh between the following pair of compounds: Methylamine and Dimethylamine
(C) How will you convert Benzene to Aniline?
(D) Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.
OR
(A) Account for the following:
(i) N-ethylethanamine boils at 329.3K and butanamine boils at 350.8K, although both are isomeric in nature.
(ii) Acylation of aniline is carried out in the presence of pyridine.
(B) What is the best reagent to convert nitrile to primary amine?
(C) Why does acetylation of -NH₂ group of aniline reduce its activating effect?
(D) Under what reaction conditions (acidic/basic) the coupling reaction of aryl diazonium chloride with aniline is carried out? [5]
Answer:
(A) Correct order is (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃

Explanation:
+1 group, like R—groups, increases the electron density on nitrogen which increases the basicity (electron rich is basic) of amine. Of the given amines, more the number of R—groups on nitrogen more is the basic nature of amine. Basicity also depends on the steric hindrance. More is the steric hindrance less is the basicity.

(B) Methylamine and dimethylamine can be distinguished by the carbylamine test. Carbylamine test: Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form foul-smelling isocyanides or carbylamines.

(D) Aniline reacts with methyl iodide to produce N, N-dimethylaniline.

With excess methyl iodide, in the presence of Na₂C0₃ solution, N, N-dimethylaniline produces N, N, N-trimethylanilinium carbonate.

(A) (i) Due to the presence of three hydrogen atoms, the intermolecular association is more in primary amines than in secondary amines as there are two hydrogen atoms available for hydrogen bond formation in it. Thus, butanamine boils at 350.8K while N-ethylethanamine boils at 329.3K.
(ii) During the acylation of aniline, stronger base pyridine is added. This done in order to remove the HCl so formed during the reaction and to shift the equilibrium to the right hand side.

(B) LiAlH₄ and sodium/alcohol are the best reagents for converting nitrile to a primary amine. The nitriles are transformed into a corresponding primary amine through reduction.

(C) The acetylation of -NH₂ group of aniline reduces its activating effect because the lone pair of electrons on the nitrogen of acetanilide interacts with oxygen atom due to resonance.

(D) Coupling reaction of aryl diazonium chloride with aniline is carried out in mildly acidic conditions, i.e. pH = 4-5.

Question 34.
(A) Identify the compound that on hydrogenation produces an optically active compound from the following compounds, giving reason.

(B) Out of chlorobenzene and benzyl chloride, identify the compound which gets hydrolysed by aqueous NaOH and why?
(C) Give the IUPAC name of the product formed when:
(i) 1- methyl cyclohexene is treated with HI.
(ii) chloroethane is treated with silver nitrite
OR
(A) Distinguish enantiomer from racemic mixture.
(B) The following compounds are given to you:
2-Bromopentane, 2-Bromo-2-methylbutane, 1-Bromopentane
(i) Write the compound which is most reactive towards S_N¹ reaction.
(ii) Write the compound which is optically active
(iii) Write the compound which is most reactive towards (3-elimination reaction.
(C) Arrange the following in decreasing order of reactivity:
R-Cl, R-Br, R-l [5]
Answer:
(A) Any compound is said to be optically inactive when:

It can show enantiomerism-lt forms non-super impossible mirror images of each other.
It has a chiral carbon-A chiral carbon is one which is bonded to four different molecules.

Hydrogenation of (I):

The product formed by the hydrogenation of compound (I) does not have a chiral carbon. Therefore, it is not optically active. Hydrogenation of (II):

The product formed by the hydrogenation of compound (II) has a chiral carbon. Therefore, it is optically active.

(B) Benzyl chloride gets easily hydrolysed by aqueous NaOH as chlorobenzene has partial double bond character in the Cl-C bond. The lone pairs delocalised in the ring strengthens Cl-C bond reducing its reactivity. Whereas benzyl chloride undergoes S_N¹ reaction to form stable benzyl carbocation.

(A) A racemic mixture is a 50:50 mixture of two enantiomers. Because they are mirror images, each enantiomer rotates plane-polarized light in an equal but opposite direction and is optically inactive. If the enantiomers are separated, the mixture is said to have been resolved.

(B) (i) 2-Bromo-2-methyl butane will be most reactive towards SJF due to maximum hindrance and formation of stable carbocation. Tertiary alkyl halides are most reactive for reaction.
(ii) 2-Bromopentane has a chiral carbon in it. So, this compound is optically active.

(iii) 2-Bromo-2-methyl pentane give a stable elimination product Most stable alkene i.e.. more substituted)

(C) R-l > R-Br > R-Cl
Explanation: The reactivity of alkyl halides mostly corresponds to bond energy and electro negativity of the halide, if we consider R-F here the bond energy is very high and the F atom is highly electronegative thus proving to be a weak leaving group, on the other hand R-l has a relatively less bond energy thus proving to be a good leaving group, that’s why it is highly reactive.

Question 35.
(A) The resistance of a conductivity cell containing 0.001M KCl solution at 298K is 1500 Ohm. What is the cell constant if conductivity of 0.001M KCl solution at 298K is 0.146 × 10^-3 S cm^-1.
(B) The cell in which the following reaction occurs:
2Fe³⁺_(aq) + 2l^–_(aq) → 2Fe²⁺ (aq) + I₂(s) has
E°_cell = 0.236 Volt at 298 K
Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction. [5]
Answer:
(A) k = \(\frac{1}{R}\) cell constant
k = 0.146 × 10^-3 s cm^-1
R = 1500 Ohm
0.146 × 10^-3 = \(\frac{1}{1500}\) ×
cell constant = 0.146 × 10^-3 × 1500
= 219 × 10^-3 = 0.219 cm^-1

(B) ∆_rG⁰ = – nFE⁰_cell
= – 2 × 96500 × 0.236
= – 45.548 kl/mol

or k_c = + antilog 8.0
k_c = 10⁸