Important Questions – Page 3

PSEB 11th Class Biology Important Questions in Punjabi English Medium

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PSEB 11th Class Chemistry Important Questions in Punjabi English Medium

August 16, 2024August 14, 2024 by sachin

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PSEB 12th Class Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

August 14, 2024August 13, 2024 by Prasanna

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Very short answer type questions

Question 1.
Write the product in the following reaction:

Answer:
CH₃– CH = CH-CH₂– CHO Pent-3-en-1-al

Question 2.
Complete the following reaction sequence: (NCERT Exemplar]
PSEB 12th Class Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 2
Answer:

Question 3.
Illustrate the following name reaction: Clemmensen reduction.
Answer:
Clemmensen reduction: The carbonyl group of aldehydes and ketones is reduced to CH₂ group on treatment with zinc amalgam and concentrated hydrochloric acid.
PSEB 12th Class Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 4

Question 4.
Name the electrophile produced in the reaction of benzene with benzoyl chloride in the presence of anhydrous AlCl₃.
Name the reaction also. [NCERT Exemplar]
Answer:
PSEB 12th Class Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 5

Question 5.
Which acid of each pair shown here would you expect to be stronger?
(i) F-CH₂ -COOH or Cl-CH₂ -COOH
PSEB 12th Class Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 6
Answer:
(i) F-CH₂-COOH > Cl-CH₂-COOH, because of stronger -ve-I effect of F than Cl.

Question 6.
Identify the compounds A, B, and C in the following reaction. [NCERT Exemplar]
PSEB 12th Class Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 8
Answer:

Question 7.
Write the product(s) in the following reaction:
PSEB 12th Class Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 10
Answer:

Question 8.
Name the aldehyde which does not give Fehling’s solution test.
Answer:
Benzaldehyde.

Question 9.
Give the name of the reagent that brings the following transformation: But-2-ene to ethanal.
Answer:
O₃/H₂O-Zn dust

Question 10.
Write TUPAC names of the following structures: (NCERT Exemplar)
PSEB 12th Class Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 12
Answer:

Question 11.
(CH₃) ₃C-CHO does not undergo aldol condensation comment.
Answer:
Because no α-H is present.

Short answer type questions

Question 1.
Write the reactions involved in the following:
(i) Etard reaction
(ii) Stephen’s reduction
Answer:
(i) Etard reaction: Chromyl chloride oxidizes toluene to chromium complex which on hydrolysis gives benzaldehyde.
PSEB 12th Class Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 14

(ii) Stephen reductIon: Nitriles are reduced to corresponding imines with SnCl₂ in the presence of HCl, which on hydrolysis give the corresponding aldehyde.
SnCl₂ + 2HCl → SnCl₄ + 2[H]
PSEB 12th Class Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 15

Question 2.
Oxidation of ketones involves carbon-carbon bond cleavage. Name the products formed on oxidation of 2,5-dimethyl hexane-3-one. [NCERT Exemplar]
PSEB 12th Class Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 16
Answer:

Question 3.
Carboxylic acids contain carbonyl group but do not show the nucleophilic addition reaction like aldehydes or ketones. Why? [NCERT Exemplar]
Answer:
Carboxylic acids contain, carbonyl group but do not show nucleophilic addition reaction like aldehyde and ketone. Due to resonance as shown below the partial positive charge on the carbonyl carbon atom is reduced.
PSEB 12th Class Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 18

Question 4.
Give reasons:
(i) The α-hydrogen atoms of aldehydes and ketones are acidic in nature.
(ii) Propanone is less reactive than ethanal toward addition of HCN
(iii) Benzoic acid does not give Friedel Crafts reaction.
Answer:
(i) The acidity of a-hydrogen atom of carbonyl carbon is due to the strong withdrawing effect of the carbonyl group and resonance stabilization of the conjugate base.
PSEB 12th Class Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 19
(ii) This is due to steric and electronic reasons. Sterically, the presence of two methyl groups in propanone hinders more the approach of nucleophile to carbonyl carbon than in ethanal having one methyl group. Electronically two methyl groups reduce the positivity of the carbonyl carbon more effectively in propanone than in ethanal.

(iii) Benzoic acid does not give Friedel Craft reaction because:

the carboxyl group is strongly deactivating.
the catalyst AlCl₃ which is a lewis acid gets bonded to the carboxyl group strongly.

Question 5.
An organic compound T having molecular formula C₄H₈O gives orange-red ppt. with a 2,4-DNP reagent.
It does not reduce Tollen’s reagent but gives yellow ppt. of iodoform on heating with NaOI.
Compound X on reduction with LiAlH₄ gives compound T’ which undergoes dehydration reaction on heating with -cone. H₂SO₄ to form but-2-ene. Identify the compounds X and Y.
Answer:
PSEB 12th Class Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 20

Reaction involved:
PSEB 12th Class Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 21

Long answer type questions

Question 1.
(a) Account for the following:
(i) Cl – CH₂COOH is a stronger acid than CH₃COOH.
(ii) Carboxylic acids do not give reactions of carbonyl group.
(b) Out of CH₃CH₂ – CO – CH₂ – CH₃ and CH₃CH₂ – CH₂ –
CO – CH₃, which gives iodoform test?
Answer:
(a) (i) Because of -I effect of Cl atom in ClCH₂COOH and +I effect of CH₃ group in CH₃COOH the electron density in the O-H bond in ClCH₂COOH is much lower than CH₃COOH.
As a result O-H bond in ClCH₂COOH is much weaker than in CH₃COOH therefore loses a proton more easily than CH₃COOH.
Hence ClCH₂COOH acid is stronger acid than CH₃COOH.

(ii) Carboxylic acids are resonance hybrid of the following structures:
PSEB 12th Class Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 22
Similarly, a carbonyl group of aldehydes and ketones may regarded as resonance hybrid of following structures :

Because of contribution of structure (IV), the carbonyl carbon in aldehydes and ketones is electrophilic. On the other hand, electrophilic character of carboxyl carbon is reduced due to contribution of structure (II). As carbonyl carbon of carboxyl group is less electropositive than carbonyl carbon in aldehydes and ketones, therefore, carboxylic acids do not give nucleophilic addition reactions of aldehydes and ketones.
(b) CH₃ – CH₂ – CH₂ – COCH₃.

Question 2.
Write the products formed when ethanal reacts with the following reagents:
(i) CH₃MgBr and then H₃O⁺
(ii) Zn-Hgyconc.HCl
(iii) C₆H₅CHO in the presence of dilute NaOH
Answer:
PSEB 12th Class Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 24

Question 3.
(a) Write the products of the following reactions :
PSEB 12th Class Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 25
(b) Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Benzaldehyde and Benzoic acid
(ii) Propanal and Propanone
Answer:

(iii) Cl-CH₂-COOH

(b)
(i) NaHCO₃ test
(ii) Iodoform test or Fehling test or Tollen’s test.

Question 4.
(a) Account for the following:
(i) CH₃CHO is more reactive than CH₃COCH₃ towards reaction with HCN.
(ii) 2-Fluorobutanoic acid is a stronger acid than 3-Fluorobutanoic acid.

(b) Write the chemical equations to illustrate the following name reactions:
(i) Etard reaction.
(ii) Rosenmund’s reaction.
(c) Give the mechanism of cyanohydrin formation when carbonyl compounds react with HCN in the presence of alkali.
Answer:
(a)

CH₃CHO is more reactive than CH₃COCH₃ towards reaction with HCN due to steric and electronic factors.
Because the inductive effect decreases with distance and hence the conjugate base of 2-Fluorobutanoic acid is more stable.

(b)
PSEB 12th Class Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 27

PSEB 12th Class Chemistry Important Questions Chapter 3 Electrochemistry

August 14, 2024August 13, 2024 by Prasanna

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 3 Electrochemistry Important Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 3 Electrochemistry

Very short answer type questions

Question 1.
Why on dilution the ∧_m of CH₃COOH increases drastically while that of CH₃COONa increases gradually? [NCERT Exemplar]
Answer:
In the case of CH₃COOH, which is a weak electrolyte, the number of ions increase on dilution due to an increase in degree of dissociation.
CH₃COOH + H₂o → CH₃COO^– + H₃O⁺

Question 2.
Why is alternating current used for measuring resistance of an electrolytic solution? [NCERT Exemplar]
Answer:
Alternating current is used to prevent electrolysis so that concentration of ions in the solution remains constant.

Question 3.
Define electrochemical series.
Answer:
The arrangement of elements in the increasing or decreasing order of their standard reduction potentials is called electrochemical series.

Question 4.
What are secondary cells?
Answer:
Secondary cells are those cells which are rechargeable, i.e., the products can be changed back to reactants.

Question 5.
What is the necessity to use a salt bridge in a Galvanic cell?
Answer:
To complete the inner circuit and to maintain the electrical neutrality of the electrolytic solutions of the half-cells we use a salt bridge in a Galvanic cell.

Question 6.
Why does a dry cell become dead after a long time even if it has not been used?
Answer:
Even though not in use, a dry cell becomes dead after some time because the acidic NH4C1 corrodes the zinc container.

Question 7.
What does the negative sign in the expression \(\boldsymbol{E}_{\mathbf{Z n}^{2+} / \mathbf{Z n}}[latex] = -0.76 V mean?[NCERT Exemplar]
Answer:
It means that Zn is more reactive than hydrogen. When zinc electrode will be connected to SHE, Zn will get oxidised and H⁺ will get reduced.

Question 8.
Write the Nemst equation for the cell reaction in the Daniel cell. How will the £cell be affected when concentration of Zn2+ ions is increased? [NCERT Exemplar]
Answer:
Zn + Cu²⁺ > Zn²⁺ + Cu
E_cell = [latex]E_{\text {cell }}^{\ominus}\) – \(\frac{0.059}{2}\) log \(\frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}\)
E_cellcell decreases when concentration of Zn²⁺ ions, [Zn²⁺ ] increases.

Question 9.
What does the negative value of \(E_{\text {cell }}^{\ominus}\) indicate?
Answer:
Negative \(E_{\text {cell }}^{\ominus}\) value means Δ_rG^Θ will be +ve, and the cell will not work.

Question 10.
Can \(E_{\text {cell }}^{\ominus}\) or Δ_rG^Θ for cell reaction ever be equal to zero? [NCERT Exemplar]
Answer:
No.

Short answer type questions

Question 1.
State Kohlrausch law of independent migration of ions. Why does the conductivity of a solution decrease with dilution?
Answer:
Kohlrausch law states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of cation and anion of the electrolyte.

In general, if an electrolyte on dissociation gives v₊ cations and v_– anions then its limiting molar conductivity is given by
\(\Lambda_{m}^{\ominus}=v_{+} \lambda_{+}^{0}+v_{-} \lambda_{-}^{0}\)

Where, \(\lambda_{+}^{0}\) and \(\lambda_{-}^{0}\) are the limiting molar conductivities of cations and anions respectively.

Conductivity of a solution decreases with dilution. This is due to the fact that the number of ions per unit volume that carry the current in a solution decreases with dilution.

Question 2.
Define the following terms:
(i) Fuel cell
(ii) Limiting molar conductivity (\(\))
Answer:
(i) A fuel cell is a device which converts the energy produced during the combustion of fuels like hydrogen, methanol, methane etc. directly into electrical energy. One of the most successful fuel cell is H₂ – O₂fuel cell.

(ii) When concentration approaches zero, the molar conductivity is known
as limiting molar conductivity. It is represented by \(\Lambda_{m}^{\ominus}\).
\(\Lambda_{m}^{\ominus}\) (∧b)when → c

Question 3.
(i) Write two advantages of H₂ – O₂ fuel cell over ordinary cell.
(ii) Equilibrium constant (K_c) for the given cell reaction is 10. Calculate \(\boldsymbol{E}_{\text {cell }}^{\ominus}\).
PSEB 12th Class Chemistry Important Questions Chapter 3 Electrochemistry 1
Solution:
(i) The two main advantages of H₂ – O₂ fuel cell over ordinary cell are as follows:

It has high efficiency of 60%-70%.
It does not cause any pollution.

PSEB 12th Class Chemistry Important Questions Chapter 3 Electrochemistry 2

Question 4.
Following reactions occur at cathode during the electrolysis of aqueous silver chloride solution:
Ag⁺(aq) + e^– → Ag(s) E^Θ = + 0.80 V
H⁺ (aq)+ e^– → \(\frac{1}{2}\)H₂(g) E^Θ = 0.00 V
On the basis of their standard reduction electrode potential (E^Θ) values, which reaction is feasible at the cathode and why?
Answer:
The reaction, Ag⁺ (aq) + e^– → Ag(s) is feasible at cathode as
cathodic reaction is one which has higher standard reduction electrode potential (\(E_{\text {red }}^{\ominus}\)).

Question 5.
Calculate ∆G and log K_c for the following reaction at 298 K:
2Cr(8) + 3Fe²⁺(aq) > 2Cr³⁺ (aq) + 3Fe(s)
Given: \(\boldsymbol{E}_{\text {cell }}^{\ominus}\) = 0.30 V
Solution:
PSEB 12th Class Chemistry Important Questions Chapter 3 Electrochemistry 3

Question 6.
Calculate the emf of the following cell at 298 K:
Cr(s)/Cr³⁺ (0.1M)//Fe<>2+ (0.01M)/(Fe(s) [Given: \(\boldsymbol{E}_{\text {cell }}^{\ominus}\) = + 0.30 V]
Answer:
The cell reaction is as follows :
2Cr(s) + 3Fe²⁺(aq) > 3Fe(s) + 2Cr³⁺(aq)
For this reaction, n = 6
Now,
E_cell = \(\boldsymbol{E}_{\text {cell }}^{\ominus}\) – \(\frac{2.303 R T}{n F}\) log \(\frac{\left[\mathrm{Cr}^{3+}\right]^{2}}{\left[\mathrm{Fe}^{2+}\right]^{3}}\)
E_cell = 0.30 – \(\frac{0.059}{6}\) log \(\frac{\left[10^{-1}\right]^{2}}{\left[10^{-2}\right]^{3}}\)
E_cell = 0.26V

Question 7.
The conductivity of 10^-3 mol/L acetic acid at 25°C is 4.1 x 10^-5 S cm^-1. Calculate its degree of dissociation, if \(\Lambda_{m}^{0}\) for acetic acid at 25°C is 390.5 S cm² mol^-1.
Answer:
We know that ∧_m = \(\frac{1000 \mathrm{~K}}{\mathrm{C}}\)
∧_m = \(\frac{1000 \times 4.1 \times 10^{-5}}{10^{-3}}\)
= 41 S cm² mol^-1
Now, α = \(\frac{\Lambda_{m}^{c}}{\Lambda_{m}^{0}}\)
= \(\frac{41}{390.5}\) = 0.105 390.5

Question 8.
(i) Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The limiting molar conductivity of ‘B’ increases 1.5 times while that of ‘A’ increases 25 times. Which of the two is a strong electrolyte? Justify your answer.
(ii) The products of electrolysis of aqueous NaCl at the respective electrodes are:
Cathode: H₂
Anode: Cl₂ and not 0₂. Explain.
Answer:
(i) ‘B’ is a strong electrolyte.
Because a strong electrolyte is already dissociated into ions, but on dilution inter ionic forces are overcome, ions are free to move. So, there is slight increase in molar conductivity on dilution.

(ii) On anode water should get oxidised in preference to Cl^– but due to overvoltage/over potential Cl” is oxidised in preference to water.

Long answer type questions

Question 1.
(a) The conductivity of 0.20 mol L^-1 solution of KC1 is 2.48 x 10^-2 S cm^-1. Calculate its molar conductivity and degree of dissociation (α). Given λ⁰ (K⁺) = 73.5 S cm² mol^-1 and λ⁰ (Cl^–) = 76.5 S cm² mol^-1.

(b) What type of battery is mercury cell? Why is it more advantageous than dry cell?
Answer:
(a) Substituting K =2.48 x 10^-2 S cm^-1, M = 0.20 molL^-1 in the expression ∧_m = \(\frac{K \times 1000}{M}[latex] , we get
PSEB 12th Class Chemistry Important Questions Chapter 3 Electrochemistry 4
Substituting ∧_m = 124 S cm² mol^-1, [latex]\Lambda_{m}^{\ominus}\) = 150 S cm² mol^-1 in the expression α = \(\frac{\Lambda_{m}}{\Lambda_{m}^{\ominus}}\), we get
Degree of dissociation, α = \(\frac{124 \mathrm{~s} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}}{150 \mathrm{~s} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}}\) = 82666
α = 82.67%

(b) Primary cell. Mercury cell is more advantageous than dry cell because its cell potential remains constant during its life as the overall reaction does not involve any ion in solution whose concentration can change during its life period.

Question 2.
(i) The conductivity of 0.001 mol L^-1 solution of CH₃COOH is 3.905 x 10^-5 S cm^-1. Calculate its molar conductivity and degree of dissociation (α).
Given λ⁰ (H⁺) = 349.6 S cm² mol^-1 and λ⁰ (CH₃COO ) = 40.9 S cm² mol^-1

(ii) Define electrochemical cell. What happens if external potential applied becomes greater than \(\boldsymbol{E}_{\text {cell }}^{\ominus}\) of electrochemical cell?
Solution:
PSEB 12th Class Chemistry Important Questions Chapter 3 Electrochemistry 5

(b) A device which is used to convert chemical energy produced in a redox reaction into electrical energy is called an electrochemical cell.
If external potential applied becomes greater than \(\boldsymbol{E}_{\text {cell }}^{\ominus}\) of
electrochemical cell, the reaction gets reversed and the electrochemical cell function as an electrolytic cell.

Question 3.
Calculate e.m.f and ΔG for the following cell at 298 K:
Mg(s) | Mg²⁺ (0.01 M) || Ag⁺ (0.0001 M) | Ag(s)
Given: \(\underset{\left(\mathrm{Mg}^{2+} / \mathrm{Mg}\right)}{\boldsymbol{E}^{\circ}}\) = -2.37V, \(\boldsymbol{E}^{\ominus}{\left(\mathbf{A g}^{+} / \mathbf{A g}\right)}\) = + 0.80V
Solution:
PSEB 12th Class Chemistry Important Questions Chapter 3 Electrochemistry 6
E_cell = 3.17 – 0.0295 log 10⁶
E_cell = 3.17 – 0.177 V = 2.993 V
E_cell = 2.993 V
Substituting n = 2, F = 96500 C mol^-1, E_cell = 2.993 V in the
expression, ΔG = – nFE_cell, we get
ΔG = nFE_cell = -2 x 96500 C mol^-1 x 2.993V
ΔG = – 577649 J mol^-1 = – 577.649 kJ mol^-1

PSEB 12th Class Chemistry Important Questions Chapter 13 Amines

August 14, 2024August 13, 2024 by Prasanna

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 13 Amines Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 13 Amines

Very short answer type questions

Question 1.
CH₃NH₂ is more basic than C₆H₅NH₂ why?
Answer:
Aliphatic amines (CH₃NH₂) are stronger bases than aromatic amines (C₆H₅NH₂) because due to resonance in aromatic amines, the lone pair of electrons on the nitrogen atom gets delocalized over the benzene ring and thus is less easily available for protonation.

Question 2.
What is the role of pyridine in the acylation reaction of amines? [NCERT Exemplar]
Answer:
Pyridine and other bases are used to remove the side product, L e., HCl from the reaction mixture.

Question 3.
A primary amine, RNH₂ can be reacted with CH₃ – X to get secondary amine, R-NHCH₃ but the only disadvantage is that 3° amine and quaternary ammonium salts are also obtained as side products. Can you suggest a method where RNH₂ forms only 2° mine? [NCERT Exemplar]
Answer:
PSEB 12th Class Chemistry Important Questions Chapter 13 Amines 1
Primary amines show carbylamine reaction in which two H-atoms attached to N-atoms of NH₂ are replaced by one C-atom. On catalytic reduction, isocyanide (formed) produces secondary amine and not tertiary or quaternary salts.

Question 4.
What is Hinsberg reagent? (NCERT Exemplar)
Answer:
Benzene sulphonyl chloride (C₆H₅SO₂Cl) is known as Hinsberg reagent. It is used to distinguish between primary, secondary and tertiary amine.

Question 5.
The conversion of primary aromatic amines into diazonium salts is known as……
Answer:
Diazotisation.

Question 6.
Rearrange the following in an increasing order of their basic strengths:
C₆H₅NH₂,C₆H₅N(CH₃)₂,(C₆H₅)₂NH and CH₃NH₂
Answer:
(C₆H₅)₂NH < C₆H₅NH₂ <C₆H₅N(CH₃)₂ <CH₃NH₂

Question 7.
What is the best reagent to convert nitrile to primary amine? [NCERT Exemplar]
Answer:
Reduction of nitriles with sodium/alcohol or LiAlH₄ gives primary amine.

Question 8.
Suggest a route by which the following conversion can be accomplished: [NCERT Exemplar]
PSEB 12th Class Chemistry Important Questions Chapter 13 Amines 2
Answer:

Question 9.
What is the role of HNO₃ in the nitrating mixture used for the nitration of benzene? [NCERT Exemplar]
Answer:
HNO₃ acts as a base in the nitrating mixture and provides the electrophile NO₂.

Question 10.
Why is benzene diazonium chloride not stored and is used immediately after its preparation? [NCERT Exemplar]
Answer:
Benzene diazonium chloride is very unstable.

Short answer type questions

Question 1.
Write the structures of A, B and C in the following:
PSEB 12th Class Chemistry Important Questions Chapter 13 Amines 4
Answer:

Question 2.
Give a chemical test to distinguish between C₆H₅CH₂NH₂ and C₆H₅NH₂.
Answer:
C₆H₅CH₂NH₂ reacts with HNO₂ at 273-278 K to give diazonium salt, which being unstable decomposes with brisk evolution of N₂ gas.
PSEB 12th Class Chemistry Important Questions Chapter 13 Amines 6
whereas, C₆H₅NH₂ reacts with HNO₂ at 273-278 K to form stable benzene diazonium chloride, which upon treatment with an alkaline solution of f3-naphthol, gives an orange dye.

Question 3.
Complete the following reaction: PSEB 12th Class Chemistry Important Questions Chapter 13 Amines 8 [NCERT Exemplar]
Answer:
The reaction exhibits an azo-coupling reaction of phenols. Benzene diazonium chloride reacts with phenol in such a manner that the para position of phenol is coupled with diazonium salt to form p-hydroxy azobenzene.
PSEB 12th Class Chemistry Important Questions Chapter 13 Amines 9

Question 4.
A solution contains 1 g mol. each of p-toluene diazonium chloride and p-nitrophenyl diazonium chloride. To this lg mol. of alkaline solution of phenol is added. Predict the major product. Explain your answer. [NCERT Exemplar]
Answer:
The above-stated reaction is an example of electrophilic aromatic substitution. In alkaline medium, phenol generates phenoxide ion which is more electron-rich than phenol and more reactive for electrophilic attack. The electrophile in this reaction is aryldiazonium cation. As we know, stronger the electrophile faster is the reaction. p Nitrophenyldiazonium cation is a stronger electrophile than p-toluene diazonium cation. So, nitrophenyl diazonium chloride couples preferentially with phenol
PSEB 12th Class Chemistry Important Questions Chapter 13 Amines 10

Question 5.
Under what reaction conditions (acidic, basic) the coupling reaction of aryl diazonium chloride with aniline is carried out? [NCERT Exemplar]
Answer:
In strongly basic conditions, benzene diazonium chloride is converted, into diazohydroxide and diazoate as both of which are not electrophilic and do not couple with aniline.
PSEB 12th Class Chemistry Important Questions Chapter 13 Amines 11
Similarly, in highly acidic conditions, aniline gets converted into an anilinium ion. From this, result aniline is no longer nucleophilic acid and hence will not couple with diazonium chloride. Hence, the reaction is carried out under mild conditions, i.e., pH 4-5
PSEB 12th Class Chemistry Important Questions Chapter 13 Amines 12

Question 6.
An organic aromatic compound ‘A’ with the molecular formula C₆H₇N is sparingly soluble in water. ‘A’ on treatment with dil. HCl gives a water-soluble compound ‘B’ ‘A’ also reacts with chloroform in presence of alcoholic KOH to form an obnoxious smelling compound ‘C’. ‘A’ reacts with benzene sulphonyl chloride to form an alkali-soluble compound ‘D’ ‘A’ reacts with NaNO₂ and HCl to form a compound ‘E’ which on reaction with phenol forms an orange-red dye ‘F’ Elucidate the structures of the organic compounds from ‘A’ to ‘F’
Answer:
PSEB 12th Class Chemistry Important Questions Chapter 13 Amines 13

Long answer type questions

Question 1.
Predict the reagent or the product in the following reaction sequence : [NCERT Exemplar]
PSEB 12th Class Chemistry Important Questions Chapter 13 Amines 15
Answer:
A correct sequence can be represented as follows including all reagents:

Hence,

(v) 5 = H₃PO₂/H₂O

Question 2.
A hydrocarbon ‘A’ (C₄H₈) on reaction with HC1 gives a compound ‘B’, (C₄H₉Cl), which on reaction with 1 mol of NH₃ gives compound ‘C’, (C₄H₁₁N). On reacting with NaNO₂ and HCl followed by treatment with water, compound ‘C’ yields an optically active alcohol, ‘D’ Ozonolysis of ‘A’ gives 2 mols of acetaldehyde. Identify compounds ‘A’ to ‘D\ Explain the reactions involved. [NCERT Exemplar]
Answer:
(i)Addition of HCl to compound A shows that compound A is alkene.
Compound ‘B’ is C₄H₉Cl.
(ii) Compound‘B’reacts with NH₂. It forms amine‘C’.

(iii) ‘C’ gives diazonium salt with NaNO₂/HCl, which yields an optically active alcohol.
So, ‘C’ is aliphatic amine.
(iv) ‘A on ozonolysis produces 2 moles of CH₃CHO. So, A is CH₃– CH =CH-CH₃ (But-2-ene).
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PSEB 12th Class Chemistry Important Questions Chapter 13 Amines 19