Bihar Board Class 12 Chemistry Solutions Chapter 13 Amines

Bihar Board Class 12 Chemistry Solutions Chapter 13 Amines Textbook Questions and Answers, Additional Important Questions, Notes.

BSEB Bihar Board Class 12 Chemistry Solutions Chapter 13 Amines

Bihar Board Class 12 Chemistry Amines Intext Questions and Answers

Question 1.
Classify the following amines as primary, secondary or tertiary?

(iii) (C₂H₅)₂ CHNH₂,
(iv) (C₂H₅)₂ NH.
Answer:
(i) Primary amine,
(ii) tertiary amine,
(iii) primary amine,
(iv) secondary amine.

Question 2.
(i) Write structures of different isomeric amines corresponding to the molecular formula C₄H₁₁N.
(ii) Write IUPAC names of all isomers.
(iii) What type of isomerism is exhibited by different pairs of amines?
Answer:
(i) C₄H₁₁N stands for the following isomeric amines
(a) Primary amines

(b) Secondary amines

(c) Tertiary amines

(ii) IUPAC names of all isomeric amines have been given in brackets along with their structures.
(iii) (a) Chain isomerism -CH₃-CH₂-CH₂-CH₂NH₂ and

(b) Position isomerism-CH₃-CH₂-CH₂-CH₂ NH₂ and,

(c) Metamerism-CH₃-CH₂-NH-CH₂-CH₃ and CH₃-NH-CH₂-CH₂-CH₃ and

(d) Functional Isomerism-Primary, Secondary and tertiary amines having the same molecular formula show the functional isomerism among themselves.
CH₃-CH₂-CH₂-CH₂-NH₂; CH₃-NH-CH₂-CH₃,

Question 3.
How will you convert (i) Benzene into aniline.
(ii) Benzene into N, N-dimethylaniline.
(iii) Cl-(CH₂)₄-Cl into hexane-1,6-diamine?
Answer:
(i) Benzene into aniline-

(ii) Benzene into N, N-dimethylaniline

(iii) Cl(CH₂)₄ Cl into hexane 1,6-diamine

Question 4.
Arrange the following in increasing order of their basic strength.
(i) C₂H₅NH₂, C₆H₅NH₂, NH₃, C₆H₅CH₂NH₂, and (C₂H₅)₂NH.
(ii) C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N, C₆H₅NH₂.
(iii) CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, C₆H₅NH₂, C₆H₅CH₂NH₂.
Answer:
(i) C₆H₅NH₂ < NH₃ < C₆H₅CH₂ NH₂ < C₂H₅NH₂ < (C₂H₅)₂ NH.
(ii) C₆H₅NH₂ < C₂H₅ NH₂ < (C₂H₅)₃ N < (C₂H₅)₂ NH.
(iii) C₆H₅NH₂ < C₆H₅CH₂NH₂ < (CH₃)₃ N < CH₃NH₂ < (CH₃)₂ NH.

Question 5.
Complete the following acid-base reactions and name the products:
(i) CH₃ CH₂ CH₂ NH₂ + HCl →
(ii) (C₂H₅)₃N + HCl →
Answer:
(i) CH₃ CH₂ CH₂ NH₂ + HCl →CH₃ CH₂ CH₂ NH₃Cl^– n-Propylinium chloride
(ii) (C₂H₅)₃N + HCl → [(C₂H₅)₃ Triethyl ammonium chloride

Question 6.
Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.
Answer:

Question 7.
Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.
Answer:

Question 8.
Write structures of different isomers corresponding to the molecular formula C₃H₉N. Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid.
Answer:

Out of (i), (ii), (iii) and (iv) above only (i) and (ii) being primary aliphatic amines will liberate N₂ gas on reaction with nitrous acid.
(i) CH₃ CH₂ CH₂ NH₂ IUPAC name: Propan-1-amine

CH₃CH₂CH₂NH₂ + HNO₂ → CH₃CH₂CH₂OH + N₂ + H₂O

Question 9.
Convert (i) 3-Methylaniline into 3-nitrotoluene, (ii) Aniline into 1,3,5-tribromobenzene.
Answer:
(i) 3-Methylaniline into 3-nitrotoluene-

(ii) Aniline into 1, 3,5-tribromobenzene-

Bihar Board Class 12 Chemistry Amines Text Book Questions and Answers

Question 1.
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiery amines.
(i) (CH₃)₂ CHNH₂,
(ii) CH₃ (CH₂)₂ NH₂,
(iii) CH₃NHCH(CH₃)₂,
(iv) (CH₃)₃ CNH₂,
(v) C₆H₅NHCH₃,
(vi) (CH₃CH₂)₂ NCH₃,
(vii) m- BrC₆H₄NH₂.
Answer:

Question 2.
Give one chemical test to distinguish between the following pairs of compounds.
(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylamine and aniline
(iv) Aniline and benzylamine
(v) Aniline and N-methlyaniline.
Answer:
(i) Methylamine and dimethylamine-Methylamine-a primary amine respond to Carbylamine reaction whereas dimethylamine does not.

(ii) Secondary and tertiery amines-A secondary amine reacts with benzene sulphonyl chloride to form N, N-dialkyl benzene sulphonamide which does not dissolve in KOH. A tertiary amine does not react with benzene sulphonyl chloride.

(iii) Ethylamine and aniline-Ehtylamine on reaction with nitrous acid (HNO₂) gives out N₂ gas, whereas aniline does not.

Aniline, on the other harid, reacts with nitrous acid and HCl to form benzene diazonium chloride which gives orange dye on reaction with β-Naphthol. It does not give out N₂ gas.

(iv) Aniline and benzylamine-
Nitrous acid test-Benzylamine on reaction with HNO₂ decomposes even at low temp, to liberate N₂ gas.

Aniline, on the other hand, reacts with HNO₂ + HCl to form benzene diazonium chloride which is stable at 273-278 K and hence does not decompose to give N₂ gas.

(v) Aniline and N-methyl aniline-Aniline is a primary aromatic amine, N-methyl aniline is a secondary aromatic amine. They may be distinguished by Carbylamine test.
Aniline gives phenyl isocyanide on heating with chloroform (CHCl₃) and alcoholic KOH which is extremely foul-smelling while N-methyl aniline does not give this test.

Question 3.
Account for the following:
(i) pK_b of aniline is more than that of methylamine.
(ii) Ethylamine is soluble in water, whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although amino group is o- and p- directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
(v) Aniline does not undergo Friedel-Crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Answer:
(i) Aniline is a much weaker amine than methylamine. In methylamine, the lone pair of electrons on N of is made many phones readily available for sharing due to the +1 effect exerted by – CH₃ group thus making its KJ, large or p^kb small. On the other hand, the benzene ring present in aniline withdraws electrons towards itself making the lone pair of electrons less available on N of NH₂ group. Here its p^kb is large.
(ii) Ethylamine is soluble in water because it can form H-bonds with water molecules.

On the other hand aniline is insoluble in water due to the larger hydrocarbon part [C₆H₅ as compared to C₂H₅ of ethylamine] which tends to retard the formation of H-bonds.

(iii) Methylamine like ammonia reacts with transition metal ions in aqueous solutions like Fe³⁺ [of ferric chloride] to form complex like [Fe (NH₃)₆]³⁺ as methylamine in aqueous solution is a good nucleophile and hence attacks metal ions like Fe³⁺ which are electrophiles. The iron complex in aqueous solution precipitates as hydrated ferric oxide.

(iv) Aniline, in the strongly acidic medium (like in a mixture of cone. HNO₃ and cone. H₂SO₄) is protonated to form anilinium ion C₆H₅ NH₃ which is meta-directing. That is why NH₂ group in spite of being 0- and p-directing also forms significant amount of meta-derivative during nitration in strongly acidic medium in electrophilic substitution reactions.

(v) Aniline does not undergo Friedel Crafts reaction like alkylation and acetylation because it forms salt with aluminium chloride which is used at a catalyst and is a Lewis acid. Due to this, nitrogen of – NH₂ of aniline acquires a positive charge and hence acts as a strong deactivating group for further reaction.
(vi) Diazonium salts of aromatic amines are more stable than that of aliphatic amines. It is because even at low-temperature diazonium salts of aliphatic amines decompose to give out N₂ gas.

(vii) Gabriel’s phthalimide synthesis is preferred for synthesis of primary aliphatic amines because primary amines are obtained in the pure form by this method. Aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.

Question 4.
Arrange the following:
(i) In decreasing order of the pK_b values:
C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂NH and C₆H₅NH₂
(ii) In increasing order of basic strength:
C₆H₅NH₂, C₆H₅N (CH₃)₂, (C₂H₅)₂NH and CH₃NH₂
(iii) Increasing order of basic strength
(a) Aniline, p-nitroaniline and p-toluidine
(b) C₆H₅NH₂,C₆H₅NHCH₃,C₆H₅CH₂NH₂
(iv) In decreasing order of basic strength in gas phase C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N and NH₃
(v) In increasing order of boiling point C₂H₅OH, (CH₃)₂NH, C₂H₅NH₂
(vi) Increasing order of solubility in water
C₆H₅NH₂, (C₂H₅)₂NH, C₂H₅NH₂
Answer:
(i) In decreasing order of p^kb values
C₆H₅NH₂ > C₆H₅NHCH₃ > C₂H₅NH₂ > (C₂H₅)₂ NH.

(ii) In increasing order of basic strength
C₆H₅NH₂ < C₆H₅N(CH₃)₂ < CH₃NH₂ < (C₂H₅)₂NH.

(iii) Increasing order of basic strength
(i) p-nitroaniline < aniline < p-toluidine.

(iv) Decreasing order of basic strength in gas phase.
(C₂H₅)₃N > (C₂H₅)₂NH > C₂H₅NH₂ > NH₃.

(v) Increasing order of boiling point (CH₃)₂NH<C₂H₅NH₂<C₂H₅OH.
(vi) Increasing order of solubility in water C₆H₅NH₂ < (C₂H₅)₂ NH < C₂H₅NH_2.

Question 5.
How will you convert
(i) Ethanoic acid into methenamine,
(ii) Hexanenitrtile into 1-amino pentane,
(iii) Methanol to ethanoic acid,
(iv) Ethanamine into methenamine,
(v) Ethanoic acid into propanoic acid,
(vi) Methanamine into ethanamine,
(vii) Nitromethane into dimethylamine,
(viii) Propanoic acid into ethanoic acid?
Answer:
(i) Ethanoic acid into methanamine:

(ii) Hexanenitrile into 1-amino pentane:

(iii) Methanol to ethanoic acid:

(iv) Ethanamine to methenamine:

(v) Ethanoic acid into propanoic acid-

(vi) Methanmine into ethanemine-

(vii) Nitromethane into dimethylamine

(viii) Propanoic acid into ethanoic acid-

Question 6.
Describe the method for the identification of primary, secondary and tertiery amines. Also write chemical equations of the reactions involved.
Answer:
Hinsberg’s Method-This is an excellent test for distinguishing primary, secondary and tertiery amines. The amine is shaken with benzene sulphonyl chloride in the presence of aqueous KOH solution.
(i) A primary amine gives a clear solution which an acidification gives an insoluble N-alkyl benzene sulphonamide.

(ii) A secondary amine gives an insoluble N, N-dialkyl benzene sulphonamide which remains unaffected on addition of acid.

(iii) A tertiery amine does not react at all. Therefore, it remains insoluble in the alkaline solution but dissolves on acidification to give a clear solution.

Question 7.
Write short notes on the following:
(i) Carbylamine reaction
(ii) Diazotisation
(iii) Hofmann’s bromamide reaction
(iv) Coupling reaction
(v) Ammonolysis
(vi) Acetylation
(vii) Gabriel phthalimide synthesis
Answer:
(i) Carbylamine reaction-Both aliphatic and aromatic primary amines when warmed with chloroform and an alcoholic solution of KOH produce isocyanides or carylamines which have an unpleasant odour.

In contrast, secondary and tertiery amines (both aliphatic and aromatic) do not give this test.
(ii) Diazotisation-The reaction of converting aromatic primary amines into diazonium salts with a cold (273-278 K) solution of nitrous acid is called diazotisation.

(iii) Hoffmann bromamide reaction-The reaction of a higher add amide with Br2 and KOH to give a lower amine is called Hoffman bromamide reaction. It gives an amine-containing one carbon atom less than the original amide.

(iv) Coupling Reaction-Arenediazonium salts react with highly reactive (i.e., electron-rich) aromatic compounds such as phenols and amines to form brightly coloured azo compounds. Ar-N = N-Ar.

(v) Ammonolysis-Reactions of an alkyl halide (R-X) with an alcoholic solution of ammonia giving a mixture of 1°, 2°, 3° amines and quaternary salt is called ammonolysis.

If R-X is used in excess, a mixture of 1°, 2°, 3° amines along with some quartemary ammonium salts is obtained.

It is a typical nucleophilic substitution reaction.

(vi) Acetylation-The replacement of an active hydrogen of alcohols, phenols or amines with an acyl group (RCO) to form the corresponding esters or amides is called acetylation. Acetylation is carried out in the presence of a base like pyridine, dimethylaniline etc. by acetyl chloride or acetic anhydride.

(vii) Gabriel’s phthalimide synthesis-In this reaction phthalimide is converted into its potassium salt by treating it with alcoholic KOH. Then potassium phthalimide is heated with an alkyl halide to yield on N-alkyl phthalimide which is hydrolysed to phthalic acid and a primary amine by heating with HCl or KOH solution. This synthesis is very useful for the preparation of pure aliphatic primary amines.

Question 8.
Accomplish the following conversions:
(i) Nitrobenzene to benzoic acid.
(ii) Benzene to m-bromophenol.
(iii) Benzoic acid to aniline.
(iv) Aniline to 2,4,6-tribromofluorobenzene.
(v) Benzyl chloride to 2-phenylethanolamine.
(vi) Chlorobenzene to p-chloroaniline.
(vii) Aniline to p-bromoaniline.
(viii) Benzamide to toluene (ix) Aniline to benzyl alcohol.
Answer:
(i) Nitrobenzene to benzoic acid

(ii) Benzene to m-bromophenol-

(iii) Benzoic acid into aniline-

(iv) Aniline to 2,4,6-tribromofluorobenzene-

(v) Benzyl chloride to 2-phenyl ethanamine-

(vi) Chlorobenzene to p-chloroaniline-

(vii) Aniline top-bmmoaniline-

(ix) Aniline to benzyl alcohol-

Question 9.
Give the structures of A, B and C in the following reactions:

Answer:
(i) A is CH₃CH₂CN; B is CH₃CH₂CONH₂; C is CH₃CH₂NH₂.
(ii) A = C₆H₅ CN; B = C₆H₅COOH; C = C₆H₅CONH₂.
(iii) A = CH₃ CH₂CN;B = CH₃CH₂ NH₂; C = CH₃CH₂CH₂OH
(iv) A = C₆H₅NH₂; B = C₆H₅N₂⁺ Cl^–; C = C₆H₅OH
(v) A = CH₃CONH₂; B = CH₃NH₂; C = CH₃OH
(vi) A = C₆H₅NH₂; B = C₆H₅N₂⁺ Cl^–;

Question 10.
An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br₂ and KOH forms a compound ‘C’ of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B and C.
Answer:
Since it is an aromatic compound, it has a benzene ring in it. B on heating with Br₂ and KOH [Hoffmann bromamide reaction] forms a compound ‘C’ of molecular formula C₆H₇N. Now only higher amides on Hoffman bromamide reaction [Br₂ + KOH treatment] give rise to lower amine. Therefore B is C₆H₅CONH₂ and ‘C’ with the molecular formula is C₆H₅NH₂. Since the compound C₆H₅CONH₂ is obtained from A on treatment with aqueous ammonia and heating; A has to be a carboxylic acid C₆H₅COOH. The sequence of reactions along with names (IUPAC) of A, B and C are as follows :

Question 11.
Complete the following reactions:
(i) C₆H₄NH₂ + CHCl₃ + ale. KOH
(ii) C₆H₅N₂Cl + H₃PO₂+ H₂O →
(iii) C₆H₅NH₂+ H₂SO₄ (cone.) →
(iv) C₆H₅N₂Cl + C₂H₅OH →
(v) C₆H₅NH₂ + Br₂ (aq) →
(vi) C₆H₅NH₂+(CH₃CO)₂O →

Answer:

Question 12.
Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
Answer:
Aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide, that is,

Question 13.
Wite the reaction of
(i) aromatic and (ii) aliphatic primary amines with nitrous acid.
Answer:
Aromatic primary amines react with nitrous acid in the presence of dil. HCl to form aromatic diazonium chloride.

Whereas aliphatic primary amines react with nitrous acid to give primary alcohols librating N₂ gas.

Question 14.
Give plausible explanation for each of the following:
(i) Why are amines less acidic than alcohols of comparable molecular masses?
(ii) Why do primary amines have higher boiling than tertiery amines?
(iii) Why are aliphatic amines stronger bases than aromatic amines?
Answer:
(i) The alkoxide ion left after the removal of a proton from an alcohol ROH is more stable than RNH left after the removal of a proton from an amine RNHr This is due to the fact that oxygen atom in alcohols is more electronegative than N atom in amines and thus can accommodate the negative charge better.

(ii) Due to the presence of two H-atoms on N-atom of primary amines, they undergo extensive intermolecular H-bonding while tertiery amines due to the absence of a H-atom on the N-atom, do not undergo hydrogen bonding. As a result, primary amines have higher boiling points than tertiery amines of comparable molecular mass, e.g., B.Pt of n- Butylaniline (351K) is much higher than that of tert-butylamine (319 K).

(iii) Aliphatic amines are stronger bases than aromatic amines. In aliphatic amines, alkyl groups present exert +1 effect

(electron releasing) thus make the lone pair of electrons more readily available for sharing by an acid, whereas in aromatic amines, the presence of benzene ring exerts -1 effect (electron-withdrawing) due to which the availability of lone pair of electrons on N in NH₂ decreases thus making it less basic.

Bihar Board Class 12 Chemistry Amines Additional Important Questions and Answers

Very Short Answer Type Questions

Question 1.
Name the amide which on reduction gives
Answer:

Question 2.
Complete the reaction
Answer:

Question 3.
How will you convert aniline to phenyl isothiocyanate. (PB. S.B. 1994)
Answer:

Question 4.
Write the TUPAC name of (I.T.T. 91)
Answer:
N, N Dimethyl-3-methyl pentane-3-amine.

Question 5.
How is aniline converted to sulfanilic acid. (A.I.S.B. 1992)
Answer:

Question 6.
Write the chemical test to distinguish between aniline and N-methyl aniline. (A.I.S.E. 2001)
Answer:
Aniline, being a primary amine, responds to carbylamine test.

Question 7.
Complete the reaction:
ArN₂Cl+H₃PO₂+H₂O→
Answer:

Question 8.
Write the IUPAC and trivial name of CH3NC. (A.I.S.E. 1991)
Answer:
IUPAC name is Methyl carbylamine.
Trivial name is Methyl isocyanide (Aceto isonitrile).

Question 9.
What happens when methyl cyanide reacts with H2 in the presence of Pt? (H.S.B. 2002)
Answer:

Question 10.
Arrange the following: (M.L.N.R. 1992)
Aniline, ethylamine, ethane, phenol in order of increasing basic strength.
Answer:
Phenol < ethane < aniline < ethylamine.

Question 11.
What are ambident nucleophiles. Give two examples. (A.I. S.E.2001)
Answer:
A nucleophile which can form bonds through more than one atom in it, e.g., NO₂^– and CN (through C and N).

Question 12.
Name one reagent used for the selective reduction of m- dinitrobenzene to m-nitroaniline.
Answer:
Ammonium sulphide (NH₄)₂ S.

Question 13.
Give the I.U.P.A.C. name of
Answer:
2-Methyl-l, 4-dinitrobutane.

Question 14.
– NO₂^– group is an ambident group. What are the active sites in it.
Answer:
Active sites in – NO₂^– group are oxygen and nitrogen.

Question 15.
What is a Coupling reaction?
Answer:
The reaction of diazonium salts with phenols in basic medium (pH 9-10) and with amines in acidic medium (pH 4 – 5) to give corresponding azo (- N = N -) dyes is called coupling reaction.

Question 16.
Horw will you convert aniline to iodobenzene?
Answer:

Question 17.
How will you get benzonitrile from aniline?
Answer:

Question 18.
Complete the following:
Answer:

Question 19.
What is Hoffmann Bromamide reaction?
Answer:
The conversion of a higher amide (both aliphatic as well as aromatic) into a lower amine (1°) in the presence of ale. KOH and Br₂ is known as Hoffman Bromamide reaction.

Question 20.
What is the IUPAC name of a tertiary amine-containing one methyl, one ethyl and one n-propyl group?
Answer:
N-Ethyl-N-methyl propane-l-amine.

Question 21.
What is the IUPAC name of
Answer:
3-Methylaniline or 3-Methylbenzenamine.

Question 22.
Name one chemical test to distinguish between an aromatic primary amine and an aliphatic primary amine.
Answer:
Azo dye test.

Question 23.
What is the directive influence of the amino group in arylamines?
Answer:
o, p-directing.

Question 24.
Mention two important uses of sulphanilic acid.
Answer:
It is used in the manufacture of

1. dyes,
2. sulpha drugs.

Question 25.
Give an example of a Zwitterion.
Answer:

Question 26.
How will you convert benzene into aniline?
Answer:

Question 27.
Which is more basic: CH₃NH₂ or (CH₃)₃N? (H.S.B. 2005)
Answer:
CH₃NH₂ is more basic than (CH₃)₃N due to greater stability of conjugate acid CH₃⁺ NH₃over (CH₃)₃N⁺ H due to hydrogen bonding.

Question 28.
Which is more acidic (or basic): aniline or ammonia? (H.S.B. 2005)
Answer:
Since ammonia is more basic than aniline, therefore, aniline is more acidic than ammonia.

Question 29.
Aniline gets coloured on standing in air for a long time. Why? (H.S.B. 2005)
Answer:
Due to strong electron-donating effect (+ R effect) of NH₂ group, the electron density on the benzene ring increases. As a result, aniline is easily oxidized on standing in air for a long time to form coloured products.

Question 30.
Arrange the following in the order of their increasing basicity: p-toluidine, N, N-dimethyl-p-toluidine, p-nitroaniline, aniline.
Answer:
p-Nitroaniline < aniline < p-toluidine < N, N-dimethyl-p- toluidine.

Question 31.
Arrange the following in increasing order of their acid strength: methylamine, dimethyl-p-toluidine, p-nitroaniline, aniline.
Answer:
The acid strength increases in the reverse order of their basic strength, i.e., dimethylamine < methyl amine < N-methyl aniline < aniline.

Short Answer Type Questions

Question 1.
Which one is more acidic? Explain. (IIT 2004)

Answer:
Due to powerful -I effect of the F-atom, it withdraws electrons from the N⁺H₃ group. As a result, the electron density in the N-H bond of p-fluoro anilinium ion decreases and hence release of a proton from it is much more easier than from anilinium ion. Therefore, p-fluoro- anilinium ion is more acidic than anilinium ion.

Question 2.
The following reaction gives two products. Write the structure of the products. (IIT. 1998)

Answer:
CH₃CONH group is o, p-directing. Therefore CH₃CONHC₆H₅ (acetanilide) on bromination gives a mixture of o- and p-bromo acetamides.

Question 3.
What is the major product in the following reaction? (W. Bengal JEE 2001)

Answer:
In Hoff man elimination reaction, it is the less sterically hindered (β-hydrogen that is removed and hence less substituted alkene is the major product.

Question 4.
Aniline does not undergo Friedel-Crafts reactions. Explain. (Roorkee 1999, C.B.S.E. PMT 2004)
Answer:
Aniline is a base while AlCl₃ used as a catalyst in Friedel- Crafts reaction is a Lewis acid. They combine with each other to form the salt.
C₆H₅NH₂ + AlCl₃ → C₆H₅N⁺H₂ AlCl₃^– Due to the presence of a positive charge on N-atom, the group N⁺H₂ AlCl₃^– acts as a strongly deactivating group. As a result, it reduces the electron density in the benzene ring and hence aniline does not undergo Friedel craft reaction.

Question 5.
Aniline dissolves in aqueous HCl. Why?
Answer:
Aniline dissolves in aqueous HCl due to the formation of water-soluble salt.

Question 6.
Why does AgCl dissolves in methylamine solution?
Answer:
Because it forms soluble complex
2CH₃NH₂ +AgCl → [ Ag(CH₃NH₂)₂]+Cl^– Soluble complex.

Question 7.
Identify A, B, C in the following:

Answer:

Question 8.
How will you convert aniline → benzylamine in 3 steps. (I.I.T. 2000)
Answer:

Question 9.
Identify : A,B,C

Answer:

Question 10.
Suggest a structural formula for a compound having molecular formula C₈H₁₁N (A) which is optically active, dissolves in dil. aq. HCl and releases N₂ on reaction with HNO₂.
Answer:

Question 11.
Write the structures of the reagents/organic compounds A to F in the following sequence of reactions: (A.I.S.E. 96)

Answer:

Question 12.
An optically inactive compound A having molecular formula C₄H₁₁N or treatment with HNO₂ gave an alcohol (B). B an heating at 440 K gave an alkene (C). C on treatment with HBr gave an optically active compound (D) having the molecular formula C4HgBr. Identify A, B, C and D and write down their structural formula. Also, write the equations involved. (C.B.S.E. Sample Paper)
Answer:
Since compound A is optically inactive and contains N which gives alcohol on reaction with HNO₂. It is therefore a primary amine – NH₂. The reactions are

C* is Chiral Centre. D is optically active.

Question 13.
How is classification of amines different from that of alcohols?
Answer:
Classification of amines is made on the nature of the Nitrogen atom (whether it has one, two or no hydrogen atom) to which the alkyl group is attached. The classification of alcohols, on the other hand, is made on the basis of carbon atom to which the – OH group is attached.

Question 14.
tert-Butylamine cannot be prepared by the action of NH₃ on tertbutyl bromide. Explain why?
Answer:
tert-Butyl bromide being a 3° alkyl halide on treatment with a base (i.e. NH₃) prefers to undergo elimination rather than substitution.

Question 15.
Suggest chemical reactions for the following conversions:
(i) Cyclohexanol → Cyclohexylamine
(ii) n-Hexanenitrile → 1-Aminopentane (D.S.B. 2002)
Answer:

Question 16.
Although trimethylamine and n-propylamine have the same molecular weight, the former boils at a lower temperature than the Explain.
Answer:
n-Pfopylamm (CH₃CH₂CH₂NH₂) has two H-atoms on N- atom, and hence undergoes intermolecular H-bonding thereby raising boiling point. Trimethylamine (CH₃)₃N being a 3° amine does not a H-atom on the N-atom. As a result, it does not undergo H-bonding and hence its boiling point is low.

Question 17.
Sulphanilic acid is soluble in dilute NaOH but not in dil. HCl. Explain.
Answer:
Sulphanilic acid exists as a Zwitter ion. In the presence of dil. NaOH, the weakly acidic – N⁺H₃ transfer its H⁺ ion to OH^– ion to form a soluble p-NH₂C₆H₄SO₃^– Na⁺, On the other hand, – SO₃^– is very weak base and hence does not accept a proton from dil. HCl to form p-N⁺H₃C₆H₄SO₃H and hence it does not dissolve in it.

Question 18.
Give increasing order towards electrophilic substitution of the following:

Answer:

Question 19.
Why is it. necessary to add excess of mineral aids to diazotisation of amines. (Roorkee 1999, CBSE PMT 2004)
Answer:
During diazotisation of arylamines, excess of mineral acids is used. In general, one mole of amine is treated with approx, three moles of acid, (i) One mole to dissolve amine, (ii) one mole to liberate HNO₂ from NaNO₂ and (iii) one mole to maintain proper acidity of the reaction mixture to prevent its coupling with diazotised salt.

Question 20.
Convert benzene to m-chloroaniline.
Answer:

Long Answer Type Questions

Question 1.
Two isomeric compounds A and B have same molecular formula C₃H₅N. Predict the structure A and B as the basis of following information.
(i) A and B do not react with HNO₂ or CH₃COCl.
(ii) On refluxing with dil. HCl A gives C and B gives D, C and D being the monobasic carboxylic acids.
(iii) Molecular mass of D = 74.
(iv) C gives Tollen’s test positive, whereas D does not Predict the structure of C and D. Also give the reduction products of A and B.
Answer:
(i) Since A and B do not react with HNO₂ or CH₃COCl it implies that they are not amines.
(ii) Since their hydrolytic products are monobasic carboxylic acids (C and D) it means that A and B may be cyano compounds.
(iii) Let the formula of D be C_nH_2n+1 – COOH.
Its molecular mass = 74.
74-46
Now 14n + 46 = 74; n = \(\frac{74-46}{14}\) =2.
∴D is C₂H₅COOH i.e., Propionic acid.
(iv) C gives Tollen’s test positive i.e., reduces ammonical AgNO₃. The only reducing carboxylic acid is formic acid.
∴ C is HCOOH i.e., Formic acid.
Since D (CH₃CH₂COOH) is a hydrolytic product of B. Therefore, B should be propane nitrile having formula CH₃CH₂CN.
Similarly, A which is an isomer of B gives formic acid (HCOOH) as hydrolytic product, therefore, A should be ethyli sonitrile with formula
CH₃CH₂CN=C.

Question 2.
Bring out the following conversions:
(i) Nitrobenzene to benzoic acid
(ii) Benzene to 3-bromophenol
(iii) Benzene to m-nitroaniline
(iv) Propanenitrile to ethanamine
(v) m-nitroaniline to m-chloriodobenzene (Roorkee 1988)
(vi) 4-nitroaniline to 1,2,3-tribromobenzene. (I.I.T.J.E.E. 1990)
Answer:
(i) Nitrobenzene to benzoic acid

(ii) Benzene to 3-bromophenol

(iii) Benzene to m – nitroaniline

Na₂S₂, being a very weak reducing agent, is effective in carrying out selective reduction of only one – NO₂ group.
(iv) Propanenitri le to ethanamine

(v) m-nitroaniline to m- chloriodobenzene

(vi) 4- nitroaniline to 1,2,3 – tribromobenzene.