Q: Nitration of aniline in strong acidic medium also gives m-nitroaniline because :

−H⊕3 is m-directing, hence besides para (51%) and ortho (2%), meta product (47%) is also formed in significant yield.

Q: The product formed by the reaction of an aldehyde with a primary amine is:

> ? = ? + ? − NH2 → > ? = ? − ? ??ℎ??? ????

Q: In the reaction A is:

H3PO2 and H2O works as a reducing agent.

Q: Which of the following reactions is appropriate for converting acetamide to methanamine?

CH3−C−NH2+Br2+4NaOH⟶ΔCH3−NH2+2NaBr+Na2CO3+3H2O This is Hoffmann Bromamide reaction.

Question 1

Which of the following amine will give the carbyle test?

Accepted Answer

Aliphatic and aromatic primary es give carbylamine reaction.

Question 2

The correct order of the basic strength of methyl substituted amines in aqueous solution is :

Accepted Answer

In aqueous solution, electron donating inductive effect, solvation effect (H-bonding) and steric hindrance all together affect basicstrength of substituted aminesBasic character :(CH3​)2​NH>CH3​NH2​>(CH3​)3​N 			2° 					1° 					3°

Question 3

Nitration of aniline in strong acidic medium also gives m-nitroaniline because :

Accepted Answer

−H⊕​3​  is m-directing, hence besides para (51%)  and ortho (2%),  meta product (47%)  is also formed in significant yield.

Question 4

The product formed by the reaction of an aldehyde with a primary amine is:

Accepted Answer

> ? = ? + ? − NH2​  	→ 	> ? = ? − ?											??ℎ??? ????

Question 5

The correct statement regarding the basicity of aryl amines is:

Accepted Answer

Here lone pair is in conjugation with double bond so basic strength decreased R −NH2​ (No conjugation)

Question 6

Which of the following will be most stable diazonium salt RN2+​X− ?

Accepted Answer

Primary aliphatic amines form highly unstable alkyldiazonium salts. Primary aromatic amines form arene diazonium salts which are stable for a short time in solution at low temperature (273−278 K) . The stability of arenediazonium can be explained on the basis of resonance.

Question 7

In the reaction A is:

Accepted Answer

H3​PO2​  and H2​O  works as a reducing agent.

Question 8

Which of the following reactions is appropriate for converting acetamide to methanamine?

Accepted Answer

CH3​−C−NH2​+Br2​+4NaOH⟶Δ​CH3​−NH2​+2NaBr+Na2​CO3​+3H2​O This is Hoffmann Bromamide reaction.

Question 9

The following reactionis known by the name:

Accepted Answer

Benzoylation of aniline is an example of Schotten-Bauman reaction.

Question 10

Method by which Aniline cannot be prepared is:

Accepted Answer

Due to resonance C – Cl bond acquires double bond character.

Question 11

Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.(i) (CH3​)2​CHNH2​(ii) CH3​(CH2​)2​NH2​(iii) CH3​NHCH(CH3​)2​(iv) (CH3​)3​CNH2​(v) C6​H5​NHCH3​(iv) (CH3​CH2​)2​NCH3​(vii) m−BrC6​H4​NH2​.

Accepted Answer

The IUPAC names along with the classification are given below:(i) 1-Methylethanamine (primary amine)(ii) Propan-1-amine (primary amine)(iii) N-Methyl-2-methylethanamine (secondary amine)(iv) 2-Methylpropan-2-amine (primary amine)(v) N-methylbenzenamine or N-methylaniline (secondary amine)(vi) N-Ethyl-N-methylethanamine (tertiary amine)(vii) 3-Bromobenzenamine or 3-bromoaniline (primary amine)

Question 12

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-methylaniline.

Accepted Answer

(i) Methylamine and dimethylamine :             Carbylamine test. Methylamine, on heating with alc. KOH solution and chloroform form foul-smelling methyl isocyanide. This test is not given by dimethylamine.(ii) Secondary and tertiary amines :Secondary amines give Libermann nitrosoamine test. On heating with nitrous acid (prepared in situ), they give yellow coloured oily N-nitrosoamine.Tertiary amines do not give such test.(iii) Ethylamine and aniline :                          Azo dye test. Aniline on diazotization ( ice cold nitrous acid solution) followed by coupling with 2-naphthol (in alkaline solution) forms brilliant orange or red dye. Ethylamine will not form dye. It will give brisk effervescence (due to liberation of nitrogen gas) but solution remains clear.(iv) Aniline and benzylamine :Azo dye test.Aniline on diazotization ( ice cold nitrous acid solution) followed by coupling with 2-naphthol (in alkaline solution) forms brilliant orange or red dye. Benzylamine will not give such test.(v) Aniline and N-methylaniline :Carbylamine test. Aniline, on heating with alc. KOH solution and chloroform form foul-smelling methyl isocyanide. This test is not given by N-methyl aniline.

Question 13

Account for the following:(i) pKb​ 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.

Accepted Answer

(i) pKb​ of aniline is more than that of methylamine.In aniline, the lone pair of electrons on N atom is in resonance with benzene ring. Hence, it cannot be easily donated to an acid. This decreases its basicity. In methyl amine, the +I effect of methyl group increases the electron density on N atom so that the lone pair of electrons on N atom can be easily donated to an acid. Hence, methylamine is more basic than aniline. Higher is the basicity, lower is the pKb and vice versa. (ii) Ethylamine is soluble in water whereas aniline is not. With increase in the molecular weight, the solubility decreases. Aniline has higher molecular weight than ethylamine.(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide. Due to the +I effect of −CH3​ group, methylamine is more basic than water. Therefore, in water, methylamine produces OH− ions by accepting H+ ions from water. OH− ions react with ferric chloride to precipitate hydrated ferric oxide.

Question 14

Arrange the following:(i) In decreasing order of the pKb​ values: C2​H5​NH2​,C6​H5​NHCH3​,(C2​H5​)2​NH  and C6​H5​NH2​(ii) In increasing order of basic strength: C6​H5​NH2​,C6​H5​N(CH3​)2​,(C2​H5​)2​NH and CH3​NH2​(iii) In increasing order of basic strength:    (a) Aniline, p-nitroaniline and p-toluidine

Accepted Answer

(i) Stronger base has lower pKb value.C6H5NH2> C6H5NHCH3>C2H5NH2>(C2H5)2NH(ii) Increasing order of basicity is as follows: C6H5NH2Et2NH>EtNH2>NH3The stabilization of conjugate acids due to hydrogen bonding is absent.(v) Increasing order of boiling pointMe2NH

Question 15

Describe a method for the identification of primary, secondary and tertiary amines. Also, write chemical equations of the reactions involved.

Accepted Answer

Hinsberg's test is used for the identification of primary, secondary, and tertiary amines. Hinsberg's reagent is benzenesulphonyl chloride (C6​H5​SO2​Cl).It reacts differently with primary, secondary, and tertiary amines. (i) Hinsberg's reagent reacts with primary amines to form N− alkylbenzenesulphonyl amide which is acidic in nature and soluble in alkali.Note: N− alkylbenzenesulphonyl amide contains a strong electron-withdrawing sulphonyl group. Due to this, the H− atom attached to nitrogen can be removed easily. Hence, it is acidic.(ii) Hinsberg's reagent reacts with secondary amines to form a sulphonamide which is insoluble in alkali.Note: As there is no hydrogen atom attached to the N atom in the sulphonamide, it is not acidic and insoluble in alkali.(iii) Hinsberg's reagent does not react with tertiary amines.

Question 16

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.

Accepted Answer

(i) Carbylamine reaction: Both aliphatic and aromatic primary amines when warmed with chloroform and an alcoholic solution of KOH, produces isocyanides or carbylamines which have very unpleasant odours. This reaction is called carbylamine reaction.R−NH2​+CHCl3​+3KOH( alc )⟶R−N≡C+3KCl+3H2​O (ii) Diazotisation: The process of conversion of a primary aromatic amino compound into a diazonium salt, is known as diazotisation. This process is carried out by adding an aqueous solution of sodium nitrite to a solution of primary aromatic amine (e.g., aniline) in excess of HCl at a temperature below 5∘C . I’amine ArNH2​​+NaNO2​+2HX⟶273−278 K​ Arene tiazonium ​Arsalt​+X−+NaX+2H2​O (iii) Hoffmann's bromamide reaction: When an amide is treated with bromine in alkali solution, it is converted to a primary amine that has one carbon atom less than the starting amide. This reaction is known as Hoffinann's bromamide degradation reaction.C6​H5​CONH2​⟶Br2​+NaOH​C6​H5​NH2​ (iv) Coupling reaction: In this reaction, arene diazonium salt reacts with aromatic amino compound (in acidic medium) or a phenol (in alkaline medium) to form brightly coloured azo compounds. The reaction generally takes place at para position to the hydroxy or amino group. If para position is blocked, it occurs at ortho position and if both ortho and para positions are occupied, than no coupling takes place.(v) Ammonolysis: It is a process of replacement of either halogen atom in alkyl halides (or aryl halides) or hydroxyl group in alcohols (or phenols) by amino group. The reagent used for ammonolysis is alcoholic ammonia. Generally, a mixture of primary, secondary and tertiary amine is formed.ROH+NH3​ High temp, Pressure N2​O3​,Δ​⟶RNH2​+H2​O (vi) Acetylation: The process of introducing an acetyl (CH3​CO−) group into molecule using acetyl chloride or acetic anhydride is called acetylation.(vii) Gabriel phthalimide synthesis: It is a method of preparation of pure aliphatic and aralkyl primary amines. Phthalimide on treatment with ethanolic KOH gives potassium phathalimide which on heating with a suitable alkyl Or aralkyl halides gives N -substituted phthalimides, which on hydrolysis with dil HCl or with alkali give primary amines.

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