SaraNextGen.Com
Updated By SaraNextGen
On March 11, 2024, 11:35 AM

Page No 400: - Chapter 13 Amines Exercise Solutions class 12 ncert solutions Chemistry - SaraNextGen [2024]


Question 13.1:

Write IUPAC names of the following compounds and classify them into primary,

secondary and tertiary amines.

(i) (CH3)CHNH(ii) CH3(CH2)2NH2

(iii) CH3NHCH(CH3)(iv) (CH3)3CNH2

(v) C6H5NHCH3 (vi) (CH3CH2)2NCH3

(vii) m−BrC6H4NH2

Answer:

(i) 1-Methylethanamine (10 amine)

(ii) Propan-1-amine (10 amine)

(iii) N−Methyl-2-methylethanamine (20 amine)

(iv) 2-Methylpropan-2-amine (10 amine)

(v) N−Methylbenzamine or N-methylaniline (20 amine)

(vi) N-Ethyl-N-methylethanamine (30 amine)

(vii) 3-Bromobenzenamine or 3-bromoaniline (10 amine)

Question 13.2:

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-methylaniline.

Answer:

(i) 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. Methylamine (being an aliphatic primary amine) gives a positive carbylamine test, but dimethylamine does not.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5791/NS_01-12-08_Utpal_12_Chemistry_13_14_html_m517262ec.gif

(ii) Secondary and tertiary amines can be distinguished by allowing them to react with Hinsberg’s reagent (benzenesulphonyl chloride, C6H5SO2Cl).

Secondary amines react with Hinsberg’s reagent to form a product that is insoluble in an alkali. For example, N, N−diethylamine reacts with Hinsberg’s reagent to form N, N−diethylbenzenesulphonamide, which is insoluble in an alkali. Tertiary amines, however, do not react with Hinsberg’s reagent.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5791/NS_01-12-08_Utpal_12_Chemistry_13_14_html_m589211d5.jpg

(iii) Ethylamine and aniline can be distinguished using the azo-dye test. A dye is obtained when aromatic amines react with HNO2 (NaNO2 + dil.HCl) at 0-5°C, followed by a reaction with the alkaline solution of 2-naphthol. The dye is usually yellow, red, or orange in colour. Aliphatic amines give a brisk effervescence due (to the evolution of N2 gas) under similar conditions.

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/4%20(1)(2).png ​

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5791/NS_01-12-08_Utpal_12_Chemistry_13_14_html_m176e96fc.gif

(iv) Aniline and benzylamine can be distinguished by their reactions with the help of nitrous acid, which is prepared in situ from a mineral acid and sodium nitrite. Benzylamine reacts with nitrous acid to form unstable diazonium salt, which in turn gives alcohol with the evolution of nitrogen gas.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5791/NS_01-12-08_Utpal_12_Chemistry_13_14_html_4522e3e2.jpg

On the other hand, aniline reacts with HNO2 at a low temperature to form stable diazonium salt. Thus, nitrogen gas is not evolved.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5791/NS_01-12-08_Utpal_12_Chemistry_13_14_html_723261b9.gif

(v) Aniline and N-methylaniline can be distinguished using the Carbylamine test. Primary amines, on heating with chloroform and ethanolic potassium hydroxide, form foul-smelling isocyanides or carbylamines. Aniline, being an aromatic primary amine, gives positive carbylamine test. However, N-methylaniline, being a secondary amine does not.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5791/NS_01-12-08_Utpal_12_Chemistry_13_14_html_3aefe8f7.gif

Question 13.3:

Account for the following:

(i) pKof 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 op− 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) pKof aniline is more than that of methylamine:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5792/NS_01-12-08_Utpal_12_Chemistry_13_14_html_20e86793.jpg

Aniline undergoes resonance and as a result, the electrons on the N-atom are delocalized over the benzene ring. Therefore, the electrons on the N-atom are less available to donate.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5792/NS_01-12-08_Utpal_12_Chemistry_13_14_html_1e024b12.jpg

On the other hand, in case of methylamine (due to the +I effect of methyl group), the electron density on the N-atom is increased. As a result, aniline is less basic than methylamine. Thus, pKb of aniline is more than that of methylamine.

(ii) Ethylamine is soluble in water whereas aniline is not:

Ethylamine when added to water forms intermolecular H−bonds with water. Hence, it is soluble in water.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5792/NS_01-12-08_Utpal_12_Chemistry_13_14_html_m2fb454bd.jpg

But aniline does not undergo H−bonding with water to a very large extent due to the presence of a large hydrophobic −C6H5 group. Hence, aniline is insoluble in water.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5792/NS_01-12-08_Utpal_12_Chemistry_13_14_html_m7ef2598c.jpg

(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5792/NS_01-12-08_Utpal_12_Chemistry_13_14_html_1402fdcc.jpg

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.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5792/NS_01-12-08_Utpal_12_Chemistry_13_14_html_1b5790f.gif

Ferric chloride (FeCl3) dissociates in water to form Fe3+ and Cl ions.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5792/NS_01-12-08_Utpal_12_Chemistry_13_14_html_m4dcf4bbe.gif

Then, OH ion reacts with Fe3+ ion to form a precipitate of hydrated ferric oxide.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5792/NS_01-12-08_Utpal_12_Chemistry_13_14_html_50f4191c.gif

(iv) Although amino group is o,p− directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline:

Nitration is carried out in an acidic medium. In an acidic medium, aniline is protonated to give anilinium ion (which is meta-directing).

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5792/NS_01-12-08_Utpal_12_Chemistry_13_14_html_6b643617.jpg

For this reason, aniline on nitration gives a substantial amount of m-nitroaniline.

(v) Aniline does not undergo Friedel-Crafts reaction:

A Friedel-Crafts reaction is carried out in the presence of AlCl3. But AlCl3 is acidic in nature, while aniline is a strong base. Thus, aniline reacts with AlCl3 to form a salt (as shown in the following equation).

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5792/NS_01-12-08_Utpal_12_Chemistry_13_14_html_2928dfb9.jpg

Due to the positive charge on the N-atom, electrophilic substitution in the benzene ring is deactivated. Hence, aniline does not undergo the Friedel-Crafts reaction.

(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines:

The diazonium ion undergoes resonance as shown below:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5792/NS_01-12-08_Utpal_12_Chemistry_13_14_html_m73db1057.jpg

This resonance accounts for the stability of the diazonium ion. Hence, diazonium salts of aromatic amines are more stable than those of aliphatic amines.

(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines:

Gabriel phthalimide synthesis results in the formation of 1° amine only. 2° or 3° amines are not formed in this synthesis. Thus, a pure 1° amine can be obtained. Therefore, Gabriel phthalimide synthesis is preferred for synthesizing primary amines.

Question 13.4:

Arrange the following:

(i) In decreasing order of the pKbvalues:

C2H5NH2, C6H5NHCH3, (C2H5)2NH and C6H5NH2

(ii) In increasing order of basic strength:

C6H5NH2, C6H5N(CH3)2, (C2H5)2NH and CH3NH2

(iii) In 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) In increasing order of solubility in water:

C6H5NH2, (C2H5)2NH, C2H5NH2.

Answer:

(i) In C2H5NH2, only one −C2H5 group is present while in (C2H5)2NH, two −C2H5 groups are present. Thus, the +I effect is more in (C2H5)2NH than in C2H5NH2. Therefore, the electron density over the N-atom is more in (C2H5)2NH than in C2H5NH2. Hence, (C2H5)2NH is more basic than C2H5NH2.

Also, both C6H5NHCH3 and C6H5NHare less basic than (C2H5)2NH and C2H5NH2 due to the delocalization of the lone pair in the former two. Further, among C6H5NHCH3 and C6H5NH2, the former will be more basic due to the +T effect of −CH3 group. Hence, the order of increasing basicity of the given compounds is as follows:

C6H5NH2 < C6H5NHCH3 < C2H5NH2 < (C2H5)2NH

We know that the higher the basic strength, the lower is the pKb values.

C6H5NH2 > C6H5NHCH3 > C2H5NH2 > (C2H5)2NH

(ii) C6H5N(CH3)2 is more basic than C6H5NH2 due to the presence of the +I effect of two −CH3 groups in C6H5N(CH3)2. Further, CH3NH2 contains one −CH3 group while (C2H5)2NH contains two −C2H5 groups. Thus, (C2H5)2 NH is more basic than C2H5NH2.

Now, C6H5N(CH3)2 is less basic than CH3NH2 because of the−R effect of −C6H5 group.

Hence, the increasing order of the basic strengths of the given compounds is as follows:

C6H5NH2 < C6H5N(CH3)2 < CH3NH< (C2H5)2NH

(iii) (a)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5793/NS_01-12-08_Utpal_12_Chemistry_13_14_html_m3c560074.jpg

In p-toluidine, the presence of electron-donating −CH3 group increases the electron density on the N-atom.

Thus, p-toluidine is more basic than aniline.

On the other hand, the presence of electron-withdrawing

−NO2 group decreases the electron density over the N−atom in p-nitroaniline. Thus, p-nitroaniline is less basic than aniline.

Hence, the increasing order of the basic strengths of the given compounds is as follows:

p-Nitroaniline < Aniline < p-Toluidine

(b) C6H5NHCH3 is more basic than C6H5NH2 due to the presence of electron-donating −CH3 group in C6H5NHCH3.

Again, in C6H5NHCH3, −C6H5 group is directly attached to the N-atom. However, it is not so in C6H5CH2NH2. Thus, in C6H5NHCH3, the −R effect of −C6H5 group decreases the electron density over the N-atom. Therefore, C6H5CH2NH2 is more basic than C6H5NHCH3.

Hence, the increasing order of the basic strengths of the given compounds is as follows:

C6H5NH2 < C6H5NHCH3 < C6H5CH2NH2.

(iv) In the gas phase, there is no solvation effect. As a result, the basic strength mainly depends upon the +I effect. The higher the +I effect, the stronger is the base. Also, the greater the number of alkyl groups, the higher is the +I effect. Therefore, the given compounds can be arranged in the decreasing order of their basic strengths in the gas phase as follows:

(C2H5)3N > (C2H5)2NH > C2H5NH2 > NH3

(v) The boiling points of compounds depend on the extent of H-bonding present in that compound. The more extensive the H-bonding in the compound, the higher is the boiling point. (CH3)2NH contains only one H−atom whereas C2H5NH2 contains two H-atoms. Then, C2H5NH2 undergoes more extensive H-bonding than (CH3)2NH. Hence, the boiling point of C2H5NH2 is higher than that of (CH3)2NH.

Further, O is more electronegative than N. Thus, C2H5OH forms stronger H−bonds than C2H5NH2. As a result, the boiling point of C2H5OH is higher than that of C2H5NH2 and (CH3)2NH.

Now, the given compounds can be arranged in the increasing order of their boiling points as follows:

(CH3)2NH < C2H5NH2 < C2H5OH

(vi) The more extensive the H−bonding, the higher is the solubility. C2H5NH2 contains two H-atoms whereas (C2H5)2NH contains only one H-atom. Thus, C2H5NH2 undergoes more extensive H−bonding than (C2H5)2NH. Hence, the solubility in water of C2H5NH2 is more than that of (C2H5)2NH.

Further, the solubility of amines decreases with increase in the molecular mass. This is because the molecular mass of amines increases with an increase in the size of the hydrophobic part. The molecular mass of C6H5NH2 is greater than that of C2H5NH2 and (C2H5)2NH.

Hence, the increasing order of their solubility in water is as follows:

C6H5NH2 < (C2H5)2NH < C2H5NH2

Also Read : Page-No-401:-Chapter-13-Amines-Exercise-Solutions-class-12-ncert-solutions-Chemistry

SaraNextGen