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Page No 402: - Chapter 13 Amines Exercise Solutions class 12 ncert solutions Chemistry - SaraNextGen [2024]


Question 13.10:

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:

It is given that compound ‘C’ having the molecular formula, C6H7N is formed by heating compound ‘B’ with Br2 and KOH. This is a Hoffmann bromamide degradation reaction. Therefore, compound ‘B’ is an amide and compound ‘C’ is an amine. The only amine having the molecular formula, C6H7N is aniline, (C6H5NH2).

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5799/NS_01-12-08_Utpal_12_Chemistry_13_14_html_139c1078.jpg

Therefore, compound ‘B’ (from which ’C’ is formed) must be benzamide, (C6H5CONH2).

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5799/NS_01-12-08_Utpal_12_Chemistry_13_14_html_m2acbd204.jpg

Further, benzamide is formed by heating compound ‘A’ with aqueous ammonia. Therefore, compound ‘A’ must be benzoic acid.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5799/NS_01-12-08_Utpal_12_Chemistry_13_14_html_96caac5.jpg

The given reactions can be explained with the help of the following equations:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5799/NS_01-12-08_Utpal_12_Chemistry_13_14_html_66bf633a.jpg

Question 13.11:

Complete the following reactions:

(i) https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5800/NS_01-12-08_Utpal_12_Chemistry_13_14_html_3de866fd.gif

(ii) https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5800/NS_01-12-08_Utpal_12_Chemistry_13_14_html_m708f17dd.gif

(iii) https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5800/NS_01-12-08_Utpal_12_Chemistry_13_14_html_56505600.gif

(iv) https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5800/NS_01-12-08_Utpal_12_Chemistry_13_14_html_656f92eb.gif

(v) https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5800/NS_01-12-08_Utpal_12_Chemistry_13_14_html_413fd36a.gif

(vi) https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5800/NS_01-12-08_Utpal_12_Chemistry_13_14_html_6e1f407a.gif

(vii) https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5800/NS_01-12-08_Utpal_12_Chemistry_13_14_html_m9dbf284.gif

Answer:

(i)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5800/NS_01-12-08_Utpal_12_Chemistry_13_14_html_72cd3afe.jpg

(ii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5800/NS_01-12-08_Utpal_12_Chemistry_13_14_html_2b0b8084.gif

(iii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5800/NS_01-12-08_Utpal_12_Chemistry_13_14_html_30135938.gif

(iv)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5800/NS_01-12-08_Utpal_12_Chemistry_13_14_html_m778f8639.gif

(v)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5800/NS_01-12-08_Utpal_12_Chemistry_13_14_html_536142b2.jpg

(vi)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5800/NS_01-12-08_Utpal_12_Chemistry_13_14_html_2cc0d2aa.jpg

(vii)

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5800/NS_01-12-08_Utpal_12_Chemistry_13_14_html_786e5e66.gif

Question 13.12:

Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?

Answer:

Gabriel phthalimide synthesis is used for the preparation of aliphatic primary amines. It involves nucleophilic substitution (SN2) of alkyl halides by the anion formed by the phthalimide.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5801/NS_01-12-08_Utpal_12_Chemistry_13_14_html_m64e1bb09.jpg

But aryl halides do not undergo nucleophilic substitution with the anion formed by the phthalimide.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5801/NS_01-12-08_Utpal_12_Chemistry_13_14_html_2101c4a4.jpg

Hence, aromatic primary amines cannot be prepared by this process.

Question 13.13:

Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.

Answer:

(i) Aromatic amines react with nitrous acid (prepared in situ from NaNO2 and a mineral acid such as HCl) at 273 − 278 K to form stable aromatic diazonium salts i.e., NaCl and H2O.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5802/NS_01-12-08_Utpal_12_Chemistry_13_14_html_3e438ae8.jpg

(ii) Aliphatic primary amines react with nitrous acid (prepared in situ from NaNO2 and a mineral acid such as HCl) to form unstable aliphatic diazonium salts, which further produce alcohol and HCl with the evolution of N2 gas.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5802/NS_01-12-08_Utpal_12_Chemistry_13_14_html_e47ab88.jpg

Question 13.14:

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 point than tertiary amines?

(iii) Why are aliphatic amines stronger bases than aromatic amines?

Answer:

(i) Amines undergo protonation to give amide ion.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5803/NS_01-12-08_Utpal_12_Chemistry_13_14_html_m53cfb456.gif

Similarly, alcohol loses a proton to give alkoxide ion.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5803/NS_01-12-08_Utpal_12_Chemistry_13_14_html_m1bd576b1.gif

In an amide ion, the negative charge is on the N-atom whereas in alkoxide ion, the negative charge is on the O-atom. Since O is more electronegative than N, O can accommodate the negative charge more easily than N. As a result, the amide ion is less stable than the alkoxide ion. Hence, amines are less acidic than alcohols of comparable molecular masses.

(ii) In a molecule of tertiary amine, there are no H−atoms whereas in primary amines, two hydrogen atoms are present. Due to the presence of H−atoms, primary amines undergo extensive intermolecular H−bonding.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5803/NS_01-12-08_Utpal_12_Chemistry_13_14_html_m2b271f15.jpg

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/272/5803/NS_01-12-08_Utpal_12_Chemistry_13_14_html_m3a98e3b6.jpg

As a result, extra energy is required to separate the molecules of primary amines. Hence, primary amines have higher boiling points than tertiary amines.

(iii) Due to the −R effect of the benzene ring, the electrons on the N- atom are less available in case of aromatic amines. Therefore, the electrons on the N-atom in aromatic amines cannot be donated easily. This explains why aliphatic amines are stronger bases than aromatic amines.

Also Read : INTRODUCTION-Chapter-13-Amines-Intext-Solutions-class-12-ncert-solutions-Chemistry

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