CBSE Class 12 Chemistry Organic Chemistry Amines Worksheet Set 01

Short Answer Type Questions

Question. (i) Arrange the following in increasing order of boiling points.
\((CH_3)_3N, C_2H_5OH, C_2H_5NH_2\)
(ii) Arrange the following in increasing order of base strength in gas phase:
\((C_2H_5)_3N, C_2H_5NH_2, (C_2H_5)_2NH\)
Answer: (i) \((CH_3)_3N < C_2H_5NH_2 < C_2H_5OH\)
(ii) \(C_2H_5NH_2 < (C_2H_5)_2N < (C_2H_5)_3N\)
Base strength is the ability to donate lone pair. Due to inductive +I effect of \(C_2H_5\), the negative charge density on nitrogen atom increases. Therefore, lone pair is easily available for donation.
\(C_2H_5NH_2 < (C_2H_5)_2NH < (C_2H_5)_3N\)

Question. (i) Arrange the following in decreasing order of solubility in water:
\((CH_3)_3N, (CH_3)_2NH, CH_3NH_2\)
(ii) Arrange the following in decreasing order of the basic character:
\(C_6H_5NH_2, (CH_3)_3N, C_2H_5NH_2\)
Answer: (i) \(CH_3NH_2 > (CH_3)_2NH > (CH_3)_3N\)
(ii) \((CH_3)_3N > C_2H_5NH_2 > C_6H_5NH_2\)
\(C_6H_5NH_2\) is least basic due to presence of -I group i.e., \(– C_6H_5\) group that withdraws unshared pair of electrons present on N-atom of \(–NH_2\) group due to which it does not undergo protonation easily; whereas \((CH_3)_3N\) is most basic due to strong +I effect of \(–CH_3\) groups which release their bonding electrons towards N-atom and it undergoes protonation easily. Hence, the decreasing order of the basic character is \((CH_3)_3N > C_2H_5NH_2 > C_6H_5NH_2\).

Question. (i) Arrange the following in increasing order of \(pK_b\) values:
\(C_6H_5CH_2NH_2, C_6H_5NHCH_3, C_6H_5NH_2\)
(ii) Arrange the following in decreasing order of solubility in water:
\((C_2H_5)_2NH, C_2H_5NH_2, C_6H_5NH_2\)
Answer: (i) \(C_6H_5CH_2NH_2 < C_6H_5NHCH_3 < C_6H_5NH_2\)
Greater the basic nature of amine, lesser will be the \(pK_b\) value of amine. Since, \(C_6H_5CH_2NH_2\) is most basic, so, it possesses least \(pK_b\) value, while, \(C_6H_5NH_2\) is least basic, so, it possesses highest \(pK_b\) value.
(ii) \(C_2H_5NH_2 > (C_2H_5)_2NH > C_6H_5NH_2\)
\((C_2H_5)_2NH\) and \(C_2H_5NH_2\) are aliphatic amines and \(C_6H_5NH_2\) is aromatic amine. Lower aliphatic amines can form hydrogen bonds with water molecules. Therefore, such amines are soluble in water. When number of alkyl groups (hydrophobic) increases then molar mass of amines increases. This usually results in a decrease in its solubility in water. Hence, \(C_2H_5NH_2\) is more soluble in water than \((C_2H_5)_2NH\). Whereas, aromatic amines are insoluble in water because of large hydrocarbon part (hydrophobic) which retards the formation of hydrogen bonding.

Question. (i) Write IUPAC name of the following compound: \((CH_3)_2N — CH_2CH_3\)
(ii) Complete the following reaction equation:
\(C_6H_5N_2Cl + H_3PO_2 + H_2O \longrightarrow .......................\)
Answer: (i) N, N-Dimethylethanamine
(ii) \(ArN_2^+Cl^- + H_3PO_2 + H_2O \longrightarrow ArH + N_2 + H_3PO_3 + HCl\) (where Ar is \(C_6H_5\)) Benzene

Question. Why does acetylation of \(—NH_2\) group of aniline reduce its activating effect?
Answer: The acetyl group being electron withdrawing attracts the lone pair of electrons of the N - atom towards carbonyl group. As a result, the activation effect of \(–NH_2\) group is reduced, that is, that lone pair of electrons on nitrogen is less available for donation to benzene ring by resonance. Therefore, activating effect of \(-NHCOCH_3\) group is less than that of \(-NH_2\) group.