Give reasons for the following : (a) Haloalkanes react with KCN to form alkyl cyanides, while with AgCN they form isocyanides. (b) is more reactive towards $\mathrm{S_N2}$ displacement as compared to . (c) Grignard reagents should be prepared under anhydrous conditions.
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(a) ionic vs covalent → C- vs N-attack; (b) benzylic sp³ vs aromatic sp²; (c) water destroys the C−Mg bond.
All three parts start from aniline, $\mathrm{C_6H_5NH_2}$, and just change the reagent. Let us take them one by one.
(a) answer: To get sulphanilic acid, heat aniline with concentrated sulphuric acid. At high temperature (about 453 to 473 K) the $\mathrm{-SO_3H}$ group goes onto the ring at the para position relative to the amino group, giving sulphanilic acid (p-aminobenzenesulphonic acid):
\[ C_6H_5NH_2 + H_2SO_4 \rightarrow p\text{-}H_2N\text{-}C_6H_4\text{-}SO_3H + H_2O \]
(b) answer: To get 2,4,6-tribromoaniline, treat aniline with bromine water. The $\mathrm{-NH_2}$ group is a very strong activating group that directs incoming groups to the ortho and para positions, and it activates the ring so strongly that all three of those positions react at once. So bromine goes in at positions 2, 4 and 6:
\[ C_6H_5NH_2 + 3Br_2 \rightarrow 2,4,6\text{-tribromoaniline} + 3HBr \]
(c) answer: To get acetanilide, treat aniline with acetic anhydride. This adds an acetyl group, $\mathrm{-COCH_3}$, onto the nitrogen (acetylation of the amino group), giving acetanilide:
\[ C_6H_5NH_2 + (CH_3CO)_2O \rightarrow C_6H_5NHCOCH_3 + CH_3COOH \]
