Which of the following reactions DOES NOT involve Hoffmann bromamide degradation?
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The Correct Option is B
Solution and Explanation
Option (1): \(\mathrm{C_6H_5CH_2CONH_2}\) with Br2/NaOH gives \(\mathrm{C_6H_5CH_2NH_2}\). This is Hoffmann degradation.
Option (2): \(\mathrm{C_6H_5CH_2COCH_3}\) with Br2/NaOH undergoes the haloform reaction to give \(\mathrm{C_6H_5CH_2COOH}\); NH3/\(\Delta\) converts it to the amide \(\mathrm{C_6H_5CH_2CONH_2}\); LiAlH4 then reduces the amide to \(\mathrm{C_6H_5CH_2CH_2NH_2}\) with no loss of carbon. The amide is never treated with Br2/NaOH, so this sequence does NOT involve Hoffmann bromamide degradation.
Option (3): \(\mathrm{C_6H_5COCl \xrightarrow{NH_3} C_6H_5CONH_2 \xrightarrow{Br_2/NaOH} C_6H_5NH_2}\). The second step is Hoffmann degradation.
Option (4): \(\mathrm{C_6H_5CN \xrightarrow{KOH,H_2O} C_6H_5CONH_2 \xrightarrow{Br_2/NaOH} C_6H_5NH_2}\). The second step is Hoffmann degradation.
Answer: (2) the sequence C6H5CH2COCH3 with (i) Br2, NaOH/H+, (ii) NH3/\(\Delta\), (iii) LiAlH4/H2O
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