Choose the correct sequential order for the conversion of 3-methyl butan-1-ol to 2-methylpropanol
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → B, D, A, C, E
The alcohol has 5 carbons and the product, 2-methylpropan-1-ol (E), has 4, so one carbon has to be removed. The step that removes a carbon is the Hofmann bromamide degradation of an amide. So the route first builds up that amide from the alcohol, then degrades it to the amine and turns the amine into the alcohol.
B — first. PCC oxidises the primary alcohol only as far as the aldehyde, 3-methylbutanal.
\( \ce{(CH3)2CHCH2CH2OH} \xrightarrow{\ce{PCC}} \ce{(CH3)2CHCH2CHO} \)
D — second. Acidified \( \ce{KMnO4} \) oxidises the aldehyde to 3-methylbutanoic acid.
\( \ce{(CH3)2CHCH2CHO + [O]} \xrightarrow{\ce{KMnO4/H+}} \ce{(CH3)2CHCH2COOH} \)
A — third. The acid and ammonia form the ammonium salt, which loses water on heating to give 3-methylbutanamide.
\( \ce{(CH3)2CHCH2COOH + NH3 -> (CH3)2CHCH2COO^-NH4^+} \xrightarrow{\Delta} \ce{(CH3)2CHCH2CONH2 + H2O} \)
C — fourth. Bromine and NaOH convert the amide into an amine with one carbon atom less, 2-methylpropan-1-amine; the carbonyl carbon leaves as sodium carbonate.
\( \ce{(CH3)2CHCH2CONH2 + Br2 + 4NaOH -> (CH3)2CHCH2NH2 + Na2CO3 + 2NaBr + 2H2O} \)
E — last. Nitrous acid (from \( \ce{NaNO2 + HCl} \)) converts the primary aliphatic amine into the alcohol, 2-methylpropan-1-ol, with the loss of nitrogen gas.
\( \ce{(CH3)2CHCH2NH2 + HNO2} \xrightarrow{\ce{NaNO2 + HCl}} \ce{(CH3)2CHCH2OH + N2 + H2O} \)
NCERT (2026-27 edition):
Hoffmann Bromamide Degradation Reaction
