Question

The product 'C' in the following reaction is:

\[\ce{CH3CH2Br ->[alc. KCN] {'A'} ->[H3O^+][\Delta] {'B'} ->[i. LiAlH4][ii. H3O^+] {'C'}}\]

Options

• CH₃CH₂CHO

• CH₃CH₂OH

• CH₃CH₂CH₂OH

• CH₃COOH

Answer 1

CH₃CH₂CH₂OH

Explanation:

\[\ce{\underset{Ethyl bromide}{CH3CH2Br} ->[alc. KCN] \underset{(A)}{CH3CH2CN} ->[H3O^+][\Delta] \underset{(B)}{CH3CH2COOH} ->[i. LiAlH4][ii. H3O^+] \underset{\underset{(C)}{Propan-1-ol}}{CH3CH2CH2OH}}\]