Question

In the following sequence of reactions:

$CH_3CH_2OH \xrightarrow{KMnO_4} (X) \xrightarrow{SOCl_2/NH_3} (Y) \xrightarrow{Br_2+NaOH} (Z)$

The end product 'Z' is

• A. Acetic acid
• B. acetone
• C. methylamine
• D. Ethylamine

Answer 1

Answer: (C)

methylamine

Answer 2

The reaction proceeds as follows:

1. Oxidation: Ethanol ($CH_3CH_2OH$) is oxidized by $KMnO_4$ to acetic acid ($CH_3COOH$).

   $CH_3CH_2OH \xrightarrow{KMnO_4} CH_3COOH$

2. Conversion to Amide: Acetic acid is converted to acetyl chloride ($CH_3COCl$) using $SOCl_2$, which then reacts with ammonia ($NH_3$) to form acetamide ($CH_3CONH_2$).

   $CH_3COOH \xrightarrow{SOCl_2} CH_3COCl$

   $CH_3COCl + NH_3 \rightarrow CH_3CONH_2$

3. Hofmann Rearrangement: Acetamide undergoes Hofmann rearrangement in the presence of $Br_2$ and $NaOH$ to yield methylamine ($CH_3NH_2$).

   $CH_3CONH_2 \xrightarrow{Br_2 + NaOH} CH_3NH_2 + CO_2$

Therefore, the end product 'Z' is methylamine.