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Question 39

The product (C) in the below mentioned reaction is: $$CH_3-CH_2-CH_2-Br \xrightarrow[\Delta]{KOH_{alc}} A \xrightarrow{HBr} B \xrightarrow[\Delta]{KOH_{aq}} C$$

Solution

$$CH_3CH_2CH_2Br \xrightarrow{KOH_{alc}} CH_3CH=CH_2$$ (A = propene, elimination).

$$CH_3CH=CH_2 \xrightarrow{HBr} CH_3CHBrCH_3$$ (B = 2-bromopropane, Markovnikov addition).

$$CH_3CHBrCH_3 \xrightarrow{KOH_{aq}} CH_3CH(OH)CH_3$$ (C = propan-2-ol, substitution).

The answer is Option (4): Propan-2-ol.

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