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The major product formed in the following reaction is
$\mathrm{CH}_3 \mathrm{CH}(\mathrm{Cl}) \mathrm{CH}_2-\mathrm{CH}_2 \mathrm{OH} \stackrel{\text { aq. } \mathrm{KOH}}{\longrightarrow}$
Options:
$\mathrm{CH}_3 \mathrm{CH}(\mathrm{Cl}) \mathrm{CH}_2-\mathrm{CH}_2 \mathrm{OH} \stackrel{\text { aq. } \mathrm{KOH}}{\longrightarrow}$
- A $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_2 \mathrm{OH}$
- B $\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{OH}$
- C
- D
Solution:
1739 Upvotes
Verified Answer
The correct answer is:
Halogenated compounds on treatment with aq. $\mathrm{KOH}$ form alcohols. The halogen atom is substituted by - $\mathrm{OH}$ group.
$$
\begin{aligned}
& \mathrm{CH}_3 \mathrm{CH}(\mathrm{Cl}) \mathrm{CH}_2-\mathrm{CH}_2 \mathrm{OH} \underset{\text { aq. } \mathrm{KOH}}{\longrightarrow} \\
& \qquad \mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_2-\mathrm{CH}_2 \mathrm{OH}
\end{aligned}
$$
Here $\mathrm{OH}^{-}$acts as a nucleophile.
$$
\begin{aligned}
& \mathrm{CH}_3 \mathrm{CH}(\mathrm{Cl}) \mathrm{CH}_2-\mathrm{CH}_2 \mathrm{OH} \underset{\text { aq. } \mathrm{KOH}}{\longrightarrow} \\
& \qquad \mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_2-\mathrm{CH}_2 \mathrm{OH}
\end{aligned}
$$
Here $\mathrm{OH}^{-}$acts as a nucleophile.
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