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Question:
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The major product of the following synthetic reactions is
Options:
- A
- B $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}$
- C $\mathrm{CH}_2=\mathrm{CHCH}_2 \mathrm{OH}$
- D $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}$
Solution:
1803 Upvotes
Verified Answer
The correct answer is:
$\mathrm{CH}_2=\mathrm{CHCH}_2 \mathrm{OH}$
Chlorination of propane with chlorine in presence of light gives chloropropane.
$$
\mathrm{H}_3 \mathrm{C}-\mathrm{CH}_2-\mathrm{CH}_3+\mathrm{Cl}_2 \stackrel{h v}{\longrightarrow} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl}
$$
Dehydrohalogenation in presence of alc. $\mathrm{KOH}$ gives propene.
$$
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl} \stackrel{\text { Alc. } \mathrm{KOH}}{\longrightarrow} \mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2
$$
At high temperature terminal allylic $\mathrm{C}$ undergoes chlorination.
$$
\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2+\mathrm{Cl}_2 \stackrel{773 \mathrm{~K}}{\longrightarrow} \mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{Cl}
$$
Allylic chlorination $\mathrm{Cl}$ is substituted by $\mathrm{OH}$
$$
\begin{aligned}
& \mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{Cl}+\mathrm{AgOH}(a q) \longrightarrow \\
& \mathrm{CH}_2=\mathrm{CHCH}_2 \mathrm{OH}
\end{aligned}
$$
$$
\mathrm{H}_3 \mathrm{C}-\mathrm{CH}_2-\mathrm{CH}_3+\mathrm{Cl}_2 \stackrel{h v}{\longrightarrow} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl}
$$
Dehydrohalogenation in presence of alc. $\mathrm{KOH}$ gives propene.
$$
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl} \stackrel{\text { Alc. } \mathrm{KOH}}{\longrightarrow} \mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2
$$
At high temperature terminal allylic $\mathrm{C}$ undergoes chlorination.
$$
\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2+\mathrm{Cl}_2 \stackrel{773 \mathrm{~K}}{\longrightarrow} \mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{Cl}
$$
Allylic chlorination $\mathrm{Cl}$ is substituted by $\mathrm{OH}$
$$
\begin{aligned}
& \mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{Cl}+\mathrm{AgOH}(a q) \longrightarrow \\
& \mathrm{CH}_2=\mathrm{CHCH}_2 \mathrm{OH}
\end{aligned}
$$
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