Identify $\mathrm{Z}$ in the following serie's $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I} \stackrel{\text {…
$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I} \stackrel{\text { Alc. } \mathrm{KOH}}{\longrightarrow} \mathrm{X} \stackrel{\mathrm{Br}_{2}}{\longrightarrow} \mathrm{Y} \stackrel{\text{excess}\mathrm{KCN}}{\longrightarrow} \mathrm{Z}$
- $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CN}$
- $\mathrm{NCCH}_{2}-\mathrm{CH}_{2} \mathrm{CN}$
- $\mathrm{BrCH}_{2}-\mathrm{CH}_{2} \mathrm{CN}$
- $\mathrm{BrCH}=\mathrm{CHCN}$
Solution
$\mathrm{BrCH}_{2}-\mathrm{CH}_{2} \mathrm{Br} \stackrel{\text{excess}\mathrm{KCN}}{\longrightarrow} \mathrm{CNCH}_{2} \cdot \mathrm{CH}_{2} \mathrm{CN}$ ,
Asked in: JEE-TOPICTESTS-CHEMISTRY
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