- ✓Neohexyl chloride
- BSecondary butyl iodide
- CTertiary butyl bromide
- DIso-propyl iodide
e.g., $C{H_3} - O - {C_2}{H_5}\, + \,HI\,\xrightarrow{{373\,\,K}}C{H_3}I\, + \,{C_2}{H_5}OH$
The alkyl halide is formed from the smaller alkyl group.
However, in case of tertiary alkyl ether following reaction occurs.
$\mathop {C{H_3} - OC{{(C{H_3})}_3}}\limits_{ter - butyl\,\,methyl\,\,ether} + \,HI\,\xrightarrow{{373\,\,K}}$ ${(C{H_3})_3}C - I\, + \,C{H_3}OH$
The alkyl halide is formed from the tertiary alkyl group and the cleavage of such ethers occurs by $S_{N^1}$ mechanism as the product is controlled by the formation of more stable intermediate tertiary carbocation from orotonated ether.
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$(a)\;\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}(\mathrm{OH}) \mathrm{CH}_{3} \stackrel{\mathrm{conc.H}, \mathrm{SO}_{4}}{\longrightarrow}$
$(b)\;\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}(\mathrm{Br}) \mathrm{CH}_{3} \stackrel{\text { alc.KOH }}{\longrightarrow}$
$(c)\;\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}(\mathrm{Br}) \mathrm{CH}_{3} \xrightarrow[\text { It should be }\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CO}^{-} \mathrm{K}^{+}]{\text { given by } \mathrm{NTA}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{O}^{-} \mathrm{K}^{+}}$
$(d)\;\begin{array}{*{20}{c}}
{{{(C{H_3})}_2}C - C{H_2} - CHO} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}\xrightarrow{\Delta }$
Which of these reaction(s) will not produce Saytzeff product ?
