- ✓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.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

$Zn_{(s)} +Cu^{+2}_{(aq)} \to Zn^{2+}_{(aq)} +Cu_{(s)}$
$[E^o_{Cu^{+2 /Cu}} = 0.34\,V$ , $E^o_{Zn^{2+} /Zn} = -0.76\,V ]$