Question

Match the starting materials given in Column $I$ with the products formed by these $\left(ColumnII\right)$ in the reaction with $HI.$

	Column I		Column II
A		(i)
B		(ii)
C		(iii)
D		(iv)
		(v)
		(vi)
		(vii)

Answer 1

The reaction of ether with HI starts with the protonation of the ether molecule.


Attack of the nucleophile on the protonated ether.

Iodide is a good nucleophile. It attacks the least substituted carbon of the oxonium ion formed in step 1 and displaces an alcohol molecule.




When a primary or a secondary alkyl group is present, the lower alkyl group forms the alkyl iodide via $S_{N}2$ pathway.

In the case of a tertiary alkyl group, the tertiary alkyl group forms the alkyl iodide via $S_{N}1$ pathway.

Analysing the parts in coloumn-I

Part$\left(a\right)$


$\left(a\right)\to \left(iv\right)$

Part$\left(b\right)$


$\left(b\right)\to \left(v\right)$

Part$\left(c\right)$


This reaction will proceed via $S_{N}1$ pathway because tertiary carbocation is highly stable, and hence, it can be formed.

$\left(c\right)\to \left(ii\right)$

Part$\left(d\right)$


$\left(d\right)\to \left(i\right)$

	Column I		Column II
A		(iv)
B		(v)
C		(iv)
D		(v)