Question

Match the starting materials given in Column I with the products formed by these (Column II) in the reaction with HI.

	Column I		Column II
(i)	CH3—O—CH3	(a)
(ii)	\[\begin{array}{cc} \ce{CH3}\phantom{..................}\\ \backslash\phantom{.............}\\ \ce{CH-O-CH3}\\ /\phantom{..............}\\ \ce{CH3}\phantom{..................} \end{array}\]	(b)	\[\begin{array}{cc} \ce{CH3}\phantom{....}\\ |\phantom{.......}\\ \ce{CH3-C-I + CH3OH}\\ |\phantom{.......}\\ \ce{CH3}\phantom{....} \end{array}\]
(iii)	\[\begin{array}{cc} \ce{CH3}\phantom{.}\\ |\phantom{....}\\ \ce{H3C-C-O-CH3}\\ |\phantom{....}\\ \ce{CH3}\phantom{..} \end{array}\]	(c)
(iv)		(d)	CH3—OH + CH3—I
		(e)	\[\begin{array}{cc} \ce{CH3}\phantom{.....................}\\ \backslash\phantom{.................}\\ \ce{CH-OH + CH3I}\\ /\phantom{.................}\\ \ce{CH3}\phantom{.....................} \end{array}\]
		(f)	\[\begin{array}{cc} \ce{CH3}\phantom{.....................}\\ \backslash\phantom{.................}\\ \ce{CH-I + CH3OH}\\ /\phantom{.................}\\ \ce{CH3}\phantom{.....................} \end{array}\]
		(g)	\[\begin{array}{cc} \ce{CH3}\phantom{....}\\ |\phantom{.......}\\ \ce{CH3-C-OH + CH3I}\\ |\phantom{.......}\\ \ce{CH3}\phantom{....} \end{array}\]

Answer 1

	Column I		Column II
(i)	CH3—O—CH3	(d)	CH3—OH + CH3—I
(ii)	\[\begin{array}{cc} \ce{CH3}\phantom{..................}\\ \backslash\phantom{.............}\\ \ce{CH-O-CH3}\\ /\phantom{..............}\\ \ce{CH3}\phantom{..................} \end{array}\]	(e)	\[\begin{array}{cc} \ce{CH3}\phantom{.....................}\\ \backslash\phantom{.................}\\ \ce{CH-OH + CH3I}\\ /\phantom{.................}\\ \ce{CH3}\phantom{.....................} \end{array}\]
(iii)	\[\begin{array}{cc} \ce{CH3}\phantom{.}\\ |\phantom{....}\\ \ce{H3C-C-O-CH3}\\ |\phantom{....}\\ \ce{CH3}\phantom{..} \end{array}\]	(b)	\[\begin{array}{cc} \ce{CH3}\phantom{....}\\ |\phantom{.......}\\ \ce{CH3-C-I + CH3OH}\\ |\phantom{.......}\\ \ce{CH3}\phantom{....} \end{array}\]
(iv)		(a)

Explanation:

(i) \[\ce{CH3 - O - CH3}\] is a symmetrical ether so the products are \[\ce{CH3I}\] and \[\ce{CH2OH}\].

(ii) In \[\ce{(CH3)2CH – O – CH3}\] unsymmetrical ether, one alkyl group is primary while another is secondary. So, it follows S_N2 mechanism. 

(iii) In this case, one of the alkyl group is tertiary and the other is primary. It follows S_N1 mechanism and halide ion attacks the tertiary alkyl group and the products are \[\ce{(CH3)3 C-I}\] and \[\ce{CH3OH}\].

(iv) Here, the unsymmetrical ether is alkyl aryl ether. In this ether \[\ce{O - CH3}\] bond is weaker than \[\ce{O - C6H5}\] bond which has partial double bond character due to resonance. So, the halide ion attacks on alkyl group and the products are \[\ce{C6H5 - OH}\] and \[\ce{CH3I}\].