Question

Decreasing order of reactivity in Williamson’s ether synthesis of the following.
I. $Me_{3}CCH_{2}Br$
II. $CH_{3}CH_{2}CH_{2}Br$
III. $CH_{2}=CHCH_{2}Cl$
IV. $CH_{3}CH_{2}CH_{2}CH_{2}Cl$

• A. I > II > IV > III
• B. III > II > IV > I
• C. I >III > II > IV
• D. II > III > IV > I

Answer 1

Answer: (D)

The reactivity of alkyl halides is in the order, $CH_{3}>1^{o}>2^{o}>3^{o}$ as tendency of alkyl halides to undergo elimination is $3^{o}>2^{o}>1^{o}.$
C - Br bond length is longer than C - Cl bond. So, C - Br bond is easier to break than C - Cl bond.
$\text{CH}_{2} = \text{CH} \overset{\oplus}{\text{C}} \text{H}_{2}$ carbocation is resonance stabilized
i.e., $\left[\right.CH_{2}=CH-\overset{\oplus}{C}H_{2}\leftrightarrow \overset{\oplus}{C}H_{2}-CH=CH_{2}\left]\right.$
So, it reacts faster than n- butyl chloride