Decreasing order of reactivity in Williamson’s ether synthesis of the following.
I. $M e_{3} C C H_{2} B r$
II. $C H_{3} C H_{2} C H_{2} B r$
III. $C H_{2} = C H C H_{2} C l$
IV. $C H_{3} C H_{2} C H_{2} C H_{2} C l$

Question

Decreasing order of reactivity in Williamson’s ether synthesis of the following. I. M e 3 C C H 2 B r II. C H 3 C H 2 C H 2 B r III. C H 2 = C H C H 2 C l IV. C H 3 C H 2 C H 2 C H 2 C l

TestHub · Accepted Answer

The reactivity of alkyl halides is in the order, $C H_{3} > 1^{o} > 2^{o} > 3^{o}$ 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.
${CH}_{2} = CH \overset{\oplus}{C} H_{2}$ carbocation is resonance stabilized
i.e., $[C H_{2} = C H - \overset{\oplus}{C} H_{2} \leftrightarrow \overset{\oplus}{C} H_{2} - C H = C H_{2}]$
So, it reacts faster than n- butyl chloride

Chemistry - Halogen Derivative Question with Solution | TestHub

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