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Decreasing order of reactivity in Willia...

Decreasing order of reactivity in Williamson's ether synthesis of the following .
I. `Me_3"CC"H_2Br`
II. `CH_3CH_2CH_2Br`
III. `CH_2=CHCH_2Cl`
IV. `CH_3CH_2CH_2CH_2Cl`

A

III gt II gt IV gt I

B

I gt II gt IV gt III

C

II gt III gt IV gt I

D

I gt III gt II gt IV

Text Solution

Verified by Experts

The correct Answer is:
A
This reaction proceeds by `S_N2` mechanism. The rate of reaction 1° > 2° > 3° alkyl halide Br is better leaving group than Cl and allyl halide is most reactive.
`CH_2=CH-CH_2-Cl>CH_3 -CH_2-CH_2-Br>CH_3-CH_2-CH_2-Cl>(CH_3)_3C-CH_2-Br`.
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