1.

Decreasing order of reactivity in Williamson's ether synthesis of the following is: I. Me_(3)C CH_(2)Br II. CH_(3)CH_(2)CH_(2)Br III. CH_(3)=CHCH_(2)Cl IV. CH_(3)CH_(2)CH_(2)Cl

Answer»

IIIgtIIgtIVgtI
IgtIIgtIVgtIII
IIgtIIIgtIVgtI
IgtIIIgtIIgtIV

Solution :`C-Br` BOND is weaker than C-Cl bond, THEREFORE, ALKYL bromide (II) reacts faster than alkyl chlorides (III) and (IV). Since `CH_(2)=CH-` is electron withdrawing while `CH_(3)CH_(2)-` is electron donating, therefore, `CH_(2)` has more +ve charge

on III than on IV, in other words, nucleophilic attack occurs faster on III than on IV. further, since Williamson's synthesis occurs by `S_(N)2` mechanism, therefore, due to steric hindrance, neopentyl bromide (I) is the least REACTIVE. thus, the decreasing order of reactivity is: IIgtIIIgtIVgtI.


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