Question

Decreasing order of reactivity in Williamson synthesis of the following:
$I.{\text{ }}M{e_3}CC{H_2}Br \\\ II.{\text{ }}C{H_3}C{H_2}C{H_2}Br \\\ III.{\text{ }}C{H_2} = CHC{H_2}Cl \\\ IV.{\text{ }}C{H_3}C{H_2}C{H_2}Cl \\\$
A.) $III > II > IV > I$
B.) $I > II > IV > III$
C.) $II > III > IV > I$
D.) $I > III > II > IV$

Answer 1

This question can be solved by the concept that Williamson synthesis is a reaction which occurs by the ${S_N}2$ mechanism. In the ${S_N}2$ mechanism, the reactivity is more for less hindered groups.

Complete step by step answer:
The Williamson synthesis reaction is an ether synthesis organic reaction in which an alkyl halide is heated with sodium alkoxide or potassium alkoxide then the corresponding ether and alkyl halide is produced in the product. By Williamson ether synthesis reaction, we can prepare simple as well as mixed ethers. Here, sodium alkoxide and potassium alkoxide have the chemical formula as $R - ONa$ and $R - OK$ respectively. The Williamson Ether Synthesis reaction usually takes place by ${S_N}2$ mechanism. Here, ${S_N}2$ mechanism works only when the alkyl halide is a primary or secondary group.
As we know that in the ${S_N}2$ reaction, the less hindered group is more reactive. That is when we remove the halide group ($Cl{\text{ or Br}}$) from the alkyl halide then the formed carbocation should be more stable. As we know that $C - Br$ bond is weaker than $C - Cl$ bond therefore in $C - Br$, bromine will remove easily to form carbocation. Therefore, alkyl halide $(II)$ reacts faster than alkyl chloride $(III)$ and $(IV)$. AS the $C{H_2} = CH -$ is an electron withdrawing group therefore the stability of carbocation is more in $(III)$ than in $(IV)$. In the $(I)$, steric hindrance is most therefore, it will be least reactive.
Therefore, the correct order of reactivity of Williamson synthesis is:
$II > III > IV > I$

Hence, C.) is the correct answer.

Note: As we know that in the ${S_N}2$ mechanism the attack of the nucleophile is from the backside which reverses the configuration of the carbon. Hence, remember that in Williamson ether synthesis, the carbon configuration will also be reversed because it follows the ${S_N}2$ mechanism.