Alcohols Phenols And Ethers

Q: Decreasing order of reactivity in Williamson synthesis of the following :I.\$Me_3CCH_2Br\$ II.\$CH_3CH_2CH_2Br\$III. \$CH_2=CHCH_2Cl\$ IV. \$CH_3CH_2CH_2Cl\$

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 therefore, \$CH_2\$ has more positive charge on III than on IV. In other words, nucleophilic attack occurs faster on III than on IV. Further, since Williamson synthesis occurs by \$S_N \: ^2\$ mechanism, therefore, due to steric hindrance alkyl bromide (I) is the least reactive. Thus, the decreasing order of reactivity is \$II > III > IV > I\$. III. \$CH_2=CH \leftarrow \overset{\,\,\delta \delta +}{CH_2}-Cl\$IV. \$CH_3-CH_2\rightarrow \overset{ \delta+}{CH_2}-Cl\$

Question

Decreasing order of reactivity in Williamson 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. $C H_{3} C H_{2} C H_{2} C l$

Alcohols Phenols And Ethers

Chemistry 2932 Views

Decreasing order of reactivity in Williamson synthesis of the following : I.Me3​CCH2​Br II.CH3​CH2​CH2​Br III. CH2​=CHCH2​Cl IV. CH3​CH2​CH2​Cl

Chemistry
2932 Views

Decreasing order of reactivity in Williamson 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. $C H_{3} C H_{2} C H_{2} C l$