question_answer

Arrange the following in the decreasing order of reactivity towards \[{{S}_{N}}2\] displacement reaction and give reasons in support of your answer. (a)\[{{C}_{2}}{{H}_{5}}Br,\,\,{{C}_{2}}{{H}_{5}}I,\,\,{{C}_{2}}{{H}_{5}}Cl\] (b) \[{{(C{{H}_{3}})}_{3}}CBr,\]\[C{{H}_{3}}C{{H}_{2}}CHBrC{{H}_{3}},\]\[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Br\]

Answer:

                                 (a) The reactivity increases as the carbon-halogen bond dissociation enthalpy decreases. Since the bond dissociation enthalpy decreases in the order: \[C-Cl>C-Br>C-I,\] therefore, reactivity of the three ethyl halides decreases in the reverse order, i.e., \[{{C}_{2}}{{H}_{5}}I>{{C}_{2}}{{H}_{5}}Br>{{C}_{2}}{{H}_{5}}Cl\] (b) In \[{{S}_{N}}2\] reactions, reactivity decreases as steric hindrance around the carbon atom carrying the halogen increases. Now in \[{{(C{{H}_{3}})}_{3}}CBr,\] there are three bulky \[C{{H}_{3}}\] groups at the carbon carrying Br, in\[C{{H}_{3}}C{{H}_{2}}CHBrC{{H}_{3}},\] there two bulky alkyl groups ; one \[C{{H}_{3}}\] and one \[C{{H}_{3}}C{{H}_{2}}\] at the carbon carrying Br but in \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Br\] there is only one bulky \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}\]  group and two small H atoms. As a result, steric hindrance increases in the order : \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Br<\]\[C{{H}_{3}}C{{H}_{2}}CHBrC{{H}_{3}}<\]\[{{(C{{H}_{3}})}_{3}}\,CBr\]. Therefore, reactivity decreases in the reverse order i.e., \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Br>\]\[C{{H}_{3}}C{{H}_{2}}CHBrC{{H}_{3}}>{{(C{{H}_{3}})}_{3}}CBr.\]