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., \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Br>\]\[C{{H}_{3}}C{{H}_{2}}CHBrC{{H}_{3}}>{{(C{{H}_{3}})}_{3}}CBr.\]
You need to login to perform this action.
You will be redirected in
3 sec