| (I)\[M{{e}_{3}}CC{{H}_{2}}Br\] |
| (II)\[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br\] |
| (III)\[C{{H}_{2}}=CHC{{H}_{2}}Cl\] |
| (IV)\[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Cl\] |
A) (I) > (II) > (IV) > (III)
B) (III) > (II) > (IV) > (I)
C) (I) > (III) > (II) > (IV)
D) (II) > (III) > (IV) > (I)
Correct Answer: D
Solution :
The reactivity of alkyl halides is in the order, \[C{{H}_{3}}>{{1}^{o}}>{{2}^{o}}>{{3}^{o}}\]as tendency of alkyl halides to undergo elimination is\[{{3}^{o}}>{{2}^{o}}>{{1}^{o}}.\] C-Br bond length is longer than C-Cl bond. So, C-Br bond is easier to break than C-Cl bond. \[C{{H}_{2}}=CH\overset{\oplus }{\mathop{C}}\,{{H}_{2}}\] Carbocation is resonance stabilized. i.e., \[[C{{H}_{2}}=CH-\overset{\oplus }{\mathop{C}}\,{{H}_{2}}\leftrightarrow \overset{\oplus }{\mathop{C}}\,{{H}_{2}}-CH=C{{H}_{2}}]\] So, it reacts faster than n-propyl choride.
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