A) Toluene
B) Benzene
C) n-Pentane
D) 2, 2-dimethylpropane
Correct Answer: A
Solution :
Toluene More stable the free radical formed upon homolytic fusion of C - H bond, lesser is the bond dissociation energy. Since stability of the radicals formed from toluene, benzene, n-pehtane, 2, 2-dimethylpropane follows the order: \[{{\operatorname{C}}_{6}}{{H}_{5}} \overset{\bullet }{\mathop{C}}\,{{H}_{2}}> {{\left( C{{H}_{3}} \right)}_{3}} \overset{\bullet }{\mathop{C}}\,> C{{H}_{3}} \overset{\bullet }{\mathop{C}}\,HC{{H}_{2}}C{{H}_{2}}\] \[{{\operatorname{CH}}_{3}} > {{\overset{\bullet }{\mathop{C}}\,}_{6}}{{H}_{5}}\] Therefore C-H bond of toluene has lowest bond dissociation energy.
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