(I) \[PHC{{H}_{2}}C{{H}_{2}}Br\] and \[PhC{{D}_{2}}C{{H}_{2}}Br\] towards dehydrohalogenation by strong base. |
(II) Conversion of: ![]() |
(III) Conversion of: ![]() |
(IV) Conversion of benzene to![]() |
A) I, II
B) I, II, III
C) II, III, IV
D) III, IV
Correct Answer: A
Solution :
[a] For \[1{}^\circ \] isotope effect \[{{K}_{H}}/{{K}_{D}}>1\] For \[2{}^\circ \] isotope effect \[{{K}_{H}}/{{K}_{D}}<1\] Since both are \[1{}^\circ RX,\] so dehydrohalogenation reaction proceeds via \[E2\] mechanism and this depends on the breaking of the \[(C-H)\] or \[(C-D)\]bond. It shows \[1{}^\circ \] isotope effect.You need to login to perform this action.
You will be redirected in
3 sec