question_answer

Which of the following examples have \[{{K}_{H}}/{{K}_{D}}\](ratio of rate constant) is greater than I?

(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. Since in reaction (I) and \[(C-H)\] bond is broken and in (II) the \[(C-D)\] bond is broken, the \[(C-H)\] bond is broken more easily than the \[(C-D)\] bond. So the rate of (I) is faster than that of (II). Hence, \[{{K}_{H}}/{{K}_{D}}>1\]. So,   \[{{K}_{H}}/{{K}_{D}}>1\] In both the reactions neither the \[(C-H)\] nor the \[(C-D)\] bond break and rate of both the reaction is: Rate \[=K\,[C{{l}^{\oplus }}]\]  [Benzene]; so \[{{K}_{H}}/{{K}_{D}}=1.\] IV. In the nitration to intermediate no \[(C-H)\] bond is broken in R.D.S. \[\therefore \,\,\,\,{{K}_{H}}/{{K}_{D}}=1\]