question_answer

Which of the following reaction shows secondary isotope effect?

A) An alkene prepared by \[{{E}_{2}}\] mechanism               

B) An alkene prepared by \[{{E}_{1}}\] mechanism

C) Dehydrohalogenation of \[PhC{{H}_{2}}C{{H}_{2}}Br\] by strong base. 

D) Dehydrohalogenation of \[PhC{{D}_{2}}C{{H}_{2}}Br\] by strong base.

Correct Answer: B

Solution :

[b] In \[E2,\] the rate of reaction depends on the breaking of the\[(C-H)\] or \[(C-D)\] bond, hence it shows \[1{}^\circ \] isotope effect. (2) In \[E\,1,\]the reaction takes place by the formation of \[3{}^\circ \]\[{{C}^{\oplus }}\]in the R.D.S. \[{{(C{{H}_{3}})}_{3}}C-X\xrightarrow{R.D.S.}{{(C{{H}_{3}})}_{3}}{{C}^{\oplus }}+{{X}^{\Theta }}\] \[(C{{D}_{3}})C-X\xrightarrow{R.D.S.}{{(C{{D}_{3}})}_{3}}{{C}^{\oplus }}+{{X}^{\Theta }}\] Here, \[(C-H)\] or \[(C-D)\] bond is not broken, but the rate of reaction depends on the nature of the surrounding groups at \[(C-X)\]or the rate of reaction is slowed down if D is incorporated; hence it shows \[2{}^\circ \]isotope effect and 4 Since both are \[1{}^\circ \] RX, so dehydrohalogenation reaction proceeds via \[E\,2\]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, \[{{K}_{H}}/{{K}_{D}}>1\].