$C H_{3} - H | C | H - D + C l_{2} \to C H_{3} C D | H - C l + H C l] 93 % + C H_{3} - C H_{2} - C l + D C l] 7 %$
What is the relative abstraction of H and D?

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Solution

i. A number of H atoms abstracted by D = 5 and the percentage of the compound formed = 93%.
Therefore, reactivity factor for replacement of H atoms = $93 / 5$ = $18.6$
ii. Number of D abstracted by Cl = 1 and percentage of compound formed = 7%.
Therefore, reactivity factor for D replacement = $7 / 1$ = 7
iii. Relative reactivity of H : D abstration = $18.6 / 7$ = $2.7 : 1$ , i.e.,
each H is abstrated 2.7 times faster than D.

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CH3−H|C|H−D+Cl2→CH3CD|H−Cl+HCl]93%+CH3−CH2−Cl+DCl]7%What is the relative abstraction of H and D?

$C H_{3} - H | C | H - D + C l_{2} \to C H_{3} C D | H - C l + H C l] 93 % + C H_{3} - C H_{2} - C l + D C l] 7 %$
What is the relative abstraction of H and D?