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For the following reactions: where k_{s} and k_{e} are respectively, the rate constants for substitution and elimination, and \mu = \frac{k_{s}}{k_{e}}, the correct option is ___________.
Option: 1 \mu _{A}>\mu _{B}and k_{e}(A)>k_{e}(B)
Option: 2 \mu_{B}>\mu _{A}and k_{e}(A)>k_{e}(B)
Option: 3 \mu_{A}>\mu_{B}and k_{e}(B)>k_{e}(A)
Option: 4 \mu_{B}>\mu_{A}and k_{e}(B)>k_{e}(A)
 

Answers (1)

best_answer

 

This is williamson synthesis.

 

Substitution is better for prrimary Z means A and Elimination tertiary or secondary Z means B.

\mathrm{k_e(B)>k_e(A)} will come.

Now, \mathrm{\mu = \frac{k_s}{k_e}}

A favours substitution and B favour elimination.
So, for A,  ks  will be greater  than ke.

then  \mathrm{\mu_A > \mu_B} will right.

Therefore, Option(3) is correct.

Posted by

Ritika Jonwal

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