Question

For the reaction $R-X+O{H}^{-}\to ROH+X^{-}$ The Rate is given by
$Rate=5.0\times {10}^{-5}[R-X]\left[O{H}^{-}\right]+0.20\times {10}^{-5}[R-X].$ What percentage of R - X reaction is by $S_{N}2$ mechanism when $\left[O{H}^{-}\right]=1.0\times {10}^{-2}M?$

• A. 96.1 %
• B. 3.9 %
• C. 80 %
• D. 20 %

Answer 1

The correct option is D 20 %

$Rate\left(S_{N}2\right)=5.0\times {10}^{-5}\times {10}^{-2}[R-X]=5.0\times {10}^{-7}[R-X]$
$Rate\left(S_{N}1\right)=0.20\times {10}^{-5}[R-X]$
%$of{S}_{N}2=\frac{5\times {10}^{-7}[R-X]\times 100}{5\times {10}^{-7}[R-X]+0.20\times {10}^{-5}[R-X]}=20$