2-Chloro pentane on reaction with NaOH in 60% ethanol and 40% water at 250C undergoes nucleophilic substitution reaction. The rate law for the reaction is r=4.5×10−2[2−chloro pentane][OH−] +5×10−3[2−chloro pentane]What will be the percentage of front side attack of OH− if concentration of OH− is 1M ?(consider ideal SN1 and SN2 reactions)

% Of front side attack % of SN12% of SN1=5×10−34.5×10−2+5×10−3×100 % of SN1=10% Of front side attack = 5

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2-Chloro pentane on reaction with $N a O H$ in 60% ethanol and 40% water at $25^{0} C$ undergoes nucleophilic substitution reaction. The rate law for the reaction is $r = 4.5 \times 10^{- 2} [2 - chloro pentane] [O H^{-}] + 5 \times 10^{- 3} [2 - chloro pentane]$
What will be the percentage of front side attack of $O H^{-}$ if concentration of $O H^{-}$ is $1 M$ ?
(consider ideal $S N^{1}$ and $S N^{2}$ reactions)

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answer is 5.

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$%$ Of front side attack $\frac{% o f S N^{1}}{2}$

$% o f S N^{1} = \frac{5 \times 10^{- 3}}{4.5 \times 10^{- 2} + 5 \times 10^{- 3}} \times 100 % o f S N^{1} = 10$

$%$ Of front side attack = 5

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