Question

For reaction $RX + OH^{-} \to ROH + X^{-},$ rate expression is $R = 4.7 \times 10^{-5} \left[RX\right]\left[OH^-\right]+2.4\times10^{-5}\left[RX\right]$ What % of reactant react by $S_N2$ mechanism when $\left[OH^{-}\right]=0.001$ molar?

• A. $1.9$
• B. $66.2$
• C. $95.1$
• D. $16.4$

Answer 1

Answer: (A)

Fraction of reactant decomposing by $S_N2$ mechanism
$=\frac{4.7\times10^{-5}\left[RX\right]\left[OH^{-}\right]}{4.7\times10^{-5}\left[RX\right]\left[OH^{-}\right]+2.4\times10^{-5}\left[RX\right]}$
Put $\left[OH-\right]=0.001 M$