Question

A partially racemised (+)-2-bromo-octane (2${}^o$ RX) on reaction with aq. NaOH in acetone gives an alcohol with 80% inversion and 20% racemisation. What is the rate expression of the reaction?

• A. Rate $=K_1\left[2^{o}RX\right]\left[O{H}^{-}\right]$
• B. Rate $=K_2\left[2^{o}RX\right]$
• C. Rate $=K_1\left[2^{o}RX\right]\left[O{H}^{-}\right]+K_2\left[2^{o}RX\right]$
• D. Rate $=K_1\left[O{H}^{-}\right]$
• E. Rate $=K_2[2^{o}RX{]}^2$

Answer 1

The correct option is C Rate $=K_1\left[2^{o}RX\right]\left[O{H}^{-}\right]+K_2\left[2^{o}RX\right]$

It is the case of simultaneous reaction where reactant (+)-2-bromo-octane $\left(2^{o}RX\right)$ is undergoing both $S{N}^1$ and $S{N}^2$ simultaneously .
The first term $K_1$[$2^{o}RX$][$\stackrel{⊝}{O}H]$ is the rate equation of $S{N}^2$ reaction which depends on both reactant and substrate and the second term $K_2$[$2^{o}RX$] is the rate equation of $S{N}^1$ reaction which depends only on substrate.