When secondary alkyl halide undergoes SN reaction, it follows SN1 (first-order) and SN2 (second-orde

Question

When secondary alkyl halide undergoes SN reaction, it follows SN1 (first-order) and SN2 (second-order) pathway simultaneously. The dominant pathway depends on large number of factors. An alkyl halide undergoes both pathway with rate constants 3 × 10−4 s−1 and 5 × 10−4 M−1 s−1, respectively. Which of the following is/are correct information for the reaction?

Accepted Answer

For SN1 path: r1 = (3 × 10–4 s–1) [RX]For SN2 path: r2 = (5 × 10–4 M–1 s–1) [RX] (a) = 0.1 M, then r1 > r2(b) = 1.0 M, then r1 < r2(c) = 0.6 M, then r1 = r2(d) = 0.4 M, then $\frac{r_{1}}{r_{2}} = \frac{2}{3}$∴ Percentage product by SN1 = $\frac{2}{2 + 3} 	imes 100\% = 40\%$

Question

Options

Solution

More Chemical Kinetics and Nuclear Chemistry Questions