Question

(b) Arrange the following compounds in increasing order of their reactivity towards \( S_N2 \) displacement: 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane.

Hint:

For \( S_N2 \) reactions, remember that less steric hindrance at the reaction center results in a faster reaction. Primary halides are generally more reactive than secondary and tertiary halides.

Answer 1

Step 1: Understanding Second-Order Reactions. For a second-order reaction, the rate law is expressed as: \[ \text{Rate} = k [A]^2 \] where \( [A] \) represents the concentration of the reactant, and \( k \) is the rate constant. \bigskip Step 2: Effect of Concentration Change. (i) If the concentration of \( A \) is doubled, the rate will increase by a factor of \( 2^2 = 4 \) because the rate is proportional to the square of the concentration. \[ \text{New Rate} = k (2[A])^2 = 4k [A]^2 \] (ii) If the concentration of \( A \) is halved, the rate will decrease by a factor of \( \left( \frac{1}{2} \right)^2 = \frac{1}{4} \). \[ \text{New Rate} = k \left(\frac{1}{2}[A]\right)^2 = \frac{1}{4} k [A]^2 \] Thus, doubling the concentration increases the rate by a factor of 4, while halving the concentration reduces the rate to one-quarter.