See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: The SN2 reaction proceeds via a bimolecular mechanism where the nucleophile attacks the electrophilic carbon from the back side, forming a transition state in which the nucleophile, the carbon, and the leaving group are all partially bonded simultaneously. Step 1 - Steric effects in SN2: The rate of an SN2 reaction is highly sensitive to steric hindrance at the carbon bearing the leaving group. As the degree of substitution increases (methyl < primary < secondary < tertiary), the transition state becomes more sterically crowded, raising the activation energy and dramatically reducing the reaction rate. Step 2 - Expected reactivity order: CH3-Br (methyl, no steric hindrance, fastest) > CH3CH2-Br (primary, slight hindrance) > CH3CH(CH3)Br (secondary, significant hindrance) > (CH3)3CBr (tertiary, severe hindrance, essentially does not undergo SN2). This gives a monotonically decreasing rate as steric bulk increases. Step 3 - Matching to the graphs: Option (a) shows a monotonically decreasing curve of log(rate) as the substrate goes from methyl bromide through primary, secondary, to tertiary bromide. This perfectly matches the known SN2 reactivity order. Step 4 - Why other options fail: - Option (b) shows a constant rate, which is incorrect because steric effects strongly differentiate the substrates. - Option (c) shows an increasing rate with increasing substitution, which is the opposite of SN2 behavior (this might loosely resemble SN1 carbocation stability trends, but even that is not an exponential increase in this way). - Option (d) shows an oscillating pattern with no chemical basis in substitution nucleophilic reactions. Therefore, the correct answer is A.