See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: Reaction classification depends on the nature of the attacking reagent and the type of bond-making/breaking process. Step 1 - Identify the reagents and products: C6H13Br (a primary or secondary alkyl halide) reacts with OH- (hydroxide ion) to give C6H13OH (an alcohol) and Br- (bromide ion). Step 2 - Identify the attacking species: OH- is a negatively charged species rich in electrons (lone pairs). Any species that donates electrons to form a new bond is called a nucleophile. Therefore, OH- is a nucleophile. Step 3 - Identify the type of bond change: The Br atom is replaced by the OH group. One group leaves (Br-) and another group takes its place (OH-). This is a substitution reaction, not an addition reaction (no pi bond is broken to accommodate both groups). Step 4 - Combine the two observations: The attacking species is a nucleophile and the reaction type is substitution, so this is a Nucleophilic Substitution reaction (SN1 or SN2 depending on substrate). Step 5 - Eliminate other options: (a) Nucleophilic addition requires a pi bond (e.g., C=O) to be broken; none exists here. (c) Electrophilic addition also requires a pi bond and an electrophilic attacker; neither applies. (d) Electrophilic substitution involves an electrophile attacking (e.g., aromatic rings); OH- is not an electrophile. (e) Free radical substitution involves homolytic cleavage initiated by radicals; OH- reacts via heterolytic (ionic) mechanism. Therefore, the correct answer is B.