See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the substrate: The starting material is a tertiary alkyl bromide (the Br is attached to a bridgehead carbon that bears three ring carbon substituents, making it a tertiary carbon in a rigid tricyclic framework). Step 2 - Evaluate Method 1 (Grignard / CO2 route): Formation of a Grignard reagent (RMgBr) from a tertiary alkyl halide is possible because Grignard formation is not an SN2 process; it proceeds via a radical or oxidative-addition pathway at the magnesium surface. Tertiary alkyl halides can form Grignard reagents. Once RMgBr is formed, reaction with CO2 followed by H3O+ workup gives RCO2H directly. No elimination competes with Grignard formation under these mild conditions (Mg, Et2O, 0 °C or RT). Therefore Method 1 works well for this tertiary substrate. Step 3 - Evaluate Method 2 (SN2 nitrile route): The NaCN step is an SN2 reaction. SN2 reactions require a backside attack on the carbon bearing the leaving group. A tertiary carbon is completely blocked by three substituents (and here the bridgehead geometry of the rigid tricyclic system makes backside approach even more sterically impossible). SN2 at tertiary (especially bridgehead) carbons does not occur. The reaction would fail at the first step (RBr + NaCN → RCN), so Method 2 is not appropriate. Step 4 - Compare with options: Only option (c) states that Method 1 works but Method 2 does not, which matches the analysis above. Why other options fail: - (a) Both work: incorrect because Method 2 fails at the SN2 step on a tertiary bridgehead carbon. - (b) Neither works: incorrect because Method 1 (Grignard) does work. - (d) Method 2 works, Method 1 does not: incorrect; it is the reverse of what is true. Therefore, the correct answer is C.