See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the stereoisomers: In 4-t-butylcyclohexyl iodide, the bulky t-butyl group locks the ring in a conformation where t-butyl is equatorial. Structure (A) has the I127 in the equatorial position at C1 (trans to t-butyl if t-butyl is equatorial), and structure (B) has I127 in the axial position at C1. Step 2 - SN2 mechanism requirement: SN2 reactions require back-side attack by the nucleophile (128I-) along the axis opposite to the leaving group (127I-). For a cyclohexane system, back-side attack is most favorable when the leaving group is in the axial position, because the nucleophile can approach axially from the opposite face without steric interference from the equatorial substituents. Step 3 - Reactivity analysis: In structure (B), I127 is axial. The nucleophile 128I- can approach from the axial direction on the opposite face (also axial), which is relatively unhindered. This is the preferred geometry for SN2 on a cyclohexane ring. In structure (A), I127 is equatorial. For SN2 to occur, back-side attack must come from an equatorial direction, but this trajectory is severely hindered by the axial hydrogens on adjacent carbons (1,3-diaxial-type interactions in the transition state). Equatorial leaving groups are much less reactive in SN2. Step 4 - Thermodynamic stability vs. reactivity: Structure (A) with equatorial iodide is the more stable isomer (equatorial substituent preferred). Structure (B) with axial iodide is less stable. However, for SN2 reactivity, it is the axial isomer (B) that reacts faster because the geometry allows unhindered back-side attack. Step 5 - Transition state energy: Because the reaction is an identity exchange (127I- replaced by 128I-, same element), the transition state for both reactions involves identical atoms and symmetrical geometry. The transition states are of the same energy in terms of the nucleophile-carbon-leaving group arrangement. The difference in rate arises purely from the accessibility of the back-side attack trajectory. Step 6 - Why other options fail: (a) Incorrect - greater stability of A does not make it react faster in SN2; in fact it is less reactive. (b) Incorrect - A does not react faster; the product stability argument is irrelevant for an identity SN2 reaction. (c) Incorrect - for A (equatorial I), approach of 128I- is actually hindered, not unhindered, by axial H atoms on neighboring carbons. (d) Correct - B (axial iodide, less stable isomer) reacts faster because the axial leaving group allows unhindered back-side attack by 128I-, and the transition states for both reactions are of the same energy (identity reaction with same nucleophile and leaving group). Therefore, the correct answer is D.