See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the reaction type: The reagent is sodium ethanethiolate (Na+ -SCH2CH3), a strong nucleophile and weak base. The substrate is a secondary alkyl bromide (1-bromo-1-methylcyclopentane... actually looking at the structure, the carbon bearing Br also bears H, CH3, and is part of the cyclopentane ring, making it a secondary carbon). The conditions (strong nucleophile, no bulky base, secondary substrate) favor SN2 reaction. Step 2 - SN2 stereochemical outcome: SN2 proceeds with inversion of configuration (Walden inversion) at the electrophilic carbon. Step 3 - Analyze the starting material stereochemistry: In the starting material, at C1 of cyclopentane: CH3 is on a plain (up) bond, H is on a dashed bond (going back, behind the plane), and Br is on a bold wedge (coming forward, in front of the plane). Step 4 - Apply inversion: The nucleophile (-SCH2CH3) attacks from the back side (opposite to Br). After inversion, the groups that were 'back' come 'forward' and vice versa. So the SCH2CH3 replaces Br but from the opposite face. If Br was on a bold wedge (front), SCH2CH3 will end up on a dashed bond (back), and H which was on a dashed bond (back) will now be on a bold wedge (front). Step 5 - Identify the product structure: After inversion, C1 has: CH3 (up, plain bond, unchanged), SCH2CH3 (dashed, going back), H (bold wedge, coming forward). This matches option (c), which shows CH3 up (plain), SCH2CH3 on dashed bond (back), H on bold wedge (front). Step 6 - Why other options fail: - (a) shows retention of configuration (SCH2CH3 on bold wedge like Br was), which would be SN1 or neighboring group participation, not SN2. - (b) and (d) show an additional methyl group on the adjacent ring carbon that is not present in the starting material, making them incorrect structures. - (d) also shows the same incorrect additional methyl and the connectivity is the same as (c) but with extra methyl. Therefore, the correct answer is C.