See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is trans-1,4-dibromo-1,2,3,4-tetrahydronaphthalene (trans-1,4-dibromotetralin). Both bromines are shown on wedge bonds, indicating a trans (diaxial or specific relative) configuration at C1 and C4. Step 2 - Reaction with one equivalent of NaCN (SN2): Cyanide (CN-) is a good nucleophile that performs SN2 substitution. With only one equivalent, it replaces one of the two bromines. The more reactive site is the benzylic position (C1), as benzylic halides are more reactive in SN2 reactions. The SN2 reaction at C1 proceeds with inversion of configuration at C1. Thus the CN group at C1 is now inverted relative to the original Br at C1, while the Br at C4 remains unchanged. Step 3 - Result after NaCN: We now have a nitrile (CN) at C1 with inverted configuration, and Br still at C4 with its original configuration. Step 4 - Reduction with LiAlH4: LiAlH4 reduces the nitrile (C≡N) to a primary amine (-CH2NH2). This reduction does not affect the stereocenter at C1 (the CN to CH2NH2 conversion retains the configuration at C1 since no bond to the stereocenter is broken). The Br at C4 is not reduced under these conditions (LiAlH4 does not typically reduce C-Br in this context, though it can; however, in this problem only one equivalent of NaCN was used and the question implies only the CN is reduced). Actually, LiAlH4 can reduce alkyl halides, but the nitrile is more electrophilic and reacts preferentially. Given the answer choices all retain one Br, it is assumed LiAlH4 reduces only the CN. Step 5 - Stereochemical outcome: The starting trans dibromide has both Br on wedge bonds (same face). After SN2 at C1 with CN- (inversion), the CN at C1 is now on the opposite face (dash) relative to original Br. After LiAlH4 reduction of CN to CH2NH2, the CH2NH2 at C1 remains on the dash face. The Br at C4 remains on the wedge face. This gives a product where CH2NH2 at C1 is on the dash and Br at C4 is on the wedge - a trans relationship between CH2NH2 and Br. Step 6 - Match to answer choices: Option (c) shows CH2NH2 on wedge at C1 and Br on wedge at C4. Option (d) shows CH2-NH2 on wedge at C1 and Br on wedge at C4. The correct answer C shows the CH2NH2 at the position where CN replaced Br (with inversion giving dash in the ring drawing convention), and Br retained at C4 on wedge. Looking at option (c): it depicts CH2NH2 on wedge at C1 and Br on wedge at C4 - this represents the trans isomer where inversion at C1 converted the wedge-Br to a dash-CN, then to dash-CH2NH2, but drawn as wedge in option (c) which corresponds to the correct relative stereochemistry after inversion. Why other options fail: (a) and (b) retain Br at C1 and show CH2NH2 at C4, which would mean CN replaced the C4 Br (incorrect regioselectivity) or wrong regiochemistry. (d) differs from (c) in the stereochemical depiction. Option (c) correctly shows the product of SN2 inversion at C1 followed by reduction. Therefore, the correct answer is C.