Haloalkanes and HaloareneshardMCQ SINGLE

See imageHaloalkanes and Haloarenes Chemistry Question

Question

See image

Chemistry diagram for: See image
Answer: B

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is 2,3-butanediol derivative. Looking at the Fischer projection-style drawing, we have a four-carbon chain: CH3 at top, then C2 bearing H (left) and Br (right), then C3 bearing H (left) and OH (right), then CH3 at bottom. This is (2R,3R)- or (2S,3S)-2-bromo-3-butanol (erythro or threo configuration). From the drawing with both H atoms on the same side (left) and Br and OH on opposite sides, this represents the threo (anti) diastereomer of 2-bromo-3-butanol. Step 2 - Reaction with HBr to give X: The OH group reacts with HBr. The OH is protonated and leaves as water via an SN2 mechanism. In the presence of HBr, the OH at C3 is converted to Br via SN2, which proceeds with inversion at C3. Starting with the threo (anti) isomer of 2-bromo-3-butanol, SN2 at C3 gives inversion, producing the meso-2,3-dibromobutane (since the two stereocenters become opposite in configuration, giving the meso compound). Alternatively, if the starting material is erythro, SN2 gives the (2R,3R) or (2S,3S) dibromobutane. Step 3 - Reaction of X with NaI in acetone: NaI in acetone is a classic condition for E2 elimination (actually NaI is a source of iodide as nucleophile/base for SN2, but with dihalides, elimination via E2 can occur). More precisely, NaI in acetone promotes E2 elimination. For E2 elimination from a vicinal dibromide, the two bromine atoms must be anti-periplanar. Step 4 - Determining stereochemistry of alkene product: If X is meso-2,3-dibromobutane, the anti-periplanar arrangement for E2 elimination requires the two Br atoms to be anti to each other. In meso-2,3-dibromobutane, when the two Br atoms are anti, the two CH3 groups are also anti to each other. Anti elimination from this conformation gives trans-2-butene as the major product. Step 5 - Why other options fail: cis-2-butene would result from syn elimination or from the (±) dibromobutane isomer. 1-butene and isobutene are not reasonable products of elimination from a 2,3-dibromobutane. Therefore, the major product Y is trans-2-butene. Therefore, the correct answer is B.

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