An optically active compound A (C 9 H 11 Br) reacts with sodium ethoxide in ethanol to
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An optically active compound A (C9H11Br) reacts with sodium ethoxide in ethanol to give an optically inactive hydrocarbon B (NMR spectrum in Fig. P16.58). Compound B undergoes hydrogenation over a Pd/C catalyst at room temperature to give a compound C, which has the formula C9H12. Give the structures of A, B, and C.
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The formula of compound B would be useful in solving this problem Consider the integral in the NMR s...View the full answer
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