Explain when compound a is treated with pbr3 in pyridine

Reference no: EM13702927

Problem- Compound A, (C6H14O) has two asymmetric carbons, both with the S configuration. When compound A is treated with PBr3 in pyridine, compound B is formed which has two asymmetric carbons, one of which is R while the other is S. When compound B is treated with sodium hydroxide, compound C is formed. When B is treated with potassium tert-butoxide (KOC(CH3)3), compound D is formed. Compounds C and D are isomers with the formula C6H12. Compound C has no chiral center, while compound D contains one chiral center and is optically active. Catalytic hydrogenation (H2/Pt) of either C or D gives the same product, compound E. Determine the structures of compounds A-E, and show stereochemistry where relevant.

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Reference no: EM13702927

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