The reaction of with H₂/Ni forms: A)2-methylheptane R)-2-methyl-5-heptanol S)-6-methyl-3-heptanol R)- and ( S)-6-methyl-3-heptanol E)Achiral 6,6-dimethyl-3-hexanol

Question

Quizplus · Accepted Answer

The reaction of 11eaa4bc_3d9b_f84c_9180_6b1265d84974_TB5902_00 with H₂/Ni results in a reduction reaction, with the Ni acting as a catalyst. The product formed will have the same number of carbons as the starting material, but with a different functional group. In this case, the starting material is a ketone with a chiral center. The reduction reaction will result in an alcohol, and the stereochemistry at the chiral center will be determined by the nature of the reducing agent. H₂/Ni is a mild reducing agent that will reduce the ketone selectively at the carbonyl group. The reduction will result in the formation of both the (R)- and (S)-enantiomers of the alcohol. The stereochemistry at the new chiral center will be opposite to that of the ketone, since the reducing agent adds a hydride ion to the carbonyl group from the opposite side of the molecule. Therefore, the product of the reaction will be a mixture of the (R)- and (S)-enantiomers of 6-methyl-3-heptanol, which corresponds to choice D.

The Reaction of with H2/Ni Forms

The Reaction of with H₂/Ni Forms