Which of the following would produce enantiom | vedclass

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(A) The reaction of a carbonyl compound $R-CO-R'$ with methyl magnesium iodide $(CH_3MgI)$ followed by hydrolysis yields a tertiary or secondary alcoh

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Benzaldehyde

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(A) The reaction of a carbonyl compound $R-CO-R'$ with methyl magnesium iodide $(CH_3MgI)$ followed by hydrolysis yields a tertiary or secondary alcohol with a chiral center if the four groups attached to the central carbon are different.For the product to be chiral (enantiomeric), the central carbon must be bonded to four distinct groups: $R$, $R'$, $CH_3$, and $H$ (or $OH$ group).$1$. Benzaldehyde $(C_6H_5CHO)$: Reacts to form $1$-phenylethanol, which has a chiral center $(C_6H_5, CH_3, H, OH)$. Thus, it produces enantiomers.$2$. Propiophenone $(C_6H_5COCH_2CH_3)$: Reacts to form $2$-phenylbutan-$2$-ol. Here, the groups are $C_6H_5, CH_3, CH_2CH_3, OH$. All four are different, so it forms enantiomers.$3$. Acetone $(CH_3COCH_3)$: Forms $2$-methylpropan-$2$-ol, which is achiral.$4$. Acetaldehyde $(CH_3CHO)$: Forms propan-$2$-ol, which is achiral.Given the options, Benzaldehyde is the standard example for producing a chiral secondary alcohol.

Which of the following would produce enantiomeric products when reacted with methyl magnesium iodide?

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