Which of the following gives two isomers of 3 | vedclass

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(B) When $4$-methylcyclohexanone is treated with phenyl magnesium bromide $(PhMgBr)$,the nucleophilic $Ph^-$ attacks the carbonyl carbon. Since the mo

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$4-$methylcyclohexanone

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(B) When $4$-methylcyclohexanone is treated with phenyl magnesium bromide $(PhMgBr)$,the nucleophilic $Ph^-$ attacks the carbonyl carbon. Since the molecule has a chiral center at the $C-4$ position (due to the methyl group),the attack of $PhMgBr$ on the planar carbonyl group creates a new stereocenter at the $C-1$ position. This results in the formation of two diastereomers (geometrical isomers) because the $Ph$ group can approach from either the same side or the opposite side of the methyl group at $C-4$. Thus,$4$-methylcyclohexanone gives two isomers of $3^o$ alcohol.

Column-$I$ (Example)	Column-$II$ (Reaction)
$(A)$ $CH_3COCl + H_2 \xrightarrow{Pd-C/BaSO_4} CH_3CHO$	$(1)$ Friedel-Crafts acylation
$(B)$ $C_6H_5CHO \xrightarrow{NaOH} C_6H_5CH_2OH + C_6H_5COONa$	$(2)$ $HVZ$ reaction
$(C)$ $C_6H_6 + CH_3COCl \xrightarrow{AlCl_3} C_6H_5COCH_3$	$(3)$ Aldol condensation
$(D)$ $R-CH_2COOH \xrightarrow{Br_2/Red P} R-CH(Br)COOH$	$(4)$ Cannizzaro's reaction
$(E)$ $CH_3-CN \xrightarrow{SnCl_2/HCl, H_2O/H^+} CH_3-CHO$	$(5)$ Rosenmund's reduction
$(F)$ $2CH_3CHO \xrightarrow{NaOH} CH_3-CH=CH-CHO$	$(6)$ Stephen's reaction

Which of the following gives two isomers of $3^o$ alcohol,when treated with phenyl magnesium bromide?

Similar Questions

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