For the conversion of cyclohexanone to 1-hydr | vedclass

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Question

(A) The conversion proceeds as follows:$1$. Cyclohexanone reacts with $CH_3MgBr$ followed by $H^+$ to form $1$-methylcyclohexanol (Reagent $1$).$2$. $

Vedclass · Accepted Answer

$1 	o 2 	o 3 	o 4$

Vedclass · Answer

(A) The conversion proceeds as follows:$1$. Cyclohexanone reacts with $CH_3MgBr$ followed by $H^+$ to form $1$-methylcyclohexanol (Reagent $1$).$2$. $1$-methylcyclohexanol undergoes acid-catalyzed dehydration with $H^+ / \Delta$ to form $1$-methylcyclohexene (Reagent $2$).$3$. $1$-methylcyclohexene reacts with cold dilute $KMnO_4$ (Baeyer's reagent) to undergo syn-dihydroxylation,forming $1$-methylcyclohexane-$1,2$-diol (Reagent $3$).$4$. Finally,$CrO_3$ selectively oxidizes the secondary alcohol group to a ketone,yielding $1$-hydroxy-$1$-methylcyclohexan-$2$-one (Reagent $4$).Thus,the correct order is $1 	o 2 	o 3 	o 4$.

For the conversion of cyclohexanone to $1-$hydroxy$-1-$methylcyclohexan$-2-$one,the correct order of reagents used is:
$(1)$ $CH_3MgBr / H^+$
$(2)$ $H^+ / \Delta$
$(3)$ $KMnO_4$ (cold dil.)
$(4)$ $CrO_3$

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For the conversion of cyclohexanone to $1-$hydroxy$-1-$methylcyclohexan$-2-$one,the correct order of reagents used is:$(1)$ $CH_3MgBr / H^+$$(2)$ $H^+ / \Delta$$(3)$ $KMnO_4$ (cold dil.)$(4)$ $CrO_3$