Product (D) in the above reaction is: | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) $1$. The starting material is $2-(hydroxymethyl)cyclohexanol$. Treatment with $PCC$ (Pyridinium chlorochromate) selectively oxidizes the primary a

Vedclass · Accepted Answer

$2-$($1$-hydroxy$-1-$methylcyclohexyl)methanol

Vedclass · Answer

(A) $1$. The starting material is $2-(hydroxymethyl)cyclohexanol$. Treatment with $PCC$ (Pyridinium chlorochromate) selectively oxidizes the primary alcohol to an aldehyde,forming $2-formylcyclohexanone$ $(A)$. $2$. Reaction with ethylene glycol in the presence of $H^+$ forms a cyclic acetal at the aldehyde group,protecting it to give $2-(1,3-dioxolan-2-yl)cyclohexanone$ $(B)$. $3$. Addition of $MeMgBr$ followed by $H_3O^+$ attacks the ketone carbonyl to form a tertiary alcohol,yielding $2-(1,3-dioxolan-2-yl)-1-methylcyclohexanol$ $(C)$. $4$. Finally,the acetal is hydrolyzed (implied in the workup or subsequent step) and the resulting aldehyde is reduced by $NaBH_4$ to a primary alcohol,resulting in $2-(1-hydroxy-1-methylcyclohexyl)methanol$ $(D)$.

Product $(D)$ in the above reaction is:

Similar Questions

$HOOC-COOH + HO^{18}CH_2-CH_2^{18}OH \xrightarrow[H^+]{\Delta} (A)$; Product $(A)$ is:

$p-$Cresol reacts with chloroform in an alkaline medium to give compound $A$,which reacts with hydrogen cyanide to form compound $Y$. Hydrolysis of $Y$ gives a chiral carboxylic acid. Identify the structure of the carboxylic acid.

The reaction of elemental sulphur with a Grignard reagent followed by acidification leads to the formation of:

$1^o, 2^o, 3^o$ alcohols are distinguished by

Dichlorocarbene is formed by chloroform and alcoholic $KOH$. Which reaction involves this carbene?

Vedclass Test Series

Exam Paper Generator

Online Exam Module