The reaction sequence is:2-text{hydroxycycloh | vedclass

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

Question

(B) The starting material is $2-(1-	ext{hydroxyethyl})	ext{cyclohexanone}$.$1$. Treatment with $I_2/NaOH$ (iodoform test) on the methyl ketone group

Vedclass · Accepted Answer

$2-$oxocyclohexanecarboxylic acid

Vedclass · Answer

(B) The starting material is $2-(1-	ext{hydroxyethyl})	ext{cyclohexanone}$.$1$. Treatment with $I_2/NaOH$ (iodoform test) on the methyl ketone group $(CH_3-CO-)$ converts it into a carboxylate group $(-COO^-)$ and produces $CHI_3$.$2$. Heating $(\Delta)$ and subsequent acidification $(H^ )$ leads to the formation of the corresponding carboxylic acid,$2-	ext{hydroxycyclohexanecarboxylic acid}$.$3$. Further heating $(\Delta)$ causes dehydration of the $\beta-	ext{hydroxy}$ acid,but in this specific structure,the product is $2-	ext{oxocyclohexanecarboxylic acid}$ or its derivative depending on the specific conditions. Given the options,the product $A$ is $2-	ext{oxocyclohexanecarboxylic acid}$.

The reaction sequence is:
$2-\text{hydroxycyclohexylacetone}$ $\xrightarrow{I_2/NaOH}$ $\xrightarrow{\Delta}$ $\xrightarrow{H^+}$ $\xrightarrow{\Delta} A$
$A$ is:

Similar Questions

The increasing order of the acidity of the following carboxylic acids is:

Which of the following compounds on hydrolysis gives reactant $(A)$?

If $(\pm)$ $2$-methylbutanoic acid is esterified by reaction with $(\pm)$ $2$-butanol,how many optically active compounds will be present in the final equilibrium reaction mixture?

Ethanoic acid $+$ $3$-methylbutan-$1$-ol $\underset{\text{traces } H_2SO_4}{\longleftrightarrow} (A)$; Compound $(A)$ is

Identify the wrong statement from the following.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The reaction sequence is:$2-\text{hydroxycyclohexylacetone}$ $\xrightarrow{I_2/NaOH}$ $\xrightarrow{\Delta}$ $\xrightarrow{H^+}$ $\xrightarrow{\Delta} A$$A$ is: