On heating CHCl_3 with aqueous NaOH,the produ | vedclass

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

Question

(B) When chloroform $(CHCl_3)$ is heated with aqueous sodium hydroxide $(NaOH)$,it undergoes hydrolysis to form an unstable intermediate,methanetriol

Vedclass · Accepted Answer

$HCOONa$

Vedclass · Answer

(B) When chloroform $(CHCl_3)$ is heated with aqueous sodium hydroxide $(NaOH)$,it undergoes hydrolysis to form an unstable intermediate,methanetriol $(CH(OH)_3)$.This intermediate loses a water molecule to form formic acid $(HCOOH)$.Since the reaction occurs in the presence of excess $NaOH$,the formic acid is neutralized to form sodium formate $(HCOONa)$.The reaction sequence is:$CHCl_3 + 3NaOH \rightarrow CH(OH)_3 + 3NaCl$$CH(OH)_3 \rightarrow HCOOH + H_2O$$HCOOH + NaOH \rightarrow HCOONa + H_2O$Overall reaction: $CHCl_3 + 4NaOH \rightarrow HCOONa + 3NaCl + 2H_2O$Thus,the final product is $HCOONa$.

On heating $CHCl_3$ with aqueous $NaOH$,the product is:

Similar Questions

The following reaction is described as:
$CH_3(CH_2)_5-CH(CH_3)-Br + OH^- \rightarrow HO-CH(CH_3)(CH_2)_5CH_3 + Br^-$

Which of the compounds will react faster in an $S_N1$ reaction?
$CH_3-CH_2-Cl$ and $C_6H_5-CH_2-Cl$

Which of the following reactions is used for the conversion of alkyl chloride to alkyl iodide?

The major product of the following reaction is:

Alcoholic solution of $KOH$ is used for

Vedclass Test Series

Exam Paper Generator

Online Exam Module