When chloroform is hydrolysed with KOH,the fi | vedclass

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

Question

(B) The hydrolysis of chloroform $(CHCl_3)$ with aqueous $KOH$ proceeds as follows:$1$. $CHCl_3 + 3KOH \rightarrow CH(OH)_3 + 3KCl$$2$. The intermedia

Vedclass · Accepted Answer

$HCOOK$

Vedclass · Answer

(B) The hydrolysis of chloroform $(CHCl_3)$ with aqueous $KOH$ proceeds as follows:$1$. $CHCl_3 + 3KOH \rightarrow CH(OH)_3 + 3KCl$$2$. The intermediate product,methanetriol $(CH(OH)_3)$,is unstable and loses a water molecule to form formic acid $(HCOOH)$: $CH(OH)_3 \rightarrow HCOOH + H_2O$$3$. Since the medium is basic ($KOH$ is present),the formic acid reacts with $KOH$ to form potassium formate $(HCOOK)$: $HCOOH + KOH \rightarrow HCOOK + H_2O$Thus,the final product is potassium formate $(HCOOK)$.

When chloroform is hydrolysed with $KOH$,the final product is:

Similar Questions

What is the decreasing order of acidic strength for the following acids?
$[I]$ Methanoic acid
$[II]$ Ethanoic acid
$[III]$ Propanoic acid
$[IV]$ Butanoic acid

Which of the following acids has the smallest dissociation constant?

What is the correct order of acidity of compounds $(I)$,$(II)$,and $(III)$?
$(I)$ $4$-Nitrobenzoic acid
$(II)$ $4$-Methoxybenzoic acid
$(III)$ Benzoic acid

The ease of hydrolysis in the compounds $CH_3COCl$ $(I)$,$(CH_3CO)_2O$ $(II)$,$CH_3COOC_2H_5$ $(III)$ and $CH_3CONH_2$ $(IV)$ is of the order:

What sized ring is formed when two carboxylic acids form a dimeric structure through intermolecular hydrogen bonding? (Consider the hydrogen bond as a bond in the ring structure.)

Vedclass Test Series

Exam Paper Generator

Online Exam Module