On heating CHCl_3 with aqueous NaOH,the produ | vedclass

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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:

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