When CHCl_3 is boiled with NaOH,it gives | vedclass

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(D) When $CHCl_3$ (chloroform) is boiled with aqueous $NaOH$,it undergoes nucleophilic substitution to form an intermediate,$CH(OH)_3$ (methanetriol o

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Sodium formate

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(D) When $CHCl_3$ (chloroform) is boiled with aqueous $NaOH$,it undergoes nucleophilic substitution to form an intermediate,$CH(OH)_3$ (methanetriol or trihydroxy methane).This intermediate is highly unstable due to the presence of three hydroxyl groups on a single carbon atom.It immediately loses a water molecule $(H_2O)$ to form formic acid $(HCOOH)$.Since the medium is basic $(NaOH)$,the formic acid reacts with $NaOH$ to form sodium formate $(HCOONa)$ and water.The overall reaction is: $CHCl_3 + 4NaOH \rightarrow HCOONa + 3NaCl + 2H_2O$.

When $CHCl_3$ is boiled with $NaOH$,it gives

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