Show how the following alcohols can be prepared by the reaction of a suitable Grignard reagent with methanal:
$(i)$ $C_6H_5CH_2OH$
$(ii)$ $CH_3CH_2CH(CH_3)CH_2OH$

(N/A) Grignard reagents $(RMgX)$ react with methanal $(HCHO)$ to form primary alcohols with one additional carbon atom.
$(i)$ To prepare $C_6H_5CH_2OH$,the Grignard reagent required is phenylmagnesium bromide $(C_6H_5MgBr)$.
$C_6H_5MgBr + HCHO$ $\rightarrow C_6H_5CH_2OMgBr$ $\xrightarrow{H_3O^+} C_6H_5CH_2OH + Mg(OH)Br$
$(ii)$ To prepare $CH_3CH_2CH(CH_3)CH_2OH$,the Grignard reagent required is sec-butylmagnesium bromide $(CH_3CH_2CH(CH_3)MgBr)$.
$CH_3CH_2CH(CH_3)MgBr + HCHO$ $\rightarrow CH_3CH_2CH(CH_3)CH_2OMgBr$ $\xrightarrow{H_3O^+} CH_3CH_2CH(CH_3)CH_2OH + Mg(OH)Br$