Which of the following reactions produce n-bu | vedclass

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(D) Let us analyze each reaction:$I$. $CH_3CH_2CH=CH_2 \xrightarrow[(ii) NaOH/H_2O_2]{(i) B_2H_6} CH_3CH_2CH_2CH_2OH$ ($n$-butanol). This is hydrobora

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$I, III$ and $IV$

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(D) Let us analyze each reaction:$I$. $CH_3CH_2CH=CH_2 \xrightarrow[(ii) NaOH/H_2O_2]{(i) B_2H_6} CH_3CH_2CH_2CH_2OH$ ($n$-butanol). This is hydroboration-oxidation,which follows anti-Markovnikov addition of water.$II$. $CH_3CH_2CHO \xrightarrow[(ii) H_2O]{(i) CH_3MgI} CH_3CH_2CH(OH)CH_3$ (butan-$2$-ol). This is a Grignard reaction forming a secondary alcohol.$III$. $CH_3CH_2CH_2CN$ $\xrightarrow[(ii) H_3O^+]{(i) SnCl_2, HCl} CH_3CH_2CH_2CHO$ $\xrightarrow{NaBH_4} CH_3CH_2CH_2CH_2OH$ ($n$-butanol). The Stephen reduction converts nitrile to aldehyde,which is then reduced to primary alcohol.$IV$. $CH_3CH_2CH_2CO_2H \xrightarrow{B_2H_6} CH_3CH_2CH_2CH_2OH$ ($n$-butanol). Diborane specifically reduces carboxylic acids to primary alcohols.Thus,reactions $I, III$ and $IV$ produce $n$-butanol.

Column - $(I)$ Product	Column - $(II)$ Preparation Process
$(A)$ Butan$-1-$ol	$(i)$ Hydration of propene in the presence of dilute sulfuric acid
$(B)$ Propan$-2-$ol	$(ii)$ Reaction of propanone with methylmagnesium bromide followed by hydrolysis
$(C)$ $2-$Methylpropan$-2-$ol	$(iii)$ Hydroboration of propene followed by reaction with $(NaOH + H_2O_2)$
$(D)$ Propan$-1-$ol	$(iv)$ Catalytic reduction of butanal

Which of the following reactions produce $n$-butanol?

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