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-$I$ (Reaction)	Column-$II$ (Product and Reaction Name)
$A$. Phenol $+$ dilute $HNO_3$	$i$. Mononitrophenol,Nitration
$B$. Phenol $+$ Bromine water	$ii$. $2,4,6$-Tribromophenol,Bromination
$C$. Phenol $+$ $Na_2Cr_2O_7/H_2SO_4$	$iii$. Benzoquinone,Oxidation
$D$. Sodium phenoxide $+$ $CO_2$ followed by $H^+$	$iv$. Salicylic acid,Kolbe reaction

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

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