Q.Complete the following reaction: RCOOH --(i) LiAlH4

Red phosphorus and HI is a powerful reducing agent that reduces the $-COOH$ group all the way to $-CH_3$, giving an alkane.

While mild reducing agents like $LiAlH_4$ reduce carboxylic acids only as far as the primary alcohol, red phosphorus + HI (a strong reducing system) removes all the oxygen from the acid, converting $-COOH$ completely into $-CH_3$:

$R-COOH \xrightarrow[\Delta]{Red\ P + HI} R-CH_3$ …

LiAlH₄ is a strong reducing agent that reduces the –COOH group of benzoic acid all the way to –CH₂OH, giving benzyl alcohol.

$C_6H_5COOH \xrightarrow[\text{(ii) H}_3O^+]{\text{(i) LiAlH}_4} C_6H_5CH_2OH$

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LiAlH4 is a strong reducing agent that reduces –COOH completely to –CH2OH, a primary alcohol.

Lithium aluminium hydride (LiAlH4) is a powerful hydride-transfer reducing agent. It reduces carboxylic acids in two steps: it first reduces –COOH to the aldehyde oxidation level and then continues reducing all the way to the primary alcohol, because the aldehyde intermediate is even more reactive toward LiAlH4 than the acid itself.

$RCOOH \xrightarrow{(i)\ LiAlH_4} RCH_2-O^-\ (alkoxide) \xrightarrow{(ii)\ H_2O} RCH_2OH$

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Carboxylic acids are reduced all the way to primary alcohols by the strong reducing agent lithium aluminium hydride (LiAlH₄); the other three reagents either do the reverse (oxidise an alcohol) or cannot reduce a carboxylic acid at all.

$LiAlH_4$ is a powerful hydride-donating reducing agent capable of reducing even the relatively unreactive carboxyl group all the way down to a primary alcohol (it reduces esters and even acids, unlike the milder $NaBH_4$):

$CH_3COOH \xrightarrow{(i)\ LiAlH_4/\text{ether}} CH_3CH_2OH$

The reaction is carried out in anhydrous ether (to avoid LiAlH₄ reacting violently with water), and the aluminium-alkoxide intermediate formed is then hydrolysed with dilute acid in a separate work-up step to liberate the free alcohol.

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