Question: When \[2,3\] -dimethyl hexanoic acid is treated with \[SOC{l_2}\] and pyridine, followed by \[LiAl{\...

Question

When $2,3$ -dimethyl hexanoic acid is treated with $SOC{l_2}$ and pyridine, followed by $LiAl{\left[ {OC{{\left( {C{H_3}} \right)}_3}} \right]_3}H$ . What would be the product?

Answer 1

Solution

Carboxylic acids are the chemical compounds that consist of a functional group $- COOH$ . When carboxylic acids were treated in presence of $SOC{l_2}$ and pyridine gave the product in which the carboxylic acid groups were converted into acyl chloride. Then treated with a reducing agent like lithium aluminium hydride gives an aldehyde.

Complete answer: :
Chemical compounds are classified into functional groups based on the groups present in it. Carboxylic acids are the functional groups that consist of $R - COOH$ , in which $R$ represents the alkyl group.
When carboxylic acids are treated with $SOC{l_2}$ and pyridine, then the $- COOH$ group converts into $- COCl$ .
When acyl chloride compounds react with reducing agents like Lithium aluminium hydride, then the acyl group converts into an aldehyde group $- CHO$ .
Given chemical compound is $2,3$ -dimethyl hexanoic acid which is a carboxylic acid when treated with $SOC{l_2}$ and pyridine forms $2,3$ -dimethyl hexanoyl chloride. Further treated with $LiAl{\left[ {OC{{\left( {C{H_3}} \right)}_3}} \right]_3}H$ which has a hydride act as a reducing agent form $2,3$ -dimethyl hexanal, which is an aldehyde. The chloride ion is replaced by hydride and forms an aldehyde.
The chemical reaction involved will be as follows:
$C{H_3}C{H_2}C{H_2}CH\left( {C{H_3}} \right)CH\left( {C{H_3}} \right)COOH\xrightarrow{{SOC{l_2}/pyridine}}C{H_3}C{H_2}C{H_2}CH\left( {C{H_3}} \right)CH\left( {C{H_3}} \right)COCl\xrightarrow{{LiAl{{\left[ {OC{{\left( {C{H_3}} \right)}_3}} \right]}_3}H}}C{H_3}C{H_2}C{H_2}CH\left( {C{H_3}} \right)CH\left( {C{H_3}} \right)CHO$
Thus, when $2,3$ -dimethyl hexanoic acid is treated with $SOC{l_2}$ and pyridine, followed by $LiAl{\left[ {OC{{\left( {C{H_3}} \right)}_3}} \right]_3}H$ the product is $2,3$ -dimethyl hexanal.

Note:
For the reduction of acyl chlorides, strong reducing agents like lithium aluminium hydride or sodium borohydride and its derivatives were used in general. Given reducing agent is $LiAl{\left[ {OC{{\left( {C{H_3}} \right)}_3}} \right]_3}H$ which has three tertiary butoxy groups in place of hydrogen. Due to the presence of one remaining hydride, it converts the acyl chloride molecule to aldehyde.