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

Hint: 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.