Question

Draw the expected product when 2,3-dimethylhexanoic acid is treated with SOCl2/pridine, followed LiAl [OC(CH3)3]3H. ???

Answer 1

Acid `rarr` Acid chloride `rarr` Aldehyde

Explanation:

`RC(=O)OH + SOCl_2 + C_6H_5N rarr RC(=O)Cl + C_6H_5N*HCl + SO_2uarr`

`RC(=O)Cl + Li[Al{OC(CH_3)}_3H] rarr R(C=O)H + Al{OC(CH_3)}_3 + LiCl`

Lithium tri(tert)butoxy aluminum hydride is a reducing agent specifically for the reduction of acid halides to aldehydes. The other common metal reducing agent, `LiAlH_4`, would likely reduce the acid back down to the alcohol (`LiAlH_4` transfers up to 4 equiv of `H^-`, whereas the butoxy reagent transfers only the 1).