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

1 Answer
anor277
Mar 21, 2016

The dialdehyde results: #H(O=C)CH(CH_3)CH(CH_3)CH_2CH_2C(=O)H#

Explanation:

#"Diacid + thionyl chloride "rarr" Diacid halide + sulfur dioxide"#

#"Diacid halide + hydride transfer reagent "rarr" Dialdehyde"#

#Li^+[HAl(OBu^t)_3]^-# is a hydride transfer reagent that delivers 1 hydride instead of the 4 that lithium aluminum hydride (#LiAlH_4#) delivers. it will react with acid halides to give the formyl species, the aldehyde:

#"Diacid halide + hydride transfer reagent"# #rarr# #"Dialdehyde + lithium chloride"#

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