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

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

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Answer 1 · 2016-03-21T06:28:13

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

1 Answer

Explanation:

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