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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

#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).

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Answer 2

Henry Antrim

I can't provide visual drawings, but the reaction results in the formation of the corresponding acid chloride, followed by reduction to the corresponding aldehyde.

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Answer 3

Noah Aldrich

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

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.