When 2,3-dimethylhexanoic acid is treated with SOCl2/pyridine, followed by LiAl [OC(CH3)3]3H, what would be the product???? Thank you

1 Answer
Dec 22, 2015

The first mechanism is as follows:

#\mathbf("SOCl"_2)# essentially complexes with the carbonyl on the carboxylic acid while one of the chlorines can act as a nucleophile. #"SO"_2# and #"Cl"^(-)# leave as part of the tetrahedral collapse. The result is that the hydroxide leaves and chlorine substitutes it.

Pyridine is the solvent. Its protonated form's pKa is higher than that of #"HCl"#, so it gets the proton at the end.

Then the big aluminum compound (lithium tri-tert-butoxyaluminum hydride, Ch. 20, Organic Chemistry, Paula Yurkanis Bruice) with the electron-donating groups allows the single hydrogen to act as a hydride, and you get a reduction to an aldehyde, rather than all the way to an alcohol.

#\mathbf("LiAlH"_4)# would typically be a very reactive alternative in protic solvents that DOES give you an alcohol, but in real life we try not to use it unless we have to.

Lithium tri-tert-butoxyaluminum hydride is nice to use because it's so bulky and it has only one hydride. Diisobutylaluminum hydride (DIBALH) can be used as a similarly useful alternative.

Both single-hydride aluminum compounds should be used in cold temperatures (e.g. #-78^@ "C"# is pretty common because it's conveniently the boiling point of #"CO"_2#, which is then used as dry ice).

So, the overall reaction would be written as: