Question

[Et is \(-\mathrm{C_2H_5}\)]
Reaction: the substrate \(\mathrm{EtO-C(=O)-CH_2-}\)(cyclopentene ring, 1,3-disubstituted, C=C in the ring)\(\mathrm{-CH_2CH_2CN}\) is treated with (i) DIBAL-H, (ii) \(\mathrm{H_2O}\).
Consider the above reaction and predict the major product.

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

DIBAL-H partially reduces both a nitrile (to an imine, hydrolysed to an aldehyde) and an ester (to an aldehyde); an isolated C=C is not touched and no carboxylic acid forms.

Answer 1

The Correct Option is A

DIBAL-H is a bulky hydride donor that reduces the polar \(\mathrm{C\equiv N}\) and C=O (ester) groups only to the aldehyde stage, and leaves the ring C=C unchanged.
Nitrile: \(\mathrm{R-C\equiv N}\to\mathrm{R-CH=N-Al(iBu)_2}\) (one hydride added); on aqueous work-up (\(\mathrm{H_2O}\)) the imine hydrolyses to \(\mathrm{R-CHO}\). So \(\mathrm{-CH_2CH_2CN}\to\mathrm{-CH_2CH_2CHO}\).
Ester: \(\mathrm{R-COOEt}\to\mathrm{R-CH(OEt)O-Al}\) hemiacetal-aluminium adduct, which on hydrolysis loses EtOH to give \(\mathrm{R-CHO}\). So \(\mathrm{EtOOC-CH_2-}\to\mathrm{OHC-CH_2-}\).
Both groups therefore end up as aldehydes. No \(\mathrm{COOH}\) can form (DIBAL-H is a reducing agent) so options (3) and (4) are out; option (2) leaves the ester untouched, but DIBAL-H also reduces esters, so it is out.
The product is the dialdehyde, option (1).
Answer: (1) OHC-CH₂-(cyclopentene)-CH₂CH₂-CHO