0

What will be the final product upon reaction of $\ce{CH2=CH-CH2-C6H4-CH=CH-CHO}$ with $\ce{LiAlH4}$ (in dry ether) followed by acidic water?

LAH will reduce the carbonyl group to alcoholic group and the alkene group which is directly attached to carbonyl and aryl group to saturated bond. But it will not reduce the isolated alkene group.

Hence, the final product must be $\ce{CH2=CH-CH2-C6H4-CH2-CH2-CH2-OH}$

Apurvium
  • 1,272
  • 11
  • 34
  • 3
    This is correct. LAH does not react with isolated double bonds. – Waylander Jun 08 '20 at 10:28
  • @Waylander What would be the product in case of $\ce{C6H5-CH=CH-CH2-CH=CH-CHO}$? Is it $\ce{C6H5-CH=CH-CH2-CH=CH-CH2-OH}$? – Apurvium Jun 08 '20 at 15:37
  • No, C6H5-CH=CH-CH2-CH2-CH2-CH2-OH. The double bond conjugated to the C=O will be reduced. – Waylander Jun 08 '20 at 15:58
  • 1
    This answer may be applicable: https://chemistry.stackexchange.com/questions/87138/why-does-lialh4-reduce-the-double-bond-of-cinnamaldehyde-whereas-nabh4-does-not/87195#87195 – user55119 Jun 09 '20 at 20:09
  • @user55119 according to the link (and the link therein), the product will be $\ce{C6H5−CH=CH−CH2−CH=CH−CH2−OH}$ in case of $\ce{LiAlH4}$ and $\ce{C6H5-CH=CH-CH2-CH2-CH2-CH2-OH}$ in case of $\ce{NaBH4}$. – Apurvium Jun 10 '20 at 07:09
  • The link only speaks to how cinnamaldehyde undergoes 1,4-reduction. Whether an alkyl group in place of a phenyl ring will give the same result is questionable. – user55119 Jun 10 '20 at 13:13
  • @user55119 Yes, that is questionable but check this link (https://chemistry.stackexchange.com/questions/6977/why-does-nabh4-reduce-double-bonds-conjugated-to-carbonyl-groups-while-lialh4-d?noredirect=1&lq=1) and my comment. – Apurvium Jun 10 '20 at 15:46

0 Answers0