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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 $\ce{CH2=CH-CH2-C6H4-CH2-CH2-CH2-OH}$

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    $\begingroup$ This is correct. LAH does not react with isolated double bonds. $\endgroup$ – Waylander Jun 8 '20 at 10:28
  • $\begingroup$ @Waylander What would be the product in case of $\ce{C6H5-CH=CH-CH2-CH=CH-CHO}$? Is it $\ce{C6H5-CH=CH-CH2-CH=CH-CH2-OH}$? $\endgroup$ – Apurvium Jun 8 '20 at 15:37
  • $\begingroup$ No, C6H5-CH=CH-CH2-CH2-CH2-CH2-OH. The double bond conjugated to the C=O will be reduced. $\endgroup$ – Waylander Jun 8 '20 at 15:58
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    $\begingroup$ This answer may be applicable: chemistry.stackexchange.com/questions/87138/… $\endgroup$ – user55119 Jun 9 '20 at 20:09
  • $\begingroup$ @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 $\ce{NaBH4}$. $\endgroup$ – Apurvium Jun 10 '20 at 7:09

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