Reduction of aldehydes & ketones by LiAlH4 & NaBH4 occurs by nucleophilic attack of hydride ion on the carbonyl group. All the four hydrogen atoms may be successively used in reduction and the alcohol is liberated from the intermediate metal alkoxides by hydrolysis.
Neither LiAlH4 nor NaBH4 reduces isolated C=C. However, in case of a¾b unsaturated aldehydes & ketones LiAlH4 usually reduces only the carbonyl group leaving the double bond intact. While with NaBH4 , the situation is somewhat different. As aldehydes are more reactive than ketones, in case of a¾b unsaturated aldehydes , the reduction of aldehydic group occurs preferentially. E.g.,
1.NaBH4,2.H3O+
CH3-CH=CH-CHO ® CH3-CH=CH-CH 2OH
However, with ketones a mixture of C=O reduction product and C=C & C=O reduction product is formed, where C=O reduction product predominates. For ex.
1.NaBH4,2.H3O+
Cyclohex-2en-1-one ® Cyclohex-2-en-1-ol + cyclohexanol
59% 41%
1.LiAlH4/dry ether,2.H3O+
Cyclohex-2en-1-one ® Cyclohex-2-en-1-ol
98%
A C=C is an electron dense species where electrophilic attack occurs. A hydride ion is an electron-dense species and thus behaves as a strong nucleophile. So a hydride ion experiences a repulsion when it comes close to a C=C and thus fails to reduce it usually. However, when attached to benzene ring, the electron withdrawing effect of an aromatic system on one side and that of the carbonyl group on the other side make the C=C less electron dense and enables the hydride ion to attack the C=C. For this reason the LiAlH4 reduces C=C as well as C=O in cinnamaldehyde.
1.LiAlH4/dry ether,2.H3O+
C6H5-CH=CH-CHO ® C6H5-CH2CH2CH2OH
Thus we cannot conclude that only NaBH4 reduces the C=C and not LiAlH4.It will depend on the compound to be reduced.
Moderation Team
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