Chemistry, asked by sapna2000, 1 year ago

ch2=ch-ch2-cho in the presence of LiAlH4 /ether and then H2o gives???

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Answered by Anonymous
3

Lithium aluminium hydride (LiAlH4) is a strong reducing agent. It cannot reduce an isolated non-polar multiple bond like C=C. However, the double bonds are reduced when a phenyl group is attached to the beta-carbon atom.

Answered by Anonymous
3
Oxidation and reduction reactions are the most complex of all reactions in organic chemistry and they are dependent on so many factors. I have tried to keep a balance here between what have been/maybe asked in IIT/AIIMS exam. The results which are not explained here are better understood by factors related to different solvents and sometimes easily explained by MOT.
Generally, neither LiAlH4 nor NaBH4 is able to reduce an isolated C=C But——
Cinnamaldehyde, C6H5-CH=CH-CHO when added slowly to an excess of the LiAlH4 dissolved in ether (normal addition), gives C6H5CH2CH2-OH
whereas in “inverse addition method” where LiAlH4/ether is added slowly to Cinnamaldehyde, so that LiAlH4 is never in excess, the product is cinnamyl alcohol, C6H5-CH=CH-CH2OH.
The exact process and products of reduction of conjugated carbonyl compounds are dependent on the nature of reductant.
In case of conjugated aldehyde or ketone, the nucleophilic attack takes place at either the β-carbon (conjugate addition) or at the carbonyl carbon (direct addition).
According to the HSAB Principle the vinylic β-carbon is a "soft" electrophilic center and prefer to react with "soft" nucleophiles whereas the carbonyl carbon is a "hard" electrophilic center and hence prefers to react with "hard" nucleophiles.
NaBH4 is a rather soft nucleophile and thus it prefers in some cases to react with 4-position or reduces conjugated >C=C< (called 1,4-addition) followed by reduction of aldehyde or ketone into the alcohol by NaBH4 in a second step.
The extent of reduction to saturated alcohol is usually greater with NaBH4 than with LiAlH4. 


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