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Which of the reagents listed below would efficiently accomplish the transformation of ethyl-3-pentenoate into 3-penten-1-ol?
a.i) lialh4; ii) h2o
b.nabh4,h2o
c.h2, pd
d.two of these choices.
e.all of these choices?

Respuesta :

transitionstate
Reactions of Ethyl-3-pentenoate with all given reagents are given below.

Reaction with H₂ / Pd:
                                     The non-polar double bond present in Ethyl-3-pentenoate is reduced to saturated chain. This reagent can not reduce the carbonyl group.

Reaction with NaBH₄:
                                   Sodium Borohydride is a weak reducing agent at compared to LiAlH₄. It can only reduce aldehydes and Ketones to corresponding alcohols.

Reaction with LiAlH₄:
                                  Lithium Aluminium hydride is a strong reducing agent. It can reduce all types of carbonyl compounds to corresponding alcohols, But, it can not reduce non-polar double bonds like alkenes and alkynes.

Result:
           The correct answer is Option-A (Highlighted RED below).
Ver imagen transitionstate

Esters form two alcohol when they undergo hydride reduction. Ethyl-3-pentenoate transforms into 3-penten-1-ol by lithium aluminum hydride and water.

What is lithium aluminum hydride?

Lithium aluminum hydride is a reducing agent that reduces esters, aldehydes, carboxylic acids, ketones, etc. It is a strong reducing agent that donates electrons.

The reaction of Ethyl-3-pentenoate with hydrogen and palladium results in the reduction of the saturated chain but the carbonyl group is not reduced.

The reaction of Ethyl-3-pentenoate with sodium borohydride cannot reduce esters, carboxylic acids, and amides, so no reaction will occur.

The reaction of Ethyl-3-pentenoate with lithium aluminum hydride reduces them to alcohol to produce 3-penten-1-ol.

Therefore, option a.  [tex]\rm LiAlH_{4}[/tex] and [tex]\rm H_{2}O[/tex] transforms ethyl-3-pentenoate into 3-penten-1-ol.

Learn more about lithium aluminum hydride here:

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