Question

CH3 -CH=CH-CH2CHO in LiAlH4 and H3O gives ?? ​

Answer 1

Answer:

Explanation:

pent - 3 ene - 1 ol

Answer 2

Final answer: The product formed is $CH_{3}-CH=CH-CH_{2}-CH_{2}OH$ (pent-3-enol).

Given that: We are given $CH_{3} -CH=CH-CH_{2}-CHO$ in $LiAlH_{4}$ and $H3O^{+}$.

To find: We have to find the product formed.

Explanation:

• Given compound: $CH_{3} -CH=CH-CH_{2}-CHO$

The IUPAC name of this compound is pent-3-enal.

It contains an isolated non-polar $> C = C <$ and an aldehyde ($-CHO$) group.

• $LiAlH_{4}$ is a strong reducing agent (add hydrogen).

It is a nucleophilic reducing agent, which reduces aldehyde ($R-CHO$) to alcohol ($R-CH_{2}OH$).

• But here the given compound is enal.

• $LiAlH_{4}$ cannot reduce isolated non-polar multiple bonds.

It selectively reduces only the $-CHO$ group in pent-3-enal.

• So first given compound (pent-3-enal) react with $LiAlH_{4}$ to form an adduct by eliminating $AlH_{3}$. Which further reacts with $H_{3}O^{+}$ to yield pent-3-enol by eliminating $H_{2}O$.

• Pent-3-enol is also written as pent-3-en-1-ol.

Chemical formula: $CH_{3} -CH=CH-CH_{2}-CH_{2}OH$

• The chemical reaction:

                                                 $(i) LiAlH_{4}$

$CH_{3} -CH=CH-CH_{2}-CHO$ ──→ $CH_{3} -CH=CH-CH_{2}-CH_{2}OH$

                                                  $(ii) H_{3}O^{+}$              $Pent-3-enol$

Hence, $LiAlH_{4}$ and $H_{3}O^{+}$ reduced $CH_{3} -CH=CH-CH_{2}-CHO$ (given) to $CH_{3}-CH=CH-CH_{2}-CH_{2}OH$.

To know more about the concept please go through the links

https://brainly.in/question/1502552

https://brainly.in/question/2765193

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