Question

Complete the following reaction
D RCN + 4H --------LiAlH4
​

Answer 1

$\huge\underline\mathfrak\pink{ᴀɴSᴡᴇʀ}$

Acetamide on treatment with LiAlH

4

forms ethyl amine.

CH

3

CONH

2

(1)LiAlH

4

/diethylether

(2)H

3

O

+

CH

3

CH

2

−NH

2

Lithium aluminium hydride reduces amide group to form corresponding amine.

Answer 2

The complete reaction is :

$RCN + 4H$ ----- $LiAlH_{4}$ -----> $RCH_{2} NH_{2}$

The product formed in Reduction is Alkyl Amine.

Chemical Reaction Details :

• The reactant RCN is alkyl nitrile having the Cyanide i.e. CN as its a functional group
• RCN has a triple strong bond between the Carbon and the Nitrogen atom
• $LiAlH_{4}$ i.e. lithium aluminum hydride is a strong reducing agent i.e. helps in the removal of oxygen or the addition of hydrogen
• Here, the lithium aluminum hydride enables the addition of hydrogen within the alkyl nitrile by reducing it and breaking the triple bond between Carbon and Nitrogen.
• As the triple bond breaks, due to reduction, the Hydrogen atoms attach to both Carbon and Nitrogen of the cyanide group
• This converts alkyl to amine i.e. alkyl nitrile to alkyl amine
• Carbon i.e. C now has 2 bonds free to which the Hydrogen attaches. The other two are for bonds to Nitrogen and 'R'.
• The Nitrogen i.e. N will also have 2 Hydrogen bonds added to it.
• Thus, the final product formed i.e. $RCH_{2} NH_{2}$ is Alkyl Amine.

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