Question

How will you carry out the following reactions? Write relevant balanced equations and name the reactions.
A.Acetyl chloride to acetaldehyde
B.Acetamide to methyl-amine

Answer 1

We must know the reagents used for the conversion to the given products. The reagents used are hydrogen with palladium and potassium hydroxide with bromine respectively for conversion to aldehyde and amine.

Complete step by step answer:
A. The conversion of acetyl chloride to acetaldehyde is done by reducing it with hydrogen adsorbed over palladium metal and barium sulphate as reagents. Whenever we use Hydrogen as a reducing agent, we usually take metals such as Nickel, platinum or palladium, to adsorb hydrogen over their surface and the reaction can proceed. The reaction is as follow:
${\text{C}}{{\text{H}}_3}{\text{COCl}}\xrightarrow[{{\text{BaS}}{{\text{O}}_4}}]{{{{\text{H}}_2}/{\text{Pd}}}}{\text{C}}{{\text{H}}_3}{\text{CHO}}$
The above reaction is known as Rosenmund reduction reaction or simply Rosenmund reaction in which acid chlorides are reduced to aldehydes with the help of above mentioned reagents.
B. Now the conversion of Acetamide to methyl-amine is done with the help of Hoffmann bromamide degradation reaction. The amine thus formed has one less carbon than that of amide. The reaction proceeds with a mixture of halogen, a strong alkali and water. Only primary amines are produced in this reaction and no tertiary or secondary amines are there.
The above conversion occurs as:
${\text{C}}{{\text{H}}_3}{\text{CON}}{{\text{H}}_2}\xrightarrow[{{\text{B}}{{\text{r}}_2}}]{{{\text{KOH}}}}{\text{C}}{{\text{H}}_3}{\text{N}}{{\text{H}}_2}$
As we can clearly see that methyl amine has one less carbon than that of the acetamide.

Note: Acetaldehyde is a very important aldehyde. It is produced in a very large amount and occurs in a huge amount in nature. It is used in the production of acetaldehyde and many other organic compounds. Methylamine is colourless gas that is basically substituting one hydrogen atom from the ammonia molecule.