What are P and Q? Name the reaction occuring in step (1);
(1) CH3CONH2 →Br2/NaOH→P
(2) P→NaNO2.HCL—0^0C→Q
This is an important question for 3 marks in Class 12 chemistry.
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Given: (1)CH₃CONH₂ +Br₂/NaOH→P
(2)P + NaNO₂.HCl (at 0° C) →Q
To Find: P and Q
Name reaction (1)
Solution:
(1) CH₃CONH₂ + Br₂ + 4NaOH → CH₃NH₂
(methanamide) (methanamine)
- This reaction is called Hoffman Bromamide Reaction.
- In Hoffman Bromamide Reaction, an alkali such as sodium hydroxide is used to attack an amide, leading to its deprotonation.
- It is used to convert an amide into an amine with 1 Carbon atom less than amide.
- P here is CH₃NH₂ (methanamine).
(2) P + NaNO₂.HCl (at 0° C) →Q
From (1) P is CH₃NH₂
CH₃NH₂ + NaNO₂.HCl (at 0°C) → CH₃OH + H₂O + N₂↑
(methanamine) (methanol)
- NaNO₂ and HCl at 0°C react to form nitrous acid (HNO₂).
- Primary aliphatic amine reacts with nitrous acid to give diazonium salt.
- Diazonium salt thus prouduced is very unstable, so it further reacts to form primary alcohol and nitrogen gas is evolved.
- Hence, we can say Q is CH₃OH (methanol).
Therefore,
P is CH₃NH₂ (methanamine) and the reaction ocurring in step (1) is Hoffman Bromamide Reaction.
Compound Q is CH₃OH (methanol).
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