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Chloromethane on treatment with excess of ammonia yields mainly
Options:
- A
N, N-dimethylmethanamine
- B
$\mathrm{N}$-methylmethanamine $\left(\mathrm{CH}_3-\mathrm{NH}-\mathrm{CH}_3\right)$
- C
methanamine $\left(\mathrm{CH}_3 \mathrm{NH}_2\right)$
- D
mixture containing all these in equal proportion
Solution:
1468 Upvotes
Verified Answer
The correct answer is:
methanamine $\left(\mathrm{CH}_3 \mathrm{NH}_2\right)$
methanamine $\left(\mathrm{CH}_3 \mathrm{NH}_2\right)$
$\mathrm{CH}_3 \mathrm{Cl}+\mathrm{NH}_3 \rightarrow \mathrm{CH}_3 \mathrm{NH}_2+\mathrm{HCl}$
Excess Methanamine
However, if the two reactants are present in the same amount, then the mixture of amines(i.e., primary, secondary and tertiary) are obtained.
Excess Methanamine
However, if the two reactants are present in the same amount, then the mixture of amines(i.e., primary, secondary and tertiary) are obtained.
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