# Doesn't methyl amine give carbylamine test?

A compound $$\ce X$$ of formula $$\ce{C2H5NO2}$$ on treatment with $$\ce {HNO2}$$ $$(\ce{NaNO2}+\ce{HCl})$$ gives off an inactive gas $$\ce Y$$ and forms $$\ce Z$$. When heated with soda lime, $$\ce X$$ gives $$\ce{A(CH5N)}$$ soluble in acid. Find the true statements :

1. $$\ce A$$ answers carbylamine test.
2. $$\ce X$$ is an amino acid.
3. $$\ce Y$$ is $$\ce{CO2}$$.
4. $$\ce Z$$ is a hydroxyl acid.

My attempt :

$$\ce X$$ is glycine.

$$\ce A$$ is methyl amine.

$$\ce Y$$ is dinitrogen.

$$\ce Z$$ is 2-hydroxy ethanoic acid.

Hence, correct options should be : $$1,2,4$$. But the answer given is $$2,4$$.

Can anyone help me to understand if and where I went wrong?

You are indeed correct.

$$\ce{HOOC-CH2-NH2(X) ->[HNO2] HOOC-CH2-OH (Z) + N2(Y)}$$

$$\ce{HOOC-CH2-NH2(X) ->[soda lime] CH2-NH2(A)}$$

Also, $$\ce{(A) + CHCl3 + 3KOH ->[Heat] CH3-NC + 3KCl + 3H2O}$$.

Methylamine does give the carbylamine test.

So clearly, the answer given must be wrong.

The correct options are 1,2 and 4.