Question

Which among the following are the correct statements?
Statement 1: Methyl-amine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
Statement 2: Methylamine is more basic than ammonia.

Answer 1

Hint: We should be aware of, which is the methyl group which is more basic in water. On the basis of this fact, we can determine whether the statements are correct or not.

Complete step-by-step answer:
Let us analyse the first statement. The first statement states that methyl-amine in water reacts with ferric chloride to precipitate hydrated ferric oxide. Due to the +I effect of the methyl group, the Methylamine is more basic in water. Methylamine reacts with water to produce hydroxide ions. The reaction is given below:

$\text{C}{{\text{H}}_{\text{3}}}-\text{N}{{\text{H}}_{\text{2}}}\text{ + }{{\text{H}}_{\text{2}}}\text{O }\xrightarrow{{}}\text{ C}{{\text{H}}_{\text{3}}}-\text{N}{{\text{H}}_{\text{3}}}^{\text{+}}\,\text{+ O}{{\text{H}}^{-}}$
Hydroxide ions react with ferric chloride to precipitate hydrated ferric oxide. The formula of ferric oxide is ${\text{F}}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}} \cdot {\text{3}}{{\text{H}}_{\text{2}}}{\text{O}}$.
The reaction is given below:
$2\text{FeC}{{\text{l}}_{\text{3}}}\text{ + 6O}{{\text{H}}^{-}}\text{ }\xrightarrow{{}}\text{ F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}\cdot \text{3}{{\text{H}}_{\text{2}}}\text{O + 6C}{{\text{l}}^{-}}$
The statement mentioned in I is correct.

Now moving on to the statement II. We know that an alkyl group is less electron-withdrawing than a hydrogen group. This means that R- group shows the +I effect, and tends to give electrons. This increases the overall electron density on the molecule, and the molecule is able to donate an electron. This makes the Methylamine more basic as compared to Ammonia.

The statement mentioned in II is correct.
So, both the statements in I and II are correct.

Note: The +I effect, or the inductive effect that is mentioned in the answer refers to the transmission of unequal sharing of the bonding electron through a chain of atoms in a molecule. This results in the formation of the permanent dipole in a bond.
The electron-withdrawing inductive effect is also known as the -I effect.