Compounds Containing Nitrogen

$ \text { The complete reaction is } $

$ \mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}_2^{+} \mathrm{Cl}^{-} \longrightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{Cl}+\mathrm{N}_2 $

Hence, reagents are: $\mathrm{NaNO}_2 / \mathrm{HCl}$, $273-278 \mathrm{~K} ; \mathrm{Cu}_2 \mathrm{Cl}_2 / \mathrm{HCl}$.

The complete reaction sequence is as follows

Molar mass of $Z$ i.e.,

$ \begin{aligned} \mathrm{C}_7 \mathrm{H}_7 \mathrm{NO}_2 & =(7 \times 12)+(7 \times 1)+(1 \times 14) \\ & =137 \mathrm{u} \end{aligned} $

Now \% of carbon in $Z$

$ \begin{aligned} & =\frac{\text { Mass of carbon }}{\text { Molar mass of } Z} \times 100 \\ & =\frac{84}{137} \times 100=61.3 \% \end{aligned} $

The complete reaction sequence is as follows

The complete reaction is as follows

Among the given options, Benzylamine can be prepared by reaction given in option (d)

$ \text { The complete reaction sequence is as follows } $

$ \text { The complete reaction sequence is as follows } $

Statement I

In the nitration of aniline, more amount of m-nitroaniline is formed than expected.

Normally, –NH₂ is an activating, ortho/para-directing group.

Hence, we expect mainly o- and p-nitroaniline.

However, nitration is carried out in strongly acidic medium (H₂SO₄/HNO₃), and under these conditions, aniline gets protonated to form the anilinium ion (C₆H₅NH₃⁺).

The –NH₃⁺ group is an electron-withdrawing, deactivating, meta-directing group.

So the nitration mixture now contains protonated aniline, which gives meta nitro product.

As a result, the actual nitration product mixture contains a higher proportion of m-nitroaniline than we would expect based on the activating nature of –NH₂ alone.

✅ Hence, Statement I is correct.

Statement II

In the presence of a strongly acidic medium, aniline is protonated to form anilinium ion, which is meta-directing.

As discussed above, this is true — in a strongly acidic medium, aniline is converted to the anilinium ion (C₆H₅NH₃⁺).

The anilinium ion is meta-directing because the positively charged nitrogen withdraws electrons from the ring, deactivating it, especially at ortho and para positions.

✅ Hence, Statement II is also correct.

✅ Final Answer

Option A: Both Statement I and Statement II are correct.

The reaction of benzene diazonium chloride with Cu and HCl is called as Gattermann reaction to form chlorobenzene.

The sequence of reagent required is, $\mathrm{NaNO}_2 / \mathrm{HCl} / 273-278 \mathrm{~K} ; \mathrm{CuCN} / \mathrm{KCN} ; \mathrm{H}_3 \mathrm{O}^{+}$ The reaction is,

$ \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2+\mathrm{NaNO}_2+2 \mathrm{HCl} \xrightarrow{273-278 \mathrm{~K}} \mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}_2^{+} \mathrm{Cl}^{-}+\mathrm{NaCl}+2 \mathrm{H}_2 \mathrm{O} $

$ \begin{aligned} & \mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}_2^{+} \mathrm{Cl}^{-}+\mathrm{CuCN} \longrightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{CN}+\mathrm{CuCl}+\mathrm{N}_2 \\ & \mathrm{C}_6 \mathrm{H}_5 \mathrm{CN}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}+\mathrm{NH}_3 \end{aligned} $

The reaction mentioned is a Fitting reaction, The complete reaction is as follow,

$ \text { The complete reaction sequence is as follow } $

Total three amines with formula $\mathrm{C}_3 \mathrm{H}_9 \mathrm{~N}$ can react with benzene sulphonyl chloride.

(i) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{NH}_2$ : Propylamine

(ii) $\left(\mathrm{CH}_3\right)_2 \mathrm{CHNH}_2$ : Isopropylamine

(iii) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NHCH}_3$ :

N -methylethylamine

$ \text { The complete reaction is as follows } $

Statement I:

$ \mathrm{CH_3NH_2} $ is more basic than $ \mathrm{NH_3} $ but $ \mathrm{C_6H_5NH_2} $ is less basic than $ \mathrm{NH_3} $.

Explanation:

• Methylamine (CH₃NH₂):

  The –CH₃ group is electron donating (+I effect), which increases the electron density on nitrogen, making it more basic than ammonia.

• Aniline (C₆H₅NH₂):

  The lone pair on nitrogen gets delocalized into the benzene ring through resonance, reducing its availability to accept a proton.

  So aniline is less basic than ammonia.

✅ Hence, Statement I is correct.

Statement II:

The order of basic strength of amines in aqueous phase follows the order

$ (\mathrm{C_2H_5})_3\mathrm{N} > (\mathrm{C_2H_5})_2\mathrm{NH} > \mathrm{C_2H_5NH_2} $.

Let's check this:

In the aqueous phase, basic strength is influenced by two factors:

1. +I effect (electron donating alkyl groups increase basicity)

2. Solvation effect (stabilization of the conjugate ammonium ion)

• In aqueous solution, the solvation effect is strong for primary > secondary > tertiary amines, because primary amines can form more hydrogen bonds.

• However, the +I effect increases as we move from primary → secondary → tertiary.

The combined effect in aqueous phase gives the observed order:

$ \textbf{Secondary amine} > \textbf{Primary amine} > \textbf{Tertiary amine} $

That is:

$ (\mathrm{C_2H_5})_2\mathrm{NH} > \mathrm{C_2H_5NH_2} > (\mathrm{C_2H_5})_3\mathrm{N} $

But the statement says the order is:

$ (\mathrm{C_2H_5})_3\mathrm{N} > (\mathrm{C_2H_5})_2\mathrm{NH} > \mathrm{C_2H_5NH_2} $

❌ This is incorrect in aqueous medium.

✅ Correct Answer:

Statement I is correct but Statement II is not correct.

✅ Option C

The complete reaction sequence is as follows

Among the given options, $p$-methyl benzene nitrile can be prepared by

$ \text { The reaction involved is } $

Conversion of benzene diazonium chloride to chlorobenzene is called as Sandmeyer reaction.

Image The correct sequence of reaction involved is the given coversion is

So, the processes involved in given sequence are bromination, reduction and carbylamine reaction.

The correct order of relative basic strength will be

In $\mathrm{NH}_3$, the lone pair is completely available for donation, hence is most basic whereas $\mathrm{NH}_2-\mathrm{NH}_2$ and $\mathrm{NH}_2-\mathrm{OH}$ can be considered as derivatives of $\mathrm{NH}_3$. Here, H is replaced by $-\mathrm{NH}_2$ forming $\mathrm{NH}_2-\mathrm{NH}_2$ and $-$OH forming $\mathrm{NH}_2-\mathrm{OH}$ being highly electron withdrawing due to $-$O will decrease the electron density most on $-$N .

$\therefore$ Least basic electron withdrawing $-I$-effect of $-\mathrm{NH}_2$ is less as compared to $-$OH .

Hence correct order of basic strength is

$\mathrm{NH}_3>\mathrm{NH}_2-\mathrm{NH}_2>\mathrm{NH}_2-\mathrm{OH}$

The boiling point of $(\mathrm{A})$ alcohol is highest due to H -bonding.

Out of primary (B), secondary (D) and tertiary amines (C), the boiling point in decreasing order is Primary amine $>$ Secondary amine $>$ Tertiary amine

as intermolecular association decreases in same order, being highest in primary amine and least in tertiary amine due to steric hindrance.

$\therefore$ The correct decreasing order of boiling point in given compound is

$\mathrm{A}>\mathrm{B}>\mathrm{D}>\mathrm{C} \text {. }$

Due to presence of electron withdrawing nitro group, delocalisation of lone pair of electron is improved. This effect observed in para and ortho position.

In ortho position, very close inductive effect of nitro group is also observed. Hence, o-nitroaniline is less basic than p-nitroaniline. In m-nitroaniline, only inductive effect is present.

Order of basicity is as follows:

aniline > m-nitroaniline > p-nitroaniline > o-nitroaniline.

$25 \mathrm{~mL}$ of $1 \mathrm{M} \mathrm{~H}_2 \mathrm{SO}_4$ corresponds to $25 \mathrm{~m} \mathrm{~mol}$.

It will neutralise $50 \mathrm{~mol}$ of $\mathrm{NH}_3$.

weight of $\mathrm{N}=\frac{14 \times 50}{1000} \mathrm{~g}=0.70 \mathrm{~g}$

Thus $\mathrm{l} \mathrm{g}$ of sample of organic compound contains $0.70 \mathrm{~g}$ of nitrogen.

$\%$ of nitrogen in organic sample is $\frac{0.70}{1} \times 100=70 \%$