Compounds Containing Nitrogen

Nitrating mixture is mixture of conc. HNO₃ and a conc. H₂SO₄.

When aniline is reacted with nitrating mixture ortho, meta and para nitroanilines are obtained.

Aniline being basic in nature, reacts with acids to form anilinium ion which is meta directing.

(a) If KNO₂ (Q)-on reacting with PhCH₂Br(P) gives PhCH₂ONO which on reduction with LiAlH₄ does not give PhCH₂NH₂.

(b) If product R formed is benzylamine not benzene amine. KCN reacts with PhCH₂ Br to give PhCH₂CN which does not have foul smelling.

(c) If AgNO₂ (Q) reacts with PhCH₂ Br to give PhCH₂NO₂ which on reduciton give PhCH₂NH₂ (Phenyl methanamine or benzyl amine).

(d) If amide reacts with LiAlH 4 (W) to give amine. PhCH₂ Br reacts with AgCN to give PhCH₂NO₂ .

PhSNa replaces F from o- and p-dinitro- fluorobenzene via nucleophilic aromatic substitution.

PhSNa replaces Cl from compound D via nucleophilic bimolecular substitution reaction.

In case of compound B and C, nucleophilic substitution is not possible neither by S_NAr or S_N2 . because, the electron withdrawing group is present at meta-position and halogen is attached to sp²-carbon.

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 $\frac{0.70}{1} \times 100=70 \%$

Epoxy ethane reacts with ammonia forming a mixture of mono, di-, tri-ethanol amines, it presence of small amount of water as follows.

The Gabriel phthalimide synthesis is a method used to prepare primary aliphatic and aromatic amines. The process begins with phthalimide being treated with ethanolic potassium hydroxide (KOH), forming the potassium salt of phthalimide. This salt is then heated with an alkyl halide (RX), and subsequent treatment with either alkaline hydrolysis or hydrazinolysis (using hydrazine, $ \mathrm{H}_2\mathrm{N} \cdot \mathrm{NH}_2 $) yields the corresponding primary amine. However, this method cannot be used to synthesize primary aromatic amines that have electron-withdrawing groups (EWGs) at the ortho- and para-positions, such as the nitro group ($ -\mathrm{NO}_2 $). This limitation arises because aryl halides (ArX) do not undergo nucleophilic substitution ($\mathrm{S}_{\mathrm{N}}$) reactions with the anion formed by phthalimide.

$ \text { In the given reaction, the final product is as follows : } $

In step (i), we get diazonium product, which go for coupling reaction to give the final product as shown in the reaction. Hence, option (d) is the correct answer.

$ \text { The major product in the following reaction is } $

(i) $\mathrm{Sn} / \mathrm{HCl}$ show reduction.

(ii) $\mathrm{Br}_2$ show addition

(iii) Show formation of diazonium salt.

(iv) Show, the final product.

Hence, option (c) is the correct answer.

Step I Amide react with bromine and aqueous KOH to form amine.

Step II Amine react with $\mathrm{NaNO}_2, \mathrm{HCl}$ followed by hydrolysis form alcohol.

The given reactions take place as follows

The reaction take place as follows

The correct answer is Option B: C₆H₅SO₂Cl.

Hinsberg's reagent is benzenesulfonyl chloride (C₆H₅SO₂Cl). It is used to differentiate primary, secondary, and tertiary amines.

Here's why the other options are incorrect :

• Option A: C₆H₅COCl is benzoyl chloride, a reagent used in Friedel-Crafts acylation reactions.


• Option C: SOCl₂ is thionyl chloride, a reagent used to convert alcohols into alkyl chlorides.


• Option D: (COCl)₂ is oxalyl chloride, a reagent used for the conversion of carboxylic acids into acid chlorides.

All the given compounds are acid derivatives, thus contain carbonyl group in them. LiAlH₄ reduces these compounds through nucleophilic substitution via addition elimination (S_NAE) reaction. The rate of reaction depends upon the following factors:

(i) Size of alkyl group.

(ii) Steric hinderance around the > C == O group.

(iii) (+) ve charge on the C-atom of > C == O group.

The alkyl groups are more or less same in the given compounds. Thus, the reactivity order of given compounds depends upon 2nd and 3rd factor written above. The cumulative effect of these two factors results to leaving group ability (LGA) of the substituents in the following order :

This leaving group ability (weak conjugate base) corresponds directly to the reactivity order.

Hence, the correct reactivity order is:

The complete pathway of reactions is as follows :