Organic Compounds Containing Nitrogen Formula Sheet — JEE Main Chemistry

Q: What are the most important Organic Compounds Containing Nitrogen formulas for JEE Main?

This Organic Compounds Containing Nitrogen formula sheet covers all the high-yield Chemistry formulas, definitions and theorems you need for JEE Main, across General methods of preparation, properties, reactions and uses, Amines: Nomenclature, classification, structure, Basic character and identification of primary, secondary and tertiary amines, Diazonium Salts: Importance in synthetic organic chemistry — each shown with the key result and, where useful, a worked example.

Q: Is this Organic Compounds Containing Nitrogen formula sheet free?

Yes — the full chapter formula sheet is free to read online, no login or payment required.

Q: How should I revise Organic Compounds Containing Nitrogen formulas?

Blurt the Organic Compounds Containing Nitrogen formulas from memory, then check against this sheet to find your gaps — and practise a few previous-year questions on the chapter to make sure you can apply them under time pressure.

Every key Organic Compounds Containing Nitrogen formula, definition and theorem for JEE Main Chemistry in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNTA · 4 sub-topics

General methods of preparation, properties, reactions and uses
Amines: Nomenclature, classification, structure
Basic character and identification of primary, secondary and tertiary amines
Diazonium Salts: Importance in synthetic organic chemistry

Amines — Classification, Nomenclature & Structure

Amines: Derivatives of ammonia in which one, two or all three H atoms of

N H_{3}

are replaced by alkyl/aryl groups. The nitrogen keeps an

unshared electron pair

in a fourth

s p^{3}

orbital, making amines pyramidal and the key to their behaviour as

bases and nucleophiles

The pyramidal (sp^3) nitrogen of an amine, carrying the lone pair that makes it a base and nucleophile. — The pyramidal ( $s p^{3}$ ) nitrogen of an amine, carrying the lone pair that makes it a base and nucleophile.

Structure	Common	IUPAC
$C H_{3} N H_{2}$	methylamine	methanamine
$C H_{3} NHC H_{2} C H_{3}$	ethylmethylamine	N-methylethanamine
$(C H_{3})_{3} N$	trimethylamine	N,N-dimethylmethanamine
$H_{2} N (C H_{2})_{2} N H_{2}$	ethylenediamine	ethane-1,2-diamine
$C_{6} H_{5} N H_{2}$	aniline	aniline / benzenamine

Classifying by number of C on N

▸ $Primary$ ( $1^{\circ}$ ): $R - N H_{2}$ or $Ar - N H_{2}$ — one C on N.
▸ $Secondary$ ( $2^{\circ}$ ): $R_{2} NH$ — two C on N.
▸ $Tertiary$ ( $3^{\circ}$ ): $R_{3} N$ — three C on N.
▸ $Quaternary$ : $R_{4} N^{+} X^{-}$ (ammonium salt, surfactant).
▸ $simple$ = all R same; $mixed$ = R groups differ.

★ Remember · Naming: common vs IUPAC

Common

= alkyl group(s)

+

'amine' (

C H_{3} N H_{2}

= methylamine;

(C H_{3})_{3} N

= trimethylamine).

IUPAC

= alkan

amine

(replace the 'e' of the alkane):

C H_{3} N H_{2}

= methanamine. Substituents on N take the

N -

locant.

🎯 Exam · Diamines & arylamines keep the 'e'

H_{2} N (C H_{2})_{2} N H_{2}

= ethane-1,2-diamine (the 'e' is kept before a consonant). The simplest arylamine

C_{6} H_{5} N H_{2}

= aniline (also benzenamine); substituted ones are named as derivatives, e.g. 4-bromoaniline.

★ Remember · Why the N is pyramidal

N is

s p^{3}

with three bonds plus a lone pair, so the

C - N - E

angle is

\approx 10 8^{\circ}

(e.g. trimethylamine) — a little less than the tetrahedral

109. 5^{\circ}

because the bulky lone pair compresses it.

🚫 Examiner Trap · Examiner traps

(1) Count the C atoms ON N to assign

1^{\circ} / 2^{\circ} / 3^{\circ}

— NOT the carbon that bears

- N H_{2}

(opposite to alcohols/halides). (2) N is pyramidal

s p^{3}

with a lone pair (C

- N -

\approx 10 8^{\circ}

)

\Rightarrow

base AND nucleophile. (3) Quaternary

R_{4} N^{+}

has NO lone pair

\Rightarrow

not basic/nucleophilic (just a salt). (4) Aniline

=

aminobenzene; don't read it as a

2^{\circ}

amine.

Amines — Preparation (Six Routes)

★ Remember · Reduction of nitro & ammonolysis

Ar - N O_{2} H_{2} /Pd or Sn/HCl Ar - N H_{2}

(Fe/HCl preferred — HCl regenerated).

Ammonolysis

R - X + N H_{3} \to RN H_{2} \to R_{2} NH \to R_{3} N \to R_{4} N^{+} X^{-}

⚠️ Watch out · Ammonolysis gives a mixture

Because each amine formed is still nucleophilic, ammonolysis yields all of

1^{\circ}

2^{\circ}

3^{\circ}

and the quaternary salt. The

1^{\circ}

amine is the major product only with a

large excess of

N H_{3}

. Reactivity of halides:

RI > RBr > RCl

Carbon-count is the giveaway

▸ $Nitrile$ ( $R - CN LiAl H_{4} RC H_{2} N H_{2}$ ): $ascent$ — one C $more$ .
▸ $Amide$ ( $R - CON H_{2} LiAl H_{4} RC H_{2} N H_{2}$ ): same carbons.
▸ $Hoffmann bromamide$ ( $R - CON H_{2} + B r_{2} + 4 NaOH$ ): one C $less$ .
▸ $Gabriel$ : clean $1^{\circ}$ amine — but $no aryl amines$ (aryl halides resist the phthalimide anion).

🎯 Exam · Gabriel vs Hoffmann — pick by need

Use

Gabriel

when you need a

pure

1^{\circ}

aliphatic amine (no

2^{\circ} / 3^{\circ}

contamination). Use

Hoffmann bromamide

to step

down

a carbon (

R - CON H_{2} \to R - N H_{2}

); it also makes aromatic

1^{\circ}

amines, which Gabriel cannot.

🚫 Examiner Trap · Examiner traps

(1) Ammonolysis gives a MIXTURE (

1^{\circ} / 2^{\circ} / 3^{\circ} /

quaternary); excess N

H_{3}

favours the

1^{\circ}

amine. (2)

Gabriel

makes

1^{\circ}

amines ONLY and

cannot make aryl amines

(phthalimide N won't do S

_{N}

Ar). (3) Nitrile reduction ADDS one C (ascent); Hofmann bromamide LOSES one C (amide

\to

amine with one less C) — opposite carbon counts. (4) Reduce

- N O_{2}

with Fe/HCl or

H_{2}

/Ni, NOT with Sn/HCl if selectivity matters.

Basicity of Amines

Amine	$p K_{b}$ (smaller = stronger)
$(C H_{3})_{2} NH$	3.27
$C H_{3} N H_{2}$	3.38
$N H_{3}$	4.75
aniline	9.38

🎯 Exam · Gas phase is clean; water is not

Gas phase

(pure

+ I

3^{\circ} > 2^{\circ} > 1^{\circ} > N H_{3}

Aqueous

(also solvation of the cation, which favours smaller ions):

(C H_{3})_{2} NH > C H_{3} N H_{2} > (C H_{3})_{3} N > N H_{3}

— the bulky

3^{\circ}

cation is poorly solvated, so it drops.

★ Remember · Aniline is a weak base

In aniline the

- N H_{2}

lone pair is conjugated into the ring (aniline = hybrid of

5

resonance forms; anilinium has only

2

). The lone pair is less available, so aniline (

p K_{b} 9.38

) is far weaker than ammonia (4.75) or any alkylamine (

\sim 3 - 4

🎯 Exam · Substituents on aniline

Electron-

releasing

groups (

- OC H_{3}

- C H_{3}

) raise basicity; electron-

withdrawing

groups (

- N O_{2}

- S O_{3} H

- COOH

- X

) lower it. So p-toluidine

>

aniline

> p

-nitroaniline.

⚠️ Watch out · Don't quote one order for everything

There is no single 'basicity order' — it depends on phase (gas vs water) and on the alkyl group (the

ethyl

series gives

(C_{2} H_{5})_{2} NH > (C_{2} H_{5})_{3} N > C_{2} H_{5} N H_{2} > N H_{3}

, slightly different from the methyl series).

🚫 Examiner Trap · Examiner traps

(1) GAS phase basicity is pure

+

3^{\circ} > 2^{\circ} > 1^{\circ} > N H_{3}

; in WATER the order scrambles (solvation

+

steric) — e.g.

(C H_{3})_{2} N H > C H_{3} N H_{2} > (C H_{3})_{3} N

. Quote the right environment! (2) Aromatic amines are MUCH weaker bases (lone pair delocalised into ring). (3) EDG on aniline

\Rightarrow

stronger base; EWG (

- N O_{2}

)

\Rightarrow

weaker. (4) Smaller p

K_{b}

=

stronger base.

Reactions of Amines I — Acylation, Carbylamine & Nitrous Acid

★ Remember · Basic character & salt formation

R - N H_{2} + HX ⇌ R - N H_{3}^{+} X^{-}

(water-soluble salt), regenerated by NaOH. This dissolve-in-acid / recover-with-base cycle is how amines are

separated

from neutral organic compounds.

🎯 Exam · Acylation → amides (1° and 2° only)

R - N H_{2} + R^{'} COCl base (pyridine) R - NHCO R^{'}

. The base removes HCl and drives it forward; benzoylation with

C_{6} H_{5} COCl

gives an N-substituted benzamide.

3^{\circ}

amines have no N–H, so they cannot be acylated.

★ Remember · Carbylamine — a test for 1° amines

R - N H_{2} + CHC l_{3} + 3 KOH Δ R - NC

(foul-smelling isocyanide)

+ 3 KCl + 3 H_{2} O

. Only

primary

amines (aliphatic or aromatic) react;

2^{\circ}

and

3^{\circ}

give nothing.

Amine	Product
$1^{\circ}$ aliphatic	alcohol $+ N_{2} ↑$ (via unstable diazonium)
$1^{\circ}$ aromatic	stable arenediazonium salt (273–278 K)
$2^{\circ}$ (any)	N-nitrosamine (yellow oil)
$3^{\circ}$ (any)	soluble nitrite salt

🎯 Exam · Quantitative N₂ is useful

Because

1^{\circ}

aliphatic amines release

N_{2}

quantitatively with

HN O_{2}

, the volume of gas is used to

estimate amino acids and proteins

. The brisk

N_{2}

evolution also flags a

1^{\circ}

amine.

🚫 Examiner Trap · Examiner traps

(1)

Carbylamine

- N H_{2}

+ CHC

l_{3}

+ KOH

\to

foul isocyanide) is a test for

1^{\circ}

amines ONLY (

2^{\circ} / 3^{\circ}

give no reaction). (2) With HN

O_{2}

1^{\circ}

aliphatic

\to

alcohol + N

_{2} ↑

(unstable diazonium);

1^{\circ}

aromatic

\to

stable arenediazonium (273

-

278 K);

2^{\circ} \to

N-nitrosamine (yellow oil);

3^{\circ} \to

salt. (3) Acetylation TAMES the ring before nitration. (4) Acylation needs

1^{\circ} / 2^{\circ}

(an N

-

H to replace).

Reactions of Amines II — Hinsberg Test & Aniline Ring Substitution

Amine	Result
$1^{\circ}$	N-alkyl sulphonamide; acidic N–H ⇒ soluble in alkali
$2^{\circ}$	N,N-dialkyl sulphonamide; no N–H ⇒ insoluble in alkali
$3^{\circ}$	no reaction (no N–H to replace)

🎯 Exam · Why alkali solubility separates 1° from 2°

The sulphonyl group is strongly electron-withdrawing, so the remaining N–H in the

1^{\circ}

product is acidic ⇒ it dissolves in KOH. The

2^{\circ}

product has

no

N–H ⇒ stays insoluble. This both distinguishes and separates the three classes.

★ Remember · Aniline: –NH₂ is a strong o/p-activator

Bromination with

B r_{2}

water gives 2,4,6-tribromoaniline (white ppt) at once — too reactive to stop at mono. To get a monosubstituted product,

protect

- N H_{2}

as acetanilide (acetylation), substitute, then hydrolyse back.

⚠️ Watch out · Nitration & Friedel-Crafts traps

In acidic nitration, aniline protonates to anilinium (a

meta

-director), so

\sim

47% m-nitroaniline also forms — protect first for clean p. Aniline gives

no Friedel-Crafts

AlC l_{3}

bonds the N lone pair, leaving a deactivating

^{+} N

. Sulphonation

\to

sulphanilic acid (a zwitterion).

🚫 Examiner Trap · Examiner traps

(1)

Hinsberg

1^{\circ} \to

sulphonamide soluble in alkali (acidic N

-

H);

2^{\circ} \to

insoluble (no N

-

H);

3^{\circ} \to

no reaction — separates all three. (2) Aniline

+

r_{2}

/water can't stop at mono

\Rightarrow

2,4,6-tribromoaniline; ACETYLATE first to get p-product. (3) No Friedel

-

Crafts on aniline (AlC

l_{3}

binds the lone pair, deactivating the ring). (4) Sulphanilic acid exists as an inner salt (zwitterion).

Diazonium Salts — Preparation & Replacement of N₂

Arenediazonium salt:

Ar - N_{2}^{+} X^{-}

(X = Cl, Br,

HS O_{4}

B F_{4}

). The

- N_{2}^{+}

(diazonium) group is resonance-stabilised by the ring, so

aromatic

diazonium salts survive briefly at 273–278 K —

aliphatic

ones are far too unstable and decompose to alcohol

+ N_{2}

★ Remember · Diazotisation

C_{6} H_{5} N H_{2} + NaN O_{2} + 2 HCl 273 - 278 K C_{6} H_{5} N_{2}^{+} C l^{-} + NaCl + 2 H_{2} O

. The unstable salt is used at once; the fluoroborate is the one stable-at-RT exception.

Replacing N₂ (it is a great leaving group)

▸ $Sandmeyer$ : $CuCl/HCl \to ArCl$ ; $CuBr/HBr \to ArBr$ ; $CuCN/KCN \to ArCN$ .
▸ $Gattermann$ : Cu powder $+ HX \to ArX$ (yield $<$ Sandmeyer).
▸ $Iodide$ : $+ KI \to ArI$ (no Cu). $Fluoride$ (Balz–Schiemann): $+ HB F_{4} \to Ar N_{2} B F_{4} Δ ArF$ .
▸ $H$ (deamination): $H_{3} P O_{2}$ / $C_{2} H_{5} OH \to ArH$ . $OH$ : warm $H_{2} O \to ArOH$ . $NO_{2}$ : $NaN O_{2} /Cu$ .

🎯 Exam · Sandmeyer beats Gattermann

Both put Cl/Br on the ring, but

Sandmeyer

(

C u_{2} X_{2}

/HX) gives better yields than

Gattermann

(Cu powder/HX). For

Ar - I

no copper is needed — just warm with KI.

⚠️ Watch out · Keep it cold, use it fast

Benzenediazonium chloride decomposes if warmed (it hydrolyses to phenol at

\sim

283 K) or dried. Always prepare and react it

in situ

at 273–278 K; the only storable form is the fluoroborate.

🚫 Examiner Trap · Examiner traps

(1) Diazotisation works for AROMATIC

1^{\circ}

amines (273

-

278 K) — aliphatic diazonium is too unstable (decomposes to alcohol +

N_{2}

). (2)

Sandmeyer

needs Cu(I) salt (CuCl/CuBr/CuCN); Gattermann uses Cu powder (lower yield). (3)

-

I and

-

F on a ring are best installed via diazonium (

-

I directly with KI;

-

F via Balz

-

Schiemann HB

F_{4}

), NOT by direct halogenation. (4) Keep cold and use at once — warming hydrolyses it to phenol.

Diazonium Salts — Coupling Reactions & Synthetic Importance

★ Remember · Coupling retains the diazo group → azo dyes

Here

- N_{2}^{+}

kept

: it acts as a weak electrophile and couples at the

para

position of an activated ring.

+

phenol (mild

O H^{-}

)

\to

p-hydroxyazobenzene (orange dye);

+

aniline

\to

p-aminoazobenzene (yellow dye).

🎯 Exam · The –N=N– bridge is why they are coloured

Azo compounds join two aromatic rings through

- N = N -

, giving an extended conjugated system that absorbs visible light — the basis of the whole class of

azo dyes

Why diazonium salts are so useful

▸One $Ar N_{2}^{+}$ opens the door to $- F$ , $- Cl$ , $- Br$ , $- I$ , $- CN$ , $- OH$ , $- N O_{2}$ on the ring.
▸Aryl $- F$ and $- I$ can't be made by direct halogenation — go via diazonium.
▸ $- CN$ won't replace ring Cl, but $Ar N_{2}^{+} \to ArCN$ works (then $\to$ acid/amine).
▸Place a group, then $deaminate$ ( $H_{3} P O_{2}$ ) to reach substitution patterns blocked by direct routes.

🎯 Exam · The –NH₂ → –N₂⁺ → anything strategy

Many JEE syntheses run: aromatic compound

\to

nitro

\to

amine

\to

diazonium

\to

desired group. Mastering the replacement table lets you build aromatics that direct substitution simply cannot.

🚫 Examiner Trap · Examiner traps

(1)

Coupling RETAINS

N_{2}

(azo

- N = N -

dye); replacement reactions LOSE

N_{2}

— opposite fates of the diazonium group. (2) Coupling happens at the PARA position with electron-rich arenes (phenol, aniline). (3) Phenol coupling needs mildly alkaline medium; aniline coupling needs mildly acidic — get the pH right. (4) Azo products are coloured (extended conjugation)

\Rightarrow

used as dyes/indicators.

More JEE Main Chemistry formula sheets

Frequently Asked Questions

What are the most important Organic Compounds Containing Nitrogen formulas for JEE Main?

This Organic Compounds Containing Nitrogen formula sheet covers all the high-yield Chemistry formulas, definitions and theorems you need for JEE Main, across General methods of preparation, properties, reactions and uses, Amines: Nomenclature, classification, structure, Basic character and identification of primary, secondary and tertiary amines, Diazonium Salts: Importance in synthetic organic chemistry — each shown with the key result and, where useful, a worked example.

Is this Organic Compounds Containing Nitrogen formula sheet free?

Yes — the full chapter formula sheet is free to read online, no login or payment required.

How should I revise Organic Compounds Containing Nitrogen formulas?

Blurt the Organic Compounds Containing Nitrogen formulas from memory, then check against this sheet to find your gaps — and practise a few previous-year questions on the chapter to make sure you can apply them under time pressure.

Also useful: all formula sheets · JEE Main previous-year papers · most important chapters.