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Amines · Class 12 Chemistry Notes

Class 12 Chemistry notes on Amines: structure and classification, nomenclature and isomer counting, six preparation routes and what each does to the carbon count, hydrogen bonding and boiling point, the basicity order and the three effects behind it, carbylamine, Hinsberg and nitrous acid, aniline and ring protection, the nine products of a diazonium salt, and a JEE Advanced tier on mechanisms and synthesis strategy.

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In short

Follow the nitrogen's lone pair and the chapter follows. That lone pair makes an amine both basic and nucleophilic, so basicity order, acylation, carbylamine and Hinsberg are all the same electron pair doing different jobs. In aniline the pair is delocalised into the ring, which is why aniline is a far weaker base and why the ring is protected before nitration.

Contents

·How to Read This Set
1Structure and Classification a pyramid with a lone pair
2Names and Isomers common, IUPAC, and how to count
3Preparation six routes and the carbon count
4Physical Properties hydrogen bonds decide boiling point
5Basicity three effects, one order
6Reactions of the Amino Group acylation, carbylamine, HNO₂, Hinsberg
7Aniline and the Ring too reactive, so protect it
8Diazonium Salts one salt, nine products
9Basicity Beyond the Basics JEE Advanced tier begins
10Mechanisms Hofmann, carbylamine, diazotisation
11Synthesis Strategy use N as a temporary tool
12Advanced Worked Problems five problems, five techniques
13Practice Set fifteen questions with traps explained
·Practice Set: Answers and Traps
★Amines · Fact Sheet

How to Read This Set

Sections 1 to 8 cover the full syllabus for NEET and JEE Main. Sections 9 to 12 are the JEE Advanced layer. Section 13 is a practice set across all three exams.

The whole chapter in one line

Follow the nitrogen's lone pair.

Basicity and nucleophilic reactions come from that one pair of electrons. So do the reactivity of aniline's ring and the diazonium salt.

Think it through

What this chapter is worth

It appears in every NEET and JEE paper, often as a conversion or a basicity order.
Diazonium conversions are a favourite for two-step and three-step questions.
The tests here (carbylamine, Hinsberg, nitrous acid) also come up in board practicals.

Trap alert

The habit that fixes this chapter

Before ranking or predicting, ask where the lone pair is.
Is it free on an sp³ N? Shared with a benzene ring? Shared with a C=O? Part of an aromatic ring?
That single answer settles most questions before any memorised order.

Structure and Classification

An amine is ammonia with one or more H atoms replaced by carbon groups. The nitrogen is sp³. Three corners hold bonds and the fourth holds the lone pair.

The lone pair is the business end of every amine. In aniline it leaks into the ring.

Type	General formula	Example
primary (1°)	R-NH₂	CH₃NH₂, methanamine
secondary (2°)	R₂NH	(CH₃)₂NH, N-methylmethanamine
tertiary (3°)	R₃N	(CH₃)₃N, N,N-dimethylmethanamine
quaternary salt	R₄N⁺X⁻	(CH₃)₄N⁺I⁻
aryl amine	N on a benzene ring	C₆H₅NH₂, aniline

Think it through

Why most amines are not optically active

An amine with three different groups and a lone pair looks chiral.
But the nitrogen flips inside out very fast, like an umbrella in the wind.
The two mirror forms swap too fast to separate. Quaternary salts cannot flip, so they can be resolved.

Trap alert

1°, 2°, 3° mean something different here

For amines, count the carbons on N, not on the carbon carrying N.
(CH₃)₃C-NH₂ is a primary amine, though its carbon is tertiary. For alcohols it would be called tertiary.

Names and Isomers

IUPAC names replace the final -e of the alkane with -amine. Groups on N get the prefix N-.

Formula	Common name	IUPAC name
CH₃CH₂NH₂	ethylamine	ethanamine
CH₃CH₂CH₂NH₂	n-propylamine	propan-1-amine
(CH₃)₂CHNH₂	isopropylamine	propan-2-amine
CH₃NHC₂H₅	ethylmethylamine	N-methylethanamine
(CH₃)₃N	trimethylamine	N,N-dimethylmethanamine
H₂N-CH₂CH₂-NH₂	ethylenediamine	ethane-1,2-diamine
C₆H₅NHCH₃	N-methylaniline	N-methylbenzenamine
CH₃-C₆H₄-NH₂ (para)	p-toluidine	4-methylaniline

Same atoms, four molecules. Two are primary, one secondary, one tertiary.

Worked example

How many amines have the formula C₄H₁₁N? Sort them by type.

Primary: butan-1-amine, butan-2-amine, 2-methylpropan-1-amine, 2-methylpropan-2-amine. That is 4.

Secondary: N-methylpropan-1-amine, N-methylpropan-2-amine, N-ethylethanamine. That is 3.

Tertiary: N,N-dimethylethanamine. That is 1.

Answer: 8 amines, or 9 if you count both mirror forms of butan-2-amine

Trap alert

Metamers are still isomers

N-methylpropan-1-amine and N-ethylethanamine are different molecules.
The groups on N differ in size. Such isomers are called metamers. Count each one.

Preparation

Six routes lead to amines. Choose by two things: the starting material you have, and whether the carbon count must change.

Each route keeps, adds or removes carbon. Questions on conversions turn on exactly that.

C₆H₅NO₂ + Fe/HCl (or Sn/HCl) → C₆H₅NH₂

R-X + NH₃ → RNH₂ → R₂NH → R₃N → R₄N⁺X⁻

R-X + KCN → R-C≡N →(LiAlH4)→ R-CH₂-NH₂

phthalimide →(KOH)→ N⁻K⁺ →(R-X)→ N-R →(NaOH aq)→ R-NH₂

R-CONH₂ + Br₂ + 4 NaOH → R-NH₂ + Na₂CO₃ + 2 NaBr + 2 H₂O

Route	Good for	Watch out
nitro reduction	aryl amines like aniline	Fe/HCl is preferred: the FeCl₂ formed regenerates HCl
ammonolysis of R-X	quick, cheap	gives a mixture; excess NH₃ favours 1°
nitrile reduction	adding one C to a halide	the C of CN becomes CH₂
Gabriel phthalimide	pure 1° amines	fails for aryl amines: aryl halides resist SN2
Hoffmann bromamide	dropping one C	the C=O carbon leaves as carbonate

Worked example 1

Convert ethanoic acid into methanamine.

The target has one carbon fewer. That points to the Hoffmann bromamide route.

CH₃COOH → (NH₃, heat) → CH₃CONH₂ → (Br₂, KOH) → CH₃NH₂.

Answer: make the amide, then use Br₂ and KOH

Worked example 2

Convert bromoethane into propan-1-amine.

The target has one carbon more. A cyanide adds exactly one.

CH₃CH₂Br → (KCN) → CH₃CH₂CN → (LiAlH₄) → CH₃CH₂CH₂NH₂.

Answer: KCN, then LiAlH₄

Trap alert

Gabriel cannot make aniline

The Gabriel route needs an SN2 step. An aryl halide will not do SN2, so no aryl amine forms.
For aniline and its relatives, reduce the nitro compound instead.

Physical Properties

Primary and secondary amines can hydrogen bond through their N-H bonds. Tertiary amines have none to give.

Same mass, same shape of chain. The number of N-H bonds sets the boiling point.

Property	What happens	Why
boiling point	1° > 2° > 3° for isomers	more N-H, more hydrogen bonds
against alcohols	amines boil lower	N is less electronegative than O, so N-H...N is weaker
solubility in water	small amines dissolve; it falls with size	H-bonds to water, but a larger carbon chain repels it
smell	small amines smell fishy	volatile and basic
colour of aniline	colourless when pure, darkens in air	slow oxidation in air

Trap alert

Tertiary amines still dissolve in water

A 3° amine cannot donate an H-bond, but its lone pair can accept one from water.
So trimethylamine is soluble. It just cannot H-bond to itself, so it boils low.

Basicity

An amine is a base because its lone pair can take a proton. The more available the lone pair, the stronger the base.

Measuring base strength

RNH_2+H_2O\rightleftharpoons RNH_3^{+}+OH^{-}

K_b=\dfrac{[RNH_3^+][OH^-]}{[RNH_2]}

pK_b=-\log K_b

A lower pK_b means a stronger base.

Three effects, scored for each methylamine ion. No single effect predicts the order. The balance does.

Basicity orders feel random?

Watch the three effects compete.

Add CH₃ groups one at a time. See the push grow, the hydrogen bonding fall, and the order flip from gas to water.

Play this concept in the app

Series	Order in water	pK_b values
methyl	2° > 1° > 3° > NH₃	3.27, 3.38, 4.22, 4.75
ethyl	2° > 3° > 1° > NH₃	3.00, 3.25, 3.29, 4.75
gas phase, any series	3° > 2° > 1° > NH₃	only the +I effect counts

Structures II to IV use the lone pair inside the ring. The proton finds it less available.

Aliphatic amines sit near pK_b 3 to 5. Aromatic amines sit near 9 to 13.

Group on the ring	Effect on aniline's basicity	Example
-CH₃, -OCH₃ at para	raises it: they push electrons in	p-toluidine 8.92, p-anisidine 8.66
-Cl at para	lowers it: -I wins	p-chloroaniline 10.0
-NO₂ at para	lowers it a lot: -M and -I	p-nitroaniline 13.0
-NO₂ at meta	lowers it less: only -I reaches	m-nitroaniline 11.5
CH₂ between ring and N	behaves like an aliphatic amine	benzylamine 4.70

Worked example

Arrange in increasing basicity in water: NH₃, CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, C₆H₅NH₂.

Aniline's lone pair is shared with the ring, so it is the weakest.

Among the rest, use the water order for methyl amines: 2° > 1° > 3° > NH₃.

Answer: C₆H₅NH₂ < NH₃ < (CH₃)₃N < CH₃NH₂ < (CH₃)₂NH

Think it through

Salts let you separate amines

Aniline does not dissolve in water. It does dissolve in dilute HCl, as anilinium chloride.
Add NaOH and the free aniline comes back out.
This is how amines are pulled out of a mixture with neutral compounds.

Trap alert

Water order and gas order differ

Gas phase: 3° > 2° > 1°. In water: 2° > 1° > 3° for methyl. Read which the question asks.
Do not use the methyl order for ethyl amines. With ethyl groups, 3° beats 1°.

Reactions of the Amino Group

The lone pair makes N a nucleophile. Most reactions here start with N attacking something.

Acylation: C₂H₅NH₂ + CH₃COCl →(pyridine)→ C₂H₅NHCOCH₃ + HCl

Benzoylation: C₆H₅NH₂ + C₆H₅COCl →(NaOH)→ C₆H₅NHCOC₆H₅ + HCl

Carbylamine: R-NH₂ + CHCl₃ + 3 KOH →(heat)→ R-N≡C + 3 KCl + 3 H₂O

Acylation needs an N-H, so it works for 1° and 2° amines only. Pyridine soaks up the HCl and pushes the reaction forward. The carbylamine test gives a foul-smelling isocyanide. Only primary amines give it.

Amine	With nitrous acid (NaNO₂ + HCl)	What you see
1° aliphatic	RNH₂ → ROH + N₂	bubbles of N₂
1° aromatic	C₆H₅NH₂ → C₆H₅N₂⁺Cl⁻ at 0 to 5 °C	no gas when cold; the salt couples to dyes
2°, either kind	R₂NH → R₂N-N=O	yellow oily nitrosamine
3° aliphatic	forms a soluble salt	no visible change
3° aromatic	C₆H₅N(CH₃)₂ → para-nitroso product	green solid

One reagent and one base give three different outcomes. The difference is the N-H left behind.

Worked example

Three bottles hold aniline, N-methylaniline and N,N-dimethylaniline. Tell them apart with nitrous acid.

Aniline is 1° aromatic. Cold HNO₂ gives a diazonium salt. Added to alkaline 2-naphthol, it gives a red dye.

N-methylaniline is 2°. It gives a yellow oily N-nitroso compound.

N,N-dimethylaniline is 3° aromatic. It gives a green p-nitroso compound.

Answer: red dye, yellow oil, green solid

Trap alert

Two tests, two traps

Only 1° amines give the carbylamine smell. A 2° amine gives nothing, not a weaker smell.
In the Hinsberg test, 3° amines do not react at all. They do dissolve in acid, which tells them apart from 2° products.

Aniline and the Ring

The -NH₂ group pushes its lone pair into the ring. That makes aniline react with electrophiles very fast, at ortho and para. Often too fast.

C₆H₅NH₂ + 3 Br₂ (water) → 2,4,6-tribromoaniline (white precipitate) + 3 HBr

Acetylation slows the ring down. After the substitution, the acetyl group comes off and NH₂ returns.

In strong acid, aniline and anilinium compete. Each directs to different positions.

Reaction	What happens	Reason
bromine water	2,4,6-tribromoaniline at once	the ring is strongly activated
nitration, direct	51% para, 47% meta, 2% ortho; some oxidation	anilinium ion forms and directs meta
sulphonation	sulphanilic acid on heating at 453 to 473 K	forms a zwitterion, H₃N⁺-C₆H₄-SO₃⁻
Friedel-Crafts	does not work	N's lone pair binds AlCl₃, and the ring is deactivated

Worked example

Make p-nitroaniline from aniline in good yield.

Direct nitration gives a mixture and oxidation. So protect N first.

Aniline → ((CH₃CO)₂O) → acetanilide → (conc. HNO₃, H₂SO₄) → p-nitroacetanilide → (H⁺, heat) → p-nitroaniline.

Answer: acetylate, nitrate, then hydrolyse

Trap alert

Friedel-Crafts fails, and so does direct nitration

Aniline is a Lewis base. It grabs AlCl₃ and turns into a deactivated salt.
In nitrating acid it is protonated. Expect lots of meta product, not pure para.

Diazonium Salts

A primary aromatic amine with cold nitrous acid gives a diazonium salt. This one salt can be turned into almost anything.

C₆H₅NH₂ + NaNO₂ + 2 HCl →(273 to 278 K)→ C₆H₅N₂⁺Cl⁻ + NaCl + 2 H₂O

Aryl diazonium ions are stable in the cold because the positive charge spreads into the ring. Alkyl diazonium ions have no ring to help, so they fall apart at once to N₂ and a carbocation.

Eight routes replace the N₂ group. In coupling, the N=N stays and joins two rings.

Too many reagents to remember?

Spin the diazonium wheel.

Pick a product and the reagent lights up. Then try it the other way round, from reagent to product.

Play this concept in the app

Too acidic and the partner loses its push. Too basic and the diazonium salt is destroyed.

C₆H₅N₂⁺Cl⁻ + C₆H₅OH →(OH⁻)→ p-hydroxyazobenzene (orange)

C₆H₅N₂⁺Cl⁻ + C₆H₅NH₂ →(mild acid)→ p-aminoazobenzene (yellow)

Worked example 1

Make fluorobenzene from benzene.

Fluorine cannot be put on directly. Go through the diazonium salt.

Benzene → (HNO₃, H₂SO₄) → nitrobenzene → (Sn, HCl) → aniline → (NaNO₂, HCl, 0 °C) → diazonium salt → (HBF₄, heat) → fluorobenzene.

Answer: four steps, ending with the Balz-Schiemann reaction

Worked example 2

Make benzoic acid from aniline.

Diazotise, then swap N₂ for CN with CuCN. Then hydrolyse the nitrile.

C₆H₅NH₂ → C₆H₅N₂⁺ → (CuCN/KCN) → C₆H₅CN → (H₃O⁺, heat) → C₆H₅COOH.

Answer: diazotise, CuCN, then hydrolyse

Trap alert

Three diazonium slips

Diazotise at 0 to 5 °C. Warm, the salt turns into phenol before you can use it.
Iodine needs only KI. Fluorine needs HBF₄ and heat. Neither is a Sandmeyer reaction.
In coupling, the N₂ is not lost. The product keeps -N=N-.

Tier 2

The JEE Advanced layer

Everything so far is complete for NEET and JEE Main. From here each problem needs a route you build yourself. No single formula reaches the answer.

Basicity Beyond the Basics

JEE Advanced mixes nitrogen types that the simple rules do not cover. One question settles all of them: where does the lone pair sit, and is it busy?

Same atom, four situations. The pK_b spans from 0.4 to almost no basicity at all.

Effect	What it does	Example
more s-character	lone pair held closer to N, so less basic	pyridine (sp²) 8.75 against piperidine (sp³) 2.88
lone pair in an aromatic ring	not available at all	pyrrole is almost non-basic
lone pair on C=O	shared with oxygen	amides are almost neutral
cation spread over 3 N	the ion is very stable	guanidine, pK_b about 0.4
ortho effect	any ortho group weakens an aniline	o-toluidine 9.56, weaker than aniline 9.38
steric inhibition of resonance	ortho groups twist NMe₂ out of the ring plane	N,N-dimethyl-o-toluidine is stronger than N,N-dimethylaniline

Worked example

Arrange in increasing basicity: acetamide, aniline, pyridine, piperidine, guanidine.

Acetamide's lone pair is shared with C=O. It is the weakest.

Aniline (9.38) is weaker than pyridine (8.75). Pyridine's sp² lone pair is free; aniline's is shared with the ring.

Piperidine (2.88) is a normal sp³ amine. Guanidine (0.4) forms a cation spread over three N atoms.

Answer: acetamide < aniline < pyridine < piperidine < guanidine

Trap alert

Pyridine is stronger than aniline

Most students rank pyridine last because of sp². But aniline's lone pair is busy in the ring.
Pyridine's lone pair points out, in the plane, and is not part of the aromatic sextet.

Mechanisms

JEE Advanced asks why a product forms, not just what forms. Four mechanisms cover this chapter.

R moves from carbon to nitrogen in one step. That is why a chiral R keeps its configuration.

Reaction	Key intermediate	What it explains
Hofmann bromamide	isocyanate, R-N=C=O	one C lost; configuration kept
carbylamine	dichlorocarbene, :CCl₂, from CHCl₃ and base	only 1° amines can lose two H to form an isocyanide, R-NC
diazotisation	nitrosonium ion, NO⁺, from HNO₂ and acid	needs N-H bonds to lose, so only 1° gives N₂⁺
Gabriel	phthalimide anion, an N nucleophile	SN2, so it fails with aryl and 3° halides

A bulky, positive leaving group sends the base to the most accessible hydrogen.

Worked example

N,N,N-trimethylpentan-2-aminium hydroxide is heated. Which alkene forms mainly? What would 2-bromopentane give with ethoxide?

The leaving group is the bulky -N(CH₃)₃⁺. The base takes H from the CH₃ end, C1, which is easiest to reach.

2-bromopentane with ethoxide follows Saytzeff and gives mostly pent-2-ene.

Answer: pent-1-ene from the ammonium salt; pent-2-ene from the bromide

Trap alert

Hofmann rule versus Saytzeff rule

A quaternary ammonium leaving group gives the less substituted alkene.
A halide leaving group with a small base gives the more substituted alkene. Same skeleton, opposite answer.

Synthesis Strategy

In ring synthesis, the nitrogen group is a tool. You put it on to steer, then change it or remove it.

Tool	Use
-NH₂, then remove with H₃PO₂	a strong ortho/para director that can vanish afterwards
-NO₂, then reduce	a meta director that later becomes NH₂
-NHCOCH₃	a milder ortho/para director; gives one substitution, at para
the diazonium salt	places F, I, CN, OH or H where no direct reaction can

Worked example 1

Make m-bromophenol from nitrobenzene.

OH and Br must be meta. NO₂ is a meta director, so brominate first.

Nitrobenzene → (Br₂, FeBr₃) → m-bromonitrobenzene → (Sn, HCl) → m-bromoaniline.

→ (NaNO₂, HCl, 0 °C) → diazonium salt → (H₂O, warm) → m-bromophenol.

Answer: brominate, reduce, diazotise, then warm with water

Worked example 2

Make 1,3,5-tribromobenzene from aniline.

The three Br atoms sit meta to each other. No single group directs like that.

But NH₂ sends Br to 2, 4 and 6, which are meta to each other. Then remove the NH₂.

Aniline → (Br₂ water) → 2,4,6-tribromoaniline → (NaNO₂, HCl, 0 °C) → (H₃PO₂) → 1,3,5-tribromobenzene.

Answer: brominate, diazotise, then remove N with H₃PO₂

Trap alert

Order matters

Decide the order by which group directs the next step. Swap two steps and the product changes.
In Example 1, reducing first would make NH₂ the director, and Br would go ortho and para.

Advanced Worked Problems

Each problem uses a different technique. None of them can be solved by recalling one reaction.

Problem 1 · mechanism and stereochemistry

(S)-2-methylbutanamide is treated with Br₂ and KOH. Name the product, with its configuration.

What it demands: knowing the R group migrates with retention, then checking that the CIP label does not change

Hofmann bromamide removes the C=O carbon. The group CH₃CH₂CH(CH₃)- moves from C to N. So the product is butan-2-amine.

The group moves with its bonds intact, so the arrangement in space is kept.

Now re-check the label. Before, -CONH₂ ranked first at the stereocentre. After, -NH₂ ranks first, in the same place. The other three groups keep their ranks.

Answer: (S)-butan-2-amine

Problem 2 · two chapters

0.59 g of a saturated primary amine gives 224 mL of N₂ at STP with nitrous acid. The alcohol formed gives a yellow precipitate with I₂ and NaOH. Identify the amine. Assume one N₂ per NH₂ group.

What it demands: mole counting from a gas volume, then an alcohol test to pick the isomer

Moles of N₂ $=\dfrac{224}{22400}=0.01$ . One amine gives one N₂, so the molar mass is $\dfrac{0.59}{0.01}=59$ g/mol.

A saturated amine C_nH_2n+3N with mass 59 has n = 3. So it is propan-1-amine or propan-2-amine.

The iodoform test needs a CH₃-CH(OH)- group. Only propan-2-ol has it.

Answer: propan-2-amine

Problem 3 · deduction

An aromatic amine A, C₇H₉N, is diazotised in the cold. The product gives a red dye with alkaline 2-naphthol. With bromine water, A gives C₇H₆Br₃N. Identify A.

What it demands: using the dye test to fix the amine type, then turning a bromine count into a position count

A stable diazonium salt means A is a primary aromatic amine. That rules out benzylamine and N-methylaniline. So A is a toluidine.

Three Br atoms need three free positions ortho and para to NH₂: 2, 4 and 6.

In o-toluidine CH₃ blocks 2. In p-toluidine it blocks 4. Only in m-toluidine are all three free.

Answer: m-toluidine (3-methylaniline)

Each round of methylation and elimination breaks one C-N bond. Two rounds free the nitrogen.

Problem 4 · working backwards

Amine X, C₅H₁₁N, has no C=C. It takes 2 mol of CH₃I to become a quaternary salt. Two rounds of Hofmann elimination, with one more CH₃I between them, give penta-1,4-diene and trimethylamine. Identify X.

What it demands: counting CH3I to find how many C-N bonds N had, then rebuilding the ring from the product

C₅H₁₁N has one degree of unsaturation and no C=C. So X has one ring.

Two CH₃ groups to reach N⁺ means N had two C-N bonds. X is a secondary amine.

All five carbons end up in one unbranched chain with a C=C at each end. So they formed a ring, joined to N at both ends.

Answer: piperidine

Problem 5 · design a separation

A mixture holds aniline, N-methylaniline and N,N-dimethylaniline. Plan how to get each one back, pure.

What it demands: using the acidity of one product and the basicity of another to split three similar compounds

Add benzenesulphonyl chloride and aqueous KOH. Aniline's sulphonamide still has an acidic N-H, so it dissolves as a salt. N-methylaniline's sulphonamide has no N-H and stays as a solid. N,N-dimethylaniline does not react.

Separate the alkaline solution and acidify it. Aniline's sulphonamide comes out. Heat it with acid to get aniline back.

Shake the rest with dilute HCl. N,N-dimethylaniline dissolves as its salt; NaOH frees it again. The solid left is N-methylaniline's sulphonamide; acid hydrolysis frees it.

Answer: Hinsberg reagent, then acid for the 3° amine, then hydrolyse each sulphonamide

Practice Set

Questions 1 to 9 are NEET and JEE Main level. Questions 10 to 15 are JEE Advanced, in its real formats. Try each one before reading the answers.

NEET and JEE Main

Q1NEET · Level 2 · single correct

The correct order of basic strength in water is:

(A) (CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃(B) CH₃NH₂ > (CH₃)₂NH > (CH₃)₃N > NH₃(C) (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃(D) NH₃ > CH₃NH₂ > (CH₃)₂NH > (CH₃)₃N

Q2JEE Main · Level 2 · single correct

Which pair can be told apart by the carbylamine test?

(A) ethylamine and aniline(B) methylamine and dimethylamine(C) aniline and benzylamine(D) propan-1-amine and propan-2-amine

Q3JEE Main · Level 2 · single correct

Which amide gives propan-2-amine with Br₂ and KOH?

(A) 2-methylpropanamide(B) butanamide(C) propanamide(D) 2-methylbutanamide

Q4JEE Main · Level 2 · single correct

An amine C₄H₁₁N gives a Hinsberg product that does not dissolve in KOH. The amine could be:

(A) butan-1-amine(B) 2-methylpropan-2-amine(C) N,N-dimethylethanamine(D) N-ethylethanamine

Q5NEET · Level 2 · single correct

Direct nitration of aniline gives about 47% meta product because:

(A) -NH₂ is a meta director(B) the anilinium ion formed in acid is a meta director(C) the ortho positions are blocked by bulk(D) NO₂⁺ attacks N first, then moves to meta

Q6JEE Main · Level 2 · numerical

Methylamine has pK_b = 3.38. Find the pH of a 0.1 M solution at 25 °C. Give 10 × pH, rounded off to the nearest integer.

Q7NEET · Level 2 · single correct

The Gabriel phthalimide synthesis can be used to make:

(A) aniline and benzylamine(B) ethylamine and N-methylethanamine(C) benzylamine and ethylamine(D) aniline and ethylamine

Q8JEE Main · Level 2 · single correct

Which sequence turns aniline into benzonitrile?

(A) NaNO₂/HCl at 0 to 5 °C, then CuCN/KCN(B) CuCN/KCN directly(C) NaNO₂/HCl at 40 °C, then CuCN/KCN(D) NaNO₂/HCl at 0 to 5 °C, then warm water

Q9NEET · Level 2 · single correct

Benzenediazonium chloride couples best with phenol at:

(A) pH 1 to 2(B) pH 4 to 5(C) pH 13 to 14(D) pH 9 to 10

JEE Advanced

Q10JEE Advanced · Level 3 · one or more correct

Which of the following are correct?

(A) pyrrolidine is a stronger base than pyrrole(B) p-anisidine is a stronger base than aniline(C) o-toluidine is a stronger base than p-toluidine(D) guanidine is a stronger base than ethylamine

Q11JEE Advanced · Level 3 · numerical

8.7 g of butanamide is fully converted to propan-1-amine by Br₂ and KOH. What mass of KOH, in grams, does the balanced reaction use? Round off to one decimal place. (K = 39, O = 16, H = 1)

Q12JEE Advanced · Level 3 · single correct

Aniline is treated in turn with (CH₃CO)₂O; Br₂ in CH₃COOH; H₃O⁺ and heat; NaNO₂/HCl at 0 °C; then H₃PO₂. The final product is:

(A) bromobenzene(B) 1,4-dibromobenzene(C) 1,3,5-tribromobenzene(D) 4-bromophenol

Q13JEE Advanced · Level 3 · single correct

An aromatic amine C₈H₁₁N does not react with benzenesulphonyl chloride. With nitrous acid it gives a green solid. It is:

(A) N-ethylaniline(B) N,N-dimethylaniline(C) 2,4-dimethylaniline(D) N-methylbenzylamine

Q14JEE Advanced · Level 3 · one or more correct

About coupling of benzenediazonium chloride, which of the following are correct?

(A) With phenol it is done in mild alkali, because phenoxide is more strongly activated.(B) With aniline it is done at pH 12, so that the amine stays free.(C) p-cresol still couples, at the position ortho to its OH group.(D) The diazonium ion acts as the electrophile.

Q15JEE Advanced · Level 3 · numerical

Benzylamine (pK_b 4.70) and aniline (pK_b 9.38) are each made 0.1 M. Find log₁₀ of [OH⁻]_benzylamine / [OH⁻]_aniline, to two decimal places.

Practice Set: Answers and Traps

Each answer shows the reasoning, then what each wrong option gets wrong. If you chose a wrong option, find your mistake there.

Q1 · (C). In water, +I, hydrogen bonding of the ion and crowding all count. For methyl amines the balance gives 2° > 1° > 3° > NH₃. Traps: (A) is the gas-phase order, +I alone. (B) counts only hydrogen bonding. (D) counts only crowding.

Q2 · (B). Only 1° amines give the isocyanide smell. Methylamine gives it; dimethylamine does not. Traps: (A), (C) and (D) are pairs of 1° amines. Both give the smell, so the test cannot split them.

Q3 · (A). Hoffmann removes the C=O carbon. Putting it back on propan-2-amine's N gives (CH₃)₂CH-CONH₂. Traps: (B) gives propan-1-amine. (C) gives ethylamine. (D) gives butan-2-amine.

Q4 · (D). A sulphonamide that will not dissolve in KOH has no N-H. So the amine is 2°. N-ethylethanamine is the only 2° amine listed. Traps: (A) and (B) are 1°: their products dissolve. (C) is 3°: it gives no product at all.

Q5 · (B). Strong acid protonates much of the aniline. The -NH₃⁺ group has no lone pair and pulls electrons, so it directs meta. Traps: (A) is false: NH₂ directs ortho and para. (C) explains the small ortho share, not the meta share. (D) invents a mechanism.

Q6 · 118. For a weak base, pOH = ½(pK_b - log C) = ½(3.38 + 1) = 2.19. So pH = 14 - 2.19 = 11.81. Traps: Using pK_b as pOH gives 10.62. Forgetting the ½ gives 9.62.

Q7 · (C). Gabriel needs SN2 on an alkyl halide and gives only 1° amines. Benzylamine and ethylamine both qualify. Traps: (A) and (D) include aniline: aryl halides do not do SN2. (B) includes a 2° amine, which Gabriel cannot make.

Q8 · (A). First make the diazonium salt in the cold. Then CuCN swaps N₂ for CN. Traps: (B) skips diazotisation. (C) is too warm: the salt becomes phenol. (D) makes phenol on purpose.

Q9 · (D). Phenol must be ionised to phenoxide, which needs mild alkali. Too much alkali destroys the diazonium salt. Traps: (B) is aniline's window. (A) is too acidic for phenol to react. (C) turns the salt into an unreactive diazotate.

Q10 · (A), (B) and (D). Pyrrolidine has a free sp³ lone pair; pyrrole's is in the ring. OCH₃ at para pushes electrons in (8.66 against 9.38). Guanidine (0.4) is far stronger than ethylamine (3.29). Traps: (C) is false: the ortho effect makes o-toluidine (9.56) weaker than p-toluidine (8.92).

Q11 · 22.4. RCONH₂ + Br₂ + 4 KOH → RNH₂ + K₂CO₃ + 2 KBr + 2 H₂O. Butanamide is 87 g/mol, so 8.7 g is 0.1 mol. KOH used = 0.4 mol × 56 g/mol. Traps: Using 1 KOH per amide gives 5.6. Using 2 gives 11.2. The carbonate and the two KBr need all four.

Q12 · (A). The acetyl group lets one Br in, at para. Hydrolysis gives p-bromoaniline. Diazotising and H₃PO₂ then remove the N. Traps: (B) forgets that H₃PO₂ gives H, not Br. (C) assumes the ring was not protected. (D) assumes water replaced N₂.

Q13 · (B). No reaction with the Hinsberg reagent means 3°. A green p-nitroso product means 3° aromatic. C₆H₅N(CH₃)₂ fits C₈H₁₁N. Traps: (A) and (D) are 2°: they react, and give yellow oils with HNO₂. (C) is 1°: it would form a diazonium salt.

Q14 · (A), (C) and (D). Phenoxide is a stronger electron donor than phenol. If para is blocked, as in p-cresol, coupling goes ortho. The diazonium ion is the electrophile. Traps: (B) is false: at pH 12 the diazonium salt is destroyed. Aniline couples in mildly acidic solution.

Q15 · 2.34. For each weak base, [OH⁻] = √(K_bC). At equal C, the log of the ratio is ½(9.38 - 4.70). Traps: Forgetting the square root gives 4.68.

★ Amines · Fact Sheet

Every rule for revision day. Print this page alone.

STRUCTURE

sp³ N, pyramid, about 108°

1°, 2°, 3° = C atoms on N.
(CH₃)₃C-NH₂ is 1°.

PREPARATION

Nitro + Fe/HCl: aniline

Gabriel: pure 1°, no aryl.
Hoffmann: one C lost.

BOILING POINT

1° > 2° > 3° for isomers

Amines boil below alcohols.
3° still dissolves in water.

BASICITY IN WATER

Methyl: 2° > 1° > 3° > NH₃

Ethyl: 2° > 3° > 1° > NH₃
Gas phase: 3° > 2° > 1°

AROMATIC AMINES

Aniline pK_b 9.38

Ring donors raise basicity.
NO₂ at para lowers it most.

CARBYLAMINE

1° only: R-NH₂ + CHCl₃ + KOH

Foul isocyanide smell.
2° and 3° give nothing.

NITROUS ACID

1° aliphatic: N₂ gas

1° aromatic: diazonium, 0 to 5 °C
2°: yellow oil. 3° aryl: green.

HINSBERG

1°: dissolves in KOH

2°: insoluble solid
3°: no reaction

ANILINE RING

Br₂ water: 2,4,6-tribromo

Nitration: 47% meta
No Friedel-Crafts. Protect N.

DIAZONIUM

CuCl, CuBr, CuCN: Sandmeyer

KI: iodo. HBF₄, heat: fluoro
H₃PO₂: H. Warm water: OH.

COUPLING

Phenol: pH 9 to 10

Aniline: pH 4 to 5
Para first; ortho if para is taken.

LONE PAIR SITE

sp³ > sp²: piperidine > pyridine

Pyrrole: almost non-basic
Guanidine: strongest, pK_b 0.4

ORTHO EFFECT

Any ortho group weakens aniline.

o-toluidine 9.56 > aniline 9.38
Twisted NMe₂ gets stronger.

HOFMANN BROMAMIDE

R moves C to N, kept intact.

Uses 4 KOH per amide.
Isocyanate in the middle.

HOFMANN ELIMINATION

Least substituted alkene.

One CH₃I round per C-N bond.
Opposite of Saytzeff.

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