JEE Mains Chemistry · Haloalkanes and Haloarenes

Nucleophiles and Ambident Reagents

A nucleophile is stronger when it is charged, when it is the stronger base among donors of the same atom, and, in a protic solvent, when its donor atom is larger; cyanide and nitrite are ambident, so the potassium and silver salts give different products.

All Haloalkanes and Haloarenes notes JEE Mains Chemistry strategy

Why this matters

Thirteen PYQs, all multiple choice, two from 2026. Five rank nucleophiles or ask how a nucleophile changes a rate or a product; eight turn on an ambident nucleophile, cyanide or nitrite, and on whether its potassium or its silver salt was used.

Concept 1 of 2: Ranking nucleophiles: charge, basicity, size and solvent

A nucleophile gives a lone pair to carbon, so the more available that lone pair, the stronger it is. Apply three checks in order. A charged species beats its neutral form. Among donors of the same atom, the stronger base is the better nucleophile. Going down a group in a protic solvent, the larger atom wins, because small anions are held tightly by hydrogen bonds.

Definition

Charge: $\mathrm{OH^- > H_2O}$ , $\mathrm{RO^- > ROH}$ , $\mathrm{NH_2^- > NH_3}$ , $\mathrm{RS^- > RSH}$ .
Same donor atom, follow basicity: $\mathrm{RO^- > C_6H_5O^- > CH_3COO^-}$ . Resonance spreads the charge of phenoxide and acetate and makes the lone pair less available.
Protic solvent (water, alcohols): nucleophilicity rises down a group: $\mathrm{I^- > Br^- > Cl^- > F^-}$ and $\mathrm{HS^- > HO^-}$ . The small fluoride ion is caged by hydrogen bonds.
Polar aprotic solvent (DMSO, DMF, acetone): anions are not caged, and the order follows basicity: $\mathrm{F^- > Cl^- > Br^- > I^-}$ .
$\mathrm{I^-}$ speeds up the hydrolysis of an alkyl chloride: it displaces chloride quickly and then leaves quickly, being both a good nucleophile and a good leaving group.
A bulky base such as $\mathrm{(CH_3)_3CO^-}$ is strong but a poor nucleophile; it removes protons instead.

Halide nucleophilicity in protic and in aprotic solvents

\text{protic: } \mathrm{I^- > Br^- > Cl^- > F^-} \qquad \text{aprotic: } \mathrm{F^- > Cl^- > Br^- > I^-}

Worked example

Arrange for reaction with

\mathrm{CH_3CH_2CH_2Br}

in water:

\mathrm{HS^-}

\mathrm{OH^-}

\mathrm{CH_3COO^-}

\mathrm{H_2O}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 1 · Q34Moderate

Example 1 · Haloalkanes and Haloarenes · Nucleophiles and Ambident Reagents

The correct order of reactivity of

CH_{3}Br

in methanol with the following nucleophiles is

F^{-},I^{-},C_{2}H_{5}O^{-}

and

C_{6}H_{5}O^{-}

The solvent reverses the halide order

In water or an alcohol, iodide is the best halide nucleophile and fluoride the worst. In DMSO or DMF the order follows basicity and fluoride is the best. Check the solvent before ranking.

Basicity ranks only donors of the same atom

Hydroxide is a stronger base than hydrogen sulfide ion, yet

\mathrm{HS^-}

is the better nucleophile in water. Use basicity to compare two oxygen nucleophiles, not an oxygen with a sulfur.

A strong base can be a poor nucleophile

Potassium tert-butoxide is too bulky to reach a carbon from behind. It takes a β-hydrogen instead and gives an alkene.

Drill 4 more on ranking nucleophiles: charge, basicity, size and solvent

Concept 2 of 2: Ambident nucleophiles and the reagent-to-product table

Cyanide and nitrite each have two atoms that can attack carbon, so they are called ambident. The metal decides which atom is free to attack. A potassium salt gives a free anion, which attacks through carbon (cyanide) or oxygen (nitrite). In a silver salt the metal holds one end, so both silver salts bond through nitrogen.

Definition

KCN (alcoholic) → nitrile, $\mathrm{R{-}C{\equiv}N}$ . AgCN → isocyanide, $\mathrm{R{-}N{\equiv}C}$ . KCN is largely ionic; AgCN is largely covalent.
KNO₂ → alkyl nitrite, $\mathrm{R{-}O{-}N{=}O}$ . AgNO₂ → nitroalkane, $\mathrm{R{-}NO_2}$ .
A carboxylate is not ambident: its two oxygens are equivalent. Silver carboxylates give esters.
In a molecule with a ring halogen and a benzylic $\mathrm{CH_2Cl}$ , only the $\mathrm{CH_2Cl}$ reacts; the aryl C–X bonds are untouched.

Reagent with R–X	Attacking atom	Product	Class of product
Aqueous NaOH or KOH	O	$\mathrm{R{-}OH}$	Alcohol
$\mathrm{NaOR'}$	O	$\mathrm{R{-}O{-}R'}$	Ether (Williamson synthesis)
NaI in acetone	I	$\mathrm{R{-}I}$	Alkyl iodide
$\mathrm{NH_3}$	N	$\mathrm{R{-}NH_2}$ , then further alkylation	Amine
KCN (alcoholic)	C	$\mathrm{R{-}C{\equiv}N}$	Nitrile (alkyl cyanide)
AgCN	N	$\mathrm{R{-}N{\equiv}C}$	Isocyanide (isonitrile)
$\mathrm{KNO_2}$	O	$\mathrm{R{-}O{-}N{=}O}$	Alkyl nitrite
$\mathrm{AgNO_2}$	N	$\mathrm{R{-}NO_2}$	Nitroalkane
$\mathrm{R'COOAg}$	O	$\mathrm{R'COOR}$	Ester
$\mathrm{LiAlH_4}$	H (hydride)	$\mathrm{R{-}H}$	Alkane

Both silver salts bond through nitrogen: AgCN gives the isocyanide and AgNO₂ the nitroalkane.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 13 April 2023 · Q138Moderate

Example 2 · Haloalkanes and Haloarenes · Nucleophiles and Ambident Reagents

Match List I with List II. 1-Bromopropane is reacted with the reagents in List I to give the products in List II.

	List I (Reagent)		List II (Product)
(A)	$KOH$ (alc)	(I)	Nitrile
(B)	$KCN$ (alc)	(II)	Ester
(C)	$AgNO_{2}$	(III)	Alkene
(D)	$H_{3}CCOOAg$	(IV)	Nitroalkane

Choose the correct answer from the options given below:

Potassium nitrite gives the nitrite, silver nitrite the nitro compound

\mathrm{KNO_2}

→

\mathrm{R{-}O{-}N{=}O}

and

\mathrm{AgNO_2}

→

\mathrm{R{-}NO_2}

. This pair is easy to swap; remember that both silver salts bond through nitrogen.

AgCN is not ionic

KCN is largely ionic, which leaves carbon free to attack. AgCN is largely covalent, which is why it gives the isocyanide. A reason stating that both salts are highly ionic is false.

Aryl halogens survive while the side chain reacts

When a benzene ring carries both ring halogens and a

\mathrm{CH_2Cl}

group, a nucleophile such as cyanide replaces only the benzylic chlorine. Aryl C–X bonds do not undergo ordinary substitution.

Drill 7 more on ambident nucleophiles and the reagent-to-product table

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Ranking nucleophiles: charge, basicity, size and solvent
Halide nucleophilicity in protic and in aprotic solvents
$\text{protic: } \mathrm{I^- > Br^- > Cl^- > F^-} \qquad \text{aprotic: } \mathrm{F^- > Cl^- > Br^- > I^-}$

Reference tables (1)

Ambident nucleophiles and the reagent-to-product table10 rows

Reagent with R–X	Attacking atom	Product	Class of product
Aqueous NaOH or KOH	O	$\mathrm{R{-}OH}$	Alcohol
$\mathrm{NaOR'}$	O	$\mathrm{R{-}O{-}R'}$	Ether (Williamson synthesis)
NaI in acetone	I	$\mathrm{R{-}I}$	Alkyl iodide
$\mathrm{NH_3}$	N	$\mathrm{R{-}NH_2}$ , then further alkylation	Amine
KCN (alcoholic)	C	$\mathrm{R{-}C{\equiv}N}$	Nitrile (alkyl cyanide)
AgCN	N	$\mathrm{R{-}N{\equiv}C}$	Isocyanide (isonitrile)
$\mathrm{KNO_2}$	O	$\mathrm{R{-}O{-}N{=}O}$	Alkyl nitrite
$\mathrm{AgNO_2}$	N	$\mathrm{R{-}NO_2}$	Nitroalkane
$\mathrm{R'COOAg}$	O	$\mathrm{R'COOR}$	Ester
$\mathrm{LiAlH_4}$	H (hydride)	$\mathrm{R{-}H}$	Alkane

Both silver salts bond through nitrogen: AgCN gives the isocyanide and AgNO₂ the nitroalkane.

Watch out for (6)

The solvent reverses the halide order
→ Ranking nucleophiles: charge, basicity, size and solvent
Basicity ranks only donors of the same atom
→ Ranking nucleophiles: charge, basicity, size and solvent
A strong base can be a poor nucleophile
→ Ranking nucleophiles: charge, basicity, size and solvent
Potassium nitrite gives the nitrite, silver nitrite the nitro compound
→ Ambident nucleophiles and the reagent-to-product table
AgCN is not ionic
→ Ambident nucleophiles and the reagent-to-product table
Aryl halogens survive while the side chain reacts
→ Ambident nucleophiles and the reagent-to-product table

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