Identify A and B in the following reactions : (a) CH₃-CH(OH)-CH₃ xrightarrowPCl₅ A xrightarrowAgCN B (b) CH₃CH

Question

Q. Identify A and B in the following reactions :

(a)

CH_3-CH(OH)-CH_3 \xrightarrow{PCl_5} A \xrightarrow{AgCN} B

(b)

CH_3CH_2CH_2Cl \xrightarrow[ethanol]{KOH} A \xrightarrow{HBr} B

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Concept understanding — Ambident Nucleophile Reactivity

Ambident Nucleophile Reactivity

Most nucleophiles attack through a single, obvious atom — a single lone pair, a single reactive site. An ambident nucleophile is unusual: it has TWO different atoms that each carry enough electron density to act as the attacking site, so it can bond to an electrophile through either one, giving two structurally different products from the same reagent.

Why This Happens: Resonance Delocalisation

An ambident nucleophile's negative charge (or lone pair) is delocalised by resonance across more than one atom, so more than one atom is genuinely nucleophilic.

Cyanide ion, $CN^-$ : $^-C{\equiv}N: \;\leftrightarrow\; :C{=}N^-$ . Both the carbon and the nitrogen carry real electron density and can attack an electrophile.

Attack through carbon gives an alkyl cyanide (nitrile), $R{-}C{\equiv}N$ .
Attack through nitrogen gives an alkyl isocyanide (isonitrile), $R{-}N{\equiv}C$ .

Nitrite ion, $NO_2^-$ : the negative charge is shared between nitrogen and the oxygens.

Attack through oxygen gives an alkyl nitrite, $R{-}O{-}N{=}O$ .
Attack through nitrogen gives a nitroalkane, $R{-}NO_2$ .

What Decides Which End Attacks: The Counter-Ion Matters

For cyanide specifically, the identity of the metal counter-ion changes which end of $CN^-$ ends up bonded to the electrophile — this is the classic KCN-vs-AgCN contrast:

KCN is genuinely ionic: it dissociates fully to give a FREE $CN^-$ ion. The carbon end is intrinsically the more nucleophilic site (more polarisable, and it forms the stronger C–C bond with the alkyl carbon), so KCN reacts through carbon, giving the nitrile as the major product.
AgCN is covalent, with silver bonded to the CARBON of the cyanide group ( $Ag{-}C{\equiv}N$ ). With the carbon end already occupied by silver, it is the NITROGEN lone pair that is left free to attack the alkyl halide — so AgCN gives the isocyanide as the major product.

Watch out

A common mistake is to assume silver coordinates to nitrogen (since nitrogen is "more electronegative" or "harder"). It is the opposite: silver bonds to carbon, and it is precisely THAT occupation of the carbon end that forces attack to happen through nitrogen instead. …

Why this formula?

Ambident Nucleophile Reactivity: Why the Rules Hold

Ambident nucleophiles are nucleophiles that have two (or more) different atoms capable of donating a lone pair to form a bond with an electrophile. Classic examples include:

Cyanide ion ( $\ce{CN-}$ ): can attack via carbon or nitrogen
Nitrite ion ( $\ce{NO2-}$ ): can attack via oxygen or nitrogen
Enolate ions: can attack via carbon or oxygen

The key question: Why does one atom react preferentially over the other?

The Core Principle: Hard-Soft Acid-Base (HSAB) Theory

The reactivity of ambident nucleophiles is governed by HSAB theory, which states:

Hard acids prefer hard bases; soft acids prefer soft bases.

Why this holds — the reasoning:

Hard species are small, highly charged, and non-polarizable. Their interactions are dominated by ionic (electrostatic) forces.
Soft species are large, polarizable, and have diffuse electron clouds. Their interactions are dominated by covalent (orbital overlap) forces.

For an ambident nucleophile, the two attacking atoms differ in hardness/softness:

Ambident Nucleophile	Harder Atom	Softer Atom
$\ce{CN-}$	N (hard)	C (soft)
$\ce{NO2-}$	O (hard)	N (soft)
Enolate ( $\ce{CH2=CH-O-}$ )	O (hard)	C (soft)

The Key Formula(e) and Their Derivation

1. Charge Density Rule (for hard-hard interactions)

For a hard electrophile (e.g., $\ce{H+}$ , $\ce{CH3+}$ , $\ce{AlCl3}$ ):

The nucleophile attacks via the atom with higher charge density (more negative charge).

Why?

Hard-hard interactions are electrostatic. The force between charges is:

F = \frac{k \cdot q_1 \cdot q_2}{r^2}

$q_1$ , $q_2$ = charges on the species
$r$ = distance between them

A hard electrophile has a localized positive charge. The nucleophile's atom with greater negative charge density (more concentrated charge) exerts a stronger electrostatic attraction. This atom is typically the more electronegative one (e.g., O in enolate, N in cyanide).

Example:

Enolate with $\ce{CH3I}$ (hard electrophile) → O-alkylation (harder O attacks)

2. Polarizability Rule (for soft-soft interactions)

For a soft electrophile (e.g., $\ce{CH3CH2I}$ , $\ce{Hg^{2+}}$ , $\ce{Br2}$ ):

The nucleophile attacks via the atom with higher polarizability (softer atom).

Why?

Soft-soft interactions are covalent and depend on orbital overlap. The softer atom has:

Larger, more diffuse orbitals (e.g., 3p vs 2p)
Lower electronegativity
Greater polarizability — its electron cloud can distort easily to form a bond

The energy of orbital overlap is approximated by:

\Delta E \propto \frac{(\text{overlap integral})^2}{\text{energy gap}}

A softer atom has a higher-energy HOMO (closer to the electrophile's LUMO), giving a smaller energy gap and stronger interaction.

Example: …

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(a) Propan-2-ol converts to 2-chloropropane with $PCl_5$ , then to isopropyl isocyanide with $AgCN$ . (b) 1-chloropropane eliminates to propene with alcoholic $KOH$ , then adds $HBr$ via Markovnikov's rule to give 2-bromopropane.

These two sequences test your understanding of substitution, elimination, and the subtle difference between ionic and covalent reagents in nucleophilic reactions.

Reaction (a): $CH_3-CH(OH)-CH_3 \xrightarrow{PCl_5} A \xrightarrow{AgCN} B$

The starting material is propan-2-ol (isopropyl alcohol), a secondary alcohol.

Step 1: Conversion with $PCl_5$

Phosphorus pentachloride is a powerful chlorinating agent that replaces the hydroxyl group with chlorine. The mechanism involves formation of an alkyl chlorosulfite intermediate, followed by $S_N1$ or $S_N2$ displacement depending on the substrate. For a secondary alcohol, the reaction proceeds smoothly:

CH_3-CH(OH)-CH_3 + PCl_5 \longrightarrow CH_3-CHCl-CH_3 + POCl_3 + HCl

So A is 2-chloropropane (isopropyl chloride).

Step 2: Reaction with $AgCN$

Now comes the key conceptual point. Cyanide ion ( $CN^-$ ) is an ambident nucleophile—it can attack through either the carbon atom (forming a nitrile, $R-CN$ ) or the nitrogen atom (forming an isocyanide, $R-NC$ ).

The choice depends on the reagent:

Ionic cyanides like $NaCN$ or $KCN$ are "free" in solution. The carbon end is more nucleophilic (softer, more polarizable), so they predominantly give nitriles ( $R-CN$ ).
Covalent cyanides like $AgCN$ or $CuCN$ have the silver coordinated to carbon, making the nitrogen end more available for attack. These give isocyanides ( $R-NC$ ).

With $AgCN$ , the nitrogen attacks the electrophilic carbon of 2-chloropropane:

CH_3-CHCl-CH_3 + AgCN \longrightarrow CH_3-CH(NC)-CH_3 + AgCl

B is isopropyl isocyanide (2-isocyanopropane), $CH_3-CH(NC)-CH_3$ .

Tip

Remember the mnemonic: Ionic cyanide → nitrile (C-attack); Covalent cyanide → isocyanide (N-attack).

Reaction (b): $CH_3CH_2CH_2Cl \xrightarrow[ethanol]{KOH} A \xrightarrow{HBr} B$

The starting material is 1-chloropropane (n-propyl chloride), a primary alkyl halide.

Step 1: Elimination with alcoholic $KOH$

Potassium hydroxide in ethanol is a strong base in a non-polar solvent—classic conditions for elimination rather than substitution. The base abstracts a β-hydrogen, and the halide leaves, forming a double bond:

CH_3CH_2CH_2Cl + KOH_{(alc)} \longrightarrow CH_3CH=CH_2 + KCl + H_2O

According to Saytzeff's rule (Zaitsev's rule), the more substituted alkene is favored. Here, only one alkene is possible from 1-chloropropane: propene (propylene).

So A is propene, $CH_3CH=CH_2$ .

Step 2: Addition of $HBr$ to propene …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 mark
$CH_3-CH_2-Br \xrightarrow{KCN} A \xrightarrow{LiAlH_4} B$
›Reveal solutionSolution

Ethyl bromide is converted to propanenitrile by cyanide substitution, then reduced by $LiAlH_4$ to propan-1-amine — a one-carbon homologation en route to an amine.

Step 1 — formation of A

$KCN$ (predominantly ionic; $CN^-$ attacks through carbon, the less electronegative and more nucleophilic end) displaces bromide from ethyl bromide by an $S_N2$ mechanism:

$CH_3CH_2-Br + KCN \rightarrow CH_3CH_2-C\equiv N\ (A,\ \text{propanenitrile}) + KBr$

This adds a carbon atom to the chain (ethyl $\rightarrow$ propanenitrile), which is why nitrile formation followed by reduction is a standard method of chain-extending to make amines with one extra carbon than the starting haloalkane.

Step 2 — formation of B

…

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CBSE 2025Set 56/5/11 markMCQ
$Q. The treatment of ethyl bromide with alcoholic silver nitrite gives : (A) ethyl nitrite (B) nitroethane (C) nitromethane (D) ethene$
›Reveal solutionSolution
Alcoholic silver nitrite ( $\text{AgNO}_2$ ) is an ambident nucleophile — it can attack via either the oxygen or the nitrogen atom. With ethyl bromide, the major product is nitroethane (C–N bond formation) because the nitrogen centre is the more nucleophilic site in the polarisable $\text{NO}_2^-$ ion, and the silver ion helps drive the reaction via an $\text{S}_\text{N}1$ -like pathway.

Understand the reagent: ambident nucleophile

Silver nitrite ( $\text{AgNO}_2$ ) dissociates in alcohol to give $\text{Ag}^+$ and $\text{NO}_2^-$ ions. The nitrite ion has two nucleophilic sites: the nitrogen atom (with a lone pair) and the oxygen atoms (with negative charge). This is the classic example of an ambident nucleophile — it can attack an alkyl halide at either site, leading to different products.

Which site attacks?

In $\text{NO}_2^-$ , the negative charge is delocalised over the two oxygen atoms, making them harder (less polarisable). The nitrogen atom, though neutral, has a lone pair and is softer — more polarisable. Ethyl bromide is a primary alkyl halide, but the presence of $\text{Ag}^+$ changes the game. Silver ion coordinates with the bromide, weakening the C–Br bond and favouring an $\text{S}_\text{N}1$ -like mechanism (even for a primary halide) because $\text{AgBr}$ precipitates. This creates a carbocation-like transition state, which is more easily attacked by the nitrogen centre (the better nucleophile in a polarisable sense).

The major product: nitroethane

Attack by the nitrogen atom gives $\text{CH}_3\text{CH}_2\text{–NO}_2$ , which is nitroethane. This is the major product under these conditions. The oxygen attack would give ethyl nitrite ( $\text{CH}_3\text{CH}_2\text{–O–N=O}$ ), but that is a minor product here because the nitrogen centre is more nucleophilic in the ambident ion.

Why not the other options? …
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CBSE 2025Set ANNUAL1 mark
$Q. Complete the following reaction: CH3CH2Br + KCN --(aqueous ethanol)--> ?$
›Reveal solutionSolution

KCN with an alkyl halide gives the alkyl nitrile as the major product, via nucleophilic substitution through carbon.

This is a nucleophilic substitution (SN2) reaction. KCN is predominantly an ionic compound, so the cyanide ion (CN⁻, an ambident nucleophile) is free in solution. Since carbon is the more electronegative and more nucleophilic end of the CN⁻ ion, it attacks the electrophilic carbon of the alkyl halide preferentially through the carbon atom (not the nitrogen).

Reaction:

CH3CH2Br + KCN --(aq. ethanol)--> CH3CH2CN + KBr

…

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CBSE 2024Set ANNUAL1 mark
$CH_3CH_2Br + AgCN \xrightarrow{\text{Ethanol}} ?$
›Reveal solutionSolution
$AgCN$ reacts with alkyl halides through nitrogen (giving isocyanides) because silver's affinity for carbon ties up the $C$ end of the cyanide ion, exposing $N$ as the attacking site — the opposite regiochemistry to the ionic reagent $KCN$ .

The cyanide ion, $:C\equiv N:^-$ , is ambident — it can attack an electrophile through either its carbon or its nitrogen lone pair.

$KCN$ is predominantly ionic, releasing a free $CN^-$ ion in solution. Carbon is the softer, more polarizable and kinetically preferred nucleophilic site in a free cyanide ion, so $KCN$ reacts through carbon to give alkyl cyanides (nitriles). …
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CBSE 2023Set ANNUAL1 mark
$Q. How would you convert the following? Prop-1-ene to 1-nitropropane$
›Reveal solutionSolution

Peroxide-catalysed (anti-Markovnikov) addition of HBr to prop-1-ene puts $-Br$ on the terminal carbon; treating that bromide with silver nitrite then substitutes $-Br$ with $-NO_2$ (the ambident nitrite ion attacking through its more nucleophilic nitrogen atom with a covalent, largely-ionic $Ag-O$ bond), giving 1-nitropropane.

Step 1 — Anti-Markovnikov hydrobromination (Kharasch peroxide effect): In the presence of organic peroxides, addition of $HBr$ to an alkene proceeds by a free-radical chain mechanism rather than the usual ionic (Markovnikov) mechanism. The bromine radical adds first to the terminal (less substituted) carbon of the double bond, generating the more stable secondary radical at the internal carbon, and the sequence of steps places $-Br$ on the terminal carbon as the major product:

$CH_2=CH-CH_3 + HBr \xrightarrow{\text{peroxide}} CH_3-CH_2-CH_2-Br \;\;(\text{1-bromopropane, anti-Markovnikov})$

…

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CBSE 2020Set 56/1/11 mark
$Q. Read the given passage and answer the questions that follow: The substitution reaction of alkyl halide mainly occurs by SN1 or SN2 mechanism. Whatever mechanism alkyl halides follow for the substitution reaction to occur, the polarity of the carbon halogen bond is responsible for these substitution reactions. The rate of SN1 reactions are governed by the stability of carbocation whereas for SN2 reactions steric factor is the deciding factor. If the starting material is a chiral compound, we may end up with an inverted product or racemic mixture depending upon the type of mechanism followed by alkyl halide. Cleavage of ethers with HI is also governed by steric factor and stability of carbocation, which indicates that in organic chemistry, these two major factors help us in deciding the kind of product formed. Predict the stereochemistry of the product formed if an optically active alkyl halide undergoes substitution reaction by SN1 mechanism.$
›Reveal solutionSolution
An optically active alkyl halide undergoing SN1 substitution gives a racemic mixture as product because the planar carbocation intermediate allows nucleophilic attack from either face with equal probability.

The key to predicting stereochemistry in any substitution reaction lies in understanding the mechanism's intermediate. For SN1, the rate-determining step produces a carbocation — and that carbocation is planar (sp² hybridised). This flat geometry is the entire story behind the stereochemical outcome.

Let’s walk through why.

The SN1 mechanism has two distinct steps. First, the leaving group departs, creating a carbocation. This step is slow and rate-determining. Second, the nucleophile attacks this carbocation in a fast step. Because the carbocation forms before the nucleophile arrives, the nucleophile has no "memory" of which side the leaving group was on.

The carbocation intermediate is planar. A carbocation has three bonds arranged in a trigonal planar geometry (bond angles ~120°). The empty p orbital sticks out perpendicular to this plane. This means the carbocation is achiral at that carbon — it has no "handedness" left.

Nucleophilic attack can occur from either face. The nucleophile can approach the planar carbocation from above the plane or below it with equal ease. There is no steric or electronic bias (assuming the nucleophile is not itself chiral or the solvent is not chiral). So roughly half the attacks happen from one side, half from the other. …
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CBSE 2020Set NC1 mark
$Q. Write the products of the following reactions:$
(i) $C_2H_5I + KNO_2 \rightarrow$ ?
(ii) $C_2H_5I + AgCN \rightarrow$ ?
›Reveal solutionSolution

Both $NO_2^{-}$ and $CN^{-}$ are ambident nucleophiles (can attack through either of two atoms); which atom attacks depends on whether the metal salt used is predominantly ionic (favouring attack via O or C, the more electronegative/terminal end that carries more negative charge) or covalent (favouring attack via the other, more nucleophilic end).

(i) $C_2H_5I + KNO_2$ . $NO_2^{-}$ is an ambident nucleophile that can bond through either its N atom (giving a nitro compound, $R\text{–}NO_2$ ) or through an O atom (giving an alkyl nitrite, $R\text{–O–N=O}$ ). Potassium nitrite is largely ionic, so the free nitrite ion reacts as a hard/O-nucleophile preferentially, and the major product is the alkyl nitrite:

$C_2H_5I + KNO_2 \longrightarrow C_2H_5\text{–O–N=O}\ (\text{ethyl nitrite}) + KI$

…

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CBSE 2019Set 56/1/11 mark
$Q. Define ambidient nucleophile with an example.$
›Reveal solutionSolution
An ambident nucleophile is a nucleophile that has two or more different atoms with lone pairs that can act as the attacking site, leading to different products depending on the reaction conditions. A classic example is the nitrite ion ( $\text{NO}_2^-$ ), which can attack through either the nitrogen atom (forming nitro compounds) or the oxygen atom (forming nitrites).

Understanding Ambident Nucleophiles

The word "ambident" comes from Latin — ambi meaning "both" and dent meaning "tooth". So an ambident nucleophile literally has "two teeth" — two different atoms that can bite into an electrophile. This is not the same as having multiple identical attacking sites (like the two oxygens in acetate ion, which are equivalent by resonance). In an ambident nucleophile, the attacking atoms are chemically different, so the product you get depends on which atom does the attacking.

The key idea is that the nucleophile has a delocalised negative charge (or lone pair) spread over two or more different atoms. Which atom actually attacks depends on factors like:

The hardness or softness of the electrophile (HSAB principle)

The polarity of the solvent

The steric hindrance around the attacking sites

Temperature and other reaction conditions

Step-by-Step Explanation

1. Identify the defining feature of an ambident nucleophile

An ambident nucleophile must have at least two non-equivalent atoms that each possess a lone pair of electrons (or a negative charge) and can form a bond with an electrophile. The atoms are different elements, so the bond formed has different character depending on which atom attacks.

2. Understand why this matters in reactions

When an ambident nucleophile reacts with an alkyl halide or other electrophile, you can get two different products. This is called ambident reactivity. The product distribution is not random — it follows predictable patterns based on the reaction conditions.

3. Take the classic example: the nitrite ion ( $\text{NO}_2^-$ )

The nitrite ion has the following resonance structures:
$\text{O} = \text{N} - \text{O}^- \longleftrightarrow {}^-\text{O} - \text{N} = \text{O}$
The negative charge is delocalised over both oxygen atoms and the nitrogen atom. However, the nitrogen and oxygen are different elements with different properties.

4. Show the two possible attacking sites

Attack through nitrogen: The lone pair on nitrogen forms a bond with the electrophile. This gives a nitro compound ( $\text{R} - \text{NO}_2$ ).

Attack through oxygen: The lone pair on one of the oxygen atoms forms the bond. This gives an alkyl nitrite ( $\text{R} - \text{O} - \text{N} = \text{O}$ ).

$\text{R} - \text{X} + \text{NO}_2^- \rightarrow \begin{cases} \text{R} - \text{NO}_2 & \text{(nitro compound, N-attack)} \\ \text{R} - \text{O} - \text{N} = \text{O} & \text{(alkyl nitrite, O-attack)} \end{cases}$
5. Explain which product forms when — the HSAB principle in action

The nitrogen atom is a softer nucleophilic centre than the oxygen atom (nitrogen is less electronegative and its lone pair is more polarisable). Oxygen is a harder nucleophile.

With hard electrophiles (like primary alkyl halides in polar protic solvents), the harder oxygen centre tends to attack, giving alkyl nitrites. …
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Q.Identify A and B in the following reactions :

Reaction (a): CH3−CH(OH)−CH3→PCl5A→AgCNBCH_3-CH(OH)-CH_3 \xrightarrow{PCl_5} A \xrightarrow{AgCN} BCH3​−CH(OH)−CH3​PCl5​​AAgCN​B

Reaction (b): CH3CH2CH2Cl→ethanolKOHA→HBrBCH_3CH_2CH_2Cl \xrightarrow[ethanol]{KOH} A \xrightarrow{HBr} BCH3​CH2​CH2​ClKOHethanol​AHBr​B

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Reaction (a): $CH_3-CH(OH)-CH_3 \xrightarrow{PCl_5} A \xrightarrow{AgCN} B$

Reaction (b): $CH_3CH_2CH_2Cl \xrightarrow[ethanol]{KOH} A \xrightarrow{HBr} B$

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