Match the reactions given in Column I with the statements given in Column II. Column I: (i) Ammonolysis (ii) G

NCERT Exemplar · Q69

Q. Match the reactions given in Column I with the statements given in Column II. Column I:

(i) Ammonolysis

(ii) Gabriel phthalimide synthesis

(iii) Hoffmann Bromamide reaction

(iv) Carbylamine reaction
Column II:

(a) Amine with lesser number of carbon atoms

(b) Detection test for primary amines

(d) Reaction of alkyl halides with

NH_3

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Concept understanding — Nucleophilic Substitution Reactions

Nucleophilic Addition – From Intuition to Precision

Imagine you have a molecule with a carbon–oxygen double bond — a carbonyl group ( $\ce{C=O}$ ). That oxygen is greedy for electrons; it pulls them away from carbon, leaving the carbon slightly positive ( $\delta^+$ ) and the oxygen slightly negative ( $\delta^-$ ). Now, if you bring a species that is rich in electrons (a nucleophile, meaning "nucleus-loving"), it will naturally be attracted to that electron-deficient carbon. The nucleophile attacks the carbon, the $\pi$ bond breaks, and the oxygen picks up a proton (or some other electrophile) to become stable. That, in a nutshell, is nucleophilic addition.

Note

The key idea: a nucleophile adds across a polar multiple bond (usually $\ce{C=O}$ or $\ce{C≡N}$ ), breaking the $\pi$ bond and forming two new sigma bonds.

The Precise Statement

Nucleophilic addition is a reaction in which a nucleophile (an electron-rich species) forms a sigma bond with an electrophilic carbon atom of a polar multiple bond (typically a carbonyl group, $\ce{C=O}$ , or a nitrile, $\ce{C≡N}$ ), while the $\pi$ bond breaks. The resulting intermediate then captures a proton (or another electrophile) to give a neutral product.

In general form:

\ce{R2C=O + Nu^- ->[H^+] R2C(OH)Nu}

The nucleophile ( $\ce{Nu^-}$ ) attacks the carbonyl carbon; the oxygen becomes negatively charged; then a proton ( $\ce{H^+}$ ) from the medium attaches to the oxygen, yielding an alcohol.

Why It Happens – The Driving Force

The carbonyl carbon is electrophilic because:

Oxygen is more electronegative than carbon, so the $\ce{C=O}$ bond is polarised: $\ce{C^{\delta+}=O^{\delta-}}$ .
The $\pi$ bond is weaker than a $\sigma$ bond, so it can break relatively easily.

A nucleophile (like $\ce{OH^-}$ , $\ce{CN^-}$ , or $\ce{NH3}$ ) has a lone pair or a negative charge. It seeks positive centres. The attack forms a new $\sigma$ bond, and the $\pi$ electrons move entirely to oxygen, creating an alkoxide ion ( $\ce{R2C-O^-}$ ). This intermediate is then quenched by a proton.

Watch out

A common mistake: thinking the nucleophile attacks the oxygen. No — oxygen is already electron-rich; the nucleophile goes to the carbon because it is electron-deficient.

A Concrete Example – Addition of HCN to a Ketone

Take acetone ( $\ce{CH3COCH3}$ ) and hydrogen cyanide ( $\ce{HCN}$ ). In the presence of a base, $\ce{CN^-}$ (the nucleophile) attacks the carbonyl carbon:

\ce{CH3COCH3 + CN^- -> CH3C(O^-)(CN)CH3}

The alkoxide intermediate then picks up a proton from $\ce{HCN}$ (or from water) to give a cyanohydrin:

\ce{CH3C(O^-)(CN)CH3 + H^+ -> CH3C(OH)(CN)CH3}

The product is acetone cyanohydrin. Notice: two new sigma bonds formed ( $\ce{C-CN}$ and $\ce{O-H}$ ), and the $\pi$ bond is gone.

What Makes a Good Nucleophile?

Strong nucleophiles are usually negatively charged or have lone pairs:

$\ce{OH^-}$ , $\ce{CN^-}$ , $\ce{NH2^-}$ , $\ce{CH3O^-}$ , $\ce{H^-}$ (from hydride reagents like $\ce{NaBH4}$ or $\ce{LiAlH4}$ )
Neutral but polarisable: $\ce{NH3}$ , $\ce{H2O}$ (weaker, but can add under acidic conditions) …

Why this formula?

Nucleophilic Substitution Reactions: Why the Key Formulas Hold

Nucleophilic substitution reactions are a cornerstone of organic chemistry. Instead of just memorizing the rate laws, let's understand why they arise from the molecular events.

1. The Two Main Mechanisms: A Tale of Timing

The key formulas (rate laws) for nucleophilic substitution come directly from how many molecules are involved in the rate-determining step (RDS) — the slowest step that controls the overall reaction speed.

SN1: Unimolecular — The Leaving Group Goes First

The Rate Law:

\text{Rate} = k[\text{RX}]

Why?

The reaction happens in two steps:

Slow step (RDS): The C–X bond breaks spontaneously, forming a carbocation intermediate. Only the substrate ( $\text{RX}$ ) is involved.

\text{RX} \xrightarrow{\text{slow}} \text{R}^+ + \text{X}^-

Fast step: The nucleophile ( $\text{Nu}^-$ ) attacks the carbocation.

\text{R}^+ + \text{Nu}^- \xrightarrow{\text{fast}} \text{RNu}

Since the slow step depends only on the concentration of $\text{RX}$ , the rate law has no dependence on $[\text{Nu}^-]$ . The nucleophile arrives after the carbocation is formed — it cannot affect the speed of the first step.

Key insight: The rate is determined by how easily the leaving group leaves, not by how fast the nucleophile attacks.

SN2: Bimolecular — Simultaneous Attack and Departure

The Rate Law:

\text{Rate} = k[\text{RX}][\text{Nu}^-]

Why?

The reaction occurs in one concerted step:

The nucleophile attacks the carbon from one side at the same time as the leaving group departs from the opposite side.
Both $\text{RX}$ and $\text{Nu}^-$ must collide with the correct orientation and sufficient energy.

The rate depends on the frequency of productive collisions between the two molecules. This is directly proportional to the product of their concentrations:

\text{Rate} \propto [\text{RX}] \times [\text{Nu}^-]

Key insight: Both partners are involved in the transition state simultaneously — if either is missing, the reaction cannot proceed.

2. The Transition State: Why the Formulas Are Not Just "Given"

For SN2, the transition state has a pentavalent carbon (five bonds partially formed/broken). The energy barrier depends on steric hindrance — bulkier groups around carbon make it harder for the nucleophile to approach, which is why SN2 is favored at primary carbons. …

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This question tests your ability to match named reactions in organic chemistry with their correct descriptions. The key is to recall the defining feature of each reaction: Ammonolysis is the direct reaction of alkyl halides with ammonia; Gabriel phthalimide synthesis uses phthalimide to make primary amines; Hoffmann Bromamide reaction degrades an amide to an amine with one fewer carbon; Carbylamine reaction is a test for primary amines. The correct matching is (i)-(d), (ii)-(c), (iii)-(a), (iv)-(b).

Let’s go through each reaction one by one, understanding why it matches its description.

1. Ammonolysis (i) — Reaction of alkyl halides with $NH_3$

Ammonolysis is the simplest method to prepare amines. An alkyl halide ( $R-X$ ) reacts with excess ammonia ( $NH_3$ ) to give a mixture of primary, secondary, and tertiary amines, along with a quaternary ammonium salt. The primary reaction is:

R-X + NH_3 \rightarrow R-NH_2 + HX

The key point: it directly uses an alkyl halide and ammonia. So the correct match is (d).

Tip

Ammonolysis often gives a mixture because the product $R-NH_2$ is itself a nucleophile and can further react with $R-X$ . Using excess $NH_3$ favours the primary amine.

2. Gabriel phthalimide synthesis (ii) — Reaction of phthalimide with KOH and R—X

This is a method to prepare pure primary amines without contamination by secondary or tertiary amines. The steps are:

Phthalimide (a cyclic imide) is treated with alcoholic KOH to form potassium phthalimide.
This salt is then reacted with an alkyl halide ( $R-X$ ) via $S_N2$ to give N-alkylphthalimide.
Finally, hydrolysis (with aqueous NaOH or hydrazine) liberates the primary amine $R-NH_2$ .

The defining step is the reaction of phthalimide with KOH and then with $R-X$ . So the correct match is (c).

Watch out

A common mistake is to think Gabriel synthesis works for all amines. It only gives primary amines — and the alkyl halide must be primary or secondary (not tertiary, due to $S_N2$ limitations).

3. Hoffmann Bromamide reaction (iii) — Amine with lesser number of carbon atoms

The Hoffmann bromamide degradation converts an amide ( $R-CONH_2$ ) into a primary amine with one carbon atom fewer than the starting amide. The reaction uses bromine and a strong base (like NaOH):

R-CONH_2 + Br_2 + 4NaOH \rightarrow R-NH_2 + 2NaBr + Na_2CO_3 + 2H_2O

Notice: the carbonyl carbon is lost as $CO_2$ (or carbonate). So the product amine has one less carbon than the amide. This matches description (a).

R-CONH_2 \xrightarrow{Br_2/NaOH} R-NH_2

--- …

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Concept: Methods of Preparation and Detection of Amines

This question tests your understanding of named reactions in amine chemistry — specifically, which reaction produces what type of amine and which is used as a test.

Method: Reaction-to-Statement Matching by Definition

Step 1: Recall the definition of each reaction in Column I

Reaction	What it does
(i) Ammonolysis	Alkyl halide reacts with excess $NH_3$ → mixture of primary, secondary, tertiary amines & quaternary salt
(ii) Gabriel phthalimide synthesis	Phthalimide + KOH → potassium salt → reacts with R–X → hydrolysis gives pure primary amine
(iii) Hoffmann Bromamide reaction	Amide ( $RCONH_2$ ) + $Br_2$ + KOH → primary amine with one carbon less
(iv) Carbylamine reaction	Primary amine + $CHCl_3$ + KOH → foul-smelling isocyanide (test for primary amines)

Step 2: Match each with Column II statements

(i) Ammonolysis → (d) Reaction of alkyl halides with $NH_3$
(ii) Gabriel phthalimide synthesis → (c) Reaction of phthalimide with KOH and R–X …

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Here are the common mistakes students make when matching these reactions and how to avoid them.

Mistake 1: Confusing "Ammonolysis" with "Hoffmann Bromamide"

The Mistake: Students often match Ammonolysis (i) with (a) Amine with lesser number of carbon atoms. They think that because ammonia ( $NH_3$ ) has fewer carbons, the product must also have fewer carbons.
Why it's wrong: Ammonolysis is the reaction of an alkyl halide with ammonia. The product is an amine with the same number of carbon atoms as the alkyl halide (e.g., $CH_3Br \rightarrow CH_3NH_2$ ). The "lesser number of carbon atoms" refers to a reaction where the product has fewer carbons than the starting material.
How to avoid: Focus on the carbon count change.
- Ammonolysis: Starting material (alkyl halide) → Product (amine). Carbon count stays the same.
- Hoffmann Bromamide: Starting material (amide, R-CONH $_2$ ) → Product (amine, R-NH $_2$ ). Carbon count decreases by one (loss of the carbonyl carbon). This is the only reaction here that reduces the carbon chain.

Correct Match: Hoffmann Bromamide (iii) → (a) Amine with lesser number of carbon atoms.

Mistake 2: Mixing up "Gabriel Phthalimide" with "Carbylamine"

The Mistake: Students match Gabriel phthalimide synthesis (ii) with (b) Detection test for primary amines, or match Carbylamine reaction (iv) with (c) Reaction of phthalimide with KOH and R—X.
Why it's wrong:
- Gabriel phthalimide is a synthesis (preparation) of primary amines. It is not a test.
- Carbylamine reaction is a detection test (qualitative analysis) specifically for primary amines. It is not a general synthesis.
How to avoid: Remember the purpose of each reaction.
- Synthesis = Making the amine.
- Test = Identifying the amine.
- Gabriel phthalimide uses phthalimide + KOH + R-X to make a pure primary amine. This is the definition of option (c).
- Carbylamine uses a primary amine + CHCl $_3$ + KOH to give a foul-smelling isocyanide. This is the definition of option (b).

Correct Match:

Gabriel phthalimide synthesis (ii) → (c) Reaction of phthalimide with KOH and R—X.
Carbylamine reaction (iv) → (b) Detection test for primary amines.

Mistake 3: Forgetting the Reagent in "Ammonolysis" …

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Q.Match the reactions given in Column I with the statements given in Column II. Column I:

1. Ammonolysis (i) — Reaction of alkyl halides with NH3NH_3NH3​

2. Gabriel phthalimide synthesis (ii) — Reaction of phthalimide with KOH and R—X

3. Hoffmann Bromamide reaction (iii) — Amine with lesser number of carbon atoms

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Q.Match the reactions given in Column I with the statements given in Column II.
Column I:

1. Ammonolysis (i) — Reaction of alkyl halides with $NH_3$

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