Amongst the given set of reactants, the most appropriate for preparing 2° amine is __. (i) 2° R-Br + NH₃ (ii)

NCERT Exemplar · Q9

Q. Amongst the given set of reactants, the most appropriate for preparing 2° amine is ____.

(i) 2°

R{-}Br + NH_3

(ii) 2°

R{-}Br + NaCN

followed by

H_2

/Pt

(iii) 1°

R{-}NH_2 + RCHO

followed by

H_2

/Pt

(iv) 1°

R{-}Br

(2 mol) + potassium phthalimide followed by

H_3O^+

/heat

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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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The key idea is that reductive amination of an aldehyde with a primary amine selectively gives a secondary amine without over-alkylation. The correct option is (C).

To prepare a secondary amine cleanly, you need a method that avoids mixtures of primary, secondary, and tertiary products. Let's see why each option works or fails.

Nucleophilic Substitution Reactions — the core concept here — involve a nucleophile attacking an electrophilic carbon. When you use an alkyl halide with ammonia, the product (a primary amine) is itself a better nucleophile than ammonia. So it immediately attacks another alkyl halide molecule, giving a cascade of over-alkylation. That's the fundamental problem with direct alkylation.

Let's examine each option:

Option (A): 2° R–Br + NH₃

Ammonia attacks the secondary alkyl bromide to give a primary amine salt. But the free primary amine formed is more nucleophilic than ammonia, so it reacts with another molecule of R–Br to give a secondary amine, then a tertiary amine, and finally a quaternary ammonium salt. You end up with a mixture — not a clean route to a secondary amine.
Option (B): 2° R–Br + NaCN followed by H₂/Pt

This gives a nitrile (R–CN) which on reduction yields a primary amine (R–CH₂–NH₂). The carbon skeleton gains one extra carbon, and the product is a primary amine, not secondary. So this is outright wrong for preparing a secondary amine.
Option (C): 1° R–NH₂ + RCHO followed by H₂/Pt

This is reductive amination. The aldehyde and primary amine first form an imine (Schiff base) by condensation. Catalytic hydrogenation then reduces the C=N bond to a C–N single bond, giving a secondary amine.

The beauty: no free alkyl halide is present to cause over-alkylation. The imine intermediate is less nucleophilic than the starting amine, so further reaction is suppressed. This is the most controlled, high-yield method.
Option (D): 1° R–Br (2 mol) + potassium phthalimide followed by H₃O⁺/heat …

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Here are the common mistakes students make on this question, along with how to avoid each.

Mistake 1: Confusing the product of the Hoffmann Ammonolysis (Option A)

The Mistake: Students see $2^\circ R{-}Br + NH_3$ and think it directly gives a pure $2^\circ$ amine. They forget that $NH_3$ is a nucleophile and the product ( $1^\circ$ amine) is also a nucleophile.
Why it's wrong: The reaction doesn't stop. The $1^\circ$ amine product reacts with more alkyl halide to give a $2^\circ$ amine, then a $3^\circ$ amine, and finally a quaternary ammonium salt. You get a mixture.
How to avoid: Remember the Hoffmann Ammonolysis rule: Alkylation of ammonia always gives a mixture of $1^\circ$ , $2^\circ$ , $3^\circ$ amines, and quaternary salt. It is never a clean method for preparing a specific $2^\circ$ amine.

Mistake 2: Misidentifying the product of the Nitrile Route (Option B)

The Mistake: Students think $2^\circ R{-}Br + NaCN$ followed by reduction gives a $2^\circ$ amine.
Why it's wrong: The $2^\circ$ alkyl halide undergoes $S_N2$ with $CN^-$ to give a nitrile ( $R{-}CN$ ). Reduction of a nitrile ( $H_2$ /Pt) gives a $1^\circ$ amine ( $R{-}CH_2{-}NH_2$ ). The carbon skeleton increases by one, but the amine is still primary.
How to avoid: Trace the carbon chain. The $CN^-$ adds a carbon, but the $NH_2$ group ends up on that new carbon. The product is always $R{-}CH_2{-}NH_2$ — a $1^\circ$ amine, regardless of the starting halide's degree.

Mistake 3: Forgetting the Reductive Amination Mechanism (Option C)

The Mistake: Students see $1^\circ R{-}NH_2 + RCHO$ and think it forms a $2^\circ$ amine directly, or they confuse it with simple imine formation.
Why it's correct (and why students miss it): The $1^\circ$ amine reacts with the aldehyde to form an imine ( $R{-}N{=}CH{-}R'$ ). The $H_2$ /Pt then reduces the $C{=}N$ bond to a $C{-}N$ bond, giving a $2^\circ$ amine ( $R{-}NH{-}CH_2{-}R'$ ). This is reductive amination — the most reliable method.
How to avoid: Memorize the two-step logic:
1. Condensation: $1^\circ$ amine + aldehyde/ketone → imine.
2. Reduction: Imine + $H_2$ /catalyst → $2^\circ$ amine. This avoids over-alkylation (unlike option A).

Mistake 4: Misapplying the Gabriel Phthalimide Synthesis (Option D)

The Mistake: Students think using 2 moles of $1^\circ R{-}Br$ with potassium phthalimide gives a $2^\circ$ amine. …

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

AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
$\mathrm{CH_3CH_2CH_2Cl}$
›Reveal solutionSolution
Two independent routes correctly build n-propylamine (nitro reduction of 1-chloropropane, and nitrile chain-extension + reduction of chloroethane); the Hofmann degradation route instead shortens the chain by one carbon, giving ethylamine.

Concept and Intuition

Three very different strategies for making primary amines are being tested, and the key discipline is to count carbons through each transformation, since some routes preserve chain length and one (Hofmann bromamide degradation) deliberately removes a carbon.

$\mathrm{AgNO_2}$ vs $\mathrm{NaNO_2}$ : nitrite is an ambident nucleophile (can attack through O or N). With the softer, more covalent silver salt $\mathrm{AgNO_2}$ , alkylation occurs preferentially through nitrogen, giving the nitroalkane as the major product (with $\mathrm{NaNO_2}$ , the ionic character favours O-alkylation, giving the alkyl nitrite instead). Reducing a nitroalkane ( $\mathrm{Fe/HCl}$ ) gives the primary amine with the same carbon skeleton.

The cyanide route ( $\mathrm{NaCN}$ , then reduction) is a classic chain-extension: the nucleophilic $\mathrm{CN^-}$ displaces the halide, adding one carbon to the chain as a nitrile, and reducing the nitrile ( $\mathrm{H_2/Ni}$ or $\mathrm{LiAlH_4}$ ) gives a primary amine that is one carbon longer than the starting halide.

The Hofmann bromamide degradation ( $\mathrm{Br_2/OH^-}$ on an amide) is the opposite: it removes the carbonyl carbon (as it leaves via an isocyanate/carbamate intermediate that loses $\mathrm{CO_2}$ ), giving a primary amine with one carbon fewer than the starting amide.

Step-by-Step Solution

I: $\mathrm{CH_3CH_2CH_2Cl}$ (3 carbons) + $\mathrm{AgNO_2}$ → $\mathrm{CH_3CH_2CH_2NO_2}$ (1-nitropropane, N-alkylation favoured by the covalent Ag–N bond). + $\mathrm{Fe/HCl}$ (reduction) → $\mathrm{CH_3CH_2CH_2NH_2}$ = n-propylamine. Matches. …
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AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
$\mathrm{HCl + ZnCl_2}$
›Reveal solutionSolution
Tests why thionyl chloride, not the phosphorus halides, is the preferred reagent purely on the grounds of product purity.

Concept and Intuition

Several reagents convert $R-OH \to R-Cl$ , but "preferred for pure product" is a purity argument, not a yield argument. The reagent of choice is the one whose by-products leave the flask on their own (as gases), so no work-up/purification is needed to remove them from the alkyl chloride.

Step-by-Step Solution

$R-OH + HCl/ZnCl_2 \to R-Cl + H_2O$ (Lucas reagent) — works well only for tertiary/some secondary alcohols and leaves $ZnCl_2$ /water behind.

$R-OH + PCl_5 \to R-Cl + POCl_3 + HCl$ — the by-product $POCl_3$ is a liquid that must be removed by distillation, contaminating the crude product.

$R-OH + PCl_3 \to R-Cl + H_3PO_3$ — leaves phosphorous acid behind, again needing separation.

$3R-OH + SOCl_2 \to R-Cl + SO_2\uparrow + HCl\uparrow$ — both by-products are gases that simply bubble out, so the alkyl chloride is obtained directly in pure form with essentially no work-up. …
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AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
$Q. The reaction of benzene diazonium chloride with Cu and HCl is known as (A) Sandmeyer reaction (B) Etard reaction (C) Finkelstein reaction (D) Gattermann reaction$
›Reveal solutionSolution
This distinguishes the Sandmeyer reaction (uses cuprous salts, e.g. $Cu_2Cl_2/Cu_2Br_2$ ) from the closely related Gattermann reaction (uses copper powder directly with HCl/HBr) for converting a diazonium salt to an aryl halide.

Concept and Intuition

Both reactions replace the diazonium group ( $-N_2^+$ ) with a halogen via a radical/copper-mediated mechanism, and both give the same type of product (aryl halide + $N_2$ ). The distinguishing feature examiners test is the reagent form: Sandmeyer's reaction historically uses a cuprous halide salt (e.g. $Cu_2Cl_2$ ) generated in situ or added directly, while the Gattermann reaction is the simplified variant that uses copper powder (metallic Cu) together with the corresponding hydrohalic acid (HCl/HBr) — no cuprous salt needed.

Step-by-Step Solution

The reaction described is $ArN_2^+Cl^-$ treated with Cu (metal powder) and HCl.

Because the copper source is elemental copper powder (not a cuprous salt like $Cu_2Cl_2$ ), this specific reagent combination is named the Gattermann reaction, distinguishing it from the Sandmeyer reaction (which specifically uses cuprous chloride/cuprous bromide). …
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AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
$\mathrm{(CH_3)_2CHCH_2NH_2 + HCOOH}$
›Reveal solutionSolution
Isonitrile hydrolysis always gives the corresponding primary amine (with the SAME alkyl group as the isonitrile) plus formic acid — the alkyl group itself is never altered.

Concept and Intuition

An isocyanide, $R-N\equiv C$ , is hydrolysed under acidic aqueous conditions by nucleophilic attack of water at the terminal carbon, ultimately cleaving the C–N bond to release the alkyl group AS AN AMINE (with its nitrogen intact, unchanged connectivity to R) while the isocyanide carbon (originally bonded only to N and nothing else besides the lone pair/triple bond) ends up as formic acid, HCOOH. This is a key distinguishing test reaction of isocyanides vs their isomeric nitriles (which hydrolyse differently, to carboxylic acids + ammonia/amine on the OTHER side).

Step-by-Step Solution

Isobutyl isonitrile: $(CH_3)_2CHCH_2-N\equiv C$ — the isobutyl group ( $(CH_3)_2CHCH_2-$ ) is attached to the isocyanide nitrogen.

Acid-catalysed hydrolysis proceeds: $R-NC + 2H_2O \xrightarrow{H^+} R-NH_2 + HCOOH$ .

Applying this to isobutyl isonitrile: the alkyl group R = $(CH_3)_2CHCH_2-$ ends up as the primary amine $(CH_3)_2CHCH_2NH_2$ (isobutylamine), retaining the exact same carbon skeleton and nitrogen attachment. …
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AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
$\mathrm{CH_3CH_2CH_2CH_2OH}$
›Reveal solutionSolution
Aqueous KOH gives substitution (1-butanol); alcoholic KOH gives elimination, and since only C2 has beta-hydrogens, the only possible alkene is but-1-ene.

Concept and Intuition

Alkyl halides react differently depending on the solvent/base used: aqueous KOH (with water as solvent) favours $S_N2$ nucleophilic substitution because $\mathrm{OH^-}$ acts predominantly as a nucleophile in the polar protic medium, while alcoholic KOH (KOH dissolved in ethanol, with much less free water) favours $E2$ elimination, as the base character of $\mathrm{OH^-}$ /ethoxide dominates, abstracting a beta-hydrogen to form an alkene.

Step-by-Step Solution

Substrate: 1-chlorobutane, $\mathrm{CH_3CH_2CH_2CH_2Cl}$ -- a primary alkyl halide, Cl on C1.

Aqueous KOH: favours $S_N2$ substitution (primary halides are ideal $S_N2$ substrates -- unhindered backside attack). $\mathrm{OH^-}$ displaces $\mathrm{Cl^-}$ directly: product P = $\mathrm{CH_3CH_2CH_2CH_2OH}$ (1-butanol).

Alcoholic KOH: favours E2 elimination. The base removes a beta-hydrogen (a hydrogen on the carbon adjacent to the one bearing Cl).

In 1-chlorobutane, C1 bears Cl; the only carbon adjacent to C1 is C2 -- so the only beta-hydrogens available are those on C2. There is no other beta-carbon (C1 is a terminal/primary carbon), so there is only one possible elimination direction. …
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AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
$CH_3Cl \xrightarrow{KCN} (A) \xrightarrow{H^+/H_2O} (B)$
›Reveal solutionSolution
$CH_3Cl$ with $KCN$ forms methyl cyanide, and acidic hydrolysis of that nitrile gives acetic acid.

Concept and Intuition

Potassium cyanide, being an ionic (predominantly carbon-nucleophilic) source of $CN^-$ , substitutes alkyl halides via their carbon end to give alkyl cyanides (nitriles), $R$ – $C\equiv N$ , rather than isocyanides. Nitriles are the functional equivalent of a "masked carboxylic acid" — acidic (or basic) hydrolysis converts the $C\equiv N$ group all the way to $-COOH$ (via an amide intermediate).

Step-by-Step Solution

$CH_3Cl + KCN \rightarrow CH_3CN + KCl$ — this is $(A)$ , methyl cyanide (acetonitrile).

Acidic hydrolysis of a nitrile: $CH_3CN + 2H_2O \xrightarrow{H^+} CH_3COOH + NH_3$ (via the amide $CH_3CONH_2$ intermediate).

So $(B) = CH_3COOH$ (acetic acid).

Common Mistakes …
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Q.Amongst the given set of reactants, the most appropriate for preparing 2° amine is ____.

Concept understanding — Nucleophilic Substitution Reactions

Nucleophilic Addition – From Intuition to Precision

The Precise Statement

Why It Happens – The Driving Force

A Concrete Example – Addition of HCN to a Ketone

What Makes a Good Nucleophile?

Why this formula?

Nucleophilic Substitution Reactions: Why the Key Formulas Hold

1. The Two Main Mechanisms: A Tale of Timing

SN1: Unimolecular — The Leaving Group Goes First

SN2: Bimolecular — Simultaneous Attack and Departure

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

Concept: Preparation of Secondary Amines via Reductive Amination

Method Name: Reductive Amination

Steps:

Why other options fail:

Mistake 1: Confusing the product of the Hoffmann Ammonolysis (Option A)

Mistake 2: Misidentifying the product of the Nitrile Route (Option B)

Mistake 3: Forgetting the Reductive Amination Mechanism (Option C)

Mistake 4: Misapplying the Gabriel Phthalimide Synthesis (Option D)

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Common Mistakes …

Q.Amongst the given set of reactants, the most appropriate for preparing 2° amine is ____.

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