A primary amine, $RNH_2$ can be reacted with $CH_3{-}X$ to get secondary amine, $R{-}NHCH_3$ but the only disadvantage is that 3° amine and quaternary ammonium salts are also obtained as side products. Can you suggest a method where $RNH_2$ forms only 2° amine?

A primary amine, RNH₂ can be reacted with CH₃-X to get secondary amine, R-NHCH₃ but the only disadvantage is t

NCERT Exemplar · Q44

RNH_2

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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 to mask nitrogen's nucleophilicity while installing exactly one extra carbon, so a second alkylation can't happen. Formylating $RNH_2$ with formic acid gives the formamide $RNHCHO$ - its nitrogen lone pair is delocalised into the carbonyl, so the formamide cannot react further with $CH_3X$ . Reducing that one $C=O$ with $LiAlH_4$ then converts the formyl group into a methyl group, delivering $R-NHCH_3$ as the only product.

The problem you've described is a classic headache in organic chemistry: direct alkylation of a primary amine with an alkyl halide. When you treat $RNH_2$ with $CH_3X$ , the product $R-NHCH_3$ is itself a better nucleophile than the starting amine. So it reacts further - first to the tertiary amine $R-N(CH_3)_2$ , then to the quaternary ammonium salt $R-N^+(CH_3)_3 X^-$ . You end up with a messy mixture.

The trick is to install the extra carbon as a one-carbon acyl group first, and only then reduce it down to a methyl group - never by direct alkylation with $CH_3X$ .

Here's the step-by-step method:

Formylate the primary amine React $RNH_2$ with formic acid ( $HCOOH$ ) or methyl formate ( $HCOOCH_3$ ). This gives the N-substituted formamide:

RNH_2 + HCOOH \rightarrow RNHCHO + H_2O

The nitrogen is now part of an amide bond - its lone pair is tied up in resonance with the $C=O$ , so this formamide nitrogen is no longer a good nucleophile, and no over-substitution can happen at this stage.

Reduce the formamide with $LiAlH_4$ $LiAlH_4$ reduces the carbonyl of the formamide all the way to a methylene, turning the one-carbon $-CHO$ group into a $-CH_3$ group: $RNHCHO \xrightarrow{LiAlH_4} RNHCH_3$ …

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Concept: Selective Monoalkylation of Primary Amines

The problem is that direct alkylation of a primary amine with an alkyl halide is not selective — the product (secondary amine) is itself more nucleophilic than the starting material, so it reacts further to give tertiary amine and quaternary ammonium salt.

Method: Gabriel Phthalimide Synthesis (modified for secondary amines)

This method avoids over-alkylation by using a protected nitrogen that can only be alkylated once.

Steps:

Form the phthalimide salt Phthalimide ( $\ce{C6H4(CO)2NH}$ ) is treated with alcoholic KOH to give potassium phthalimide.

\ce{C6H4(CO)2NH + KOH -> C6H4(CO)2N^-K^+ + H2O}

Alkylate with the desired alkyl halide The potassium salt reacts with $\ce{CH3-X}$ via $\mathrm{S_N2}$ to give N-alkylphthalimide.

\ce{C6H4(CO)2N^-K^+ + CH3-X -> C6H4(CO)2N-CH3 + KX}

Hydrolyse to release the pure secondary amine The N-alkylphthalimide is hydrolysed (usually with aqueous NaOH or hydrazine) to give only the secondary amine and phthalic acid.

\ce{C6H4(CO)2N-CH3 + 2H2O ->[OH^-] C6H4(COOH)2 + CH3NH2}

Note: The product here is actually methylamine ( $\ce{CH3NH2}$ ), which is a primary amine. To get a secondary amine $\ce{RNHCH3}$ , you must start with an N-alkylphthalimide where the alkyl group is $\ce{R}$ (from step 2 using $\ce{R-X}$ ), then alkylate again? No — that would give tertiary.

Correction for your exact case:

You want only $\ce{RNHCH3}$ (secondary) from $\ce{RNH2}$ and $\ce{CH3X}$ .

The Gabriel method as described above gives primary amine after hydrolysis.

To get a secondary amine selectively, use:

Modified Gabriel — Alkylation of a pre-formed N-alkylphthalimide

First make N-alkylphthalimide from $\ce{R-X}$ (not $\ce{CH3X}$ ) and potassium phthalimide.
Then alkylate that with $\ce{CH3-X}$ — but this gives a tertiary product after hydrolysis.

So the correct method for your exact need is:

Hinsberg Test / Separation Method (not a synthesis, but a purification) …

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Here’s a breakdown of the common mistakes students make on this concept — alkylation of amines — and how to avoid each.

Common Mistake 1: Forgetting that amines are nucleophilic and will keep reacting

The error:

Students often think that once the secondary amine ( $R-NHCH_3$ ) forms, the reaction stops. In reality, the secondary amine is more nucleophilic than the primary amine, so it reacts further with $CH_3X$ to give tertiary amine and quaternary ammonium salt.

How to avoid:

Always remember: each alkylation makes the amine more electron-rich (more alkyl groups = more +I effect), so it becomes a better nucleophile. The reaction does not self-limit — you must actively prevent further alkylation.

Common Mistake 2: Suggesting “use excess $RNH_2$ ” without understanding the real problem

The error:

Students say “just take a large excess of primary amine” — but this only reduces the relative amount of side products, it does not eliminate them. Some secondary, tertiary, and quaternary products will still form.

How to avoid:

Understand that excess $RNH_2$ is a practical trick to favour monoalkylation, but it is not a perfect method. The question asks for a method that gives only secondary amine — so excess amine is not the answer here.

Common Mistake 3: Confusing the Hinsberg test with a synthetic method

The error:

Students recall that benzenesulfonyl chloride ( $C_6H_5SO_2Cl$ ) can distinguish primary, secondary, and tertiary amines, and think it can be used to synthesise pure secondary amine.

How to avoid:

The Hinsberg test is an analytical (identification) tool, not a preparative method. You cannot use it to make a secondary amine from a primary one — it forms sulfonamides, not the free amine.

Correct Method (for reference)

The standard exam answer is:

Use the carbylamine reaction (isocyanide formation) followed by reduction.

React $RNH_2$ with $CHCl_3$ and alcoholic $KOH$ to form an isocyanide ( $RNC$ ). …

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

TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
$\mathrm{C_6H_5N(CH_3)_2}$
›Reveal solutionSolution
The pair is a tertiary amine and a secondary amine. Only Hinsberg's reagent, $\mathrm{C_6H_5SO_2Cl}$ (option A), separates them: the secondary amine gives an alkali-insoluble sulphonamide, the tertiary amine does not react.

The concept first

Before choosing a reagent, classify the two compounds:

$\mathrm{C_6H_5N(CH_3)_2}$ ( $N,N$ -dimethylaniline): the nitrogen carries three carbon groups → tertiary amine, no N–H.

$\mathrm{(C_2H_5)_2NH}$ (diethylamine): the nitrogen carries two carbon groups and one H → secondary amine, one N–H.

So you need a test that distinguishes $2^\circ$ from $3^\circ$ — i.e. a test that keys off the presence of an N–H bond, not off basicity.

Hinsberg's test does exactly that. Benzenesulphonyl chloride acylates the nitrogen, but only if there is an N–H available:

$1^\circ$ amine → $\mathrm{C_6H_5SO_2NHR}$ . The N–H left over is made acidic by the two electron-withdrawing $\mathrm{S{=}O}$ groups → the sulphonamide dissolves in KOH/NaOH.

$2^\circ$ amine → $\mathrm{C_6H_5SO_2NR_2}$ . No N–H remains, so the sulphonamide is a solid that is insoluble in alkali.

$3^\circ$ amine → no reaction (no N–H to substitute); the amine remains as an oily layer, dissolving only on adding acid.

Step-by-step

Test option (D), $\mathrm{CHCl_3}$ + alc. KOH (carbylamine). It gives the foul-smelling isocyanide only with primary amines. Both compounds here are non-primary → both give a negative test → cannot distinguish. ✗

Test option (B), anhy. $\mathrm{ZnCl_2}$ + conc. HCl (Lucas reagent). This is a test for alcohols (distinguishing $1^\circ$ / $2^\circ$ / $3^\circ$ ROH by turbidity). Amines are not alcohols. ✗ …
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TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
$\mathrm{C_7H_9N}$
›Reveal solutionSolution
The alkali-insoluble benzenesulphonamide tells us Y must be a secondary amine, which fixes $\mathrm{Y}=\mathrm{C_6H_5NHCH_3}$ ; a secondary amine of this shape arises from catalytic reduction of an isocyanide, so $\mathrm{X}=\mathrm{H_2/Ni}$ . Option (D).

The concept: why alkali-solubility identifies the class of amine

When an amine reacts with a sulphonyl chloride (Hinsberg's reagent, here p-toluenesulphonyl chloride):

1° amine $\mathrm{RNH_2}$ gives $\mathrm{RNH{-}SO_2Ar}$ . The N–H left on nitrogen sits between an electron-withdrawing $\mathrm{SO_2}$ group and the ring, so it is acidic — the sulphonamide dissolves in NaOH forming a salt.

2° amine $\mathrm{R_2NH}$ gives $\mathrm{R_2N{-}SO_2Ar}$ . Nitrogen now has no hydrogen at all, nothing to ionise, so the product is insoluble in alkali.

3° amine does not react at all.

So "Z is insoluble in alkali" is a direct statement that Y is a secondary amine.

Step-by-step

Use the Hinsberg clue. Z insoluble in alkali $\Rightarrow$ Y has no N–H after sulphonylation $\Rightarrow$ Y is 2°.

Screen the options. $\mathrm{C_6H_5CH_2NH_2}$ (benzylamine) is a primary amine — its tosylamide $\mathrm{C_6H_5CH_2NH{-}SO_2C_6H_4CH_3}$ still has an acidic N–H and would dissolve in KOH. So options (A) and (C) are out. Y must be $\mathrm{C_6H_5NHCH_3}$ .

Now identify the reduction. A secondary amine bearing an $\mathrm{N{-}CH_3}$ group is the hallmark product of reducing an isocyanide (carbylamine): $\mathrm{C_6H_5{-}NC} + 4[\mathrm{H}] \xrightarrow{\ \mathrm{H_2/Ni}\ } \mathrm{C_6H_5{-}NH{-}CH_3}$ …
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TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
$\mathrm{NH_2OH}$
›Reveal solutionSolution
The question asks which pair of reagent (X) and product (Y) is correctly matched. The correct match is aniline ( $\mathrm{C_6H_5NH_2}$ ) with Schiff base, so option (C) is correct.

Concept & Intuition

This is a classic organic chemistry matching problem about carbonyl derivatives. Each reagent (X) reacts with a carbonyl compound (aldehyde or ketone) to give a specific nitrogen-containing derivative. The key is to recall the functional group of the product formed:

Oximes come from hydroxylamine ( $\mathrm{NH_2OH}$ ).

Hydrazones come from hydrazine ( $\mathrm{NH_2NH_2}$ ).

Semicarbazones come from semicarbazide ( $\mathrm{NH_2NHCONH_2}$ ).

Schiff bases (imines) come from primary amines ( $\mathrm{RNH_2}$ ), especially aromatic ones like aniline.

Let’s check each option step by step.

Option (A): $\mathrm{NH_2OH}$ → Hydrazone

Hydroxylamine ( $\mathrm{NH_2OH}$ ) reacts with a carbonyl to form an oxime (with a $\mathrm{C=NOH}$ group), not a hydrazone. Hydrazones come from hydrazine. So this is incorrect.

Option (B): $\mathrm{NH_2NH_2}$ → Semicarbazone

Hydrazine ( $\mathrm{NH_2NH_2}$ ) gives a hydrazone ( $\mathrm{C=NNH_2}$ ). Semicarbazones require semicarbazide ( $\mathrm{NH_2NHCONH_2}$ ). So this is incorrect.

Option (C): $\mathrm{C_6H_5NH_2}$ → Schiff base …
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TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
$\mathrm{CH_3{-}CO{-}NH_2} \xrightarrow[\mathrm{NaOH\ (aq)}]{\mathrm{Br_2}} \mathrm{X} \xrightarrow[\text{(ii) } \mathrm{H_2O}]{\text{(i) } \mathrm{NaNO_2 + HCl}} \mathrm{Y}$
›Reveal solutionSolution

Acetamide undergoes Hofmann bromamide degradation to methylamine (X); nitrous acid converts this primary aliphatic amine into an unstable diazonium ion that immediately loses $\ce{N2}$ and picks up water, giving methanol — option (C).

The concept first: why aliphatic diazonium salts die instantly

Diazotisation ( $\ce{NaNO2 + HCl}$ ) makes the same $\ce{-N2+}$ group from any primary amine. The difference is stability. In an aryl diazonium salt the $\ce{-N2+}$ is conjugated with the ring, so at 273–278 K it survives long enough to be used in coupling/Sandmeyer reactions. In an alkyl diazonium ion there is no such delocalisation, and $\ce{N2}$ is an outstanding leaving group — so the ion falls apart the moment it forms, giving a carbocation that the solvent (water) captures. That is why primary aliphatic amines simply effervesce and give alcohols with nitrous acid.

Step-by-step

Step 1 — Identify X. Acetamide with bromine in aqueous alkali is the textbook Hofmann bromamide degradation:

$\ce{CH3CONH2 + Br2 + 4NaOH -> CH3NH2 + Na2CO3 + 2NaBr + 2H2O}$ …
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TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
$\mathrm{CH_3CH_2CH(NH_2){-}COC_6H_5}$
›Reveal solutionSolution
A sulphonamide that is insoluble in alkali can only come from a secondary amine, so X is $\ce{CH3CH2NHCH3}$ ; benzoyl chloride then acylates its nitrogen to give $\ce{CH3CH2N(CH3)COC6H5}$ — option (D).

The concept first: why alkali-solubility fingerprints the amine class

p-Toluenesulphonyl chloride (a Hinsberg-type reagent) reacts with the amine's lone pair, replacing an $\ce{N-H}$ hydrogen with the bulky $\ce{-SO2Ar}$ group:

Primary amine $\ce{R-NH2}$ → $\ce{R-NH-SO2Ar}$ . One $\ce{N-H}$ remains. That hydrogen is acidic, because the resulting anion is stabilised by the strongly electron-withdrawing sulphonyl group. Hence the product dissolves in KOH/NaOH.

Secondary amine $\ce{R2NH}$ → $\ce{R2N-SO2Ar}$ . No $\ce{N-H}$ is left. There is no acidic proton to remove, so the product is insoluble in alkali. ✓

Tertiary amine $\ce{R3N}$ → no reaction (there is no $\ce{N-H}$ to substitute in the first place), so no product Y at all.

The stem says a product Y forms and is insoluble in alkali. Both facts together force X to be secondary.

Step-by-step

Step 1 — Classify X. Product formed ⇒ not tertiary. Product alkali-insoluble ⇒ not primary. Therefore X is a $2^\circ$ amine.

Step 2 — Benzoylation (Schotten–Baumann). Benzoyl chloride $\ce{C6H5COCl}$ acylates the amine nitrogen:
$\ce{R2NH + C6H5COCl -> R2N-CO-C6H5 + HCl}$
The amide nitrogen ends up carrying both original alkyl groups plus the benzoyl group — i.e. it is a tertiary (N,N-disubstituted) amide with no $\ce{N-H}$ .

Step 3 — Test each option against "benzoyl on a $2^\circ$ nitrogen".

(A) $\ce{CH3CH2CH(NH2)COC6H5}$ — the benzoyl is on carbon, and a free $\ce{-NH2}$ survives. Wrong on both counts. …
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TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
$\mathrm{Li_2O + NO_2}$
›Reveal solutionSolution
Lithium nitrate decomposes differently from other alkali metal nitrates because of the small size and high polarising power of Li⁺; it gives $\mathrm{Li_2O}$ , $\mathrm{NO_2}$ , and $\mathrm{O_2}$ , so the correct option is (B).

Concept & Intuition

Most alkali metal nitrates (like NaNO₃, KNO₃) decompose on strong heating to give the nitrite and oxygen:

$\mathrm{2\,MNO_3 \rightarrow 2\,MNO_2 + O_2}$ .

But lithium is an exception. Because Li⁺ is very small, it has a high charge density and strongly polarises the nitrate ion. This destabilises the nitrate, causing it to break down further — lithium nitrite (LiNO₂) itself is unstable at high temperature and decomposes to lithium oxide, nitrogen dioxide, and oxygen. So the final products are not simply nitrite + O₂, but oxide + NO₂ + O₂.

Step-by-step reasoning

General trend for alkali metal nitrates

For Na, K, Rb, Cs:

$\mathrm{2\,MNO_3 \xrightarrow{\Delta} 2\,MNO_2 + O_2}$

The nitrite is stable at the decomposition temperature.

Lithium’s anomaly

Li⁺ is much smaller than other alkali ions. Its high polarising power weakens the N–O bonds in the nitrate ion, so decomposition occurs at a lower temperature and proceeds further.

First step – formation of nitrite

Initially, LiNO₃ does form LiNO₂ and O₂:

$\mathrm{2\,LiNO_3 \rightarrow 2\,LiNO_2 + O_2}$

But LiNO₂ is not stable at the temperature needed for decomposition.

Second step – further decomposition of LiNO₂

Lithium nitrite decomposes to lithium oxide, nitrogen dioxide, and oxygen:

$\mathrm{2\,LiNO_2 \rightarrow Li_2O + NO_2 + NO}$

However, NO reacts immediately with O₂ to give NO₂: …
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