Q.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?

The key idea is to use acylation followed by reduction instead of direct alkylation. Direct alkylation with $CH_3X$ is an uncontrolled $S_N2$ process — the product $RNHCH_3$ is more nucleophilic than $RNH_2$, so it reacts further, giving tertiary amine and quaternary salt.

To install exactly one methyl group on nitrogen, use formylation (a one-carbon acyl group) followed by reduction:

Steps:

1. Formylate the primary amine with formic acid ($HCOOH$) or methyl formate ($HCOOCH_3$) to form the N-substituted formamide:

   $RNH_2 + HCOOH \rightarrow RNHCHO + H_2O$

2. Reduce the formamide with a strong reducing agent like $LiAlH_4$. This cleaves the C=O bond and adds two hydrogens, converting the formyl group ($-CHO$) into a methyl group and giving only the secondary amine:

   $RNHCHO \xrightarrow{LiAlH_4} RNHCH_3$

   (Acetylation with $CH_3COCl$ would instead introduce an ethyl group on reduction, giving $RNHCH_2CH_3$ — so the one-carbon formyl group is the right choice for an N-methyl 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:

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

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.

2. 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: $RNHCHO \xrightarrow{LiAlH_4} RNHCH_3$ …

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:

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

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

3. 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.

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

1. First make N-alkylphthalimide from $\ce{R-X}$ (not $\ce{CH3X}$) and potassium phthalimide.
2. 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) …

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 ($RNC$). …