Q.Write structures of different isomers corresponding to the molecular formula, $C_3H_9N$. Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid.

Concept: Nucleophilic Substitution Reactions (specifically, the reaction of amines with nitrous acid).

Reasoning:

1. The molecular formula $C_3H_9N$ corresponds to a saturated amine (no rings or double bonds). There are four isomers: two primary amines (propan-1-amine and propan-2-amine), one secondary amine (N-methylethanamine), and one tertiary amine (N,N-dimethylmethanamine, i.e. trimethylamine).
2. When treated with nitrous acid ($HNO_2$), only primary aliphatic amines liberate nitrogen gas ($N_2$) via an unstable diazonium salt that decomposes. …

Concept First: Why Amines React with Nitrous Acid

Nitrous acid ($HNO_2$) is a versatile reagent that reacts differently with primary, secondary, and tertiary amines. The key reaction for primary aliphatic amines is diazotization — the amine is converted into an unstable alkyldiazonium salt, which immediately decomposes to give nitrogen gas ($N_2$) and a carbocation. This carbocation then reacts with water (or other nucleophiles present) to form an alcohol, an alkene, or other products.

Secondary amines form stable N-nitrosamines (yellow oils), while tertiary amines simply form salts. Neither of these liberate $N_2$ gas. So the question reduces to: which isomers of $C_3H_9N$ are primary amines?

Step-by-Step Solution

1. Determine the degree of unsaturation

For $C_3H_9N$, the saturated acyclic formula for a compound with one nitrogen is $C_nH_{2n+3}N$. Here $n=3$, so $2n+3 = 9$. The formula matches exactly — zero degrees of unsaturation. All isomers must be saturated, acyclic amines (no rings, no double/triple bonds).

2. Enumerate all structural isomers

Amines are classified by the number of carbon groups attached to nitrogen:

• Primary (1°): $R-NH_2$ — one alkyl group, two hydrogens on N.
• Secondary (2°): $R_2NH$ — two alkyl groups, one hydrogen on N.
• Tertiary (3°): $R_3N$ — three alkyl groups, no hydrogen on N.

With three carbons total, we systematically vary the carbon skeleton and the position of the nitrogen.

Isomer 1: Propan-1-amine (1°)

Straight chain, amine at the end: $CH_3CH_2CH_2NH_2$

Isomer 2: Propan-2-amine (1°)

Branched chain, amine on the middle carbon: $CH_3CH(NH_2)CH_3$

Isomer 3: N-Methylethanamine (2°)

Ethyl group + methyl group on nitrogen: $CH_3CH_2NHCH_3$

Isomer 4: N,N-Dimethylmethanamine (3°)

Three methyl groups on nitrogen: $(CH_3)_3N$

3. Identify which isomers are primary amines

From the list:

• Propan-1-amine — primary (1°) …

Method: Reaction with Nitrous Acid (HNO₂) — Distinguishing 1°, 2°, and 3° Amines

This method uses the nitrous acid test (reaction with $HNO_2$) to identify which isomers liberate $N_2$ gas.

Step 1: Write all structural isomers of $C_3H_9N$

The molecular formula $C_3H_9N$ corresponds to saturated amines (no rings/double bonds). The isomers are:

Primary amines (1°)

• Propan-1-amine: $CH_3CH_2CH_2NH_2$
• Propan-2-amine: $CH_3CH(NH_2)CH_3$

Secondary amine (2°)

• N-Methylethanamine: $CH_3CH_2NHCH_3$

Tertiary amine (3°)

• N,N-Dimethylmethanamine: $(CH_3)_3N$

Total isomers = 4

Step 2: Understand the reaction with nitrous acid

Nitrous acid ($HNO_2$) is prepared in situ by reacting $NaNO_2$ with dilute $HCl$ or $H_2SO_4$.

The key reaction that liberates $N_2$ gas is:

This happens only with primary aliphatic amines (1°).

Step 3: Identify which isomers liberate $N_2$ gas

This is a classic exam trap that tests both structural isomerism and chemical reactivity in one go. Let’s break it down.

🧪 Step 1: Understand the molecular formula

$C_3H_9N$ — this is an amine (or possibly a quaternary ammonium salt, but here we focus on neutral amines).

Degree of unsaturation = $0$ → all saturated compounds.

🧩 Step 2: All possible isomers of $C_3H_9N$

There are four structural isomers:

1. Primary amines (—NH₂ group)

• Propan-1-amine (n-propylamine) $CH_3CH_2CH_2NH_2$
• Propan-2-amine (isopropylamine) $(CH_3)_2CHNH_2$

2. Secondary amine (—NH—)

• N-Methylethanamine (ethylmethylamine) $CH_3CH_2NHCH_3$

3. Tertiary amine (—N<)

• N,N-Dimethylmethanamine (trimethylamine) $(CH_3)_3N$

⚠️ Common Mistake #1: Forgetting the tertiary amine

Mistake: Students list only three isomers (the two primary + one secondary) and miss trimethylamine.

Why it happens: They think “amine” means at least one N–H bond. But tertiary amines have no N–H and are still neutral amines.

How to avoid: Always check all three classes: 1°, 2°, and 3° amines. For $C_nH_{2n+3}N$, the number of isomers = number of ways to attach alkyl groups to N.

⚠️ Common Mistake #2: Writing wrong IUPAC names

Mistake: Calling $(CH_3)_2CHNH_2$ “isopropylamine” instead of propan-2-amine in IUPAC.

How to avoid: In IUPAC, the amine is named as alkanamine — the longest carbon chain gets the “-amine” suffix, with the position number. Common names are not accepted in many board exams unless asked.

⚠️ Common Mistake #3: Confusing the reaction with nitrous acid ($HNO_2$)

Key fact: Only primary aliphatic amines liberate $N_2$ gas with $HNO_2$.

• Primary amine: $RNH_2 + HNO_2 \rightarrow ROH + N_2 + H_2O$
• Secondary amine: gives a yellow oily N-nitrosamine (no $N_2$)
• Tertiary amine: forms a salt (no $N_2$) …