Q.(i) Write structures of different isomeric amines corresponding to the molecular formula, $C_4H_{11}N$.

The key idea is that amines with the formula $C_4H_{11}N$ can be primary (1°), secondary (2°), or tertiary (3°), leading to chain, position, and functional group isomerism.

Step 1 — Identify all possible skeletons.

A four-carbon chain can be straight (butane) or branched (isobutane). The nitrogen can attach at different carbons or be inserted between carbons.

Step 2 — List all distinct isomers.

• Primary amines (—NH₂): Butan-1-amine (1-aminobutane) Butan-2-amine (2-aminobutane) 2-Methylpropan-1-amine (isobutylamine) 2-Methylpropan-2-amine (tert-butylamine)
• Secondary amines (—NH—): N-Methylpropan-1-amine N-Methylpropan-2-amine N-Ethylethanamine (diethylamine)
• Tertiary amine (—N<): N,N-Dimethylethanamine (N-ethyldimethylamine)

Step 3 — Identify isomerism types. …

1. The concept: What does $C_4H_{11}N$ tell us?

A saturated acyclic hydrocarbon with 4 carbons would be $C_4H_{10}$. Replacing one $H$ with an $NH_2$ group adds one $N$ and removes one $H$, giving $C_4H_{11}N$. So the molecule is a saturated amine — no rings, no $\pi$ bonds. The nitrogen can be primary ($1^\circ$), secondary ($2^\circ$), or tertiary ($3^\circ$), depending on how many carbon groups are attached to it.

The key is to systematically vary the carbon skeleton and the position of the nitrogen.

2. Step-by-step enumeration of all isomers

Step 1: Primary amines ($1^\circ$) — $R-NH_2$

Here the $NH_2$ group is at the end of a carbon chain. The carbon skeleton can be:

• Butane skeleton (straight chain): $CH_3CH_2CH_2CH_2NH_2$ — butan-1-amine

  Also, $CH_3CH_2CH(NH_2)CH_3$ — butan-2-amine (the $NH_2$ on carbon 2)

• Isobutane skeleton (branched): $(CH_3)_2CHCH_2NH_2$ — 2-methylpropan-1-amine

  Also, $(CH_3)_3CNH_2$ — 2-methylpropan-2-amine (the $NH_2$ on the tertiary carbon)

That gives four primary amines.

Step 2: Secondary amines ($2^\circ$) — $R_2NH$

Here the nitrogen is bonded to two alkyl groups. The total carbons in the two groups must sum to 4. Possible pairs:

• Methyl + propyl: $CH_3NHCH_2CH_2CH_3$ — $N$-methylpropan-1-amine

  Also, $CH_3NHCH(CH_3)_2$ — $N$-methylpropan-2-amine

• Ethyl + ethyl: $CH_3CH_2NHCH_2CH_3$ — $N$-ethylethanamine (diethylamine)

That gives three secondary amines. (Methyl + isopropyl is distinct from methyl + propyl; ethyl + ethyl is symmetric.)

Step 3: Tertiary amines ($3^\circ$) — $R_3N$

Three alkyl groups, total carbons = 4. The only possibility is:

• Methyl + methyl + ethyl: $(CH_3)_2NCH_2CH_3$ — $N,N$-dimethylethanamine

That gives one tertiary amine.

3. IUPAC names of all eight isomers

4. Type of isomerism exhibited …

Here is a clear, concept-first solution method for the given problem.

Method: Systematic Isomer Enumeration & Classification

This method relies on identifying all possible carbon skeletons and then placing the nitrogen atom (as an amine group, $-\text{NH}_2$, $-\text{NH}-$, or $-\text{N}<$) at every distinct position. Finally, we classify the isomers by functional group and chain/position.

Step 1: Determine the Degree of Substitution (1°, 2°, 3°)

The formula $C_4H_{11}N$ has one nitrogen and no rings or double bonds (saturated). The nitrogen can be:

• Primary (1°): $R-\text{NH}_2$ (one alkyl group on N)
• Secondary (2°): $R_2\text{NH}$ (two alkyl groups on N)
• Tertiary (3°): $R_3\text{N}$ (three alkyl groups on N)

We will generate all possible carbon skeletons for each type.

Step 2: Draw All Primary Amines ($R-\text{NH}_2$)

Treat the $-\text{NH}_2$ group as a substituent on a 4-carbon chain.

1. Straight chain (butane): Place $-\text{NH}_2$ at C1 and C2.

   • 1-Butanamine (C1)
   • 2-Butanamine (C2)

2. Branched chain (2-methylpropane): Place $-\text{NH}_2$ on the main chain.

   • 2-Methyl-1-propanamine (on the CH$_2$ group)
   • 2-Methyl-2-propanamine (on the central carbon)

Result: 4 primary amines.

Step 3: Draw All Secondary Amines ($R_2\text{NH}$)

The nitrogen is bonded to two alkyl groups (total 4 carbons). List all possible pairs of alkyl groups whose carbons sum to 4.

Result: 3 secondary amines.

Step 4: Draw All Tertiary Amines ($R_3\text{N}$)

The nitrogen is bonded to three alkyl groups (total 4 carbons). List all combinations of three alkyl groups whose carbons sum to 4.

Only one combination works: two methyls + one ethyl.

Result: 1 tertiary amine.

Step 5: Write IUPAC Names for All Isomers

Here is the complete list with IUPAC names:

Primary Amines (4):

• Butan-1-amine
• Butan-2-amine
• 2-Methylpropan-1-amine
• 2-Methylpropan-2-amine

Secondary Amines (3): …

Here are the common mistakes students make on this question about amines with formula $C_4H_{11}N$, along with clear strategies to avoid each.

(i) & (ii) Common Mistakes in Writing Structures and IUPAC Names

1. Missing isomers (especially secondary and tertiary amines)

• The Mistake: Students often list only the primary amines (like butan-1-amine, butan-2-amine, 2-methylpropan-1-amine, 2-methylpropan-2-amine) and forget that secondary and tertiary amines are also functional isomers of the same formula.
• Why it happens: The formula $C_4H_{11}N$ satisfies the rule for a saturated amine ($C_nH_{2n+3}N$). Students assume “amine” means only primary, but the nitrogen can be bonded to 1, 2, or 3 carbon groups.
• How to avoid: Systematically vary the degree of substitution at nitrogen:
  • Primary (1°): $R-NH_2$ (4 isomers)
  • Secondary (2°): $R_2NH$ (3 isomers)
  • Tertiary (3°): $R_3N$ (1 isomer)
  • Total = 8 isomers (not 4).

2. Incorrect IUPAC naming for secondary/tertiary amines

• The Mistake: Writing names like “diethylamine” or “ethylmethylamine” without following IUPAC rules (e.g., using “N-” prefixes).
• Why it happens: Common names are easier, but the question explicitly asks for IUPAC names.
• How to avoid: For IUPAC, treat the largest carbon chain as the parent alkane. Name the other alkyl groups as substituents on nitrogen, prefixed by “N-”.
  • Example: $CH_3CH_2NHCH_2CH_3$ → N-ethylethanamine (not diethylamine).
  • Example: $(CH_3)_2NCH_2CH_3$ → N,N-dimethylethanamine (not ethyldimethylamine).

3. Forgetting to check for optical isomerism

• The Mistake: Assuming all isomers are achiral.
• Why it happens: Students don’t check for chiral carbon atoms in the alkyl chains.
• How to avoid: Look for a carbon with four different groups. For $C_4H_{11}N$, butan-2-amine ($CH_3CH_2CH(NH_2)CH_3$) has a chiral centre at C-2. This means it exists as a pair of enantiomers — but the question asks for structural isomers, so you only need to write one structure (the racemic mixture is implied).

(iii) Common Mistakes in Identifying Isomerism Type

4. Confusing functional isomerism with chain/position isomerism

• The Mistake: Saying “all pairs show chain isomerism” or “all show position isomerism”.
• Why it happens: Students don’t realise that primary, secondary, and tertiary amines are functional isomers of each other (different functional group arrangement around N).
• How to avoid: Classify pairs based on what differs:
  • Chain isomerism: Same functional group, different carbon skeleton (e.g., butan-1-amine vs 2-methylpropan-1-amine).
  • Position isomerism: Same functional group, different position of $-NH_2$ on the same chain (e.g., butan-1-amine vs butan-2-amine).
  • Functional isomerism: Different class of amine (1° vs 2° vs 3°) — e.g., butan-1-amine (1°) vs N-methylpropan-1-amine (2°).

5. Mishandling metamerism

• The Mistake: Either flatly claiming “amines cannot show metamerism”, or stretching the word to cover 1°-vs-2°-vs-3° pairs. …