Q.Write the isomers of the compound having formula $C_4H_9Br$.

The key idea is structural isomerism — specifically, chain and position isomerism for a monosubstituted butane.

Step 1: The formula $C_4H_9Br$ corresponds to a butane chain ($C_4H_{10}$) with one hydrogen replaced by bromine. The number of isomers equals the number of distinct carbon skeletons and distinct positions for Br.

Step 2: Butane has two carbon skeletons:

• n-butane (straight chain): $CH_3-CH_2-CH_2-CH_3$ — Br can attach at C1 or C2, giving two isomers. …

Why this approach works

Structural isomerism (also called constitutional isomerism) arises when molecules have the same molecular formula but different connectivity — different arrangements of atoms. For $\mathrm{C_4H_9Br}$, the bromine atom can be attached to different carbon atoms, and the carbon chain itself can be straight or branched. The trick is to first sketch all possible carbon skeletons for 4 carbons, then place the bromine at every chemically distinct carbon position. Two placements that look different on paper but are actually the same molecule (by symmetry or rotation) must be counted only once.

Step-by-step reasoning

1. Draw all carbon skeletons for 4 carbons.

   There are two distinct skeletons:

   • A straight chain of 4 carbons: $\mathrm{C-C-C-C}$ (butane skeleton).
   • A branched chain with a methyl group on the middle carbon: $\mathrm{C-C(C)-C}$ (isobutane skeleton). No other skeletons exist for 4 carbons.

2. Place bromine on the straight chain skeleton.

   Number the carbons 1, 2, 3, 4 from one end.

   • Bromine on carbon 1: $\mathrm{CH_3-CH_2-CH_2-CH_2Br}$ (1-bromobutane).
   • Bromine on carbon 2: $\mathrm{CH_3-CHBr-CH_2-CH_3}$ (2-bromobutane).
   • Bromine on carbon 3: This is the same as carbon 2 by symmetry (flipping the chain end-to-end gives the same molecule). So no new isomer.
   • Bromine on carbon 4: Same as carbon 1. So from the straight chain, we get 2 isomers.

3. Place bromine on the branched skeleton.

   The only branched skeleton for $\mathrm{C_4}$ is isobutane, $\mathrm{CH_3-CH(CH_3)-CH_3}$: a central tertiary carbon (C2) bonded to one hydrogen and three other carbons — two methyl groups (C1, C3) plus the branch methyl (C4). Counting confirms 4 carbons total (C1, C2, C3, C4). The skeleton is:

   C1 - C2 - C3
        |
        C4
   

   where C2 is the central carbon.

   Now place bromine at each distinct carbon:

   • Bromine on C1 (a primary carbon): $\mathrm{CH_2Br-CH(CH_3)-CH_3}$ (1-bromo-2-methylpropane).
   • Bromine on C2 (the tertiary carbon): $\mathrm{CH_3-CBr(CH_3)-CH_3}$ (2-bromo-2-methylpropane).
   • Bromine on C3: This is equivalent to C1 by symmetry (the two methyl groups on the ends are identical). So no new isomer. …

Method: Carbon Skeleton + Functional Group Placement Method

This method systematically generates all structural isomers by:

1. First drawing all possible carbon skeletons (straight and branched chains)
2. Then placing the bromine atom at every unique carbon position

Step 1: Draw all possible carbon skeletons for 4 carbons

There are two distinct skeletons:

• Straight chain (butane):

  $C - C - C - C$

• Branched chain (isobutane / 2-methylpropane):

  $\begin{matrix} C - C - C \\ | \\ C \end{matrix}$

Step 2: Place the bromine atom at each unique carbon position

Skeleton 1: Straight chain ($n$-butane backbone)

Number the carbons:

$C_1 - C_2 - C_3 - C_4$

• Br at $C_1$ → 1-bromobutane
• Br at $C_2$ → 2-bromobutane
• Br at $C_3$ → same as $C_2$ (symmetry)
• Br at $C_4$ → same as $C_1$ (symmetry)

Two isomers from this skeleton.

Skeleton 2: Branched chain (isobutane backbone)

Structure:

$\begin{matrix} C_1 - C_2 - C_3 \\ | \\ C_4 \end{matrix}$

• Br at $C_1$ → 1-bromo-2-methylpropane
• Br at $C_2$ → 2-bromo-2-methylpropane
• Br at $C_3$ → same as $C_1$ (symmetry)
• Br at $C_4$ → same as $C_1$ (symmetry)

Two isomers from this skeleton.

Step 3: Count total distinct isomers

| Skeleton | Isomers |

|----------|---------| …

This is a classic exam favourite. Let's break it down.

🧠 The Core Concept First

The formula $C_4H_9Br$ has zero degrees of unsaturation — it is fully saturated, just like $C_4H_{10}$ with one H replaced by Br ($DoU = \frac{2(4)+2-9-1}{2} = 0$). So all isomers will be structural isomers — no rings or double bonds.

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

✓ The Correct Isomers (for reference)

There are 4 structural isomers:

1. 1-Bromobutane (straight chain, Br at end)

   $CH_3CH_2CH_2CH_2Br$

2. 2-Bromobutane (straight chain, Br on C2)

   $CH_3CH_2CHBrCH_3$

3. 1-Bromo-2-methylpropane (branched chain, Br at end)

   $(CH_3)_2CHCH_2Br$

4. 2-Bromo-2-methylpropane (branched chain, Br on tertiary carbon)

   $(CH_3)_3CBr$

✗ Common Mistakes & How to Avoid Them

1. Counting the same isomer twice (due to symmetry)

• Mistake: Writing both $CH_3CH_2CHBrCH_3$ and $CH_3CHBrCH_2CH_3$ as different isomers.
• Why it's wrong: The carbon chain is symmetric — numbering from either end gives the same molecule.
• How to avoid: Always number the carbon chain from the end nearest the Br (or functional group). If the Br is on carbon 2 from both ends, it's the same compound.

2. Forgetting branched-chain isomers

• Mistake: Only writing straight-chain isomers (1-bromobutane and 2-bromobutane).
• Why it's wrong: $C_4$ can form a branched skeleton (isobutane).
• How to avoid: Always draw all possible carbon skeletons first — for $C_4$, there are two: n-butane and isobutane (2-methylpropane). Then place Br on each unique carbon.

3. Creating a "new" isomer by rotating the molecule

• Mistake: Writing $CH_3CHBrCH_2CH_3$ and $CH_3CH_2CHBrCH_3$ as different.
• Why it's wrong: These are the same molecule (2-bromobutane) — just drawn backwards.
• How to avoid: If you can rotate the molecule in space to match another structure, it's not a new isomer. Use systematic naming to check.

4. Adding a branch that creates a longer chain

• Mistake: Writing $CH_3CH(Br)CH_2CH_3$ as a "branched" isomer.
• Why it's wrong: That's just 2-bromobutane — the longest chain is still 4 carbons.
• How to avoid: Always identify the longest continuous carbon chain first. If it's 4 carbons, it's a butane derivative, not a branched one.

5. Missing the tertiary bromide isomer

• Mistake: Writing only 1-bromo-2-methylpropane but forgetting 2-bromo-2-methylpropane. …