Q.Identify all the possible monochloro structural isomers expected to be formed on free radical monochlorination of $(CH_3)_2CHCH_2CH_3$.

The key idea is structural isomerism — specifically, constitutional isomers formed by replacing different hydrogen atoms with chlorine.

Reasoning:

1. The given compound is 2-methylbutane. Its carbon skeleton is:

   • C1 (primary, end CH$_3$), C2 (tertiary, CH), C3 (secondary, CH$_2$), C4 (primary, end CH$_3$), plus a methyl branch (primary CH$_3$) on C2.

2. Free radical chlorination substitutes H atoms at different carbon positions. Each distinct carbon type yields a different monochloro product.

3. The possible products are:

   • 1-chloro-2-methylbutane (Cl on C1)
   • 2-chloro-2-methylbutane (Cl on C2 — tertiary)
   • 2-chloro-3-methylbutane (Cl on C3)
   • 1-chloro-3-methylbutane (Cl on C4) …

Why This Approach Works

Structural isomerism in monochlorination arises from replacing a hydrogen atom on different carbon atoms of the parent alkane. The free radical mechanism is unselective — chlorine radicals abstract hydrogens from all available positions, though with different rates depending on the type of carbon (primary, secondary, tertiary). But for structural isomers, we only care about which carbon gets the chlorine, not how fast. So the task reduces to: count the distinct carbon atoms in the molecule, then consider if chlorination at each gives a unique product.

The molecule given is $(CH_3)_2CHCH_2CH_3$ — that's 2-methylbutane (isopentane). Let's draw it properly.

Number the carbon chain for clarity:

1. Carbon 1: the $CH_3$ at the left end (attached to the branch point)
2. Carbon 2: the $CH$ (the branch point, with a methyl group attached)
3. Carbon 3: the $CH_2$ in the main chain
4. Carbon 4: the $CH_3$ at the right end
5. Carbon 5: the $CH_3$ branch on carbon 2

Now, each distinct carbon environment can yield a different monochlorinated product. But careful: some carbons are equivalent by symmetry.

Step-by-Step Reasoning

1. Identify all unique carbon environments.

Look at the molecule: carbon 1 is a primary carbon (attached to one other carbon). Carbon 4 is also primary, but is it the same as carbon 1? No — carbon 1 is attached to carbon 2 (a tertiary carbon), while carbon 4 is attached to carbon 3 (a secondary carbon). So they are in different chemical environments. What about carbon 5 (the branch methyl)? It is attached to carbon 2, exactly like carbon 1 — the two methyls of the $(CH_3)_2CH-$ group are equivalent by symmetry. So there are only two distinct primary environments: the C1/C5 pair, and C4.

Carbon 2 is tertiary (attached to three other carbons). Carbon 3 is secondary (attached to two other carbons). So in total, there are four distinct carbon environments — and hence four different types of hydrogen atoms — in the molecule: the C1/C5 pair, C2, C3 and C4.

2. Consider chlorination at each carbon atom.

When a hydrogen is replaced by chlorine on a given carbon, we get a structural isomer. Let's name each product systematically:

• Chlorination at carbon 1 (the left-end $CH_3$): gives $CH_2Cl-CH(CH_3)-CH_2-CH_3$. The IUPAC name is 1-chloro-2-methylbutane.

• Chlorination at carbon 2 (the tertiary $CH$): gives $(CH_3)_2CCl-CH_2-CH_3$. That's 2-chloro-2-methylbutane.

• Chlorination at carbon 3 (the $CH_2$ in the chain): gives $(CH_3)_2CH-CHCl-CH_3$. That's 2-chloro-3-methylbutane (note: numbering starts from the end nearer the chlorine, so the methyl is on carbon 3, chlorine on carbon 2).

• Chlorination at carbon 4 (the right-end $CH_3$): gives $(CH_3)_2CH-CH_2-CH_2Cl$. That's 1-chloro-3-methylbutane.

• Chlorination at carbon 5 (the branch methyl): gives $(CH_3)(CH_2Cl)CH-CH_2-CH_3$, i.e. $CH_3-CH(CH_2Cl)-CH_2-CH_3$.

3. Check for duplicates. …

Method: Free Radical Substitution – Product Analysis via Carbon Skeleton & Symmetry

This method identifies all structural isomers by:

1. Drawing the parent alkane’s carbon skeleton.
2. Classifying each unique carbon environment (1°, 2°, 3°).
3. Replacing one H on each distinct carbon with Cl — each gives a different structural isomer.

Step 1: Identify the parent alkane

The given compound is:

This is 2-methylbutane (isopentane). Its carbon skeleton:

   C
   |
 C-C-C-C

Number the chain for clarity:

      C1
      |
 C2-C3-C4-C5

Step 2: Identify unique carbon environments

• C1 (methyl, attached to C3) → 1° carbon
• C2 (methyl, attached to C3) → 1° carbon (identical to C1 by symmetry)
• C3 (CH, attached to C1, C2, C4) → 3° carbon
• C4 (CH₂, attached to C3 and C5) → 2° carbon
• C5 (methyl, attached to C4) → 1° carbon (different from C1/C2 because it’s on a different branch)

So we have 4 distinct carbon types:

• 1° (type A): C1 and C2 (equivalent)
• 1° (type B): C5 (different)
• 2°: C4
• 3°: C3

Step 3: Replace one H on each distinct carbon with Cl

Each unique carbon gives one structural isomer of monochlorinated product:

| Carbon type | Position | Product name |

|-------------|----------|--------------| …

This is a classic exam trap in free radical halogenation and structural isomerism. Let’s break down the common mistakes and how to avoid them.

🧪 The Reaction

We are monochlorinating:

This is 2-methylbutane (isopentane). Free radical chlorination replaces one H with Cl.

✗ Mistake #1: Forgetting that all non-equivalent hydrogens give different products

Why it happens

Students often count only the carbon skeleton and miss that hydrogens on the same carbon but in different chemical environments can yield different isomers.

✓ How to avoid

• Identify all unique types of hydrogen atoms in the molecule.
• Draw the structure and label each distinct H environment.

Structure of 2-methylbutane:

      CH3
       |
CH3 — C — CH2 — CH3
       |
      H

Number the carbons:

1. C1: CH₃ (attached to C2)
2. C2: CH (tertiary, attached to C1, C3, C4)
3. C3: CH₂ (attached to C2 and C4)
4. C4: CH₃ (attached to C3)
5. C5: CH₃ (attached to C2, the branch)

Hydrogen types:

So 4 structural isomers are possible.

✗ Mistake #2: Counting the same product twice

Why it happens

C1 and C5 are both primary methyl groups attached to C2 — they are equivalent by symmetry.

✓ How to avoid

• Always check for symmetry in the molecule.
• If two carbons are in identical environments, they give the same product.

Here, C1 and C5 are equivalent → only one product from both.

✗ Mistake #3: Ignoring optical isomerism when asked only for structural isomers

Why it happens

Some students list optical isomers (chiral centres) as separate structural isomers.

✓ How to avoid

• Structural isomers = different connectivity (constitutional isomers).
• Stereoisomers (optical, geometrical) are not structural isomers.
• In this reaction, 2-chloro-3-methylbutane has a chiral centre, but its enantiomers are not counted as separate structural isomers.

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