Q.Among the isomeric alkanes of molecular formula $C_5H_{12}$, identify the one that on photochemical chlorination yields

The key idea is Structural Isomerism in alkanes — specifically, how the symmetry of a molecule determines the number of distinct monochlorination products.

Reasoning:

1. The three structural isomers of $C_5H_{12}$ are:

   • n-pentane (straight chain)
   • isopentane (2-methylbutane)
   • neopentane (2,2-dimethylpropane)

2. Photochemical chlorination replaces one H with Cl. The number of distinct monochlorides equals the number of chemically different types of hydrogen atoms in the molecule.

3. Count distinct H-environments:

   • Neopentane: all 12 H atoms are equivalent (four identical methyl groups) → 1 monochloride. …

Photochemical chlorination of alkanes is a free-radical substitution reaction. At room temperature or with UV light, chlorine molecules break into chlorine radicals, which then abstract a hydrogen atom from the alkane. The resulting alkyl radical quickly reacts with another chlorine molecule to form the alkyl chloride.

The crucial point for isomer counting: each distinct type of hydrogen atom in the starting alkane gives a different monochloride product. Two hydrogen atoms are "equivalent" if replacing either one leads to the same alkyl chloride — this happens when they are attached to the same carbon, or to carbons that are symmetric in the molecule.

So the problem reduces to: for each $C_5H_{12}$ isomer, how many chemically distinct hydrogen environments exist?

Let’s examine the three structural isomers of pentane.

1. n-Pentane (straight chain)

   Structure: $CH_3-CH_2-CH_2-CH_2-CH_3$

   Label the carbons: C1, C2, C3, C4, C5.

   • C1 and C5 are equivalent (both end methyl groups). Their 6 hydrogens are all identical.
   • C2 and C4 are equivalent (both are $CH_2$ groups next to an end). Their 4 hydrogens are identical to each other.
   • C3 is the central $CH_2$ group. Its 2 hydrogens are distinct from the others. So there are 3 types of hydrogen. Chlorination gives 3 monochlorides: 1-chloropentane, 2-chloropentane, and 3-chloropentane.

2. Isopentane (2-methylbutane)

   Structure:

       CH3
        |
   CH3-CH-CH2-CH3
   

   Number the carbons:

   • C1: the $CH_3$ attached to the branch point (top methyl).
   • C2: the branch point ($CH$).
   • C3: the $CH_2$ next to the branch.
   • C4: the terminal $CH_3$ on the straight chain.
   • C5: the other terminal $CH_3$ (the one on the branch). Now check equivalence:
   • The three methyl groups are not all equivalent. The one on C4 is attached to a $CH_2$, while the one on C1 and C5 are attached to a $CH$ (the branch point). So C4 methyl is different from C1 and C5.
   • Are C1 and C5 equivalent? Yes — both are methyls on the same branch carbon (C2). Their 6 hydrogens are identical.
   • C3 ($CH_2$) has 2 hydrogens, distinct from all methyls.
   • C2 ($CH$) has 1 hydrogen, distinct from everything. That gives 4 types: (i) C1/C5 methyl, (ii) C4 methyl, (iii) C3 $CH_2$, (iv) C2 $CH$. So chlorination yields 4 monochlorides.

3. Neopentane (2,2-dimethylpropane)

   Structure:

        CH3
        |
   CH3-C-CH3
        | …
   

Method: Systematic Carbon Skeleton Analysis + Symmetry Check

This method uses the principle that the number of monochlorination products depends on the number of non-equivalent hydrogen atoms in the alkane.

Step 1: Draw all structural isomers of $\mathbf{C_5H_{12}}$

There are three chain isomers:

Step 2: Count non-equivalent hydrogen types in each

• n-Pentane:

  • 3 types of H (end $\mathrm{CH_3}$, middle $\mathrm{CH_2}$, central $\mathrm{CH_2}$)
  • → 3 monochlorides

• Isopentane:

  • 4 types of H (two different $\mathrm{CH_3}$ groups, one $\mathrm{CH_2}$, one $\mathrm{CH}$)
  • → 4 monochlorides

• Neopentane:

  • All 12 H atoms are equivalent (all on $\mathrm{CH_3}$ groups attached to same carbon)
  • → 1 monochloride

Step 3: Match to the question …

The Core Idea

For an alkane $C_5H_{12}$, photochemical chlorination replaces one H atom with a Cl atom. The number of different monochlorides depends on how many distinct types of hydrogen atoms are present in the molecule.

• Equivalent H atoms (same chemical environment) give the same monochloride.
• Non-equivalent H atoms give different monochlorides.

So the question is really: How many different kinds of H atoms does each isomer have?

The Three Isomers of $C_5H_{12}$

Now let’s count distinct H types for each.

(i) Single monochloride → Neopentane (C)

• Structure: central C bonded to 4 identical methyl groups.
• All 12 H atoms are equivalent.
• Chlorination gives only one product: $(CH_3)_3C-CH_2Cl$ (but all positions same).

✓ Answer: Neopentane

(ii) Three isomeric monochlorides → n-Pentane (A)

• Structure: $CH_3-CH_2-CH_2-CH_2-CH_3$
• Types of H:
  • C1 and C5 (end CH₃) — equivalent → 1 type
  • C2 and C4 (CH₂) — equivalent → 1 type
  • C3 (middle CH₂) — unique → 1 type
• That gives 3 distinct H types → 3 monochlorides.

✓ Answer: n-Pentane

(iii) Four isomeric monochlorides → Isopentane (B)

• Structure: $CH_3-CH(CH_3)-CH_2-CH_3$
• Types of H:
  • 1° H on C1 (3 H)
  • 1° H on C4 (3 H) — different from C1 (C1 is next to the CH, C4 is next to the CH₂)
  • 2° H on C3 (2 H)
  • 3° H on C2 (1 H)
• That gives 4 distinct H types → 4 monochlorides.

✓ Answer: Isopentane

Summary of Correct Answers

Common Mistakes & How to Avoid Them

✗ Mistake 1: Counting only carbon skeletons, ignoring symmetry

• Why it happens: Students list all possible C₅H₁₂ isomers but don’t analyze H equivalence.
• How to avoid: For each isomer, draw the full structure and label each H atom’s environment. Use symmetry: identical groups attached to same carbon → equivalent H.

✗ Mistake 2: Assuming all terminal methyl groups are equivalent

• Why it happens: In branched alkanes, methyl groups on different carbons are not equivalent.
• How to avoid: Check the neighbouring atoms. A CH₃ attached to a CH is different from a CH₃ attached to a CH₂. …