Q.A hydrocarbon $C_5H_{10}$ does not react with chlorine in dark but gives a single monochloro compound $C_5H_9Cl$ in bright sunlight. Identify the hydrocarbon.

The key idea is structural isomerism — specifically, the hydrocarbon must be a cycloalkane (since alkenes react with chlorine in the dark via addition, but this compound does not). In bright sunlight, it undergoes free-radical substitution, and giving a single monochloro product means all hydrogen atoms are equivalent.

Reasoning:

1. No reaction with chlorine in the dark rules out alkenes (which add chlorine). So the compound is a saturated cyclic hydrocarbon: $C_5H_{10}$ fits a cycloalkane. …

Why this approach works

The problem tests two key ideas: structural isomerism and reactivity patterns of hydrocarbons.

First, the molecular formula $C_5H_{10}$ fits either an alkene (with one double bond) or a cycloalkane (saturated ring). Both have the same degree of unsaturation: one ring or one $\pi$‑bond.

Second, the clue about chlorine:

• In the dark, chlorine adds across a double bond (electrophilic addition) but does not react with a saturated hydrocarbon.
• In bright sunlight, chlorine undergoes free‑radical substitution with alkanes/cycloalkanes, replacing a hydrogen atom with chlorine.

So the hydrocarbon must be a cycloalkane (no reaction in dark) that gives only one monochloro product (all hydrogens are equivalent).

Step‑by‑step reasoning

1. Identify the possible isomers of $C_5H_{10}$

   The formula has one degree of unsaturation. The main families are:

   • Alkenes (e.g., pent‑1‑ene, pent‑2‑ene, 2‑methylbut‑1‑ene, etc.)
   • Cycloalkanes (cyclopentane, methylcyclobutane, ethylcyclopropane, dimethylcyclopropanes)

   Alkenes would react with chlorine in the dark (addition across the double bond), so they are ruled out.

2. Focus on cycloalkanes

   Cycloalkanes do not react with chlorine in the dark — they need UV light for substitution. So the hydrocarbon must be a cycloalkane.

3. Check which cycloalkane gives a single monochloro product

   In free‑radical substitution, the number of monochloro isomers equals the number of distinct types of hydrogen atoms in the molecule.

   • Cyclopentane ($\ce{C5H10}$): a regular pentagon. All 10 hydrogens are equivalent (each carbon has two identical hydrogens, and all carbons are identical). → Only one monochloro product.
   • Methylcyclobutane: has a methyl group and ring hydrogens — at least 3 different types of H → multiple products.
   • Ethylcyclopropane: even more distinct H types.
   • Dimethylcyclopropanes: also give multiple isomers.

   Only cyclopentane satisfies the “single monochloro compound” condition.

Method: Degree of Unsaturation + Symmetry Analysis

Step 1: Find the degree of unsaturation

For a hydrocarbon $C_nH_m$, the degree of unsaturation (DU) is:

For $C_5H_{10}$:

So the molecule has one double bond or one ring.

Step 2: Interpret the chemical clues

• Does not react with chlorine in dark → No alkene (alkenes add chlorine in dark). So the unsaturation is not a double bond — it must be a ring.
• Gives a single monochloro compound in sunlight → Sunlight causes free radical substitution. Only one type of hydrogen is present (all hydrogens are equivalent).

Step 3: Identify the structure …

This is a classic exam trap that tests your understanding of structural isomerism, reactivity, and symmetry. Let’s break it down.

Step 1: Understand the clues in the question

• Formula: $C_5H_{10}$ → could be an alkene or a cycloalkane (both have one degree of unsaturation).
• Does not react with chlorine in dark → this rules out alkenes (alkenes add chlorine in the dark). So the compound is not an alkene.
• Gives a single monochloro compound in bright sunlight → this is a free radical substitution reaction (like alkanes). Only one product means all hydrogen atoms are equivalent (or the molecule is highly symmetric).

So the hydrocarbon must be a cycloalkane with all hydrogens identical.

Common Mistake #1: Forgetting that cycloalkanes also have formula $C_nH_{2n}$

• The error: Students immediately think “alkene” because of $C_5H_{10}$.
• Why it’s wrong: Cycloalkanes also satisfy $C_nH_{2n}$. The “no reaction in dark” clue is the key to eliminate alkenes.
• How to avoid: Always check both possibilities — alkenes and cycloalkanes — when you see $C_nH_{2n}$. Then use the reactivity clue.

Common Mistake #2: Ignoring symmetry when counting products

• The error: Picking a cycloalkane like cyclopentane (which gives one product) but also considering methylcyclobutane (which gives multiple products).
• Why it’s wrong: The question says single monochloro compound. Methylcyclobutane has different types of hydrogens (primary, secondary, tertiary) → would give multiple products.
• How to avoid: Draw the structure and check if all hydrogen atoms are equivalent. Only cyclopentane has all 10 hydrogens identical.

Common Mistake #3: Confusing “single product” with “major product”

• The error: Thinking that if one product is formed in large amount, it counts as “single”.
• Why it’s wrong: In free radical chlorination, if multiple products are possible, you get a mixture — even if one is major. “Single” means only one possible monochlorinated product. …