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Q6.5:
A hydrocarbon ( C_{5}H_{10} ) does not react with chlorine in dark but gives a single monochloro compound ( C_{5}H_{9}Cl ) in bright sunlight. Identify the hydrocarbon.

Solution :

Step 1: Analyzing Molecular Formula and Unsaturation

Given the molecular formula $C_5H_{10}$, we calculate the Degree of Unsaturation (Double Bond Equivalent):

$DU = C + 1 - \frac{H}{2} = 5 + 1 - \frac{10}{2} = 1$

[Per the Degree of Unsaturation formula]. A $DU$ of 1 indicates the presence of either one double bond or a ring structure.

Step 2: Evaluating Reaction Conditions

The hydrocarbon does not react with $Cl_2$ in the dark, which suggests the absence of an alkene (which would undergo rapid electrophilic addition). However, it reacts in bright sunlight (UV light) to form a monochloro compound. This indicates a Free Radical Substitution reaction typical of alkanes or cycloalkanes [Per the mechanism of Free Radical Chlorination].

Step 3: Deductive Reasoning for Molecular Structure

For a hydrocarbon to yield only a single monochloro compound, all hydrogen atoms in the molecule must be chemically equivalent. If there were different types of hydrogen atoms, multiple isomeric monochloro products would form.

  • Linear/Branched Alkenes: Excluded due to the lack of reactivity in the dark (absence of $C=C$ addition).
  • Acyclic Alkanes: Must be $C_n H_{2n+2}$, but we have $C_5H_{10}$.
  • Cycloalkanes: The cyclic structure satisfies $C_n H_{2n}$. Among $C_5H_{10}$ isomers (cyclopentane, methylcyclobutane, ethylcyclopropane, etc.), we seek the one with maximum symmetry.

Step 4: Identifying the Isomer

Cyclopentane ($C_5H_{10}$) consists of a five-membered ring where each carbon atom is equivalent. Consequently, all ten hydrogen atoms are equivalent [By molecular symmetry analysis].

C C C C C Cyclopentane

Step 5: Verification of Reaction

The reaction proceeds via free radical mechanism:

$C_5H_{10} + Cl_2 \xrightarrow{hv} C_5H_9Cl + HCl$

Because all positions in cyclopentane are identical, the substitution of any hydrogen atom results in the same product: Chlorocyclopentane.

Candidate Isomer Number of equivalent H sets Monochloro products
Cyclopentane 1 1
Methylcyclobutane 4 4
1,1-Dimethylcyclopropane 3 3

Final Solution: The hydrocarbon is Cyclopentane ($C_5H_{10}$).


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