Consider the isomers of hydrocarbon with molecular formula C$$_5$$H$$_{10}$$. These isomers do not decolourise KMnO$$_4$$ solution. These isomers are subjected to chlorination with chlorine in presence of light to give monochloro compounds. The total number of monochloro compounds (structural isomers only) formed is __________.

The degree of unsaturation (or DBE) for $$C_5H_{10}$$ is
$$\frac{2(5)+2-10}{2}=1$$.

A single degree of unsaturation can arise from either one C=C double bond or one ring. The question states that the compounds do not decolourise alkaline $$KMnO_4$$, so they contain no double bond. Hence every isomer must be a cycloalkane with exactly one ring and no multiple bonds.

List all unique cycloalkane skeletons that give the formula $$C_5H_{10}$$ (only constitutional isomers are needed, stereochemistry is ignored).

Case 1: Five-membered ring - cyclopentane  Skeleton: $$\text{cyclopentane}$$

Case 2: Four-membered ring with one carbon outside the ring - methylcyclobutane  Skeleton: $$\text{methylcyclobutane}$$

Case 3: Three-membered ring with two carbons outside the ring - three possibilities  
(a) 1,1-dimethylcyclopropane (gem-dimethyl)  
(b) 1,2-dimethylcyclopropane (two methyls on adjacent ring carbons)  
(c) 1-ethylcyclopropane (an ethyl group on one ring carbon)

Thus there are in total five structural isomers that match the conditions:

1. cyclopentane
2. methylcyclobutane
3. 1,1-dimethylcyclopropane
4. 1,2-dimethylcyclopropane
5. 1-ethylcyclopropane

Next, count the number of distinct monochloro products each isomer can give on free-radical chlorination (light). Only connectivity is considered; stereochemical differences are ignored.

Isomer 1: cyclopentane
All five ring carbons are equivalent, so only one monochloro product is possible.
Number of monochloro isomers = $$1$$

Isomer 2: methylcyclobutane
Identify symmetry-independent positions:
- C1 = ring carbon bearing CH3
- C2 (and C4) = ring carbons adjacent to C1 (equivalent to each other)
- C3 = ring carbon opposite C1
- CH3 group itself
Therefore four different sites give four constitutional monochloro products.
Number of monochloro isomers = $$4$$

Isomer 3: 1,1-dimethylcyclopropane
C1 (bearing the two methyls) has no H, so it cannot be substituted.
- C2 and C3 (ring CH2 groups) are equivalent → one product.
- The two methyl groups are equivalent → substitution on either gives the same product.
Number of monochloro isomers = $$2$$

Isomer 4: 1,2-dimethylcyclopropane
The molecule has a mirror plane that exchanges C1 and C2; C3 is unique.
Independent sites:
- C1 (≡ C2) → one product
- C3 → second product
- Either methyl group (equivalent) → third product
Number of monochloro isomers = $$3$$

Isomer 5: 1-ethylcyclopropane
Symmetry exchanges C2 and C3; C1 is unique.
Independent sites:
- C1 (ring carbon bearing ethyl)
- C2 (≡ C3)
- α-carbon of ethyl group (CH2 attached to ring)
- β-carbon of ethyl group (terminal CH3)
Number of monochloro isomers = $$4$$

Finally, add the monochloro products from all isomers:

$$1 + 4 + 2 + 3 + 4 = 14$$

Hence, the total number of distinct monochloro structural isomers that can be formed is 14.