Total Monochloro Structural Isomers Formed from C5H10 Cycloalkanes

To determine the total number of monochloro structural isomers formed from the given hydrocarbon, we analyze the problem in the following steps:

Step 1: Degree of Unsaturation and Functional Group Identification

The molecular formula is $\mathrm{C_5H_{10}}$. The degree of unsaturation (DU) is calculated as: $\text{DU} = C + 1 - \frac{H}{2} = 5 + 1 - \frac{10}{2} = 1$

A degree of unsaturation of $1$ indicates either an alkene (presence of a double bond) or a cycloalkane (presence of a ring).

Since the isomers do not decolourise $\mathrm{KMnO_4}$ solution, they do not contain any carbon-carbon double bonds (alkene functionality). Therefore, all such isomers are saturated cyclic hydrocarbons, i.e., cycloalkanes.

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Step 2: Identification of Cycloalkane Isomers of $\mathrm{C_5H_{10}}$

There are $5$ cyclic structural isomers for $\mathrm{C_5H_{10}}$:

1. Cyclopentane
2. Methylcyclobutane
3. Ethylcyclopropane
4. 1,1-Dimethylcyclopropane
5. 1,2-Dimethylcyclopropane

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Step 3: Determining Monochloro Structural Isomers for Each Cycloalkane

1. Cyclopentane

All $10$ hydrogen atoms are chemically equivalent due to molecular symmetry.

• Monochloro product formed:

  1. $1\text{-chlorocyclopentane}$

• Subtotal = $1$

2. Methylcyclobutane

Methylcyclobutane has $4$ sets of non-equivalent hydrogen atoms:

• Substitution at $\text{C-1}$ (ring carbon with methyl attached): $1\text{-chloro-1-methylcyclobutane}$

• Substitution at $\text{C-2/C-4}$ (ring carbons adjacent to $\text{C-1}$): $1\text{-chloro-2-methylcyclobutane}$

• Substitution at $\text{C-3}$ (ring carbon opposite to $\text{C-1}$): $1\text{-chloro-3-methylcyclobutane}$

• Substitution at methyl group: $(\text{chloromethyl})\text{cyclobutane}$

• Subtotal = $4$

3. Ethylcyclopropane

Ethylcyclopropane has $4$ sets of non-equivalent hydrogen atoms:

• Substitution at $\text{C-1}$ of the ring: $1\text{-chloro-1-ethylcyclopropane}$

• Substitution at $\text{C-2/C-3}$ of the ring: $1\text{-chloro-2-ethylcyclopropane}$

• Substitution at $-\mathrm{CH_2}-$ of ethyl group: $1\text{-(1-chloroethyl)cyclopropane}$

• Substitution at $-\mathrm{CH_3}$ of ethyl group: $1\text{-(2-chloroethyl)cyclopropane}$

• Subtotal = $4$

4. 1,1-Dimethylcyclopropane

1,1-Dimethylcyclopropane has $2$ sets of non-equivalent hydrogen atoms ($\text{C-1}$ has no hydrogens):

• Substitution at $\text{C-2/C-3}$ of the ring: $1\text{-chloro-2,2-dimethylcyclopropane}$

• Substitution at either methyl group: $1\text{-(chloromethyl)-1-methylcyclopropane}$

• Subtotal = $2$

5. 1,2-Dimethylcyclopropane

1,2-Dimethylcyclopropane has $3$ sets of non-equivalent hydrogen atoms:

• Substitution at $\text{C-1/C-2}$ of the ring: $1\text{-chloro-1,2-dimethylcyclopropane}$

• Substitution at $\text{C-3}$ of the ring: $3\text{-chloro-1,2-dimethylcyclopropane}$

• Substitution at either methyl group: $1\text{-(chloromethyl)-2-methylcyclopropane}$

• Subtotal = $3$

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Step 4: Total Calculation

Summing all the monochloro structural isomers: $\text{Total} = 1 + 4 + 4 + 2 + 3 = 14$

The total number of monochloro compounds (structural isomers only) formed is 14.