Chemistry Q75 – JEE-MAIN 2025

Home Chemistry Organic Chemistry – Some Basic Principles and TechniquesQuestion #468

The total number of structural isomers possible for the substituted benzene derivatives with the molecular formula

C_{9} H_{12}

is _______.

Enter Calculated Numerical Value:

Clue 1 of 3: Degree of Unsaturation

Calculate the degree of unsaturation (DU) for the given molecular formula to determine the presence of rings and/or multiple bonds.

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Step 1: Calculate Degree of Unsaturation (DU)✦ Active

For the molecular formula $C_{9} H_{12}$ , the degree of unsaturation (DU) is calculated using the formula $D U = C + 1 - \frac{H}{2}$ . Substituting the values:

D U = 9 + 1 - \frac{12}{2} = 10 - 6 = 4

A DU of 4 indicates the presence of a benzene ring (which has 3 double bonds and 1 ring, totaling 4 degrees of unsaturation). This means there are no other rings or multiple bonds outside the benzene ring.

Step 2: Identify Possible Substitution Patterns○ Expand

A benzene ring has the formula $C_{6} H_{6}$ . For $C_{9} H_{12}$ , the remaining atoms for substituents are $C_{3} H_{6}$ . We need to find combinations of alkyl groups that sum up to $C_{3} H_{6}$ and can be attached to the benzene ring.

Possible substitution patterns:

1. Monosubstituted: One $C_{3} H_{7}$ group (e.g., $C_{6} H_{5} - C_{3} H_{7}$ ). This requires removing 1 H from benzene, leaving $C_{6} H_{5}$ . Total H = $5 + 7 = 12$ . This is valid.

2. Disubstituted: One $C H_{3}$ and one $C_{2} H_{5}$ group (e.g., $C_{6} H_{4} - (C H_{3}) (C_{2} H_{5})$ ). This requires removing 2 H from benzene, leaving $C_{6} H_{4}$ . Total H = $4 + 3 + 5 = 12$ . This is valid.

3. Trisubstituted: Three $C H_{3}$ groups (e.g., $C_{6} H_{3} - (C H_{3})_{3}$ ). This requires removing 3 H from benzene, leaving $C_{6} H_{3}$ . Total H = $3 + 3 \times 3 = 12$ . This is valid.

💡 Teacher's Secret Hint

Ensure the total number of carbons and hydrogens matches the molecular formula after substitution.

Step 3: Enumerate Isomers for Each Pattern○ Expand

1. **Monosubstituted ( $C_{6} H_{5} - C_{3} H_{7}$ ):** The $C_{3} H_{7}$ group can be n-propyl or isopropyl.

- n-Propylbenzene

- Isopropylbenzene (Cumene)

(2 isomers)

2. **Disubstituted ( $C_{6} H_{4} - (C H_{3}) (C_{2} H_{5})$ ):** The methyl and ethyl groups can be in ortho, meta, or para positions relative to each other.

- 1-Ethyl-2-methylbenzene (o-ethyltoluene)

- 1-Ethyl-3-methylbenzene (m-ethyltoluene)

- 1-Ethyl-4-methylbenzene (p-ethyltoluene)

(3 isomers)

3. **Trisubstituted ( $C_{6} H_{3} - (C H_{3})_{3}$ ):** Three methyl groups can be arranged in three distinct ways on the benzene ring.

- 1,2,3-Trimethylbenzene

- 1,2,4-Trimethylbenzene

- 1,3,5-Trimethylbenzene (Mesitylene)

(3 isomers)

The total number of structural isomers is the sum of isomers from all possible substitution patterns: $2 + 3 + 3 = 8$ .

💡 Teacher's Secret Hint

Remember to consider all possible positional isomers for di- and tri-substituted benzene rings.

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Chemistry Question 75 – JEE-MAIN 2025

Clue 1 of 3: Degree of Unsaturation

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