JEE MAIN - Chemistry (2025 - 29th January Evening Shift - No. 25)

Isomeric hydrocarbons → negative Baeyer’s test

(Molecular formula C9H12)

The total number of isomers from above with four different non-aliphatic substitution sites is -

Answer
2

Explanation

Molecular formula of isomeric hydrocarbon $\to C_9H_{12}$ - Negative Baeyer's test

Baeyers test is given by compounds that contain readily active carbon-carbon double bond.

So, in a negative Baeyer's test, there is no readily active c = c.

Compounds that give negative Baeyer's test are alkanes and aromatic compounds.

C$_9$H$_{12}$ compound is not an alkane

(Alkane general formula C$_n$H$_{2n+2}$)

Hydrocarbons with the formula C$_9$H$_{12}$ are considered as aromatic and isomers of substituted benzene rings.

The possible isomers of C$_9$H$_{12}$ are

JEE Main 2025 (Online) 29th January Evening Shift Chemistry - Hydrocarbons Question 12 English Explanation 1

From these, total number of the isomers with four different non-aliphatic substitution sites are 2

JEE Main 2025 (Online) 29th January Evening Shift Chemistry - Hydrocarbons Question 12 English Explanation 2 JEE Main 2025 (Online) 29th January Evening Shift Chemistry - Hydrocarbons Question 12 English Explanation 3 JEE Main 2025 (Online) 29th January Evening Shift Chemistry - Hydrocarbons Question 12 English Explanation 4
Four positions (1,2,3,4) are different. So, it contain four different non-aliphatic substitution sites. Four position (1,2,3,4) are different. So, it contain four different non-aliphatic substitution sites. Four position (1,2,3,4) are not different. So, it contain four different non-aliphatic substitution sites.

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