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