Q.72. Consider the isomers of hydrocarbon with molecular formula C₈H₁₀. These isomers do not decolou

Q: What is the correct answer for this exam Chemistry Q72?

The correct answer is 14. Sol. The molecular formula is C₈H₁₀. The degree of unsaturation (DBE) = C + 1 - H/2 = 8 + 1 - 10/2 = 9 - 5 = 4. This indicates a benzene ring (DBE=4). "These isomers do not decolourise KMnO₄ solution" means they do not have any C=C double bonds (alkenes) or C≡C triple bonds (alkynes) or benzylic hydrogens that can be oxidized by KMnO₄. This implies that all alkyl side chains are fully saturated a

Q: Q.72. Consider the isomers of hydrocarbon with molecular formula C₈H₁₀. These isomers do not decolourise KMnO₄ solution. These isomers are subjected to chlorination with chlorine in presence of light…

The correct answer is 14. Sol. The molecular formula is C₈H₁₀. The degree of unsaturation (DBE) = C + 1 - H/2 = 8 + 1 - 10/2 = 9 - 5 = 4. This indicates a benzene ring (DBE=4). "These isomers do not decolourise KMnO₄ solution" means they do not have any C=C double bonds (alkenes) or C≡C triple bonds (alkynes) or benzylic hydrogens that can be oxidized by KMnO₄. This implies that all alkyl side chains are fully saturated a

Question

Q.72. Consider the isomers of hydrocarbon with molecular formula C₈H₁₀. These isomers do not decolourise KMnO₄ 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 _________.

FirstInTest · Accepted Answer

Sol. The molecular formula is C₈H₁₀. The degree of unsaturation (DBE) = C + 1 - H/2 = 8 + 1 - 10/2 = 9 - 5 = 4. This indicates a benzene ring (DBE=4).

"These isomers do not decolourise KMnO₄ solution" means they do not have any C=C double bonds (alkenes) or C≡C triple bonds (alkynes) or benzylic hydrogens that can be oxidized by KMnO₄. This implies that all alkyl side chains are fully saturated and do not have benzylic hydrogens that are easily oxidized (e.g., no methyl or ethyl groups directly attached to the ring if the oxidation is to be avoided, but this is usually interpreted as no oxidizable groups like alkenes/alkynes).
However, the common interpretation for "do not decolorize KMnO₄" in the context of C₈H₁₀ isomers usually means no side chains that can be easily oxidized, i.e., no alkyl groups with benzylic hydrogens that can be oxidized to carboxylic acids. This is a bit tricky. If it means no benzylic hydrogens, then only tert-butylbenzene would fit, but C₈H₁₀ doesn't allow for tert-butylbenzene.

A more common interpretation for C₈H₁₀ and KMnO₄ test is that it refers to the absence of alkenes/alkynes. All aromatic C₈H₁₀ isomers are alkylbenzenes, which *do* have benzylic hydrogens and *can* be oxidized by strong KMnO₄. However, if the question implies a *mild* KMnO₄ test, then it might be referring to the absence of alkenes/alkynes.

Let's list the aromatic isomers of C₈H₁₀:
1.  Ethylbenzene: C₆H₅-CH₂CH₃
2.  o-Xylene: 1,2-dimethylbenzene
3.  m-Xylene: 1,3-dimethylbenzene
4.  p-Xylene: 1,4-dimethylbenzene

All these compounds have benzylic hydrogens and would typically be oxidized by hot, acidic KMnO₄. If the condition "do not decolourise KMnO₄ solution" is strictly applied, then none of these would fit. This part of the question might be misleading or refer to a specific context (e.g., cold, dilute KMnO₄, which tests for alkenes/alkynes).

Assuming the question implies that we should consider all possible aromatic isomers of C₈H₁₀ and then perform monochlorination:

1.  **Ethylbenzene (C₆H₅-CH₂CH₃)**:
    *   Chlorination on the ring (ortho, meta, para): 3 isomers (o-, m-, p-chloroethylbenzene)
    *   Chlorination on the ethyl group:
        *   -CH₂Cl-CH₃ (1-chloro-1-phenylethane) - chiral center, so 2 stereoisomers (R, S)
        *   -CH₂-CH₂Cl (1-chloro-2-phenylethane)
    Total for ethylbenzene = 3 + 1 + 1 = 5 structural isomers (ignoring stereoisomers for now, as the question asks for structural isomers only).
    Let's re-evaluate the side chain chlorination. For -CH₂CH₃:
    - Chlorination at C1 (benzylic carbon): C₆H₅-CHCl-CH₃. This carbon is chiral. So, 1 structural isomer.
    - Chlorination at C2: C₆H₅-CH₂-CH₂Cl. 1 structural isomer.
    Total for ethylbenzene = 3 (ring) + 2 (side chain) = 5 structural isomers.

2.  **o-Xylene (1,2-dimethylbenzene)**:
    *   Chlorination on the ring: The ring has 2 unique positions for substitution (e.g., 3-chloro-1,2-dimethylbenzene and 4-chloro-1,2-dimethylbenzene). So, 2 isomers.
    *   Chlorination on a methyl group: Since both methyl groups are equivalent, chlorinating either one gives the same product (e.g., (2-methylphenyl)methyl chloride). So, 1 isomer.
    Total for o-xylene = 2 + 1 = 3 structural isomers.

3.  **m-Xylene (1,3-dimethylbenzene)**:
    *   Chlorination on the ring: The ring has 3 unique positions for substitution (e.g., 2-chloro-1,3-dimethylbenzene, 4-chloro-1,3-dimethylbenzene, 5-chloro-1,3-dimethylbenzene). So, 3 isomers.
    *   Chlorination on a methyl group: Both methyl groups are equivalent. Chlorinating either one gives the same product (e.g., (3-methylphenyl)methyl chloride). So, 1 isomer.
    Total for m-xylene = 3 + 1 = 4 structural isomers.

4.  **p-Xylene (1,4-dimethylbenzene)**:
    *   Chlorination on the ring: All ring positions are equivalent. So, 1 isomer (e.g., 2-chloro-1,4-dimethylbenzene).
    *   Chlorination on a methyl group: Both methyl groups are equivalent. Chlorinating either one gives the same product (e.g., (4-methylphenyl)methyl chloride). So, 1 isomer.
    Total for p-xylene = 1 + 1 = 2 structural isomers.

Total number of monochloro structural isomers = 5 (from ethylbenzene) + 3 (from o-xylene) + 4 (from m-xylene) + 2 (from p-xylene) = 14.

The diagram in the solution confirms this breakdown, showing the different positions for chlorination for each isomer and summing them up to 14.

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