Octane number is assigned to gasoline to indicate the tendency of "knocking" in the automobile's engine. The higher the octane number, the more smoothly the fuel will burn without knocking. Branched-chain aliphatic hydrocarbons have higher octane numbers than straight-chain aliphatic hydrocarbons, and aromatic hydrocarbons have the highest octane numbers. (a) Arrange these compounds in the order of decreasing octane numbers: \(2,2,4-\) trimethylpentane, toluene (methylbenzene), \(n\) -heptane, and 2 -methylhexane. (b) Oil refineries carry out catalytic reforming in which a straight-chain hydrocarbon, in the presence of a catalyst, is converted to an aromatic molecule and a useful by-product. Write an equation for the conversion from \(n\) -heptane to toluene. (c) Until 2000 , tert-butylmethyl ether had been widely used as an antiknocking agent to enhance the octane number of gasoline. Write the structural formula of the compound.

Short Answer

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(a) Toluene, \(2,2,4-\) trimethylpentane, 2 -methylhexane, \(n\) -heptane (b) \( n - \) heptane + 3H2 -> Toluene + 3 H2O (c) A central carbon forming single bonds with three methyl groups and one oxygen atom, which is bonded to a methyl group.

Step by step solution

01

Identifying the order of the compounds according to octane number

According to the information provided, the order of octane numbers is: aromatic hydrocarbons > branched-chain aliphatic hydrocarbons > straight-chain aliphatic hydrocarbons. So, the order for the given compounds is toluene (aromatic hydrocarbon) > \(2,2,4-\) trimethylpentane (branched-chain aliphatic hydrocarbon) > 2 -methylhexane (branched-chain aliphatic hydrocarbon) > \(n\) -heptane (straight-chain aliphatic hydrocarbon).
02

Writing the catalytic reforming reaction

During catalytic reforming, a straight-chain hydrocarbon is converted to an aromatic molecule. This equation can be represented as follows: \( n - \) heptane + 3H2 -> Toluene + 3 H2O.
03

Drawing the structure for tert-butylmethyl ether

The structural formula for tert-butylmethyl ether is drawn as follows: A carbon atom is at the center. It forms single bonds with three methyl (CH3) groups and one oxygen atom. The oxygen atom then forms a single bond with another carbon atom, which then forms single bonds with three hydrogen atoms (making it a CH3 as well).

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