Question

Give the structure of each of the following aromatic hydrocarbons. a. $o$ -ethyltoluene b. $p$ -di-tert-butylbenzene c. $m$ -diethylbenzene d. 1-phenyl-2-butene

Short answer

The structures of the given aromatic hydrocarbons are: a. o-ethyltoluene: $$\text{C - C^6}$$ b. p-di-tert-butylbenzene: $$\text{C^{tBu - C^6 - CtBu}}$$ c. m-diethylbenzene: $$\text{C^2 – C^6 – C^2}$$ d. 1-phenyl-2-butene: $$\text{C-C^6-(C=C)-C-C}$$

Step-by-step solution

Draw the core benzene ring structure

For all four compounds, the basic structure will be derived from benzene, which has a six-membered carbon ring with alternating single and double bonds (aromatic). The benzene ring is often represented in shorthand with a hexagon containing a circle inside, which denotes the aromatic structure. o-, m-, or p- denote the positions of substituents on the benzene ring: "o-" stands for the ortho position (1,2-disubstituted), "m-" for meta (1,3-disubstituted), and "p-" for para (1,4-disubstituted).

Draw o-ethyltoluene#a.

Toluene is a benzene derivative with a methyl group substituted on one of the carbon atoms of the benzene ring. To draw o-ethyltoluene, we first draw the benzene ring and attach a methyl group to one of the carbon atoms. Next, we add an ethyl group (CH2CH3) to the adjacent carbon atom in the ortho position.

Draw p-di-tert-butylbenzene#b.

To draw p-di-tert-butylbenzene, we begin by drawing the benzene ring. Then, we attach a tert-butyl group (CH3)3C at two opposite carbon atoms of the benzene ring, denoting the para (1,4-disubstituted) positions.

Draw m-diethylbenzene#c.

To draw m-diethylbenzene, we start by drawing the benzene ring. Then, we attach two ethyl groups (CH2CH3) to the carbon atoms in the meta (1,3-disubstituted) positions.

Draw 1-phenyl-2-butene#d.

To draw 1-phenyl-2-butene, we first draw a butene molecule, which has four carbon atoms in a chain with a double bond between carbons 2 and 3 (C=C). The 1-phenyl denotes that the benzene ring is attached to the first carbon of the butene chain. After drawing both butene and the benzene ring, we connect them by attaching the first carbon of the butene chain to one of the carbons of the benzene ring. The structures of the aromatic hydrocarbons are shown below for reference: a. o-ethyltoluene: $$C-C^6$$ b. p-di-tert-butylbenzene: $$C^{tBu-C^6-CtBu}$$ c. m-diethylbenzene: $$C^2–C^6–C^2$$ d. 1-phenyl-2-butene: $$\text{C-C^6-(C=C)-C-C}$$