Problem 38 - Chemistry The Central Science

Question

(a) One test for the presence of an alkene is to add a small amount of bromine, which is a red-brown liquid, and look for the disappearance of the red-brown color. This test does not work for detecting the presence of an aromatic hydrocarbon. Explain. (b) Write a series of reactions leading to para-bromoethylbenzene, beginning with benzene and using other reagents as needed. What isomeric side products might also be formed?

Step-by-Step Solution

Answer

(a) Bromine test does not work for aromatic hydrocarbons because they don't undergo electrophilic addition reactions like alkenes. Aromatic compounds undergo electrophilic substitution reactions, which require a catalyst such as FeBr3 or AlBr3 to form brominated products. (b) To produce para-bromoethylbenzene from benzene: 1. Benzene reacts with CH3CH2Cl, in the presence of AlCl3, to form ethylbenzene. 2. Ethylbenzene reacts with Br2, in the presence of FeBr3, to form para-bromoethylbenzene. Potential isomeric side products are ortho-bromoethylbenzene and meta-bromoethylbenzene.

Step 1: (a): Explanation for the unavailability of bromine test for aromatic hydrocarbons

Bromine reacts with alkenes to form dibromides, converting the reddish-brown bromine liquid to a colorless solution. This reaction is known as bromination. The presence of the double bond in alkenes facilitates the electrophilic addition reaction with bromine. In aromatic hydrocarbons, such as benzene, the electrons are delocalized in the ring and are not so easily available for addition. Aromatic hydrocarbons undergo an electrophilic substitution reaction with bromine, which is facilitated by a strong Lewis acid (catalyst) such as FeBr3 or AlBr3, to form bromobenzene. The direct addition reaction is impossible in aromatic hydrocarbons without the catalyst, making the bromine test unresponsive.

Step 2: (b)(i): Formation of Ethylbenzene

Benzene can be ethylated to form ethylbenzene using Friedel-Crafts alkylation. We will need an anhydrous aluminum chloride (AlCl3) catalyst and ethyl chloride (CH3CH2Cl). Reaction: Benzene + CH3CH2Cl (in presence of AlCl3) → Ethylbenzene + HCl

Step 3: (b)(ii): Formation of para-Bromoethylbenzene

Ethylbenzene can be brominated to yield para-bromoethylbenzene. We will need bromine (Br2) and a strong Lewis acid catalyst, ferric bromide (FeBr3). Reaction: Ethylbenzene + Br2 (in presence of FeBr3) → para-Bromoethylbenzene + HBr

Step 4: (b)(iii): Potential isomeric side products

While carrying out the bromination of ethylbenzene, ortho- and meta-isomers may be formed as side products due to electrophilic substitution at different positions on the benzene ring. 1. ortho-Bromoethylbenzene: If bromination occurs at the position adjacent to the ethyl group on the benzene ring, it results in the production of ortho-Bromoethylbenzene. 2. meta-Bromoethylbenzene: If bromination occurs at the position next to the ortho position or one carbon away from the ethyl group, it results in the production of meta-Bromoethylbenzene. In conclusion, while attempting to synthesize para-bromoethylbenzene from benzene, ortho- and meta-Bromoethylbenzene may be formed as isomeric side products.