Problem 34 - Chemistry: The Central Science

Question

(a) One test for the presence of an alkene is to add a small amount of bromine, 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) The bromine test works for alkenes because they undergo halogenation, causing the red-brown color to disappear. Aromatic hydrocarbons like benzene do not undergo halogenation under normal conditions due to resonance stability, so the test is not effective for them. (b) To synthesize para-bromoethylbenzene from benzene, perform the following steps: 1. Friedel Crafts Alkylation: Benzene reacts with ethyl chloride (CH3CH2Cl) and AlCl3 as a catalyst to form ethylbenzene. 2. Electrophilic Aromatic Substitution: Ethylbenzene reacts with bromine (Br2) and FeBr3 as a catalyst to form para-bromoethylbenzene. The isomeric side product formed is ortho-bromoethylbenzene.

Step 1: Part (a) - Reaction of bromine with alkenes and aromatic hydrocarbons

In the presence of an alkene, bromine undergoes a reaction called halogenation, in which the red-brown color of bromine disappears. The disappearance of this color is a useful test to check the presence of an alkene. The reaction proceeds as follows: \[ \text{Alkene} + \text{Br}_2 \rightarrow \text{Dibromoalkane} \] However, when we try to perform the same test with an aromatic hydrocarbon such as benzene, the red-brown color of bromine will not disappear. This is because aromatic hydrocarbons are less reactive than alkenes and cannot undergo halogenation under normal conditions due to the resonance stability of their structure.

Step 2: Part (b) - Synthesis of para-bromoethylbenzene and isomeric side products

To synthesize para-bromoethylbenzene from benzene, we need to perform the following steps: 1. Friedel Crafts Alkylation: Reaction of benzene with ethyl chloride (CH3CH2Cl) in the presence of a Lewis acid catalyst (AlCl3). \[ \text{Benzene} + \text{CH3CH2Cl} \xrightarrow{AlCl_3} \text{Ethylbenzene} \] 2. Electrophilic Aromatic Substitution: Bromination of ethylbenzene using bromine (Br2) and iron(III) bromide (FeBr3) as a catalyst. The major product will be para-bromoethylbenzene as the ethyl group is an activating group and directs the incoming electrophile to the para-position. It will push for the formation of para-bromoethylbenzene. There may be a small yield of ortho-bromoethylbenzene as a side product. \[ \text{Ethylbenzene} + \text{Br}_2 \xrightarrow{FeBr_3} \text{para-Bromoethylbenzene} + \text{ortho-Bromoethylbenzene (side product)} \] So, the possible isomeric side product formed in this reaction is ortho-bromoethylbenzene.