Friedel Crafts Alkylation
Let's take a look at our discussion question this week!
We want to determine which of the statements regarding Friedel Crafts alkylation of benzene are correct.
From product cumene we can deduce the reactant is 2-chloropropane, with AlCl3 as catalyst.
1. Cumene does not decolourise aqueous bromine at room temperature.
Alkyl groups are activating and make benzene slightly more reactive, but bromination will still require presence of catalyst such as FeBr3.
Only highly activating groups like -OH and -NH2 groups will make benzene reactive enough to react with bromine without the need for catalyst.
Hence statement 1 is true.
2. Aluminium chloride is acting as a Lewis acid in the synthesis of cumene.
Let's take a look at the electrophilic substitution mechanism of benzene to form cumene.
In step 1, AlCl3 acts as an electron pair acceptor from Cl- to generate the electrophile.
Hence statement 2 is true.
3. The electrophile which reacted with benzene in the synthesis of cumene was CH3CH2CH2+.
Since reactant is 2-chloropropane, the electrophile should have positive charge on second carbon instead of terminal carbon.
Hence statement 3 is false.
Therefore the answer to this question is option B.
Topic: Benzene, Organic Chemistry, A Level Chemistry, Singapore
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