If 1-pentyne were treated with 1 equiv. Br2 in CCl4 and then 1 equiv. of HBr, what is the product?Select answer from the options below2,2-dibromopentane1,1,2-tribromopentane1,1-dibromopentane1,2,2-tribromopentane1,1,2,2-tetrabromopentane
Question
If 1-pentyne were treated with 1 equiv. Br2 in CCl4 and then 1 equiv. of HBr, what is the product?Select answer from the options below2,2-dibromopentane1,1,2-tribromopentane1,1-dibromopentane1,2,2-tribromopentane1,1,2,2-tetrabromopentane
Solution
The product would be 1,1,2-tribromopentane. Here's the step-by-step process:
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1-pentyne is an alkyne, which means it has a carbon-carbon triple bond. When it is treated with 1 equivalent of Br2 in CCl4, a bromine atom will add across the triple bond, turning it into a double bond. This is a halogenation reaction, specifically a bromination reaction. The product of this step is 1,2-dibromopentene.
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Then, when 1,2-dibromopentene is treated with 1 equivalent of HBr, another bromination reaction occurs. The bromine atom from HBr will add to the carbon atom that is less substituted, which is the carbon atom at the 1-position (the carbon atom at the end of the chain). This is because of Markovnikov's rule, which states that in the addition of a protic acid HX to an alkene or alkyne, the acid hydrogen (H) becomes attached to the carbon with fewer alkyl substituents, and the halide (X) group becomes attached to the carbon with more alkyl substituents. The product of this step is 1,1,2-tribromopentane.
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