Question

Propose a mechanism for the entire reaction of pent-1-yne with 2 moles of HBr. Show why Markovnikov's rule should be observed in both the first and second additions of HBr.

Accepted Answer

Step 1: Begin by identifying the structure of pent-1-yne (CH≡C-CH2-CH2-CH3), which is a terminal alkyne. The reaction involves the addition of 2 moles of HBr, so the mechanism will proceed in two stages: the first addition and the second addition. Step 2: In the first addition of HBr, the π-electrons of the triple bond attack the proton (H⁺) from HBr, leading to the formation of a carbocation intermediate. According to Markovnikov's rule, the proton will add to the carbon of the triple bond that already has more hydrogens (the terminal carbon), resulting in a more stable secondary carbocation on the internal carbon. Step 3: The bromide ion (Br⁻) then attacks the carbocation, forming a vinyl bromide intermediate (CH2=CHBr-CH2-CH2-CH3). This completes the first addition of HBr. Step 4: In the second addition of HBr, the π-electrons of the double bond in the vinyl bromide intermediate attack another proton (H⁺) from the second molecule of HBr. Again, following Markovnikov's rule, the proton adds to the carbon of the double bond that has more hydrogens, leading to the formation of a more stable carbocation on the carbon bonded to the bromine atom. Step 5: Finally, the bromide ion (Br⁻) attacks the carbocation formed in the second addition, resulting in the final product, 2,2-dibromopentane (CH3-CBr2-CH2-CH2-CH3). Markovnikov's rule is observed in both steps because the formation of the more stable carbocation intermediate drives the regioselectivity of the reaction.

Propose a mechanism for the entire reaction of pent-1-yne with 2 moles of HBr. Show why Markovnikov's rule should be observed in both the first and second additions of HBr.

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Key Concepts

Alkyne Reactivity

Markovnikov's Rule

Reaction Mechanism