Qvault OC Question

[DogeDDS]

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Can someone explain why this is the correct answer to the question? The answer I would've chosen isn't even offered on the list.

Wouldn't this mechanism (electrophilic addition) allow a hydride shift to occur, allowing 1 chlorine to attach to the tertiary carbon that's part of the ring, and only one would be on the secondary carbon that was part of the triple bond?

 

[portlandstateofmind]

This rxn produces geminal dihalides as shown in the correct answer.

 

[1Overdrive]

In a reaction like that (H-X with alkene/alkyne), the H and X add at the same time, avoiding any carbocations, thus, avoiding any methyl/hydride shifts. It is markovnikov, so the Cl will bond with the most substituted carbon, producing a gem-dihalide.

 

[DogeDDS]

In a reaction like that (H-X with alkene/alkyne), the H and X add at the same time, avoiding any carbocations, thus, avoiding any methyl/hydride shifts. It is markovnikov, so the Cl will bond with the most substituted carbon, producing a gem-dihalide.

Thanks, I believe I found the answer. However, I think the mechanism of HX + alkene still allows for the carbocation intermediate. The main reason this isn't true (I believe) with an alkyne is because a vinyl carbocation (which only has 1 alkyl group to stabilize the developing negative charge) is highly unstable, thus, the proton and halide add at the same time

 

[emitpeels]

Break the triple bond, add a Hydrogen to the carbon that has more hydrogen already, Chlorine to the other. Rinse and repeat!

 

[larrylu]

look at it as markovnikov addition of a halide x2

 

[DogeDDS]

If the original molecule was cyclohexylethylene (double CC bond instead of triple), the Cl- would then add to the tertiary carbon, correct? In other words, would a hydride shift occur.

 

[Triflate]

yes you are correct! and your product would be 1-chloro-1-ethylcyclohexane

 

[DogeDDS]

yes you are correct! and your product would be 1-chloro-1-ethylcyclohexane

Thank you