Hept-2-yne reaction with HBr ?

[Lo.Lee.Ta.]

Okay, I am trying to determine the product of hept-2-yne with HBr.

It looks like:

CH3CH2CH2CH2C≡CCH3 + HBr

I thought this thing would follow Markovnikov's rule, making the Br attach to the most substituted carbon and the H attach to the least substituted carbon...

I thought the product would only be: CH3CH2CH2CH2C(Br)=CHCH3

In my solutions manual, there are really 2 products:

CH3CH2CH2CH2C(Br)=CHCH3 and CH3CH2CH2CH2CH=C(Br)CH3

I think this is called a racemic mixture, right?

Well, why does this form a racemic mixture?
Thank you so much! :)

 

[AGNuke]

Markonikov product is generally a major product, the other being a minor product.(like 90-10) Furthermore, the products are not enantiomers of each other, so they are not a racemic mixture.

 

[Lo.Lee.Ta.]

Thanks, AGNuke! :) Nice explanation!

So there is always a major and minor product but different levels of each for different reactions?

Well, how are we supposed to know when the minor product will be so minute not to even mention, when the minor product is worth mentioning, or when the two molecules formed will be in equal quantites?

In my solutions manuel, it shows ethynylcyclopentane + HBr to have only one product, which is (1-bromoethenyl)cyclopentane.

How come they don't show a minor product in this case?

Thanks so much! :)

 

[AGNuke]

Addition of Hydrogen Halide on Alkynes generally follows Markovnikov rule (lame explanation but bear with this, the real explanation is good and you may already know it).

Electrophilic attacks on Alkynes are already difficult to initiate as the result of unstable vinylic carbocation intermediate. In the first question, making carbocation on either carbon was feasible in major/minor ratio.

In the second question, however, there is "a hydrogen" on one of the C(triple)C Carbon, so Markovnikov rule is applied in full contest. The minor product will be negligible because of its intermediate, -CH=C⁺-H is very unstable in compared to intermediate of major (and only product reported).

Its only just like that. Just the instinct that the ratio of major/minor is appreciable or not.

 

[LudusRex]

Hello, thank you for your question. It is great to see you exploring chemical reactions and their products. The reaction between hept-2-yne and HBr is an example of an addition reaction, where the double bond of the alkyne is broken and new bonds are formed with the H and Br atoms.

You are correct in thinking that this reaction would follow Markovnikov's rule, where the H atom would attach to the least substituted carbon and the Br atom would attach to the most substituted carbon. This would result in the product CH3CH2CH2CH2C(Br)=CHCH3. However, there is also a second product formed, which is CH3CH2CH2CH2CH=C(Br)CH3.

This is because the addition of HBr to hept-2-yne can happen in two different ways, resulting in two different products. In one scenario, the H atom adds to the less substituted carbon, forming the first product. In the other scenario, the H atom adds to the more substituted carbon, forming the second product. This is known as an anti-Markovnikov addition.

The reason why this reaction forms a racemic mixture is due to the nature of the alkyne. The triple bond of the alkyne is very strong and stable, making it difficult for the HBr molecule to add to it in a specific orientation. This results in the H and Br atoms adding randomly, leading to both products being formed in equal amounts. This is known as a racemic mixture, where both enantiomers (mirror images) of the product are formed in equal amounts.

I hope this helps to clarify the reaction and its products. Keep exploring and asking questions, and never stop learning!