What is the major product of each of the following...

Key Concepts

Electrophilic Addition to Alkynes

This concept refers to the reaction mechanism in which an electrophile, such as the hydrogen ion from HBr, attacks the electron?rich triple bond of an alkyne. The ? electrons of the triple bond interact with the electrophile, initiating the addition process that ultimately converts the alkyne into a vinyl halide (an alkene with a halogen substituent).

Markovnikov's Rule

Markovnikov's rule predicts the regiochemistry of electrophilic additions to unsymmetrical multiple bonds. It states that the hydrogen atom from HX will add to the carbon that already has more hydrogen atoms, while the halide (X) attaches to the more substituted carbon. This rule is crucial for determining the major product in the addition of HBr to alkynes, as it guides which carbon will bear the bromine atom.

Transcript

00:04 This problem asks us to draw the products of each of these reactions.

00:09 So starting with a, if we have hbr with a pie bond, what's going to happen is the h will be attacked by one of those pie bonds and we'll kick the br off.

00:24 So we'll have two hs on one of the carbons, right? and we always put the h on the carbon that is less substituted so that the other one can become a carbocation because we want the more stable carbocation.

00:37 And then our bromine will come in and attack that carbocation.

00:42 So here we will get h2c of our br and then the ch3.

00:50 So that's if we have one equivalent of hbr.

00:53 If we have excess hbr or two equivalents of hbr, then we can do that to both pie bonds.

00:58 And the exact same thing will happen, but now we'll have three hydrogens on this end carbon and two bromines in the middle because again we're still going to have that more stable carbocanion on the internal carbon with br2 we have a little bit different of a reaction so what happens here is if we draw vr one of the pie bonds will come in attack one of the brs and the br will also come and attack one of the carbons and we'll kick off the other one so we'll get a transition state that looks like this we'll have that br here attached to both carbons and a it will be positive.

01:38 And then our other bromine, which is now negative, will come in and attack one of these carbons from the opposite side and we'll push this off so it's only on one.

01:48 So then we'll have ch3.

01:51 There's a double bond.

01:52 The bromines have to be opposite of each other because they add ante because of the pyramid transition state.

02:00 There's no room for it to come in and attack from this side.

02:02 So it has to come in from the opposite side.

02:05 If we've accessed hbr, we'll we'll end up with an alkan now.

02:11 And we'll just put another bromine on both of those carbons.

02:15 So now there will be two here and two here.

02:24 Now if we have an internal alkyne with an excess of hbr, it doesn't matter which carbon the br goes on because they're equal...