Draw the organic products formed when each alkyne is treated with two equivalents of HBr.

Name: Problem 8, Chapter 11: Organic Chemistry
Uploaded: 2019-05-10T01:16:29-08:00
Duration: 10 min 18 s
Channel: Nima Gharibi
Description: Draw the organic products formed when each alkyne is treated with two equivalents of HBr.

Draw the organic products formed when each alkyne is treated...

Key Concepts

Electrophilic Addition to Alkynes

This reaction involves an electrophilic addition mechanism where the alkyne, a ?-rich system, reacts with an electrophilic species—in this case, the proton from HBr—to initially form a cationic intermediate. This intermediate subsequently reacts with the bromide ion, leading eventually to the addition of two equivalents of HBr across the triple bond.

Markovnikov's Rule

Markovnikov's rule dictates that in the addition of hydrogen halides to unsymmetrical alkynes, the hydrogen atom attaches to the carbon with more hydrogen substituents, while the halogen attaches to the more substituted carbon. This rule is central to predicting the regiochemistry (placement of substituents) of the product when two equivalents of HBr are added.

Formation of Geminal Dihalides

When an alkyne reacts with two equivalents of HBr, the reaction typically proceeds through successive electrophilic additions that yield a geminal dihalide, where both bromine atoms attach to the same carbon atom. Understanding this concept is crucial for predicting product structures in multi-addition reactions involving alkynes.

Transcript

0:00 Hello, everyone.

00:01 Today we're doing chapter 11 problem 8, and this problem asks us to draw the organic products formed from each of these alkynes when you treat it with two equivalents of hbr.

00:09 So, obviously, we know that alkynes are very electron -rich.

00:14 Each of these pie bonds give two electrons, so these can be your lone pair electrons or actually your pie bond electrons used to act as nucleophiles.

00:24 And these nucleophiles can attack potential electrophiles.

00:27 So nucleophile is your electron -rich.

00:30 Electron source and an electrophile is an electron -poor atom that's being attacked.

00:35 So obviously, h versus br, hydrogen is more electron -poor than bromine.

00:39 So we know that hydrogen is going to be our electrophile in this situation.

00:43 But now we need to ask, well, there's two sides of this al -q -q -x -1 -al -kind.

00:48 We have this side where there's one hydrogen and this side that has one hydrogen and one hydrocarbon.

00:55 So one of these ch2, ch2, ch3s.

01:00 So which side is this hydrogen going to add on? well, if you remember markovnikov's rule, the rich get richer and the poor get poorer.

01:08 What does that mean? that means that the rich get richer.

01:10 So the side that has more protons is going to get your extra proton.

01:16 The side with the least number of protons is going to get your other group, which is non -proton.

01:21 So if you look at this structure, we know that the right side has one proton, well the left side has one proton, but it also has another non -proton substation.

01:30 So technically the right side has more protons.

01:35 So we know that the proton is going to be added to this side while leaving our carbocatin intermediates in a more substituted side and i'll tell you guys why that happens once we get to intermediate.

01:45 So i'm going to draw my pie bond going through the carbon nucleus my sorry my arrow starting from the source of my electrons going through the carbon nucleus suggesting that the sigma bond is going to be formed there and attacking the hydrogen atom.

02:02 When we make a bond, we need to break a bond.

02:04 So this sigma bond breaks to give us our intermediate.

02:21 And then we also have that hydrogen that we just added.

02:23 So i'll draw in black so you guys can differentiate and be our minus.

02:29 And we have the original proton here.

02:33 And then now, because there's only three substituents on this carbon, we have a carbokadine intermediate.

02:38 So why does is markov's rule? why does it exist? why do we follow my carboniculture? well, by adding the proton on the least substituted side, this generates your carbocation on the most substituted side.

02:50 So when you generate your carbocation in the most substituted side, you now have more substituents inductively donating electron density to this carbocation to stabilize.

03:00 So this generates a more stabilized and a more substituted carbocation intermediate.

03:04 So here, as you see, we have a secondary carbocation.

03:07 But now imagine if you add a proton on this carbon, well, then the carbocatine will exist.

03:11 Here and this would be a primary carbocatin.

03:14 So obviously we know that a secondary carbonyl is more stable than a primary carbonython.

03:18 So this is why markovna's rule exists.

03:21 But now we also need to continue the problem.

03:24 So we have another equivalent hbr.

03:29 So we need to do the same thing.

03:31 Following markovinaikov's rule, we know that the side with more protons is going to get the extra proton.

03:35 So we know that obviously is going to go through this carbon nucleus with the two protons to pick up the extra proton, breaking the sigma bond between hydrogen bromine.

03:43 To generate your second carbocata intermediate, but before we do that, i actually skip the step.

03:51 So let's move backwards.

03:52 And now, starting from this intermediate, we have a carbocatin intermediate at this intermediate stage.

03:59 So we actually need to neutralize that.

04:00 So the bromine that fell off is now electron rich.

04:05 It has lone pair electrons...

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