Draw the major products obtained from the reaction of one...

Key Concepts

Thermodynamic Product

The thermodynamic product is the more stable product that is favored under conditions that allow the system to reach equilibrium, such as higher temperatures or longer reaction times. Even if the formation of this product involves a higher activation energy barrier, it is more stable energetically in the long run, making it the predominant product in a reaction governed by thermodynamic control.

Electrophilic Addition Reactions

Electrophilic addition involves an electrophile, such as HBr, attacking a region of high electron density like a double or triple bond. The reaction typically proceeds via the initial attack on the unsaturated bond to form a carbocation intermediate, which is then attacked by a nucleophile. This mechanism is central to understanding how various products are formed when halogens or hydrogen halides react with organic compounds containing multiple bonds.

Carbocation Intermediates

Carbocations are positively charged intermediates formed during many electrophilic addition reactions. Their stability significantly influences the course of the reaction; more substituted carbocations are generally more stable and may lead to different regioselectivity in the addition process. The formation and rearrangement of these intermediates can determine both which product is formed and whether the reaction is under kinetic or thermodynamic control.

Kinetic Product

The kinetic product is the product that forms fastest due to the lowest activation energy barrier, even if it is not the most stable one. Under kinetic control, often realized at lower temperatures or shorter reaction times, the reaction pathway favors the route with the quickest formation, which may lead to a less thermodynamically favorable structure.

Transcript

00:05 In this question, we are asked to draw the major products from the reaction of one equivalent of hbr with each of these compounds and then to indicate the kinetic and thermodynamic products.

00:16 So these are both going to give us one two and one four addition products since both of these compounds are dianes.

00:23 And so to figure out what those are going to look like, we're going to start by numbering the dine.

00:28 So numbering just the four carbons that are participating in those double bonds.

00:32 And we're going to number them to get the most stable carbocation.

00:35 On each side.

00:36 So we can either go one, two, three, four, or one, two, three, four.

00:41 Our carbokad ions will always be on carbons two and four, so we'll take a look at what those would be.

00:46 So if we went from the left on this top one here, that would be secondary for number two and tertiary for number four.

00:53 If we went from the top, it would be tertiary for number two and tertiary for number four.

00:58 So we're going to go with the way they gives us two tertiary carbocadions.

01:01 So i'll number it like this.

01:03 And the first step is always to take that one, two double bond and grab the hydrogen with it and kick that bromine away.

01:10 And that gives us a carbocadion intermediate.

01:14 So that will look like this for this structure.

01:18 So the three and four double bond hasn't moved.

01:20 We have our carbocation on number two since that's the allelic position.

01:24 And that hydrogen went on to carbon one to give it that fourth bond that it needed.

01:29 And then we have a bromine nucleophile left over.

01:33 Now this also has a resonant structure.

01:35 That's how we get to the 1 -4 product.

01:37 So if we move this double bond like this, we get a second resonance contributor for that carbocadion that looks like this.

01:44 So now the double bond is there, the carbocadion is there.

01:47 So then we'll get two products here, and those occur just from the bromine nucleophile attacking each of the carbocadions.

01:55 So from that first carbocation that i drew, we'll get the 1 -2 product.

02:00 And that will look like this.

02:01 So this double bond is still here.

02:03 Bromine is on that carbon.

02:06 And then from the second one that i drew, we'll get the bromine over here.

02:12 And so that'll be our product there.

02:14 So this is our 1 -2 and this is our 1 -4.

02:18 And then as far as kinetic and thermodynamic products, the kinetic product is always the 1 -2 product.

02:25 So this is kinetic.

02:26 And that's because of the proximity effect...

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