- Identifying nucleophilic and electrophilic centers
- Curly arrow conventions in organic chemistry
- Intro to organic mechanisms
- Alkyl halide nomenclature and classification
- Sn1 mechanism: kinetics and substrate
- Sn1 mechanism: stereochemistry
- Carbocation stability and rearrangement introduction
- Carbocation rearrangement practice
- Sn1 mechanism: carbocation rearrangement
- Sn1 carbocation rearrangement (advanced)
- Sn2 mechanism: kinetics and substrate
- Sn2 mechanism: stereospecificity
- Sn1 and Sn2: leaving group
- Sn1 vs Sn2: Solvent effects
- Sn1 vs Sn2: Summary
Examples of using hydride and methyl shifts to form more stable carbocation.
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- Are there geometric constraints on the shifts? For example, if you compared different cycloalkanes would the differing bond angles impact how readily a shift happened or what group shifts?(5 votes)
- should the shifts be an adjacent carbon away? for example, a carbon that is two away would not be able to shift(4 votes)
- Yes these rearrangements typically move a hydride or alkyl group to an adjacent carbon.
In some cases you can get stepwise movement that results in the positive charge moving more than one carbon, but only if each step results in a more stable carbocation.(2 votes)
- even though these carbocation rearrangements render products that are significantly more energetically stable, does this entire process need energy to begin? also is the net energy change for this rearrangement endothermic or exothermic?(3 votes)
- Would the secondary carbocations form at all? Are those minor products?(2 votes)
- Carbocations are quite unstable so you should be thinking of them as intermediates rather than products.
That said, in reactions with rearrangeable carbocation intermediates, you will typically get a mixture of products:
• products resulting from more stable (rearranged) carbocation intermediates will be more abundant
• products resulting from less stable (unrearranged) carbocation intermediates will be less abundant.(2 votes)
- How is hydrogen able to carry two electrons from the bond?(2 votes)
- For a tertiary carbocation, the bulky groups around it won't hinder the incoming nucleophile?Then how it is more stable?(2 votes)
- how to solve the questions containing ring like benzene ring: please give some tips..(1 vote)
- So you mean a benzylic carbocation?
And do you mean resonance structures of a benzylic carbocation or methyl and hydride rearrangements?
Because it's a rearrangement the same rule applies: trying to form a more stable carbocation than the original. It would be the same process described in the video.(2 votes)
- If tertiary carbocation is highly stable, then how come it accept a nucleophile and undergo nucleophilic reactions?
Since carbocations are already stable, won't they be least likely to accept a group or even to undergo any reaction??(1 vote)
- It’s not highly stable...but it’s much more stable (meaning significantly more likely to form) than a primary or secondary.(2 votes)
- [Narrator] In the last video we looked at hydride shifts and methyl shifts, so let's do some carbocation rearrangement practice. Let's start with this carbocation. So, we have a plus one formal charge on this carbon, and then we have the carbon with the positive charge bonded to two other carbons, so this is a secondary carbocation. Let's think about what could possibly shift here. So, first let's pick one of the two hydrogens on this carbon here, and let's say one of them is involved in a hydride shift. So, that'll be our first attempt. So, we take this hydrogen and these two electrons in a hydride shift, and we move them over to this carbon. So, let's draw what we would form. So, I'll draw in the ring here, and let me go ahead and put in the methyl group over here on the right. We know that there was already a hydrogen, I should say, on this carbon that's marked in magenta. So, let me draw in that hydrogen. And the hydride in red, right, that's going to move onto that carbon as well. So, let me go ahead and mark this as being the red hydrogen. That leaves one hydrogen, and I'll make this one in green, on this carbon in green. So, let's say this is the carbon in green, and then there's one hydrogen left on that carbon. We took a bond away from the carbon in green, and so that carbon now has a plus one formal charge. So, plus one formal charge on this carbon. So, we still have a secondary carbocation. The carbon in green is bonded to two other carbons, so this is a secondary carbocation, which is not an improvement upon our original secondary carbocation. Now, we could also draw it like this. So, I draw out all the hydrogens, but when you do a lot of practice you don't have to draw in all of the hydrogens. You could just think about a hydride shift, and you have your methyl group right here, forming a carbocation at this carbon. So, when you're first doing this you need to draw in all the hydrogens just for practice. Alright, let's think about the possibility of a methyl shift, because that hydride shift didn't work out very well, so let's try to move this methyl group over to the carbon in magenta. So, let's draw what we would form. So, here is our ring, and we're moving our methyl group in red over to this carbon. So, let me go ahead and draw in a CH3 in red. That carbon already had a hydrogen on it, so I'll go ahead and draw in the hydrogen that was already on that carbon, and we're taking a bond away from, let me make this carbon blue, so this carbon in blue is losing that CH3, but there was a hydrogen on that carbon originally, so let me go ahead and draw it in. There was a hydrogen here, so I'll make that one in blue, and so it's still there. But we did take away a bond from the carbon in blue, which means that this carbon in blue has a plus one formal charge. This is still a secondary carbocation. The carbon in blue is directly bonded to two other carbons, so this is a secondary carbocation, which, again, is not an improvement upon our original carbocation. And also, instead of drawing your carbocation this way, with all these bonds in here, you could just go ahead and show your methyl group as being here, and then a plus one formal charge on this carbon. So finally, what's left? What possibility is left? Well, we could try a hydride shift again, but instead of trying one of the two hydrogens on the left, we could try this hydrogen on the right, the one in blue. So, lets try a hydride shift with that one. So, this hydrogen and these two electrons are gonna move over here to the carbon in magenta. So, let's draw what we would form. We have our ring, and the methyl group stays there. The hydrogen in blue just moved over to here, so actually, let me go ahead and make that blue, so we can distinguish it, so here is the hydrogen, and those two electrons that moved. There was already a hydrogen on that carbon, so I will draw in the original hydrogen here, and we took a bond away from the carbon in blue. So, here's the carbon in blue, which means that is where our carbocation is. That is a plus one formal charge. Now, let's look at this resulting carbocation. The carbon that's in blue is directly bonded to one, two, three other carbons So, this is a tertiary carbocation. And we know from the previous video that a tertiary carbocation is more stable than a secondary carbocation. So, this is the rearrangement that we would see. We're going from a secondary carbocation to a tertiary carbocation via a hydride shift. And just like the previous examples, we don't need to draw in the hydrogens. We could just show our tertiary carbocation leaving out the hydrogens. So, put a methyl group in here, plus one formal charge on this carbon. So, this tertiary carbocation is more stable than the secondary ones. Let's do another carbocation rearrangement problem. So, this one's actually a little bit easier than the previous one. So, here's our carbocation, and the carbon with the plus one formal charge is directly bonded to two other carbons, which makes this a secondary carbocation. So, let's think about what kind of shifts that we could possibly have. So first, we know that this carbon has two hydrogens on it, so we could try doing a hydride shift with one of those hydrogens. So, one of these hydrogens and these two electrons could do a hydride shift and move over here to this carbon. So, let's draw what we would make. Let's draw our ring in here, and let's put in these two methyl groups coming off of that carbon. So, the hydrogen in red is now this hydrogen. Let me go ahead and make this red, and let me highlight these two electrons. So, that's what moved in our hydride shift. And that moved to this top carbon here in magenta. We know there was already a hydrogen bonded to that magenta carbon in the beginning, right? So, let's go ahead and put in that original hydrogen on that magenta carbon. We took a bond away from this carbon, so I'll make this the carbon in green. But, we still have a hydrogen attached to that carbon. So, the carbon in green is this one, and there's still a hydrogen attached to it, but since we took a bond away from it that means that we have a plus one formal charge on the carbon in green. And the carbon in green is directly bonded to two other carbons, so we formed a secondary carbocation, which is not an improvement upon our original secondary carbocation. So, we haven't increased our stability. Remember, we could also draw this without all those hydrogens in there, just a little bit easier to see. So, we have our methyl groups with a plus one formal charge on this carbon. Let's try doing another kind of shift. So, if we look over here to this carbon in blue we have these two methyl groups. So, we could do a methyl shift with one of these methyl groups. So, let's take this CH3, we're gonna move it over here to the carbon in magenta. So, let's try a methyl shift, and let's draw in our ring. Well, one of the methyl groups is going to remain on this carbon in blue. So, let me highlight it here. So, this is the carbon in blue, and this one of the methyl groups that stayed behind. Another one moved over here to this carbon. So, let me go ahead and draw that in, CH3, and let me just go ahead and highlight that in blue. This was the methyl group that underwent our methyl shift, so it's this one right here. And again, on that carbon in magenta there was originally a hydrogen, so let's draw in that hydrogen. But notice that we took a bond away from the carbon in blue, so this carbon in blue gets a plus one formal charge, and if we think about what kind of carbocation this is, the carbon in blue is directly bonded to one, two, three other carbons. So, this is a tertiary carbocation, which we know is more stable than a secondary carbocation. So, this the rearrangement that's going to occur. We get a methyl shift to form a tertiary carbocation. Now let's go ahead and draw it without the hydrogen in there so we can see what it looks like a little bit better, so we have there's methyl groups, we have a plus one, a formal charge on this carbon.