- 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
Sn1 mechanism: stereochemistry
Stereochemistry of an Sn1 reaction and how it relates to the Sn1 mechanism. Created by Jay.
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- At5:20how do you know that the carbon is becoming sp2 hybridized, doesn't it still have all single bonds?(8 votes)
- Jessica, Jay has made previous made videos on 'general chemistry overview' which detail how to work out if an atom is sp1/2/3 hybridised based on it's stereo number (number of sigma bonds)
In short, if it has 4 sigma (single covalent) bonds, it is sp3 hybridised (1xS and 3xP), if it is has 3 sigma bonds as is the case here, it is sp2 hybridised (1xS and 2xP).
Thus in the example at5:20the carbon now has 3 sigma bonds, therefore making it SP2 hybridsed.
Hope that helps :)(20 votes)
- Maybe it's a stupid question but, the products are enantiomers "of each other", producing a racemate mixture, right?3:57(3 votes)
- Correct. SN1 reactions give racemization at the α carbon atom. If that is the only chiral centre, you get a racemic mixture. If there are other chiral centres, you get a pair of diastereomers.(9 votes)
- At0:53if the bonds of the sp2-bybridized C are all on the same plane why are the bond to 2 H's drawn as wedge and dashes? Am I missing something?(5 votes)
- They are all in the same plane. But they are all just dashes only if your eye is perpendicular to the plane. In the video, the plane is drawn as a parallelogram, which tells you that you are looking at it edge-on or just above the plane. That means that one of the bonds must be closest to your eye (a wedge) and the other must be furthest away (a dashed line).(4 votes)
- So the inversion form is preferred as the product?(3 votes)
- It is but it is very slight. You'll often hear that Sn1 reactions proceeds to a racemization of products. It is more correct to say that it proceeds to a mixture of inversion and retention of products. It's a subtle distinction but it is there.(4 votes)
- Two questions (maybe related):
1. Does it take energy to make a carbon go from an sp3 configuration to an sp2 configuration, as it does at about6:10?
2. If only one carbon goes from sp3 (tetrahedral) to sp2 (trigonal planar), how does this affect the shape of the chair configuration and the boat configuration. With only one of the carbons changing its angle within the ring, it seems to me that this would "warp" the ring and put some kind of strain on the other five carbon atoms and their bonds. Does this happen, and if it does, does it affect anything?(4 votes)
- Yes, but not because of the hybridization change. It is because you are separating a negatively charged bromide ion from a positively charged carbocation. It takes energy to separate positive and negative charges from each other.
Yes, converting one of the cyclohexyane carbon atoms to sp2 does put some strain in the ring. The other bond angles would compensate somewhat, but you would still have an overall chair or boat conformation.(2 votes)
- how do we say that propan-2-ol is achiral?(3 votes)
- to say that a molecule has an chiral carbon, it must have 4 different radicals attached to it, like the drawing (sorry) above, with R1, R2, R3 and R4 all differente
propan-2-ol is achiral because it dont have a carbon with all different radicals
- At around6:00, would it be possible to undergo an E1 reaction at one of the beta carbons to the bromine to form a cycloalkene and H2S? Why or why not?(2 votes)
- What are the methods of stereochemistry(2 votes)
- That question has a much broader answer than one can provide here. I suggest going to the subsection of videos in Organic Chemistry titled "Stereochemistry" and reviewing them. It will cover all the material you'll need to better understand sterochemistry(1 vote)
- Sn2 reaction proceeds with complete stereo-chemical inversion and a Sn1 reaction proceeds with racemisation, right ?(1 vote)
- Yes, that is one characteristic of SN1 and SN2 reactions.(3 votes)
- the negatively charged halogen will be shielding from the bottom as said in the video by jay at1:54so ,possibility of attack of nucleophile decreases from bottom but it is not necessary right that the halogen should be in the bottom it could have also been on the top ....after all ions move around in free space they wont stay in the same place right?(2 votes)
- [Narrator] In this video, we're going to look at the stereochemistry of the SN1 reaction. On the left is our alkyl halide, on the right is our nucleophile with a negative charge on the sulfur. We know that the first step of our SN1 mechanism should be loss of a leaving group. So if these electrons come off onto the bromine, we would form the bromide anion. And we're taking a bond away from the carbon in red. So the carbon in red should get a plus one formal charge. So let's draw the resulting carbocation here. So let me sketch that in. The carbon in red is this carbon. So that carbon should have a plus one formal charge. In the next step of our mechanism, our nucleophile will attack, alright? So the nucleophile attacks the electrophile and a bond will form between the sulfur and the carbon in red. But remember the geometry directly around that carbon in red, the carbons that are bonded to it, so this carbon in magenta, this carbon in magenta, and this carbon in magenta are in the same plane as the carbon in red, and so the nucleophile could attack from either side of that plane. At this point, I think it's really helpful to look at this reaction using the model set. So here's a screenshot from the video I'm going to show you in a second, and in that video I make bromine green, so here you can see this green bromine, this methyl group coming out at us in space is going to be red in the video. On the right side this ethyl group here will be yellow, and finally on the left side this propyl group will be gray. So here's our alkyl halide with our bromine going away from us, our methyl group coming out at us, our ethyl group on the right, and the propyl group on the left. So I'll just turn this a little bit so we get a different viewpoint, and we know that the first step is loss of our leaving group so I'm going to show these electrons coming off onto our bromine and leaving to form a carbocation. But that's not what the carbocation should look like. We need planar geometry around that central carbon. So here's another model which is more accurate. Now the nucleophile could attack from the left or from the right, and first let's look at what happens when the nucleophile attacks from the left. So we form a bond between the sulfur and the carbon, and let's go ahead and look at a model set of one of our products. So here's the product that results when the nucleophile attacks from the left side of the carbocation. Here's our carbocation again, and this time let's say the nucleophile approaches from the right side. So we're going to form a bond between this sulfur and this carbon. And let's make a model of the product that forms when the nucleophile attacks from the right. So here is that product. And then we hold up the carbocation so we can compare the two. Now let's compare this product with the product when the nucleophile attacked from the left side. So in my left hand I'm holding the product when the nucleophile attacked from the left, and on the right I'm holding when the nucleophile attacks from the right. So what's the relationship between these two? Well they're mirror images of each other, but if I try to superimpose one on top of the other, you can see I can't do it. So these are non-superimposable mirror images of each other. These are enantiomers. Now let's look at our products from a different viewpoint. So I'm going to take the product on the left, and I'm going to turn it so that the S-H is coming out at me in space. So here we can see the S-H coming out at us in space, the methyl group going away from us, the ethyl group on the right, and the propyl group on the left. So now let's look at our other product, and this time if we're going to keep the same carbon chain, the methyl group's coming out at us in space, the S-H is going away from us, the ethyl group is on the right, and the propyl group is still on the left. Here are the two products that we got from the video, but since you won't always have a model set let's go back to the drawings over here and pretend like we don't have a model set. We know that our nucleophile attacks our electrophile and a bond forms between the sulfur and that carbon. So if I draw in my carbon chain here I know a bond formed between the sulfur and the carbon. Let me highlight the electrons. So let's say a lone pair of electrons in magenta on the sulfur form this bond, and here's our product. But if you look at our product, notice that this carbon is a chiral center. There are four different groups attached to that carbon. So if you think about the stereochemistry of this mechanism, with the nucleophile approaching the electrophile from either side of that plane, you should get a mixture of enantiomers as your product. So if I draw in my carbon chain here, I could represent one enantiomer by putting the S-H on a wedge, so let me just draw that in here. So here's our S-H on a wedge, which means the methyl group must be going away from us in space. And if I'm going to draw the other enantiomer, I would have to show the S-H going away from us in space, which means the methyl group is coming out at us. And notice that these two products match the model sets that we drew here. Since there's an equal likelihood that the nucleophile could attack from one side or the other, we would expect to see an equal mixture of our products. So I'm going to say here approximately 50% is this enantiomer, and approximately 50% of our product is this enantiomer. Finally let's go through the hybridization states of this carbon in red one more time. So for our starting alkyl halide, the carbon in red is tetrahedral, right? It's sp3 hybridized so it has tetrahedral geometry. When we formed our carbocation, the carbon in red is now sp2 hybridized, so it has planar geometry. But for our products we're back to an sp3 hybridized carbon with tetrahedral geometry. So we have to think about the stereochemistry.