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Organic chemistry
Course: Organic chemistry > Unit 5
Lesson 5: Sn1 and Sn2- 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
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Sn2 mechanism: kinetics and substrate
The mechanism, rate law, and stereochemistry of Sn2 reactions. How the sterics of the alkyl halide affect the reaction rate. Created by Jay.
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09:00You're placing the oxygen on the same side as Br but you're saying that it's inversion. How come?
Also, I really don't understand that triangle type bond, what's that?(0 votes)
See videos on stereochemistry. This video is way too advanced if you don't understand chirality or wedge and dash representations of bonds.(38 votes)
at5:30, why couldn't Br be the leaving group and Br leaves right away, leaving carbocation + to allow OH- to come attack the carbocation and it will be a SN1 reaction instead of SN2? i am having a bit of trouble drawing definitions between SN1 and SN2 :((10 votes)
The order of stability of carbocations is 3°>2°>1°. In this case, if you lose a Br-, you will get a 2° carbocation. This is so much less stable than a 3° cation that it forms too slowly to be of any synthetic use. Instead, the strong nucleophile OH- competes successfully in the much faster SN2 displacement reaction.
The only secondary alkyl halides that react via SN1 reactions are benzylic and allylic halides, since their carbocations are stabilized by resonance.
The general rule then is that:3° halides react via sN1, while 2° and 1° halides react via SN2.(21 votes)
1. I don't think that it is particularly relevant here, but C is also pulling electrons from the three H's attached. Doesn't that affect the partial positive charge of C at00:46?
2. A rather general question: at04:38Jay determines the absolute configuration. When do we have to determine the absolute configuration in dealing with reactions? At the first example in this video, we didn't consider the absolute configuration.(2 votes)- 1. Yes, it does. Every atom wants electrons, but some want electrons more than others. Cl wants them the most, so that puts a partial negative charge on Cl and a positive charge on C. H wants electrons almost as much as C, but C wants them just a little more. So the H atoms get a very small positive charge and that on the C atom gets reduced slightly, but it still stays partially positive.
2. We can largely ignore the absolute configuration of a molecule when it reacts with achiral substances, except when we want to know the mechanism, as in this video. We absolutely must know the absolute configuration when it reacts with other chiral substances. For example, the amino acids in our bodies all have one particular configuration. The enzymes that interact with them must have the correct configuration in order to react.
As another example of the importance of absolute configuration, the drug thalidomide used to be prescribed as a racemic mixture for the prevention of morning sickness in pregnant women. It was later found that only one configuration was an effective drug, but the other configuration caused severe birth defects. The drug had to be withdrawn from the market, but only after the birth of several severely malformed children.(8 votes)
Why isn't the reaction of 2-bromobutane with NaOH an E2 reaction (the major product being an alkene rather than the enantiomer in the video)? I thought with secondary carbons, you need a weaker base, but still a good nucleophile for SN2 to be the major pathway. Or is he just showing us examples of SN2 reactions regardless of major or minor products?(3 votes)
I thought that the tertiary carbon could not proceed by SN2, only SN1 due to hinderance by the connecting C atoms. What allows for the exception?(2 votes)
At7:59he says 'for right now, just assume it's an SN2 reaction mechanism.'(2 votes)
- 7:10How can you say you always get inversion of configuration for Sn2? If the nucleophile that attacks happens to be ranked 4 instead of 1, then the absolute configuration would be unchanged.(2 votes)
- Sn2 only takes place at primary or secondary carbons. By that distinction, there will always be a hydrogen (or two) present at your electrophilic carbon, and all nucleophiles will be a higher priority than them (other than the Hydride ion [H-], but if you happened to see that undergo an Sn2 reaction the carbon would no longer be chiral). Also, generally, most nucleophiles you will see will be higher priority than the carbon substituents present. Sounds like you are connecting with the material well though, keep it up!(2 votes)
- At08:43I don't get why OH is dashed and not wedged? Please, explain it to me. I've watched all videos about stereochemistry and I don't know how this happens. Something missed me. Better, I missed something!(1 vote)
This reaction proceeds via an SN2 mechanism, which always has inversion of stereochemistry (i.e. in this case the wedged Br becomes a dashed OH). The nucleophile (OH-) approaches from the opposite side of the Br ("backside attack"). The transition state is a trigonal bipyramidal shape, with the OH on one side mid-bond-formation, and the Br on the opposite side mid-bond-breaking. The three other substituents (H, CH3, and CH2CH3) form a planar triangle with each other in the middle. Once the C-Br bond fully breaks and the C-OH bond fully forms, the three other substituents have flipped over to the other side (like an umbrella flipping inside out), resulting in an inversion of stereochemistry.
As for why the OH attacks via backside attack (opposite side of Br) rather than frontside attack (same side as Br), this has to do with molecular orbitals...don't worry if you don't understand this bit yet. When the atomic orbitals of the central carbon and the Br combine, they form two new molecular orbitals. One is located between the C and the Br and has 2 electrons in it - this forms the C-Br bond and is called the bonding orbital. The other molecular orbital is on the far side of the C and the far side of the Br (relative to the C-Br bond); it is not occupied by any electrons and is called an antibonding orbital (because if electrons were in this orbital, they would be on the far sides of the atoms, pulling them apart, rather than between the atoms, bonding them together). It would look something like this http://www.chem.ufl.edu/~itl/4411/matter/FG09_032.GIF, where the bottom right is the bonding orbital and the top right is the anti-bonding orbital. If you had electrons in both the bonding orbital (pulling the atoms together) and the antibonding orbital (pulling the atoms apart), the net result would be that there was no bond. So in the SN2 mechanism, if we want to break the C-Br bond, all we have to do is put 2 electrons in the C-Br antibonding orbital. We can use two lone pair electrons from the OH- to do this. Since the antibonding orbital is located on the far side of the C (relative to the C-Br bond), the OH- must approach from this side to get its electrons to the right place; thus, the SN2 mechanism proceeds via backside attack, which results in an inversion of stereochemistry.(3 votes)
8:00- besides the presence of a strong base, what other clues are there that it will be an SN2 rxn?(1 vote)- If the substrate is primary or secondary, the reaction is usually SN2.. A tertiary substrate reacts via SN1.(2 votes)
Whats the reason for Front-side attack not taking place in SN2 mechanism?(1 vote)
The leaving group (halide) is on the "front-side" of the molecule. Since it does not leave before the nucleophile attacks, there is no room for the nucleophile to attack the front-side so it must attack the back-side.(2 votes)
What if you have a carbanion that is bonded to 4 Cs plus a halogen? You would have a quarternary alkyl halide in that case.
Quarternary is extremely reactive as an anion because it is unstable, sort of how it is with alkali metals and halogens.
Would this undergo Sn1, Sn2, E1, or E2?
Would you if you titrated this with NaOH notice the positively charged sodium acting as a nucleophile since that alkyl halide is already negatively charged?(1 vote)- Carbon doesn't make 5 bonds, so you can't have the structure you propose.(2 votes)
09:00
Video transcript
- [Instructor] Let's look at the mechanism for an SN2 reaction. On the left we have an alkyl halide and we know that this bromine is a little bit more
electronegative than this carbon so the bromine withdraws
some electron density away from that carbon which makes this carbon
a little bit positive, so we say partially positive. That's the electrophilic center so this on the left is our electrophile. On the right, we know
that this hydroxide ion which we could get from
something like sodium hydroxide, is a negative one formal
charge on the oxygen which makes it a good nucleophile. Let me write down here. This is our nucleophile on the right and on the left is our electrophile which I'm also gonna refer to
as a substrate in this video. This alkyl halide is our substrate. We know from an earlier video that the nucleophile will
attack the electrophile because opposite charges attract. This negative charge is attracted to this partially positive charge. Lone pair of electrons on the oxygen will attack this
partially positive carbon. At the same time, the two
electrons in this bond come off onto the bromine. Let me draw the bromine over here. The bromine had three lone
pairs of electrons on it and it's gonna pick up another
lone pair of electrons. Let me show those electrons in magenta. This bond breaks and these two electrons come off onto the bromine which gives the bromine a
negative one formal charge. This is the bromide anion. And we're also forming a bond between the oxygen and this carbon and this bond comes from
this lone pair of electrons which I've just marked in blue here. Those two electrons in blue form this bond and we get our product
which is an alcohol. The SN2 mechanism is a concerted mechanism because the nucleophile
attacks the electrophile, at the same time we get
loss of a leaving groups. There's only one step in this mechanism. Let's say we did a series of experiments to determine the rate
law for this reaction. Remember from general chemistry, rate laws are determined experimentally. Capital R is the rate of the reaction and that's equal to the rate constant k times the concentration
of our alkyl halide. And it's determined experimentally this is to the first power times the concentration
of the hydroxide ion also to the first power. So what does this mean? This means if we increased
the concentration of our alkyl halide. If we increase the concentration
of our alkyl halide by a factor of two, what happens to the rates of the reaction? Well, the rate of the reaction is proportional to the concentration of the alkyl halide to the first power. Two to the first is equal to two which means the overall
rate of the reaction would increase by a factor of two. Doubling the concentration
of your alkyl halide while keeping this concentration, the hydroxide ion concentration the same should double the rate of the reaction. And also, if we kept the concentration of alkyl halide the same and we double the
concentration of hydroxide, that would also increase
the rate by a factor of two. And this experimentally
determined rate law makes sense with our mechanism. If we increase the
concentration of the nucleophile or we increase the concentration
of the electrophile, we increase the frequency of
collisions between the two which increases the overall
rate of the reaction. The fact that our rate law is proportional to the concentration of both the substrate and the nucleophile fits with our idea of
a one step mechanism. Finally, let's take a look
at where this SN2 comes from. We keep on saying an SN2 mechanism, an SN2 reaction. The S stands for substitution. Let me write in here substitution because our nucleophile is substituting for our leaving group. We can see in our final products here, the nucleophile has substituted
for the leaving group. The N stands for nucleophilic because of course it is our nucleophile that is doing the substituting. And finally the two here refers to the fact that this is bimolecular which means that the rate depends on the concentration of two things. The substrate and the nucleophile. That's different from an SN1 mechanism where the rate is dependent only on the concentration of one thing. The rate of the reaction also depends on the structure of the alkyl halide, on the structure of the substrate. On the left we have a methyl halide followed by a primary alkyl halide. The carbon bonded to our bromine is directly attached to one alkyl group followed by a secondary alkyl halide, the carbon bonded to the bromine is bonded to two alkyl groups, followed by a tertiary alkyl halide. This carbon is bonded
to three alkyl groups. Turns out that the methyl halides and the primary alkyl halide react the fastest in an SN2 mechanism. Secondary alkyl halides react very slowly and tertiary alkyl halides
react so, so slowly that we say they are unreactive
toward an SN2 mechanism. And this makes sense when
we think about the mechanism because remember, the nucleophile has to attack the electrophile. The nucleophile needs to get close enough to the electrophilic carbon
to actually form a bond and steric hindrance would
prevent that from happening. Something like a tertiary alkyl halide has this big bulky methyl groups which prevent the
nucleophile for attacking. Let's look at a video so we can see this a little bit more clearly. Here's our methyl halide with our carbon directly
bonded to a halogen which I'm seeing as yellow. And here's our nucleophile which could be the hydroxide ion. The nucleophile approaches
the electrophile for the side opposite of the leaving group and you can see with the methyl halide there's no steric hindrance. When we move to a primary alkyl halide, the carbon bonded to the halogen has only one alkyl group bonded to it, it's still easy for the
nucleophile to approach. When we move to a secondary alkyl halide, so for a secondary you can see that the carbon bonded to the halogen has two methyl groups attached to it now. It gets a little harder
for the nucleophile to approach in the proper orientation. These bulky methyl groups
make it more difficult for the nucleophile to get close enough to that electrophilic carbon. When we go to a tertiary alkyl halide, so three alkyl groups. There's one, there's
two and there's three. There's a lot more steric hindrance and it's even more difficult for our nucleophile to approach. As we saw on the video, for an SN2 reaction we need
decreased steric hindrance. So, if we look at this alkyl halide, the carbon that is directly
bonded to our halogen is attached to only one alkyl group. This is a primary alkyl halide and that makes this a good SN2 reaction. The decreased steric hindrance allows the nucleophile to
attack the electrophile.