SN2 Mechanism

Time to start learning about some of the most important reactions in all of organic chemistry. In fact, you’re never allowed to forget this one!

Concept: Drawing the SN2 Mechanism

9m
Video Transcript


All right guys, so now we're going to jump into one of the most important mechanisms in all of organic chemistry and it's a mechanism that you're never allowed to forget. What I teach you in the next 20 minutes, it's going to stick because you're going to need it for your – obviously for orgo, for your graduate exams and even in graduate school. If you're planning on going to graduate school for anything pre-health, you're still going to need to know this reaction and that's called the SN2 mechanism. Let's dive right into it.
What the – if I were to just give it a tagline and just say in one sentence what an SN2 mechanism is, what it is is that a negatively charged nucleophile – hopefully, all of that is words you should be comfortable with – negatively charged nucleophile reacts with an accessible leaving group. Now, leaving group you should know what it is. Accessible, maybe you're a little confused, but we'll define it. To produce substitution – you know what that is – in one step.
Let's go ahead. I want to just get right into it. Let's just draw this mechanism out. So I have this nucleophile that I'm just generally putting as Nu-. There's a lot of different nucleophiles out there. It doesn't really matter the identity right now. Now I have to figure out – I'm reacting it with an alkyl halide. What did I say alkyl halides were good at? Leaving. What that means is I have to figure out what is the electrophilic part of this molecule because this is my electrophile and what is it going to look like after it reacts.
So how do I find out which part's electrophilic? Does it have a positive charge already on it? No. So I'm going to have to draw the dipole. What does the dipole look like? Well, remember that halogens pretty much always pull away from whatever they're attached to, so I would have – my only major dipole is pulling away from the carbon. What I would have is a negative here, a partial positive there.
Where's my nucleophile going to want to attack? It's going to want to attack the carbon. So the electrophilic part is not the X, it's the carbon. So I'm going to start off my arrow from my nucleophile and I know I'm going to attack that carbon.
But now actually we have a choice. Because what we have is a distinct set of sides. So let's think about it this way. This is my carbon. And in the past, I haven't really worried about exactly how I draw my arrows because I haven't been very picky. But, if you think about it, there's actually two different sides to this carbon. Let's say that the X side, the one with the halogen, is called the front side. So the X has three lone pairs, one, two, three. That would be what I would consider the front side. So I'm just going to write here that's the front.
And the back side would be everything that's on the other side over here. And the back side, what it's going to have is just a hydrogen and then some alkyl groups. So this would be a methyl group and an ethyl group.
Which of these two sides, front or back, do you think is going to be the easiest for my nucleophile to approach? Let's think about it this way. We know that it wants to hit the carbon. So no matter what it's going for the carbon. But all I'm asking is is it going to try to go from the front side or the back side.
It turns out that the front side is a really bad option. Why? Because the nucleophile, remember it already has extra electrons. It's got extra electrons it's trying to get rid of. In order to go through the front side, it would need to pass through a bunch of electron clouds from the hallogen. Do you think that's going to be very easy to do? It's actually going to be almost impossible. It doesn't happen. Those electrons are going to repel each other like crazy. So front side attack is actually impossible. It's never going to happen.
What that means is that this is going to lead us to one of the most inappropriate phrases in all of science and that is back-side attack. So as messed up as that sounds, alkyl halides are totally down with it. Back-side attack is something that they're all about. And we're going to be doing this every day for the rest of organic chemistry. So I hope you guys are cool with that. Got to get used to it pretty quick.
So back-side attack is the way to go because it's the way that's basically less hindered. It's going to be a lot easier for those electrons to pass through the back side where there are not as many electrons as the front.
So now what we need to do is we have to draw the transition state of what this is going to look like because let me just ask you guys. Am I done with this mechanism? Do I need to draw any more arrows or am I done? No, we should draw some more arrows.
Why? Because remember that this is going to be a nucleophile and electrophile that does not have an empty orbital. Notice that there's no empty orbital here. This carbon already has four bonds. So this carbon already has four bonds if I make a new one that's five. So if I'm making this bond, I'm going to have to break a bond. And you guys already know what I'm going to break. I'm going to break the halide off. I'm going to break the halogen off.
So that means that I'm making a bond and I'm breaking a bond at the same time. This is going to lead to something called a transition state. A transition state is just a high-energy phase of the reaction that is very, very short lived. What it means is that it never even really happens – what I'm trying to say is that it cannot be isolated. It's a very high energy thing that must happen because we know that it must go from one state to another, but if I tried to just isolate it in a test tube, I would never be able to isolate transition states.
Let's go ahead and draw what it would look like. It would be a carbon. And it would be attached to three things for sure that are just single bonds. It would be attached to a methyl group. That's the one in the front. A hydrogen in the back and then an ethyl group. And I'll just put the ethyl group facing down because I'm going to need all the space I can get.
Now this is the interesting part. We just said that I'm making a bond and I'm breaking a bond and it's all happening in one step. That means that I'm going to have to draw partial bonds. That means that my nucleophile is partially making a bond to that carbon. And my halide is partially breaking a bond to that carbon.
On top of that, now this carbon has too many bonds. It has five instead of four, so I'm going to have to put partial negatives on these atoms. What that means is that remember that carbon wants to have four bonds, it now has five, so in this transition state, it's extremely unstable. Carbon does not like to have this many bonds. I have to indicate that it has one too many by putting negatives that are distributed.
So there we go. That's our transition state. If you ever see this double dagger, that means transition state. Like I said, this is something that it must exist for like a nanosecond, but it's not something you could isolate.
After this reaction is done happening, after it's all completed, all happens at one time, now I have to figure out what my products are going to look like. Well, what I'm going to have now is that I'm going to have my nucleophile. But now my nucleophile is going to be attached to – have a single bond. I'll draw it in blue because that indicates the arrow that was just made, to that red carbon.
And what is that red carbon going to be attached to? The same three things it was attached to before. So it's going to have that methyl group, it's going to have that hydrogen. And not the ethyl group because the nucleophile came from the back, my ethyl group is getting pushed towards the front. That's going to be important. Let's just hold on to that thought.
On top of that, is there anything else that we need to draw? Our leaving group. So our leaving group is going to get X-.
So first of all, how can I tell that a substitution reaction just took place? Well, I can tell because things substituted. Before I had a carbon with an X, now I have a carbon with a nucleophile. Before I had a nucleophile with a negative, now I have an X with a negative. See how everything perfectly swapped. So this is definitely substitution. 

Summary: A negatively charged nucleophile reacts with an accessible leaving group to produce substitution in one-step.

Concept: Understanding the properties of SN2.

12m
Video Transcript

Now what I want to do so that's basically that, OK? Now what I want to do is you guys know the mechanism now this is going to be the mechanism we use every time it's very important that you guys understand everything about this mechanism but on top of that there's a lot of facts that we need to memorize about it too because your professor is going to want you to know a lot of conceptual stuff about this as well, OK? And in fact you're really not going to understand it unless you know all the concepts behind it so now what I want to do is start breaking all these different concepts down.

Let's start off with nucleophile, OK? You can already guess what this is because I already talked about it but do you think we need a strong nucleophile or a weak nucleophile to start off this reaction? And the answer is that I'm going to say whenever something's negatively charged that is strong, OK? Because that means it is actively trying to get rid of electrons so in the description I told you guys it's always a negatively charged nucleophile, OK? So that means it's strong, alright? So that means a weak nucleophile one that is neutral or doesn't have a negative charge would not be a good candidate for SN2, good so far? Cool so then leaving group is it going to be better if it's unsubstituted or if it's highly substituted? Now just so you guys know substituted has to do with R groups, OK? That's what substituted means, unsubstituted means that you have less R groups coming off the alkyl halide the leaving group, OK? Highly substituted means that you have a lot of R groups, OK? So which one do you think is going to be better? Remember that the mechanism is backside attack so which one do you think is going to be better for backside attack? Having not a lot of stuff on the backside like just hydrogens that are really small or having these big bulky groups a bunch of them that are going to take up a lot of space? When I word it that way it sounds like the best option would be unsubstituted and that's exactly right, OK? For backside attack to be favored you need to have a lot of room on the backside, alright? Otherwise it's just going to be too crammed and you're not going to be able to get it in, alright? So now it's going to the next one so those are really big points strong and unsubstituted, OK?

Next reaction co-ordinate so if I were to draw an energy diagram of this would the highest point would it be a transition state or would it be an intermediate? Is that which one is the one that I'm passing through to get to my product? And the answer is I told you guys transition state, OK? We're going to transition state because this is all happening at one time so that means I have one molecule that is kind of in between both sides, OK? And that helps us flow into the next question is this a concerted mechanism or is it a two-step mechanism? Now just so you guys know concerted just means one step, OK? Concerted is actually a word that we can use in English it just means everything happens at once, OK? So it's not just a chemistry word it's actually just like a normal word so concerted one step? Absolutely I mean I already told you guys it's one step but so you could just looked at my little sentence but also that makes sense because there's no distinct first step and second step, right? Is they're like step 1, step 2? No, I told you you're making a bond and you're breaking a bond at the same exact time, alright? So that means it's all happening at the same time.

Cool so now we're going to get into is rate questions so all this stuff down here has to do with rate and rates we haven't really talked about too much so I'm going to explain this, alright? So is the rate going to be unimolecular or Bi-molecular? A really easy way to remember this is that SN2, the 2 in SN2 stands for Bi-molecular in fact what SN2 stands for is substitution then it says nucleophilic, why? Because a nucleophile is starting it and then bi-molecular, OK? That's what it actually means that's what SN2 means, OK? So I know the rate is bi-molecular? But what the hell does that actually mean, OK? Well what it has to do with is it's basically saying the rate, remember that rates are always based on that you can trace them back to rate constants, OK? And you can say if I increase the concentration of a certain reagent is that going to affect the rate at which I make my products, OK? And if it's bi-molecular what that means is that there's going to be not just one species that the rate depends on but it's going to be two species, OK? And the best way that I can illustrate this is I always think of an arrow and a target, OK? And I think that the arrow is the nucleophile, OK? Because it's the one that's kind of starting this all off and the target is the electrophile or what we call the leaving group, OK? Cool so I've got my nucleophile and I've got my electrophile, OK? Also just so you guys know the leaving group is also my Alkyl halide, right? Because I said that Alkyl Halides are the most common type of leaving group, OK? Doesn't have to be but in most cases, right? So now here's the question if I double the amount of arrows that I'm shooting so instead I've got a bow and arrow and now instead of shooting one arrow I've got another guy next to me and he's shooting another, OK? Is that going to increase the chances that I hit the target or that I hit a bullseye? Hell yeah, it's going to increase the chances, OK? Because now even if I missed maybe the other guy hits it, OK? Now think about it this way, molecules are not smart in fact they're super stupid in fact they're the most stupid ever they don't have brains I know that's going to come as a shock to you all they are all these reactions are dictated by random motion, OK? So that means is that I have a test tube and I have a bunch of arrows and I have a bunch of targets, guess what's happening? They're not aiming for the target they're randomly colliding they're hitting the wall they're hitting each other then they hit the frontside and they bounce off, why would they bounce off if they hit the frontside? Because remember there's electrons there and the electrons repel so they keep bouncing bouncing eventually something hits the backside, BOOM that's my reaction and I go and I make product, alright? If I double the amount of arrows I'm shooting with I'm doubling the chances that I'm randomly going to make a product, cool so far? Alright next question if I were to double the amount of my leaving group of my targets, OK? Would that double the amount or would that increase the chances of getting a bullseye? What do you guys think? I can tell you guys are really thinking about this one, yes it would be, OK? Because let's say I'm only shooting with one arrow but now there's two targets, hey I could be a really bad shot remember I'm dumb, OK? So I missed the first one but actually hit the second one because you know I was really bad, alright? And that's the way it also works so if I double the concentration of my leaving group I'm going to double the chances that I get a collision that leads to a backside attack, alright? How about if I were to double both so now I have twice as many arrows and I have twice as many targets What would that do? That would wind up quadrupling the rate of my reaction because now what would happen is that I have four times the chances of hitting a bullseye, is that making sense? And what I'm trying to say is hitting the bullseye is the equivalent of backside attack it means that you get a product, alright? So what that means is that the rate of my reaction is dependent on both the nucleophile and the leaving group or my electrophile, by the way this is the same as electrophile it's got a lot of names but it's the same thing, OK? So what that means is that this is bi-molecular and that means that my rate is going to be equal to the concentration of both reagents not just one thatÕs what bimolecular means, alright? Are you guys cool with that? So we just have to memorize that for now we're not going to do a whole lot of calculations but you do need to be able to answer questions about if I increase the rate this type of reagent would that increase, would that double, would that triple whatever, OK? Cool so then we get into one last type of question which is stereo chemistry, now remember that we had a whole chapter dedicated to Chirality, alright? The Chirality here isn't going to get that confusing but if this is a Chiral center...By the way is this a Chiral center the carbon that's red? Yeah it is that's actually a Chiral center right there that's got a green carbon now I just drew a green has four different groups on it, OK? So if you have a Chiral Center at the beginning, OK? And you do a backside attack notice that at the end two of my group swapped places, OK? Notice that my H and my methyl group are still in the same place so they're good but now my Ethel group is on the right side and my high priority group my nucleophile or what used to be an X is on the left side so what that means that these switched places, OK? Any time that you switch the orientations or the positions of two groups guess what you do? You flip the configuration, so what that means is that originally if this was an R if the first was an R.....By the way I'm not going to calculate RS here, OK? You guys are cool with that, I'm just going to say that imagine that whatever.... OK I will calculate it because it's super easy it would just be that this is 1 this is 2 this is 3 so it would be an R, OK? So if this first one is R, afterwards it's going to turn into an S, OK? And if you want to prove that you could just say "OK this is 1 this is 2 this is 3 now it's going to go this way, OK?" So if you start off with an R you going to get an S at the end at the end and guess what that's called? It's a huge huge thing very important that is called inversion of configuration, OK? So now anytime you hear the word backside attack, alright? You're going to have to get used to the fact that backside attack always means the one thing with Chirality it always means inversion of configuration, OK? Now what would retention...What would these other words mean? Retention just means that you get the same thing, R turns into R afterwards that's not this reaction so don't worry about it, OK? What does Racemic mean? Remember that Racemic means it's a combination or a perfect mixture of both enantiomers so Racemic would be that I get 50 percent of R, 50 percent of S is that what's happening here? No I'm getting 100 percent of the other enantiomers, why? Because I'm flipping the orientation of two groups, is that cool? Alright so one last thing and then I'll just do a really quick practice problem, the nickname of this reaction is going to be I've said it 10 times backside attack, OK? Whenever you hear backside attack you're always thinking SN2 and that just has to come that very naturally at this point, OK? And as the semester roles on, alright? So now I want to do a practice problem I will give you guys the chance to pause the video or to answer it then we'll go onto the next video I want you to rank the following Alkyl Halides in terms of their reactivity, OK? So figure out which of these would be the most reactive? Which would be the least reactive towards an SN2 reaction? Notice that all the leave in groups of the same in terms of the actual strength of the leaving group so there must be something else that's going to make it good or bad at reacting, OK? So go ahead and try to figure that out and then I'll answer it for you GO.

Properties of SN2 reactions:

  • Nucleophile =  Strong
  • Leaving Group =  Unsubstituted
  • Reaction coordinate = Transition State
  • Reaction = Concerted
  • Rate =  Bimolecular
  • Rate =  k[Nu][RX]
  • Stereochemistry = Inversion
  • Nickname = Back-side attack!

Example: Rank the following alkyl halides in order of reactivity toward SN2 reaction. 

5m

Problem: Predict the product of the following reaction.

3m

The product must contain inversion of configuration if the original leaving group is located on a chiral center. 

Problem: Predict the product of the following reaction:

5m

SN2 Mechanism Additional Practice Problems

For the following haloalkanes, rank them from 1-4 with respect to reactivity in an SN2 reaction, with a 1 under the LEAST REACTIVE HALOALKANE, and a 4 under the MOST REACTIVE HALOALKANE.

Watch Solution

Using line-angle ONLY, draw the MAJOR product expected from the following reaction. Be sure to show stereochemistry if appropriate. If no reaction occurs write NR.

Watch Solution

Using line-angle ONLY, draw the MAJOR product expected from the following reaction. Be sure to show stereochemistry if appropriate. If no reaction occurs write NR.

Watch Solution

Predict the organic product of the following reaction. When appropriate, be sure to indicate stereochemistry. If more than one product is formed be sure to indicate the major product. Draw all answers in skeletal form.

Watch Solution

Rank the following alkyl halides in order of increasing SN 2 reactivity. (1 – least reactive, 3 – most reactive).

Watch Solution

Draw the transition state. Be sure to include relative lengths of bonds breaking and/or forming, hybridization of the α‐ carbon, and partial charges.

Watch Solution

Predict the organic product of the following reaction. When appropriate, be sure to indicate stereochemistry. If more than one product is formed be sure to indicate the major product. Draw your answer in skeletal form. You will be graded on the product your draw from the reaction no other information is needed for this question.

Watch Solution

Circle the least reactive substrate in an SN2 reaction:

Watch Solution

For the following pairs of reactions, mark an “X” in the box on the right indicating which will go faster. The compound on top of the arrow is the solvent.

Watch Solution

Consider the following SN2 reaction below and answer the following questions: 

a. Draw the mechanism

b. What is the rate equation?

c. What would happen to the rate if the solvent changed from DMSO to ethanol?

d. Draw the energy diagram. 

e. Draw the transition state of the reaction.

 

 

Watch Solution

Stereochemical inversion; Draw the product for each of the following SN2 reactions:

 

a. (S)-2-chloropentane and NaSH

b. (R)-3-iodohexane and NaCl

c. (R)-2-bromohexane and NaOH

Watch Solution

Consider the following SN2 reaction given below.

A) Write a mechanism for the reaction. Use curved arrows to show the movement of electrons. You must include the transition state.

 

 

 

B) Identify the following:

 

substrate:

 

leaving group:

 

nucleophile:

 

spectator ion:

 

solvent:

Watch Solution

Provide the major product for the following compound.

Watch Solution

Provide the major product for the following compound.

Watch Solution

Features of SN2 Reactions

Watch Solution

Transition states have partially formed bonds whereas intermediates have fully formed bonds ( True or False )

Watch Solution

 Provide the major product for the following reaction. 

Watch Solution

Provide the major product for the following compound. 

Watch Solution

Devise a synthetic pathway to form the following product. 

(S) – 2 – chlorohexane to (S) – 2 hexanenitrile  

Watch Solution

Draw the mechanism for each of the following reactions: 

Watch Solution

Draw the product for each of the following SN2 reactions:

(S)-2-Chloropentane and NaCN

Watch Solution

The reaction below exhibits a second-order rate reaction:

What happens to the rate if the concentration of 1-iodopropane is doubled and the concentration of sodium hydroxide is tripled? 

 

Watch Solution

The reaction below exhibits a second-order rate reaction: 

Show the mechanism for the reaction above.

Watch Solution

Draw the product of the reaction between (R)-3-bromohexane and NaSH in DMSO.

Watch Solution

Provide the final reaction product or "no reaction" if you believe it is the case. Please provide the R/S stereochemistry for each chiral center in the product provided (if any), also indicate if the compound is not optical active.

Watch Solution

Provide the final reaction product or "no reaction" if you believe it is the case. Please provide the R/S stereochemistry for each chiral center in the product provided (if any), also indicate if the compound is not optical active.

Watch Solution

Construct a mechanism for the following reaction. Include the transition state in your reaction mechanism.

Watch Solution

Predict the product(s) for the following reaction.

Watch Solution

Predict the product(s) for the following reaction.

Watch Solution

Draw the major product(s) for the following reaction in the box provided. Indicate stereochemistry where appropriate. When a racemic mixture is formed, you must draw both enantiomers and write RACEMIC. The mechanism of each reaction (SN2, E2, SN1, or E1) is written below the reaction arrow. Therefore draw your product accordingly.

Watch Solution

For each of the following pairs of molecules, CIRCLE the molecule that will undergo an SN2 reaction faster. 

Watch Solution

Which rate law best depicts an S N2 type mechanism? (Note: [E] is the electrophile concentration and [N] is the nucleophile concentration).

A)  R=k[N]

B)  R=k[E]

C)  R=k[N][E]

D)  R=k[N][E]2

Watch Solution

Which compound would react fastest by an S N2 mechanism?

Watch Solution

How many atoms and electrons are directly involved in the bond-making and bond-breaking for an SN2 substitution reaction? 

(A) two atoms, two electrons 

(B) three atoms, two electrons 

(C) four atoms, two electrons 

(D) two atoms, four electrons 

(E) three atoms, four electrons

Watch Solution

In the reaction below, the configuration of the starting material is S and the configuration of the product is also S. Is this consistent with the SN2 mechanism? Explain your reasoning.

Watch Solution

Which of the following compounds will undergo an S N2 reaction most readily?

a) (CH3)2CHCH2CH2CH2I

b) (CH3)3CCl

c) (CH3)2CHCH2CH2CH2Cl

d) (CH3)2CHI

e) (CH3)3CCH2I

Watch Solution

Draw the transition state for the reaction between (S)-1-iodo-3-methylpentane and NaSCH3

Watch Solution

Identify the correct potential energy diagram for a tertiary alkyl chloride (3°RCl) versus primary alkyl chloride (1°RX) undergoing an SN2 reaction. Provide rationale to support your choices

Watch Solution

Which of the following reacts the slowest with sodium cyanide, NaCN?

Watch Solution

A pentacoordinate carbon is a transition state in the _____ mechanism.

a) SN1

b) SN2

c) E1

d) E2

Watch Solution

The rate law for the following reaction is 

 

Watch Solution

1-Chloro-4-fluorobutane is reacted with one equivalent of sodium iodide in acetone. During the reaction a precipitate forms. What is the precipitate? 

a) FCH2CH2CH2CH2l

b) ClCH2CH2CH2CH2l

c) NaCl

d) NaF

Watch Solution

In the SN2 reaction, the "2" stands for

a) two reactants in the reaction

b) two steps in the reaction

c) two intermediates in the reaction

d) bimolecular kinetics for the reaction

Watch Solution

Complete the energy diagram below for an SN2 reaction. Draw the curved line ONLY, do not indicate any other details.

Watch Solution

Provide a full arrow pushing mechanism showing all steps and the transition state formed for the following reaction. 

Watch Solution

In July 1917 British soldiers on the Western front in Ypres were attacked by the German army using mustard gas, the chemical structure of which is shown below. Use curved arrows to show mechanism and draw the product of the intramolecular nucleophilic substitution reaction that this compound undergoes in the box provided.

Watch Solution

Predict the product:

 

Watch Solution

Assuming no other changes, what effect on the rate would result from simultaneously doubling the concentrations of both butyl bromide and OH - ion?

CH3CH2CH2CH2Br  +  OH -   →   CH3CH2CH2CH2OH  +  Br -

  1. No effect
  2. It would double the rate.
  3. It would triple the rate.
  4. It would quadruple (4x) the rate.
  5. It would sextuple (6x) the rate.
Watch Solution