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.

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.

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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.

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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.

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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.

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

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

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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.

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Circle the least reactive substrate in an SN2 reaction:

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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.

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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.

 

 

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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

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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:

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Provide the major product for the following compound.

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Provide the major product for the following compound.

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Features of SN2 Reactions

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

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 Provide the major product for the following reaction. 

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Provide the major product for the following compound. 

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Devise a synthetic pathway to form the following product. 

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

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Draw the mechanism for each of the following reactions: 

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Draw the product for each of the following SN2 reactions:

(S)-2-Chloropentane and NaCN

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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? 

 

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The reaction below exhibits a second-order rate reaction: 

Show the mechanism for the reaction above.

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

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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.

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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.

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

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Predict the product(s) for the following reaction.

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Predict the product(s) for the following reaction.

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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.

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

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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

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Which compound would react fastest by an S N2 mechanism?

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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

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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.

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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

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

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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

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

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

a) SN1

b) SN2

c) E1

d) E2

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The rate law for the following reaction is 

 

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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

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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

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

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

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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.

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Predict the product:

 

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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.
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