Nucleophilic Substitution Reactions - SN1 and SN2 Mechanism, Organic Chemistry
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- Опубліковано 12 січ 2017
- This organic chemistry video tutorial explains how nucleophilic substitution reactions work. It focuses on the SN1 and Sn2 reaction mechanism and it provides plenty of examples and practice problems. The Sn2 reaction is a second order nucleophilic substitution reaction where the rate is dependent on the concentration of the substrate / alkyl halide and the nucleophile. SN2 reactions occur with inversion of configuration and work well with methyl and primary substrates. It's a concerted reaction mechanism that occurs in a single step. The rate law for the SN1 reaction is given as well. SN1 reactions proceed via a carbocation intermediate and carbocation rearrangements such as the hydride shift and the methyl shift are possible. SN1 reactions work well with tertiary alkyl halide substrates due to carbocation stability. Carbocations are stabilized by means of hyperconjugation and the inductive effect. SN1 reactions will produce an unequal racemic mixture. The stereochemistry of both reaction mechanisms are discussed in detail. SN1 reactions work well with polar protic solvents but SN2 reactions work better in polar aprotic solvents. Solvolysis reactions are sn1 reactions where the nucleophile is the same as the solvent.
Stereochemistry R/S Configuration: • Stereochemistry - R S ...
Optical Activity & Specific Rotation:
• Optical Activity - Spe...
SN1, SN2, E1, E2 Reaction Mechanisms:
• SN2 SN1 E1 E2 Reaction...
SN2 Reaction Mechanisms:
• SN2 Reaction Mechanisms
SN2 - Test Question:
• SN2 Reaction Mechanism...
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SN1 Reaction Mechanisms:
• SN1 Reaction Mechanism
Carbocation Stability - Hyperconjugation:
• Carbocation Stability ...
Carbanion Stability:
• Carbanion Stability
Protic Vs Aprotic Solvents:
• Polar Protic Solvents ...
E1 Ring Expansion:
• E1 Reaction Mechanism ...
E2 - Test Question:
• Zaitsev vs Hoffman's P...
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E2 Stereochemistry - Newman Projections:
• E2 Stereochemistry Wit...
SN1, SN2, E1, E2 - Practice Test:
• SN1 SN2 E1 E2 Reaction...
Organic Chemistry PDF Worksheets:
www.video-tutor.net/orgo-chem...
Organic Chemistry 1 Exam 2 Playlist:
bit.ly/3PKEApB
Full-Length Videos and Worksheets:
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Full-Length Exams and Worksheets: www.patreon.com/MathScienceTutor/collections
SN1 SN2 E1 E2 Reactions Test Review: bit.ly/3Bt4ghw
Organic Chemistry Final Exam Review: bit.ly/2WCJ8GP
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@1:46 you say "the SN1 reaction occurs in a single step" For anyone else thatt might be reading this, it's supposed to say "the SN2 reaction" not "SN1"
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I have to criticize a bit. Sn2 has 2 in its name not because it is second order overall but because it is a bimolecular nucleophilic substitution, emphasis on bi. Sn2 can occur as a pseudo-first order reaction in case of solvolysis(when the solvent is the nucleophile and is in excess)! But otherwise, thank you for this video.
bimolecular literally means it is a second order reaction
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In your first example, isn't iodine a better leaving group, and thus a weaker base, than bromine? So the reaction will not proceed, or will at least favor the reactants if the reactions is reversible
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Good work, but it was very confusing when you switch from sn1 to Sn2 without saying anything so i cant tell which is which
First look at the Alpha C. Is is a primary, secondary, or tertiary C? If it is Primary, it will always be Sn2 (minus a couple instances), if it is secondary you need to look at what you are reacting with and the protic/aprotic conditions along with the leaving group, if it is tertiary is always favors Sn1.
It’s implied when we discuss sn1/sn2. The biggest thing to remember (what I learned) is that sn1 is usually a 2 step process while sn2 are 1 step:) hope it helps. There’s basic rules to help you immediately see the rxn that will happen when u see the reagent with starting material. So primary, sec, and tertiary substrates helps me know whether the rxn will proceed in either sn1 or sn2. Check the Carbon:)
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@kuewayneennis9521 Yes! I was also confused because at 1:47 he says "The SN1 reaction." while he meant SN2! I hope there aren't more confusions?
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Substitution reactions prefer formation of the more stable base. Reacting iodine with an alkyl-bromide would produce no reaction. Please clarify this
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For the last molecule, would it actually have 4 steroisomers? The methyl-substituted carbon is an asymmetric center, and I assume the methyl shift could add to either side.
Why does the intermediate molecule at 14:21 let go of the hydrogen instead of the methyl group? What makes it more favorable? Why would the methanol want the hydrogen? Wouldn't it have to dissociate for the OH- to grab the H+ and become H2O, leaving a methyl ion CH3+, which is kinda unfavorable?
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correct me if I'm wrong, but I'm pretty sure that the second example can only work with E2. SN1 does not work under strong basic conditions.
hi not sure if you can see this comment since this is a pretty old video,, but im open to anyone who can answer my question: for 17:43 product, is it not going to be a racemic mixture just like the previous example since H2O is a neutral nucleophile?
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does SN2 and SN1 work for alkane, alkene and alkynes or only for alkyl halides?
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i tought I- was a better abandonat group than Br- . but in the video you use Br- as the abandonat group. i am kinda cofused. can somebody explain?
@7:48- how do we know if the Nu: attacks from the front or back?? I know that you said it should attack from the back and why, but is that always the case? Can you explain this further? Should it always attack from the back WBR the Nu: is a Br-?
how do you know if it attacks from behind or in front
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You're welcome Ana.
What type of stereoisomers are the molecules in the last example?
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Overall, this video is very helpful. However, I'd like to point out that at many points in the video, some terms are not very clearly pronounced.
What is the link to video that has 75 practice problems?
1:46 don't you mean Sn2?
Yes SN2 is Concerted single step, while SN1 is multi step.
That's what I was thinking too
Does Carbonation formation occur in SN1?
at 4:05 you were drawing 2-bromo-2-methylpropane but as you can see that you made mistake for that methyl (to the left of this molecule.) C and H suppose to be switched or else I will think that the carbon in the center of the molecule is connected to the hydrogen and that hydrogen is connected to the other carbon which that is crazy.
Why does the oxygen develop a partial positive charge instead of negative when bonding to the carbocation?
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So, what type of nucleophile will determine if it will be a SN1 reaction or a SN2 reaction?
Is this for the AS level?
U said that "since bromide repels oxygen ....inverted product will be formed"....but this is Sn1 reaction so the leaving group have already expelled ...then there will be no repulsion for oxygen right ??? Kindly answer
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do your videos have any order and organization? I don't want to jump from one video to the next.
Hello
How do you know what's front and what's back? Isn't that relative?
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Any practice question for reaction with formic acid?
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Is there any influence in percent of racemic mixtures if the bromine is put in front or at the back of carbon atom?
No because all reasoning is reversed
Isn’t iodide weaker than bromide making it unable to remove it?
There are a few strange things:
1. I is a better LG than Br so it shouldn’t react
2. Methanol is acidic and would not accept protons
What else could replace this method ? -In case it is needed.
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I don't get how the bromide affects the water if it attacks from the front of the molecule, but not the back? It's all in the same solution, right? So, why can't the bromide affect the water if it's attacking from the back?
It's not really about "front" and "back", but about "one side of the plane" and "another side of the plane". That carbocation is sp²-hybridized, so its geometry is trigonal planar, and normally the water molecule could attack it from both sides. But the negative bromide ion is still hanging around nearby, attracted by that positive carbocation, and partially blocking the way from one side for that water molecule. Therefore, that water molecule has still a better chance of attacking the carbocation from one side (the one without the bromide ion) than the other (where the bromide is still hanging around and blocking its way).
For Sn2, what if both the nucleophile and substrate are halved?
1/2 times 1/2 is .25. So multiply the rate by 0.25 or divide by 4, however you wanna conceptualize it.
At 17:23, Why isnt the other carbon chiral? Also aren't the two carbons 4° and 3° respectively? Someone help!
Well the 3° carbon is chiral and no because only carbon-carbon bonds count, so the 2° carbon is bonded with 2 other carbons thats why its 2° and the 3° is bonded with 3 carbons and so its 3°
How will you know a reaction is an SN1 or an SN2 reaction
14:22 why are we getting rid of the hydrogen? Also when we get rid of that hydrogen, why is that replaced with CH3 instead of CH3OH?
We get rid of H and replace it with a pair of electrons to return O to a neutral charge.
For your second question, I think you may be looking at it wrong. H isn't replaced by CH3, that CH3 is there from the original CH3OH. He's brought in a second CH3OH to leech off a pair of electrons. He could have just as easily brought in something else for this, like OH. He also doesn't show the result of where H goes, which I guess isn't important for the heart of what we are discussing.
Can you please reply asap. Is 1:46 right or not because I dont understand
he made a mistake. SN2 reactions are always one step while SN1 are multi-step reactions. Hope it clears it up for you
Could I have the references/source please?
Why will methanol remove the hydrogen and not the methyl Group on the oxygen? At 14:18
Is there any law on why it pick that?
The simplest answer is: we've just done a nucleophile-electrophile reaction when that oxygen attacked the carbocation, so the next step shouldn't be another nucleophile-electrophile reaction, but an acid-base reaction instead. And acid-base reactions are all about giving/taking protons (hydrogen nuclei).
Also hydrogens are relatively easy to detach, because they're very small when compared with any other atom. They can just lose one electron to the oxygen and leave when someone else offers them another electron. Bonds between carbon and oxygen are harder to break, because carbon is not very happy either when losing or gaining an electron.
Wish me luck on my Chem HL exam tomorrow :(
Thank you! But why does the carbocation forms? I mean, how come the carbon-bromide bond just splits?
you dont really need to know those but it is because of the dipole interaction between carbon and bromine.