Sn2 reactions

Sn2 Reactions

Sn2 reactions

Discussion and questions for this video
A problem on my book asks to predict wich compound will undergo the SN2 reaction fster, how can you predict that??
1) something called "steric hinderance", basically you look at the 3 groups attached to the carbon that isn't the leaving group and you see how big the substituents are. so in this example, the 3 H's offer little resistance in terms of bulk for the reaction to occur. if you have 2 H's and an R group (R denotes any C-C.. bonds like a methyl or ethyl), the rate will be less because the R group hinders the backside attack due to its size, this is also called a "primary electrophilic carbon". then there is a secondary with 2 R's and 1 H, and this goes all the way to "tertiary electrophilic carbon" where you have all 3 groups as R groups, this rate of reaction is near 0 because there is so much steric hinderance that the nucleophile simply can't reach the electrophile.

2) the more polarizable the lone pairs of the nucleophile the better. so RS- is better than RO- because the S has larger and more diffuse orbitals due to the 3s and 3p vs the 2s and 2p of O. so I->Br->Cl->F-

3) also, as a side note, charged nucleophiles are better than uncharged ones, so for example: RS->ROH or H2O

hope this helps, have a nice day
The Br leaves because the C atom cannot be bonded to five atoms at the same time (This would violate the octet rule). Something has to give. The C-O bond is stronger than a C-Br bond (360 kJ/mol vs288 kJ/mol), so C-O wins and the Br has to leave. Also, there is such a thing as momentum. If the O atom hits the back side of the carbon atom hard enough, some of that energy is transferred to the Br atom.
It is as if you were choking on a piece of meat. If I hit you behind the shoulders hard enough, the offending piece if meet will come flying out of your mouth in front.
I have found your videos to be extremely helpful. I have several instances that you clarified something that helped things "click." However, I need some guidance regarding 1,2 methyl 1,2 alkyl shift, hydride shift as well as ring expansion shift and reactions with Br2 and wather. I couldn't find a video that seemed to discuss these specifically. I appreciate any assistance that you can provide or pointing me to the correct place for instruction. Thank you. Regina
hydride shift is nothing but our apporach to make our carbocation more stable like tertiary > secondary >> primary.
What would happen if you had an R-2-bromobutane and the nucleophile was sodium hydroxide instead of pure hydroxide. Would that matter?
Pure hydroxide doesn't really exist by itself. If you dissolve NaOH in water, you'll get a Na+ ion (that doesn't do a lot since it's soluble) and an -OH ion (which is the hydroxide you want). So I would say it doesn't matter.
Greetings from Norway!
I don't know if this question has been answered, and that I've just missed it. But what exactly is "partial negativity"?
When you have a polar covalent bond then one end is more positive and one end more negative. They are not charged enough to be ionic and have a full charge. So these ends are called "partial" negative charges.
I mean, whats in the name?! - partial negative, meaning slightly negative -so, thats what it is!
A weak base means a stronger conjugate acid. Weak bases are stable. Stronger acids are more reactive, meaning they are more likely to react with a nucleophile, and the weak base is more likely to leave.
Does Khan Academy have a brother/sister site where we can practice problems rather than just watching them being solved in front of us?
is it necessary to show that eighth electron on bromide ion even when putting a -ve sign above Br????
Yes! Because if you don't put the eighth electron, Br will just be neutral.
Figured I would watch this for some help with my Organic chem workbook for dummies questions. Can anyone assist? its not homework im trying to understand the book. Im trying to understand sn2 reactions, in the first question in chapter 10 it has a carbon chain with a BR at end, then KOH as substitution, then somehow the K disappears to leave a carbon chain with OH at end, I cant understand why?
The overall reaction is
R-Br + KOH → R-OH + KBr (the "molecular" equation)
But we must remember that compounds like KOH and KBr are ionic compounds.So we really should write the reaction as
RBr + K+ + OH- → ROH + K+ + Br- (the "ionic" equation)
We see that the K+ appears on both sides of the equation. It does not really participate in the reaction. Its only function is to balance the charges on the OH- and Br- ions. We call it a _spectator_ ion.Since it does not participate in the reaction, we can cancel it from each side of the equation and get
R-Br + OH- → R-OH + Br- (the "net ionic" equation).
We often omit spectator ions from equations.
Hope this helps.
When you have a primary carbon (that the LG is attached to) a polar aprotic solvent, or a strong nucleophile (that is not a strong base). It could be any combination of these 3 variables. It just takes a lot of practice to understand it well.
So that it's easier to get the picture of what's going on. You can use condensed formulas if you like but they would be a tough nut to crack. By drawing the 3-D structure it's often easier to understand other factors concerned like steric hinderance.
the video says it is 88% complete even though I have watched it all, anyone else having this issue?
What will happen if the hydroxide anion attacks from the front side? Or is it necessary for sn2 reaction having a nucleophile attacking from the backside?
The short answer is that backside attack is necessary for SN2. As to _why_ that is the case: the orientation/geometry of the orbitals demand backside attack. Have you learned about molecular orbitals? There is a bonding molecular orbital between the carbon and the leaving group. An antibonding molecular orbital has a lobe on the side of the carbon pointing directly away from the leaving group. To break the bond between the carbon and the leaving group, we need to add electron density to the antibonding molecular orbital. Since this antibonding molecular orbital points in the opposite direction of the carbon-leaving group bonding molecular orbital, the way to do this is through backside attack. Electrons on the hydroxide add electron density to the antibonding molecular orbital. This breaks the carbon-leaving group bond. In the process, a carbon-hydroxide bond can form.
Hi there I am confused. The first Sn2 step in this video is the same as the first Sn1 step in the other video and the only difference is the different reactants. So for this Sn2 reaction we could also say this is an Sn1 reaction couldn't we? Because the first rate determining step only involves one molecule as well......
In an SN2 reaction, the leaving group leaves at the same time as the nucleophile attacks,all in one rate-determining step. Since the nucleophile and the substrate are both involved in this step, this is a second order (SN2) nucleophilic substitution.
In an SN1 reaction, the leaving group leaves all by itself in a rate-determining step. The nucleophile doesn’t attack until later..Since only the substrate is involved in the rate determining step, this is a first-order (SN1) nucleophilic substitution.
The two mechanisms are quite different.
Is it that the more electro-negative an atom is, the more likely it is to form bonds by sharing electrons? Or will it be more likely to pull electrons away, breaking bonds?
electronegative atoms are in search for electrons .. so they are more likely to form bonds by sharing ( covalents bond ) so the atom has a partial positive charge and a partial negative charge
Neither,I guess. Basic definition says that in homolytic fission/cleavage, carbon free radical (Carbon with an electron and 3 single bonds with H is called methyl free radical) is formed.
While in heterolytic fission,either carbanions or carbocations are formed(charged species).
In this example none of them were formed. Carbon atom was neutral in the end.
At 2:47, the instructor said that due to the bromine is more electronegative than carbon, so the electorn is more willing to go to bromine. And this is the main reason the neucleophile share it's charge with the carbon.
However, is this case, there are three hydrogens which have lower electronegative compared with the carbon. Why doesn't the neuclephile share it's charge with the hydrogen?
Thanks a lot!
First, hydrogen is LESS electronegative than carbon. Electro-negativity is the property of attracting electrons. So the carbon will attract the electrons more.
Secondly, the carbon does not have a completely filled orbital, which makes it unstable. Whereas in the case of the hydrogen, we can see that it has attained duplet state, i.e. it has a completely filled outer orbital.
How would you differentiate whether it is a SN2 or an SN1 reaction by just looking at the compounds?
Generally, a primary alkyl halide reacts by SN2, and a tertiary alkyl halide reacts by SN1. A secondary alkyl halide can react by either mechanism, but more frequently by SN2.
What exactly makes the E- want to to go to the carbon? Is it the partial + on that side of the molecule? It seems more intuitive to me that it would always just take the H, making an elimination reaction..
The electron is attracted to the most positive part of the molecule. The electroegativity difference between C and Br is greater than the difference between C and H. The C-Br is polar, with the positive end at the carbon atom. Although the C-H bond is slightly polar, the polarity is so small that we usually consider a C-H bond to be effectively nonpolar.
So you are correct. It is the partially positively charged carbon that becomes a target for the electrons of the nucleophile.
Why won't Br(-) now act as the nucleophile and displace oxygen to give the original reactants? Or why won't it react with the hydrogen attached to the oxygen atom as it will have a partial positive charge?
due to electronegtivity difrence.oxygen is more electrongtive then bromine that iz y it can't atrct proton(hydrogen) atched to oxygen
what is a nucleophile (Nu:) and how do you know how to find out which is a nucleophile when you have an equation?
'Nucleophile' literally means 'nucleus-loving'. Since nuclei are positively charged due to its protons, and a nucleophile is attracted to it, therefore a nucleophile must be negatively charged, partially or wholly.

The oxygen in water has a partial negative charge and can behave as a nucleophile

The chloride ion has a whole negative charge and it also can behave as a nucleophile
*Nucleophile* means "nucleus loving". A nucleophile is a chemical species that donates an electron pair to the positive nuclear charge of an electron-poor species (an electrophile) to form a covalent bond.
All molecules or ions with a lone pair of electrons or at least one π bond can act as nucleophiles. Because nucleophiles donate electrons, they are by definition Lewis bases.
Just this year, X-ray diffraction proved that the norbornyl cation is symmetrical. You can see a drawing of the structure in
The norbornyl cation is a non-classical ion. Its structure involves the sharing of two valence electrons by three atoms: C6, C1, and C2.
It is not easy to explain the stability of the norbornyl cation. A clue may come from the structure itself: the C6-C1 and C6-C2 bonds are equal in length, but the C1-C2 bond is remarkably short.
If the original positive charge was on C2, we could re-hybridize C1 and C6 to get two p orbitals. These, plus the vacant p orbital on C2, give us essentially a cyclopropenium cation, without two of the frame-work σ bonds. That may be why the cation is so stable: under this interpretation, we could consider it as a Hückel aromatic π system with n = 0. It is as if we had a π bond (as well as a σ bond) between C1 and C2, with a lobe of a p orbital from C6 sticking into the π system.
What is the origin of the bond rotation barrier in ethane, steric hindrance or hyperconjugation?
There is no consensus on this.
The two explanations are that the eclipsed state is less stable (van der Waals repulsions) or that the staggered state is more stable (hyperconjugation).
We have learned that the electron clouds surrounding H atoms repel each other. Indeed, at distances of less than 210 pm, two H atoms do repel. At longer distances, they attract. The maximum attraction occurs at about 240 pm. The distance between eclipsed hydrogen atoms in ethane is about 230 pm, so the H atoms should be attracting rather than repelling each other.
Another explanation is that the rotational barrier arises because hyperconjugation makes the staggered conformation more stable. It arises because in the antiperiplanar staggered conformation, the bonding σ orbital of one C-H bond is oriented perfectly to overlap with the σ* orbital of an antiperiplanar C-H bond on the adjacent carbon atom. This interaction leads to a lower energy situation.
We would represent this as H-CH₂-CH₂-H <—> H⁺ CH2=CH2 H⁻
In staggered ethane, there are six (equivalent) antiperiplanar interactions of this type. They are not major contributors, but they do contribute. This makes the staggered form more stable than the eclipsed form.
The simple answer is that SN2 mechanisms happen in one step because this is the lowest energy pathway to get from the starting material to the products (for reactions that follow this mechanism). If the mechanism were to be two steps (like in an SN1 reaction), then there would be a carbocation intermediate. If this carbocation intermediate is not very stable (e.g. a primary carbocation) then a one-step SN2 mechanism is more likely since it avoids forming a carbocation, thus avoiding this unstable/high-energy intermediate.
when the hydroxide molecules reacted why did bromide ion got separated why not hydrogen?
There are various methods of determining what is the best "Leaving group" in a molecule. They all lead to the same answer but depends which figures you find easier to learn.

Method 1: Bond strength,
The Carbon to Hydrogen bond is one of the strongest bonds in chemistry at roughly 420kJ per mol. As the bond strength decreases the more likely the group is to leave. Bromine a much weaker bond with hydrogen at roughly 300kJ per mol, (the weakest possible is around 200kJ per mol). So the process that requires has the lowest activation, or input energy, will happen.

Method 2; pKa of Conj. Base.
The pKa of the conjugate base of the leaving group is another more complex method for working out the best leaving group. I tend to use this one as I have a better picture in my head of pKa's rather than Bond strengths.
The conjugate base of Bromine is Hydrogen Bromide (HBr). This has a pKa value of -9. The lower the pKa value of the Conj. Base the more stable the leaving group will be. So Br- is a very good leaving group.

Use whichever method you find easier. Hope I've helped.
Hi Sanika. Yes, that would be a coordinate covalent bond, also known as a dative covalent bond. It is this type of bond because both of the electrons in the bond come from the oxygen.
why the structure of the molecule is being flipped over? sal no explain it in video?
@aashish4195: There are indeed four atoms bonded to the carbon, but there are three hydrogens and one bromine, which means there are two different types of group - the hydrogens and the bromine. Now imagine you swap *any* two of the groups bonded to the carbon round. If you then rotate the whole molecule you still end up with what you started with. This means there is no chirality (any arrangement of the molecule will always be superimposable on another).
so i have a problem to figure out why the OH, which is nucleophil, attacks the nucleus of the carbon and not the nucleus of one of the Hydrogens, when all those are much easier to attack, because they point out from the molecule. one idea is that the nucleus of the carbon is bigger, therefor more positive and that makes a higer attraction to the OH. is that right?

ps: thank you so so much for all these awesome videos!!!
C-H bonds are much stronger than O-H bonds, which you can observe from the natural autodissociation process of water. Also, C-H bonds are nonpolar. This is in contrast to the C-Br bond which gives the carbon a partial positive charge allowing it to act as an electrophile. To attack the stabilized hydrogen would require the C-H bond to be broken - an (thermodynamically) unfavorable process.
Great video! Can you create a video giving the rules of arrow pushing/electron pushing?
Does Sn2 occur spontaneously and if not what are the necessary conditions (temperature, pressure etc.) or catalysts needed?
Does the nucleophile have to have a greater electronegativity than the halide initially attached to the carbon for the nuclephile to attach to the substrate and begin the reaction?
this is confusing me. well, at first it's clear but when i listened to MCAT Osmosis talking about SN1 and SN2 reaction, they said SN1 has two reactants involved and SN2 is the fastest step. apparently in SN1 you have a tertiary carbocation making it difficult for the weak base to attack but because reactions happen simultaneously, once halogen attached to the tertiary carbocation leaves, that's when the weak base gets the chance to attack. the slow attack is the rate determining step. so now i'm left confused of which is the truth!!!
Why can't the OH- interact with Br to form BrOH instead of react with the Carbon?
They're both fully negatively charged, and would therefore repel each other. Both negative ions (anions) would be attracted to a positive ion (cation) such as a proton, H+.
Organic Chemistry by John McMurry 8th Ed. pretty pricey though you may be able to get it at your local library
? I have taken orgo chem before, and this is a really messy way to explain the reactions. You should explain what the different leaving groups generally are, and how to figure out if one group is a better LG than another, and also a list of nucleophiles and how they rank in nucleophilicity and why would be incredibly helpful- rather than one random example which is very simplistic and won't help you apply it to other sn2 reactions.
No, it only has 1 carbon so it's a methane. It's called methyl bromide, or bromomethane.
in this video when sal drew the transition state he showed carbon connected with both bromine and hydroxyl ion and the three hydrogens can sp3 hybridised carbon be pentavalent?
No, I think you might be mistaken. Sp3 hybridized carbon can only be tetravalent, which means, it has 4 free bonds.
Can a substituent with an extra pair of electrons, such as an amine or an alcohol ever act as a nucleophile in an SN2 reaction, causing the SN2 reaction to occur twice?
Is this reaction reversible? if we add more and more HBr can we produce more CH3Br
i dont think so.. The definition of electronegativity is:
the ability of an atom to attract the electron density (or a pair of electrons) in a covalent bond.

Whilst Bromine displays electronegative properties, a hydroxide ion does not. It displays polarity, where the oxygen attracts more of the electron density than the hydrogen, as oxygen is more electronegative than hydrogen, but the molecule itself is not electronegative.

So remember: atoms are electronegative. Molecules are polar (except diatomic molecules).
How does it affect the reaction(or the products) if the substrate is optically active....??
_Stereospecific_ refers to a reaction or process in which starting materials that differ _only_ in their configuration produce different stereoisomers as products.
all these video's are linked with Youtube n I am unable to use it! So how can I get benefit from these video's ? any alternate option?
Can you download the cell phone app? It's a hassle, but it worked for me.
Hello, Would you explain me the mechanism of reaction between a secondary alcohol and sodium sulfide / sulfur (Na2S / S ) in both type of solvents, protic and aprotic?
Thank you!
here a nucleophile attacks on the back of carbon and leaving group leaves forward and why cant nucleophile attacks where the leaving leaves
Picture each bond/atom as balloons. There's simply not enough room for an incoming nucleophile to occupy the same space as the leaving group.
Hey, Sal, you're confusing me a bit with the rotation of the bromomethane molecule. The way you drew the second molecule with the hydrogen coming out on the right side (instead of on the left as in the first image), it looks like a different molecule (regardless of the OH group). I tried it in a 3D viewer, and if you claim the blue hydrogen in the second molecule is the same as the blue hydrogen in the first molecule, then the bromine (or the OH group) would be in the back
If a guanidino (as nucleophile) reacts with a carbocation (as electrofile), the final adduct has a sigma-bond between both gruop (and a H lost), but how can? if the reaction is occurring between the lone pair electrons in the guanidino P orbital with the empty P orbital (e.g, carbocation) ? If so, what is the geometry of the attack, parallel P orbital? THANKS
so if the substrate or reacting organic compound to the nucleophile has 2 or 3 different functional groups which one is possibly to go out as explained in the mechanism of Sn2 reaction
I think it might be the one which is most electrophilic / electronegative As it would cause the greatest positive charge on the substrate which would result in an even faster/ stronger attack of the neucleophile on the newly formed electron deficient centre.
When do we say that a group has been added to a compound ?
Please help.
Thank you.
Nucleophilic means nucleus-loving (proton-loving) so they must be negative. A nucleophilic centre is an area of negative charge on a molecule.

This could be in the form of lone pairs (i.e., the oxygen in methanol) or a partial negative charge (i.e., one of the oxygens in nitrate)
Why do the HYDROXIDE have a negative Charge on it ?
please explain.
The more electronegative oxygen atom is attached to hydrogen ,oxygen has two bonds ,one bond with hydrogen and other another free bond which is satisfied with a negative charge.
in the transition state of SN2 reaction mechanism partial negative charges are seen to be developed on the nucleophile and on the halogen atom why? and does partial negative charge should exist on the central carbon atom if it is a methyl group? why if yes/no?
An intermediate isn't formed in an Sn2 reaction.In the transition state,the C atom is bonded to 5 atoms.The halogen atom hasn't left yet and is artially bonded to the carbon atom and thus,the partial negative charge.Similar for the nucleophile.
The C atom is artially bonded to the nucleohile and the halogen atom.So,it doesn't have a negative charge.
Well that's what I think.Hope I helped.
Which grade are you in?
I thought that when the nucleophile attacks, it does so from the back and the substrate undergoes conversion.
Why did we skip the explanation for this in the video?
SN2 is a concerted process so there is no other possibility : Nu arrive one face and the LG goes the other face
It is by opposition with SN1 where first step : LG leave then the Nu attack any faces
also, can you explain the difference between a back-side and front-side attack?
Wondering about the reason for the OH negative charge. The video talks about it at 30 seconds in. Isn't the charge only negative or positive if the number of electrons is not equal to the number of protons? Shouldn't the reason for OH having a negative charge be that H2O lost one proton without loosing any electrons?
That's exactly right. The H₂O was originally neutral. It lost an H⁺. That left it with an extra electron, so the remaining OH had a negative charge.
Sal said hydroxide had a negative charge because oxygen had (technically) an extra electron from the covalent bond with hydrogen. Hydrogen provides one electron to the covalent bond, where is the other electron coming from? or is it a hydrogen 2 isotope?
So OH- comes from water. Water is constantly in a battle between being in H2O fomat, or being in OH- and OH3+ format. It's a bit complex to explain - I've put some links down to have a look at.
How can i determine if a certain reaction will undergo sn1 given reactants only
There are exceptions, but in general this is how I approach it:
Start with recognizing what Sn1 means.
Sn1 = Nucleophilic substitution where the leaving group leaves '''before''' the nucleophile attacks.
What does that mean? That means you will form a carbocation.
What kind of carbocations will form? To answer this question, we need to know the stability of carbocations.
Tertiery carbocations are more stable than secondary carbocations and secondary carbocations are more stable than primary.

So, if a tertiary carbocation will be formed, Sn1 is much more likely.
I'm so confused,if Bromine has 8 total electrons,doesn't that mean it's octet is complete? So it should be neutral? Instead it gets a negative charge . .
The negative charge is the difference between number of protons in the atom's nucleus and the number of electrons orbiting it. So while it is happier with a full valence ring, the atom itself has more negative charge (8 electrons) than positive charge (7 protons).
SAl says the negative charge is due to the fact that oxygen has 7 electrons, one of which is one halfof the shared bond. But , if hydrogen contributes its sole valence electron to the bond, and oxygen still has its six valence electrons on it, where does the other electron of the bond come from? Thanks.
In a covalent bond, both sides of the bond have same number of electrons. So, it doesn't matter who shared more.
Also, Hydrogen can't "share" its sole electron. It can only give the last electron forming Hydronium ion.
i am not getting why there is a inversion of configuration at the stereogenic center
can anyone tell me how to convert benzene to phenol?
when i came across this question i thought it would first undergo halogenation......
then it would react with NaOH to give phenol....
i dont know if it is right or wrong.... can someone tell me please?
when you showed the transition step you had a partial negative charge on both reactions. should one be partial positive charge and one be negative?
Did he not make a mistake when he redrew the whole thing after the substitution? That hydrogen went to the other side. It's a different grouping if you will. Not sure why that's so, didn't understand the explanation. What's more, in the Sn2 stereochemistry he redraws it differently again, so I'm a bit perplexed here. Help anyone? :)
you mean the 3-D structure related to the three hydrogens ?
if yes then see,
the hydrogens are not actually at the place they should be.
because as the Br- leaves , and the OH- comes and joins with the carbon , the two hydrogens are pushed towards the space created as the Br- leaves.
atleast i think of it like that.
hope i didn't confuse you.
just think about it in 3-Dimension and you will get it better.
I understand how Addition, Elimination (E1 , E2 or E1CB) Substitution (SN1 , SN2) and Rearrangemen reactions occur..

However, i do not understand which reaction mechanism has to be followed when for what solvent, reagent or substrate...
Does it need to be mugged up?

For example,
CH3CH-Br-CH2CH3 + OH- follows SN2 mechanism , i.e. it first reaches a transition state and then we get the product which is CH3CH-OH-CH2CH3OH + Br-

But i cannot understand why does not it follow SN1 mechanism i.e. it first becomes a carbocation and then OH- bonds with it.

Similarly, i cannot understand by seeing the solvent, reagent or substrate which reaction mechanism will be followed...
Everyone has this difficulty and some people never understand the reasons. A secondary bromide like the one you show could undergo Sn1 reaction if you do not use a high concentration of good nucleophile in a protic solvent.

I'd elaborate....but there are errors in your have a diol that just a typo? Are you sure you understand how substitution occurs?
I would have liked to see what the rate determining step was. Could someone clear this up for me?

Example - Rate: [ ? ] [ ? ]
What is the nature of strong bonds between organic-sulfur (and higher chalcogen) compounds and gold?
You have portrayed this as virtually a free-radical reaction when in fact it is one which does not proceed via a free radical state but an anionic state. Why have you chosen to do it this way?
Does the Carbon in CH3Br (name?) Methyl bromine? have a partial positive charge because BR is more electronegative?
Exactly.Generally for elements commonly used in Organic Chemistry, Electronegativity increases to the top and right on the periodic table as you move towards fluorine. This isn't a perfect rule in all chemistry, but I found a really good pictorial map of electronegativities here if you want the exact pattern:
At 10:50, what's the sign you've placed at the top of the transition state parenthesis?
The sign is called a "double dagger" (‡). It is the symbol that chemists use to indicate that a structure is a transition state.
So at the end of the reaction in this example will a "bromide anion" and a molecule of "methyl alcohol" form?
Yes. I think besides the formation of methyl alcohol, bromine would we liberated as a gas with brownish vapors. But do get it confirmed.
why is it that in an sn2 reactions needs a strong base which is a good nuc but in sn1 its a poor nuc weak base.
It's a matter of relative rates. A strong nucleophile doesn't wait for the leaving group to leave. It attacks the partially positive α carbon first, forcing the leaving group to leave (SN2).
If the nucleophile is weak to attack the partially positive α carbon, the leaving group will leave first (SN1). Then the nucleophile can attack the fully positive α carbon.
Sal said that an Sn2 reaction would be fast, but why? Just that it happens in 1 step is not a reson for it. There are thousand of reactions that are 1 step but takes a hell lot of time?