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Sn1 reactions

Introduction to Sn1 reactions

Sn1 reactions

Discussion and questions for this video
Hello, I am wondering as to how you determined the rate determining step. You indicated that the "slowest" step is the rate determining step and in this case would be the Bromine leaving and becoming the Bromide anion at 11:48. Could you explain why this is the rate determining step? Does this have to do with the stability of the compound given that Bromine really wants another electron? Thus what does stability have to do with determining rate determining steps? Thanks! -College Organic Chemistry student
In this case the reason that step 1 is so slow is the relative stability of the intermediates. The 2-bromo-2-methylpropane is a fairly stable species and is generally happy to just bounce around. The tertiary carbocation is much much less happy because all it's atoms are not obeying the octet rule. So every now and again a 2-bromo-2-methylpropane molecule gets enough energy that the bromine leaves and creates the carbocation, which can then grab the hydroxol group. But there are not that many carbocations around at any one time because of the energy requirements to form the cation, so this is a really slow step. Steps that form cations are often, but not necessarily always, the rate determining step of any mechanism they are in because of that huge energy difference.
Hey sal. why do weak nucleophiles favour Sn1 reactions?
this is because weak Nucleophiles don't have the tendency to attack the partial positively charged carbon.
so it needs to wait till the leaving group leaves and when it does, it can then give its lone pair
How do you decide whether it's a Sn1 or Sn2?
It's about the rate determining (slowest) step of the reaction. Sn1 means that the slowest step only has one thing in it. In Sal's example the 2-bromo-2-methylpropane is the only thing in the step that forms the cation, which happens to be the slowest. In the Sn2 reaction the slowest step was when a water molecule got together with the bromomethane, in that step the water bonded and the bromine left in the same step. So two things in the rate determining step is Sn2.
what does "rate determining" mean?
The rate determining step essentially means the slow step.

Think of you and a friend washing dishes. You are washing, your friend is drying them. It takes you 5 seconds to wash a plate, and it takes your friend 10 seconds to dry the plate. Drying the plate is the rate determining step, because it determines how quickly you and your friend can finish the job!
As Sal said the reaction is taking place in the abundance of water so why doesn't the Hydronium ion combine with the Bromine ion to form Hydrobromic acid.
Hydrobromic acid is a strong acid, and is a much stronger acid than water. If you dissolved HBr in water, it would fully dissociate to HBr + H2O -----> H3O+ + Br-. Since this reaction lies essentially completely to the right, HBr will not reform from H3O+ and Br-.
shouldn't the tertiary carbocation intermediate be trigonal planar (that is, no methyl group coming out, or going into, the page)?
You're right, it is trigonal planar. But also, the way Sal drew it is also correct because if you imagine the carbocation he drew, the triangle is sightly pointing away/towards you.
How to identify whether the reaction is SN1 or SN2 reaction?
A reaction is Sn2 if:
1)it proceeds through a transition state i.e. intermediate of nucleophile and substrate, which is usually a penta-ordinate system.
2)It takes place in a single step i.e. it is a concerted reaction
3) and nucleophile usually attacks from opposite side of leaving group
And the reaction is Sn1 if reaction proceeds through a carbocation.
So an alcohol has an oxygen bonded with a hydrogen?
Yes, an alcohol, generally, is R-O-H
do you use any specific arrows to show how the atoms and electrons move, like a half arrow or a full arrow?
The convention is to use harpoons (half-arrows) to represent the movement of single electrons and full arrows to represent the movement of electron pairs.
Why doesn't the bromine anion takes the H atom from the water molecule that attached to the substract? Why does it have to be another water molecule instead? Shouldn't it be more stable in terms of electron charges that the bromine anion attaches to a H, forming HBr?
Thank you.
HBr is a very strong acid and is fully dissociated in water. It would be very unfavourable to re-form HBr in the presence of water since Br- is a much weaker base than H2O.
Why is a terc carbon more stable than a sec?
The extra alkyl group is able to supply a little more electron density to the 3° carbocation, so that makes it more stable.
I'm confused... Shouldn't Sal use "half-arrows", indicating that only one electron is mooving, not a whole pair? Or am I totally mistaken?
Yes, it is customary to use harpoons for single electrons and arrows for electron pairs. Most chemists consider these reactions to involve movements of electron pairs and reserve the harpoons for radical reactions.
Why is the H2O unable to attack the 2-bromo-2-methylpropane at the same time that the Br leaving group leaves, because isn't the 2nd carbon in the 2-bromo-2-methypropane slightly positive due to the halide group on it?
Yes, you're right that the 2nd carbon is slightly positive due to the electron withdrawing halide group. However, because that 2nd carbon has 3 methyl groups attached to it, it is quite sterically hindered so it is difficult for the H2O to access it. Once the Br- has left, a tertiary carbocation is formed. This carbocation is planar, so at this point the 2nd (positively charged) carbon is more accessible, and the water is able to attack.
At the end of the reaction, would the Br(-) attack the H of H3O(+)? It'd leave you with 2-methylpropan-2-ol, hydrobromic acid, and water, which seem more stable than what we ended with in the video.
Remember that HBr is a strong acid. HBr ionizes completely in water: HBr + H2O → H3O+ + Br-. If HBr is a strong acid, Br- is a weak base. It will not remove the H from H3O+.
is sn1 reaction is similar to e1 reaction
Yes, they are similar in that the first step of the reaction mechanism is the same in each.
Is it possible for hydronium (H3O+) to react with Br- to form H2O and HBr?
Think about this question from an acid-base perspective: HBr is a strong acid. It completely dissociates. The reverse (as you have it), will not occur.
Can Sn1 or Sn2 be determined from a chemical equation? If so, how? I understand what makes an Sn1 or Sn2 reaction, but what method is used to determine which a reaction will be?
You have to do experiments to determine whether a reaction is SN1 or SN2. You can make predictions based on the nature of the substrate, nucleophile, solvent, leaving group, etc, but you must always do experiments to confirm your predictions.
The 1 and 2 refer to whether the reaction is unimolecular or bimolecular. That is, does the rate of the reaction depend on the concentrations of one or both components?
Let's say that you are carrying out a reaction between NaOH and an alkyl halide RX. Is the reaction SN1 or SN2?
You might first do an experiment in which [RX] and [OH-] both equal 0.1 mol/L and measure the rate of reaction r₁.
Then you would do a second experiment in which the only change is that [RX] = 0.2 mol/L. If the new rate r₂ is twice r₁, then the rate is _first order_ in [RX].
Then you would do a third experiment in which [RX] = 0.1 mol/L and [OH-] = 0.2 mol/L.
If the new rate r₃ is the same as r₁,you know that the rate of reaction depends _only_ on [RX]: the reaction depends on the concentration of only one component, the RX. The reaction is unimolecular, so we label it SN1.
Ir r₃ had been twice the value if r₁, the reaction would be first order in [OH-]. The rate of the reaction is found to depend on the concentrations of _two_ components, so this is a bimolecular reaction, and we label it SN2.
Hope this helps.
At the end shouldn't the plus be on the hydrogen? The oxygen will take the hydrogens electron to form hydronium.

The plus should go on the atom which has a different amount of electrons than its neutral counterpart. Hydrogen usually has 1 electron in its valence shell. Oxygen usually has 6. This time oxygen has 5 - it is missing one electron. Therefore it has a +1 formal charge.
How do you know if its a fast reaction? and if its in equilibrium or not?
You can tell if a reaction is fast or not only by doing an experiment. After you have done enough experiments, you discover that tertiary substrates usually react faster in SN1 reactions than secondary substrates do, and these in turn react faster than primary substrates.
at 7:15 sal says oxonium ion,what is an oxonium ion?
Oxonium is oxygen with 3 bonds, usually in the cation form. For oxygen to be a cation, or O+, it would have to loose an electron and would have a formal charge of +1. Oxygen with formal charge of 0 would have two R-groups bonded to it, and 2 lone pair electrons. When the third bond is made, the oxygen utilizes one if its lone pairs of electrons to make the bond, resulting in sharing two electrons instead of always having 2. Since the oxygen is sharing two electrons with another molecule, you can think of it only being able to have 1 electron at any time (instead of 2 like before the third bond was made). This is what gives the oxygen at formal charge of +1, which is considered a cation, and specifically with oxygen called an oxonium ion
Around 8:16 when another water molecule comes to take the proton from the positive oxygen, could you also show the bromide anion taking the proton?
You could, but water is a much stronger base than bromide ion, and there is so much of it, that it will almost certainly be the water that accepts the proton.
Having trouble understanding why the reaction would occur. At 5:20 you mention that the water molecule is attracted to the tertiary carbocation because the water molecule has a partial negative charge around the oxygen. Yet, the bromide anion has a full negative charge. Why does the carbocation attract the partial negative charge of the water more so than the full negative charge of the bromide anion? Thanks
The bromide anion would also be attracted to the carbocation, but this would just reform your starting material (this does happen a bit in reality, but we don't usually talk about it because it results in no net reaction). This reaction would likely be done with a _lot_ of water (ie water as the solvent) as compared to the amount of your starting material. So, when the carbocation forms, you would have a little bit of bromide anion and a lot of water molecules. Since there are so many more water molecules than bromide anions, the water molecules are more likely to react with the carbocation.
hey, should'nt that at the end there, when the water grab the H+ from the molecule to form, H30+, it suppose to have a '+' isnt that? and I wan to know, why the water molecules can react just like in Sn2 where two water molecules reacts to give OH-?
Good eye, yes it is suppoed to be H3O+. For your other question remember that water is amphoteric, it can react as a base OR an acid. So, in this example the water is tearing the H+ of of the other H2O molecule i.e. the water is acting like a base.
Hi, i have never understood why a water molecule as a weak nucleophile would attack the positively charged substituent before the bromine would since it is has such a high amount of electronegativity. Is it simply a matter of the water as a solvent in the reaction being at a much higher concentration?
The carbon bromine bond is very weak. Bromine is an excellent leaving group.

What you mean by "why a water molecule as a weak nucleophile would attack the positively charged substituent"......this is a two step reaction. What positively charged substituent are you referring to...the carbocation?

Water is in a much higher concentration, that is certainly true. However, Br- is a much weaker base/nucleophile than water. Once it's left, it's surrounded by a hydration shell of water molecules making it unavailable for reaction. It has nothing to to with electronegativity here....the determining factor is nucleophilicity which follows the same trend as basicity.
(5:30) I don't understand why it is that when the tertiary carbon becomes a carbocation, the H20 doesn't just donate a hydrogen (by taking an e- from the hydrogen rather than a lone pair), turning the H20 into OH- and the substrate then staying as 2-methylpropane...
If the H2O donated an H- to the carbocation to form 2-methylpropane, you would have OH+, not OH-. OH+ is a very high-energy ion, because the O is highly electronegative, Thus this process is unlikely to happen.
Can SN1 reactions also occur with a secondary carbon substrate?
Yes, they certainly can. You always get a mixture of SN1 and SN2 reactions. For most secondary substrates, the reaction is mainly SN2. But if you have an excellent leaving group, a weak nucleophile, and a highly polar protic solvent, SN1 can predominate.
At the end of the machanism, shouldn't the proton be attacked to the bromium ion to create HBr ?
that would seem more logical to me, since H3O+ would also probably instantly react with Br-, this also depends on the solubility of HBr acid..
Can you please explain how can i determine the products of hydrolysis of Alkyl Halides? As i understood it depends from the structure of carbocation? If we would hydrolyse 1,1-dichloropropane, the product would be proprionaldehyde? Now in this case we would have C connected with Cl,Cl,H and C2H5, but if we change H with lets say CH3 radical, then the product would be propanone? I think also you can get carboxylic acid in one of these cases
8:40 i think the bromine anion should give the electron that it had so the reaction would end up with HBr rather than hydronium ion...
Instead of 2-methylpropan-2-ol , can we just call it methylpropan-2-ol because if methyl was somewhere else it would be butan-2-ol?
IUPAC naming does not drop the number designating the substituent. It does not matter if placing the methyl group would result in a different parent chain.
Can the group which attacks center carbon be the group which leaves?
Sn1 reactions use protic solvents correct?
Yes, SN1 reactions tend to work well in polar protic solvents because these solvents are good at stabilizing the forming carbocation and solvating the leaving group.
So to confirm, an SN1 reaction involves a carbocation whereas an SN2 reaction involves a transition state?
why does a tertiary alcohol undergo sn1 mechanism ?
Tertiary-substituted groups prevent nucleophilic attack due to the steric bulk around the leaving group, thus SN2 cannot occur.
how OH- is a good nucleophile??
oxygen is a electronegative element so it must be much stable to hold that negative charge...
It is all about relative energies. Yes oxygen is electronegative an can stabilize a negative charge fairly well. However, it is still in a more stable configuration when there is a neutral charge. A general rule is the neutralization of charge will create a more stable molecule. As always in organic chemistry there are a few exceptions but they are not important for learning the general trends of organic chemistry.
Hey Sal, are there weak nucleophile used in Sn2 reactions like they are in Sn1 reactions?
Does the bromine pick up a hydrogen at any point?
Technically, yes it could definitely pick up a hydrogen. Realistically, it does not happen for very long since hydrogen bromide (HBr) is highly soluble in an aqueous solution. And since this is an Sn1 reaction, the solvent will most likely be nice and polar (typically water). In solution, the bromide ion (Br-) will be relatively stable on its own since the charge is insulated by the polar solvent. A bunch of water molecules with partial positive charges on their hydrogens, in Sal's example, will surround the Br-. Their positive hydrogens will be oriented towards the Br- and their negative oxygens will be pointed away.
is this a neuclophilic addition?
It is a "substitution" rather than addition. A group leaves to be "substituted" by another, in this case the bromide is replaced (substituted) by the water. Addition reactions will typically be breaking double or triple bonds to add a group to the substrate (without any other groups having to leave).
How do u determine which step is rate determining?
among all the intermadiate reactions we consider the slowest step and that reaction determines the rate.this is as per chemical kinetics classical concepts
whats an oxonium group?
An oxonium ion is any oxygen atom that is attached to three other atoms and has a positive formal charge. A hydronium ion, H3O+, is a special example of an oxonium ion, A common example of an oxonium ion in organic chemistry is a protonated alcohol, ROH2+ or a protonated ether R2OH+.
Wouldn't the Bromide ion have a stronger attraction to the carbocation than the water molecule? Is it just because the carbocation is surrounded by water molecules so that a water molecule is more likely to react?
is it possible to give any significance of this reaction from industrial or real -life point of view?
There are sooOOO many examples.
at 0:44 shouldnt the compound be butane instead of propane coz there are 4 Carbons?
There are four carbon atoms, but the longest continuous chain contains only three carbon atoms, so it is named as a propane with a methyl group attached at C-2.
In naming the 2-Bromo-2-Methylpropane. Wouldn't the propane be an Isopropane? I'm just learning to name molecules and I'm finding all of the little quirks confusing. If it isn't an isopropane, can you explain why it is not. This looks like something that would be on a test in one form or another.
When naming an organic molecule you generally start with recognizing the longest carbon chain. In this case that is propane. Then you name the substituents attached. Here, they are the bromine and the methyl group, both attached to the 2nd carbon. Isopropane is the name of a substituent, meaning it would be branched off the longest carbon chain that it would therefore not have any carbons in common with.
help me out plz...
my textbook says we can add NaOH to any alkyl halide to give an alcohol and NaX..
R-X + NaOH---> R-OH +NaX..
my question is ... can we add h20 instead of naoh? and which mechanism is this?
is it nucleophilic substituion rxn? if yes, then is it sn1 or sn2?
why sec-alkyl halide is stable
A 2° alkyl halide is less likely to react by an SN1 mechanism because the 2° carbocation intermediate is less stable than the 3° carbocation that is formed from a 3° halide.
There are references made to Sn2 rxns in this video, which suggests substitution mechanisms have been covered elsewhere, even though I would think this would be the first video about them, given the title. Please point me to where this topic is first addressed. Thanks!
A negative bromine and a positive hydronium... I can imagine what happens there, but could you enlighten me??
For the sake of quickly selecting mechanisms, a solvent with H would be a polar solvent, so an E1 or Sn1, ? Like methanol and methoxide, correct?
shouldn't it be 2-dimethyl at about 10:33 when it is getting named?
No, because you only have one methyl group sticking out from the second carbon
I guess I'm just still confused.....I know the "difference" in SN1/SN2. However, I am not understanding HOW do I know if it's happening in one step or two? I am able to follow the flow of the electrons all the way through the reactions and end up with the correct products and I understand that the rate determining step is the slow step but how can I look at a reaction and know that the Bromine leaves at the SAME time the OH moved in on the SN1 example and left completely independently in the SN2 example (just before the first H20) in the SN2 example?
would it be safe to say that a way to look at SN2 vs SN1 is that:
SN2 requires the forcing of a leaving group vs SN1 being a situation where an opportunistic nucleophile waits to jump in on say a carbocation which has formed as the result of a reactant leaving on its own accord?
Khan should really consider in making a partnership with Kaplan MCAT prep (or make an MCAT prep play list in general), im using your video and they are not far off the test prep stuff(spot on). i actually understand much more with your videos. then again I'm more of a visual learner, and visual learners' benefit so much out of your videos. thanks Khan.
Shouldn't it be h30+ and not h30 at 8:39
Why isn't the hydrocarbon named tertbutylbromide?
The common name is tert-butyl bromide (two words, with a hyphen between tert and butyl). The systematic name is
hello i'm anny.
though the video on this topic has not been made but it comes under the topic of nucleophilic reactions.
i wanted a detailed explanation of Nucleophilic internal substitution or also known as Sni.
thank you..
In a test how will I be able to tell if a reaction is Sn1 or Sn2? And do Sn1 reactions only happen with weak nucleophiles?
You can determine whether a reaction is likely to proceed predominantly via SN1 or SN2 by looking at the nucleophile, electrophile, and solvent (and, if you already know the product and just need to determine the mechanism, the stereochemistry will tell you!). If you have a polar protic solvent, a weak nucleophile, and tertiary alkyl halide as your electrophile, your reaction will be SN1. If you have a polar aprotic solvent, a strong nucleophile, and a primary alkyl halide as your electrophile, your reaction will proceed via SN2. If you have something inbetween, it can be trickier to decide (check out http://www.masterorganicchemistry.com/2012/08/08/comparing-the-sn1-and-sn2-reactions/ for more details).

Keep in mind that BOTH reaction mechanisms proceed in most situations, they just proceed at very different rates. So yes, a weak nucleophile could react via SN2, but because it's weak, it will only manage to "kick out" the halide on the electrophile every once in a while. The halide spontaneously leaving to give a (stable) carbocation would happen much more frequently, and once the carbocation is formed (slow step), the weak nucleophile would react with it quickly. So both happen, but because in this situation SN1 proceeds so much more quickly than SN2, it is the dominant reaction mechanism.
In the first part of the video of SN1 reactions, does the BR- and H30+ not associate? After all they each have an equal and opposite charge.
I was hoping for some clarification concerning the equilibrium of this reaction. Since there are bromide ions and H3O present as products, what will occur as final products. Will there be a higher concentration of hydronium (pka -1.7) or H-Br (pka -9)? I struggled with this in Ochem last year, so I really need some good input, thank you all so much :)
sir, although i like ua lectures alot.......... bt the problem is that aftr few min. the video stops playin.......... evn wen i download it , its not workin properly..... how can i get your full video
download button to the bottom left. try that.
Sn1 reaction is commonly carried by polar protic solvents.why?
When an SN1 reaction takes place the substance brakes down into two ionic components which can only be solvated by polar substances that can "cancel" their charge with theirs'.
Can Bromine also abstract a hydrogen off of the water nucleophile that bonds onto the carbon?
The bromide _ion_ *can* abstract a hydrogen from the protonated OH group, but it is such a weak base that it loses the competition with water.
Remember that HBr is a strong acid, so Br- is a very weak base.
how can be 'O' of water a weak nucleophile?
it has two FREE PAIR electrons!!
as it has only partial negative charge
I'm a little confused around 7:30 where he introduces another water molecule just being there...where does that extra water molecule come from?
The reaction takes place in water solution. There are more molecules of water than of anything else.
In the above video , is the oxygen sharing or giving its one electron ?
Please help. Thank you.
At 7:34, if the Oxygen "gave" one of it's electrons to the Hydrogen that is sharing an electron pair with the Oxygen why would the other electron be give back to the other oxygen? I thought that Hydrogen wanted two electrons... Wouldn't it want to keep that electron so that it would have two electrons and be able to exist by itself?
In the molecule where the H is already attached, it has two electrons associated with it (since it shares 2 electrons with the oxygen). In the molecule where H goes to, it also has 2 electrons associated with it (again, 2 electrons are shared between the H and O). That is, H has two electrons in both cases so is "happy" in both of these molecules.
At 03;51, Does Leaving Br from (CH3)3C-Br occur automatically?
It is not exactly "automatic". What happens is that in the solution the molecules of R-X are continuously bombarded by collisions with other molecules in the solution (remember kinetic molecular theory). The molecules of R-Br can absorb energy from these collisions, and this energy goes into more vigourous vibrations of the bonds in the molecule. If any one collision or a series of collisions gives the molecule enough energy, the C-Br bond (the weakest bond in the molecule) will be vibrating so vigorously that the C-Br bond just breaks.
For example, if your car is hit in the rear end by a car that is going fast enough, your front bumper might fly off.
Hope this helps.
How did he get the extra oxygen in the product, where he had the propanol, the bromine, and then H3O? If the original 2-bromo-2-methyl reacted with water (H2O) which in total is 1 oxygen, where did the extra oxygen come from at the end to make H3O?
In this case, water is probably used as the solvent (or at least in excess). You're right, the way Sal has drawn out the mechanism this reaction is not balanced, but it is assumed that there are extra water molecules around to deprotonate the product and make H3O+.
Couldn't it be called 2-bromo isobutane, and couldn't its alcohol be called isobutyl alcohol or isobutanol? I'm just wondering.
Actually, the common name would be tert-bromobutane, or t-bromobutane, because the central carbon, to which the bromine is bonded, is tertiary, as it is bonded to three other carbons.
If the bromine were bonded to any of the other (primary) carbons, it would be called isobutylbromide.
R-C-(CH3)3 is a tert-butyl group, whereas
R-CH2-CH-(CH3)2 is an isobutyl group.
Why compounds with large steric hinderance favours SN1 reaction ?

Please clear my doubt soon.
What is the difference between Sn-1 and Sn-2 reaction?
An SN1 reaction happens stepwise. First, the leaving group leaves, giving you a carbocation. The nucleophile then attacks this carbocation in a second step. Since the nucleophile is equally likely to attack from the top face or the bottom face of the carbocation, you will always end up with a mixture of enantiomers (when in chiral systems).

An SN2 reaction happens _in a single step_. The nucleophile attacks the central carbon _while the leaving group is leaving_. Since the nucleophile always attacks from the side exactly opposite of where the leaving group leaves, the stereochemistry of the central carbon is inverted.

You can help favour SN1 over SN2 or vise versa by changing the nucleophile, electrophile, and solvent.
I don't understand how the bromine just came off? As far as I know, elements don't randomly just fall off.... and water seems to be fairly stable so it wouldn't just make the bromine come off, HELP PLEASE.
It needs activation energy to fall of, like every reaction does. Search around for the term activation energy, it will be better explained elsewhere than I could ever try to.
If the nucleophile is weak and the cation is strong, why is this nucleophilic substitution and not electrophilic substitution?
There is only one electrophilic substitution reaction and it occurs on benzene and other aromatic rings.

The simple answer...reactions are classified by the end result to the organic compound only (It's an electrophile). We do not consider what gets exchanged (substituted) on the inorganic nucleophile.
But I don't understand how to look at the reactants and decide whether to draw Sn1 or Sn2?? And when I do draw Sn1, how am I supposed to figure out the rate determining step by just looking at it?
If you have a large concentration of good nucleophile, a polar aprotic solvent and a relatively unhindered alkyl halide/pseudohalide the reaction will be Sn2.
Let's say you have 2-bromobutane. and above the arrow is NaOH/H2O. ...NaOH is stronger and will certainly react faster than H2O. Is HO- a good nucleophile? yes. it is negatively charged. Sn2. If the alkyl halide is tertiary, Sn2 is not possible. Then you must do an E2 reaction if possible.

In an Sn1, the first step is rate -limiting (solvolysis). It always is, (unless you are doing an acid catalyzed dehydration of and alcohol).
Why did the Br leave the 2-bromo-2-methylpropane in the first step of this reaction? Why did this not also occur in the first step of the Sn2 reaction?
Because in Sn2 you have a strong base which will go grab H first.
in the first example of ur video... when oxygen became neutral by breakng h bond ..so it means hydogen gained a positive charge.......and we know that bromide already has a negative charge.......so why doest hydrogen go and form hydrogen bromide ?
Hydrogen Bromide or HBr does exist. It is a gas usually but it forms hydrobromic acid when dissolved in water.
why doesn't the second water molecule's partial negative side bond with the (now positive) oxygen atom (it instead forms the hydronium cation)? 10:30

thank you!!
At the end, couldn't the molecule be named 2-methyl-2-propanol? this makes more sense to me, intuitively. Would it be correct?
This molecule is commonly called tert-butyl alcohol in laboratories.
Why would Cl favour SN1 and Br and mixture of SN2 and SN1??
Can someone write out the entire reaction, like in words? I understand the structures, but I just want to see it in words, with the prefixes.
Basically u created a charge out of neutral species , but we know charge can't be created nor be destroyed. this way u can create energy out of any thing . that is not possible..
Where in the video are you referring to? Are you meaning when a carbocation is formed (for example, at 3:20)? You're right that we are creating a charged species from a neutral species. However, this is okay in this case because the overall charge is still neutral: the carbocation is positively charged, and the bromide is negatively charged, so there is still no net charge.
what is concenteration of sn1 reaction
The concentration, or moles of molecules per volume, for each species in an SN1 reaction will depend on how much of each reagent you put in - there is no set number!
Why wouldn't the Br- take the extra hydrogen off the water that bonds to the substrate?
It could after the water attacks the carbocation.

However, the solvent is practically infinite in concentration (water). Water is much more basic than Br- (1x10^4 times more). It would be highly improbably that Br- would remove the H. It's already surrounded by a cage of water that stabilizes the anion....it's not even able to break out of the cage to attack the hydrogen.
Whenever we're given 2 reactants and we have to find the product according to SN1 & SN2, how to know if SN1 reaction is taking place or SN2 ?
Often, both reactions occur simultaneously, but usually one mechanism is strongly favoured over the other. It is a competition among various factors.
In general, SN2 reactions are favoured whenever you have:
a strong nucleophile
a poor leaving group
a 1° or 2° substrate
a polar aprotic solvent
SN1 reactions are favoured by:
a weak nucleophile
a good leaving group
a 3° substrate
a polar protic solvent.
You have to weigh these factors against each other and then decide which ones best fit your case.
The reaction shown above occurs in acidic conditions correct? If it were in basic conditions, the reaction would favor the dissociation of the OH group and the substitution of the Bromine anion.
You mean Sn1 doesn't stand for Sal's number one?
Sn stands for Nucleophilic substitution reaction. There are two different ways they can react, and thus we label them Sn1 and Sn2.
At 9:25 the bonded water gives a hydrogen to another water group, creating an oxonium. Why doesn't the hydrogen bond with the Br-? Isn't that easier?
I'm assuming there's just so much more water than bromide that it's more likely the hydrogen will be passed to the water.
As for the product formed in the Sn1 reaction, you wrote it as, 2-methyl propan-2-ol. Can it also be written as, 2-hydroxyl-2-methyl propane?
No, it cannot be.
That is only because it violates the IUPAC way of naming.
The hydroxy there is the functional group. And the functional group becomes the main thing about the compound and that's why it ends with its suffix.
Its all in the rules of naming organic compounds. You should try brushing up on it to be clear with it all.
Why would in the first step , bromine will leave the group on its own without being attacked by any other nucleophile as in the case of SN2 reaction ?
Please help.
Thank you.
How can I distinguish between a strong and weak nucleophile? What are the properties of them? 1:47
As far as I know, it depends on how polarised the molecule is and it's overall charge. H2O is described as a 'weak' nucleophile relative to hydroxide because a hydroxide anion has an 'extra' electron, making it more unstable, thus more likely to give up its electron to a nucleus (in this case carbon). However it really is relative to the atom/compound in question. See the lectures on nucleophilicity and nucleophilicity vs basicity
is the oxygen in the hydronium molecule now positively charged?
Since the Oxygen has 3 bonds in the hydronium molecule, it does, in fact, have a positive charge.
Can somebody please help I have a final in 40 minues! How can you determine if a reaction will undergo SN1 or E1? And then SN2 or E2 if its a primary carbocation.
why does sal draw the different methyl bond to carbon in straight, triangle and dot lines.
Dotted lines or hash marks mean the bond is going into the paper. Triangle or wedges represent bonds that are coming out of the paper toward you. Regular straight line represent bonds on the same plane of the paper.
Is the addition of oh group syn addition or anti addition>>
thank you sir you saved me
adnan humayun pakistan
why end with the hydronium ion, and not the hydrogen bromide?
Oh I didn't know why the bromide didn't react with the hydronium ion, instead of staying separate
Does the carbocation in Sn1 reactions undergo hydride and methyl shifts as it does in electrophilic addition reactions?
They certainly can!
In this example, there is no hydride or methyl shift that would produce a more stable carbocation, but carbocation rearrangements do happen in SN1 reactions, for the same reasons they do in E1 reactions.
Couldn't it be called 2-bromo isobutane, and couldn't its alcohol be called isobutyl alcohol or isobutanol? I'm just wondering.
why is the second part of the video the same as the first video...
Report a mistake in the video

At 2:33, Sal said "single bonds" but meant "covalent bonds."

Report a mistake in the video

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