E2 E1 Sn2 Sn1 Reactions Example 2. Created by Sal Khan.
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- If the carbon has a bulky group attached, making it a tertiary, won't SN1 or E1 be used? or it has do to with being aprotic or protic(6 votes)
- In our case we have an Aprotic solvent which encourages - SN2/E2. The solvent cannot change the fact that we have a bulky base. Since steric hinderance is the limiting factor for an SN2 reaction, we are forced to do an E2 reaction because the nucleophile cannot complete an attack of the alpha carbon (bulky bases are too bulky to attack).
A protic solvent would help stabilize an unstable substrate (alpha carbocation) but since we have a strong nucleophile - we are right off the bat looking for Sn2/E2.
Make sense?(14 votes)
- how do you know what base is strong and weak(5 votes)
- You can take a simple look at the pKa of it's conjugated acid. A high pKa means that you have a weak acid and vice versa, but a strong conjugated base. If you take ethyne for instance, which has an acidic H and a pKa of 25. It is therefore a very weak acid, but when the H+ is gone, you have a very good base (or nucleophile) because it has a lone pair of electrons ready to attack any electrophile. There are other more theoritical ways to find your awnser, but as i mention, you can simply look at the conjugated acid's pKa.(12 votes)
- Is there a standard used to evaluate the strength of a base? I understand that steric hindrance affects the nucleophilicity strength, but what about bases?(2 votes)
- i think bases are strongest when they have the most incentive to grab a H+ atom. so F- is a stronger base than I- because its more electronegative and really wants to react. when the halogen anion is a conjugate base of a strong acid it is a weak base. so as you move down the period the basicity decreases because HCl, HBr, and HI are strong acids, they make weaker bases. i think shielding also effects basicity as atomic radius increases basicity decreases(6 votes)
- So in short, the bulkier it is, the less nucleophilic?(3 votes)
- It depends on the solvent.
I⁻ is the strongest halide nucleophile in a polar protic solvent, but the weakest in a polar aprotic solvent.(3 votes)
- At1:40, you say that the reaction will strictly take the E2 path. But now since tert-butyl oxide ion is a weak nucleophile and DMF is a aprotic POLAR solvent, wont these conditions make way for the SN1 reaction too? Wont SN1 substitution products be obtained too?(3 votes)
- Would it be fair to say that a spectator ion would also be present in this reaction to balance the negative charge of the attacking molecule? In which case, would the attacking molecule then be potassium (or some alkali metal) tert-butoxide?(2 votes)
- You're right in saying that it serves as a counter ion to the negative charge, but we still say that the attacking molecule is tert-butoxide, because that's what's doing the actual chemistry. There is no bond between the metal ion and the oxygen. Even if you're adding tBuOK, there's still dissociation when it's in whatever solvent (usually tBuOH).(3 votes)
- From what i have heard, that Ph-CH2X undergoes sn2 mech. But in this case the carbocation is resonance stabilised, then is it not more favorable to go under sn1 mech?(2 votes)
- 1° benzylic halides react via SN2, but 2° and 3° benzylic halides react SN1.
The reason is that in the 1° compound, the transition state for the 1° cation is at a high energy level. The aromatic ring does stabilize the cation somewhat, but not enough. The TS for SN2 attack is at a lower energy level, so SN2 attack predominates.
The 2° and 3° cations are much more stable (lower energy), and the ring stabilizes them further, so that in these cases SN1 attack predominates.(3 votes)
- Why (CH3)3-C-O is not a good nucleophile? Is it because it's bigger than CH3-O?(2 votes)
- (CH3)3-C-O is bulky and presents a great amount of steric hindrance. This means that the molecule will have trouble aligning itself with the position it needs to reach in order to react.(3 votes)
- What is the steric hindrance limit for determining whether Sn2 or E2? How bulk does it have to be.(2 votes)
- At the end of the video, you said that only E2 will happen because of the bulky base. Can SN2 still happen in small amounts or not at all? In my textbook, if the substrate is a primary carbon, then even if a very bulky base is used, SN2 can still occur in small proportion, is the same true for secondary carbons?(2 votes)
What I want to do in this video is think about what type of reaction we might have if we have ingredients very similar to what we saw in the last video. But instead of our nucleophile or our base being methoxide, it's going to be something slightly more involved. So it's still going to have the O minus, but it's going to be bonded to a carbon, which is then bonded to three methyl groups: CH3, CH3, CH3, just like that. So we don't have methoxide anymore. We have this thing right over here. So just like before, we have the exact same solvent. We have dimethylformamide. It's an aprotic solvent. That by itself would put us in the Sn2 or E2 direction. But now we don't have methoxide anymore. Methoxide was both a strong base, very strong base. It's also a very small molecule, and so it can really get in there and react with the substrate. So it's also a strong nucleophile. Now, this more bulky molecule, it is still a strong base. It is still an extremely strong base. But now it's this big, bulky molecule. It would actually have trouble getting in to react with your substrate, so it is no longer a good nucleophile. This is not a good nucleophile. So by making the base more, I guess, bulky, it's now-- or I guess you could also call it the nucleophile or the thing that would act as a nucleophile, more bulky. It is no longer a strong nucleophile, so it would no longer be good for an Sn2 reaction. So just by changing the base a little bit or the nucleophile a little bit, now this one would go strictly in the E2 direction. So we wouldn't see anything like this in the last video. We would only see something like this. And obviously, the base in this example is no longer just a methoxide. It looks like this. Let me clear it. Let me do my best to clear it. Edit, clear. Let me clear it over here as well. So now instead of just being bonded to a methyl group, it's bonded to a carbon. It's bonded to a carbon that's bonded to three methyl groups. So CH3, CH3, CH3, or you could call this a tert-butyl group; this whole thing over here. So that's a carbon bonded to a CH3, a CH3 and a CH3. So the reaction occurs just like what we saw in the last video, except this base is this big, old, bulky thing, but it can still act as a strong base, so it still nabs the hydrogen or really just the proton. The hydrogen's electron that was bonded now goes to the alpha carbon. The alpha carbon will then lose an electron to the bromo group and that becomes bromide, so the same exact mechanism, different base. But that base is now not a good nucleophile, so you won't see Sn2 occurring at all. You will only see E2.