Elimination vs substitution: tertiary substrate
Determining if a tertiary substrate will undergo E1, E2, or Sn1.
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- At5:55, we have MeOH as a solvent, which is protic. So wouldn't we observe a unimolecular reaction instead of a bimolecular one?(9 votes)
- This is really tripping me up as well. A quick reference to the solvent would've been very helpful here just to reinforce the role of the solvent, especially considering in other videos here it is noted that protic solvents lend themselves to the unimolecular mechanisms. So why is it here we see a biomolecular mechanism?
My exam is today and I'm going in confused now!
Edit: the 3rd edition of orgo as a second language also says: "This trend is similar to the trend that was observed for alkyl halides (Section 7.11). Similarly, tertiary sulfonates have been observed to undergo E2 reactions when treated with strong bases and to undergo unimolecular reactions (SN1 and E1) in protic solvents."
Does this only apply to tertiary sulfonates but not tertiary alklyhalides?(5 votes)
- Why's E2 a possible mechanism to occur, why wouldn't we have to worry about steric hinderance?(7 votes)
- In E2 reactions, we do not need to worry about steric hinderance as we are mostly grabbing a hydrogen from the sides of the molecule, unline in Sn2 where we attacked the central atom (generally).(7 votes)
- So, at the5:26problem. What's the reasoning for the E2 and not Sn1? Like, in the primary carbon it's said that Sn2 is favored over E2. Why in this case is E2 favored over Sn1? Plus, it has a reference for a polar protic solvent. I'd always think it was an Sn1.(4 votes)
- At6:32, it seems to me that Sn1 may be possible too. Because a tertiary carbocation is stablized by the Polar protic solvent, as Jay mentioned in the Sn1 vs Sn2: solvent effects. Any ideas?(4 votes)
- at0:37why is an E2 reaction possible, isn't Cl^- a weak base?(1 vote)
- He is only considering the possibilities based on the molecule on the left, he crosses off E2 like 10 seconds later because Cl- is a very weak base.(2 votes)
- At6:55, both SN1 and E1 reactions are possible. But even though water can act as a nucleophile, wouldn't the reaction still be relatively slow (I remember Jay mentioning in one of the previous videos)?
If heat is used to catalyse the reaction the elimination product would be favored, so what can be used to catalyse the reaction and still get a substitution product?(1 vote)
- Sal please help me out:
The major product of the following reaction is:
treated with C2H5ONa & C2H5OH
I am unable to figure out the major product (most stable one)
Is it elimination reaction that's major?
Which carbon will have the double bond?(1 vote)
- I think the logic shown in video starting at5:24is directly applicable to the first part of your question ...
Then you have to decide which 'arm' is most favorable for the double bond. I believe that the double bond in resonance with the phenyl group will be most favorable due to them forming a conjugated system.(1 vote)
- so when you're comparing substitution and elimination, are tertiary substrates slower than the primary or secondary?(1 vote)
- I- is a better Nu than Cl-, so wouldn't the first reaction favour the opposite direction?(1 vote)
- At0:50minutes, since Cl- is a weak base and a strong nucleophile with tertiary carbon, shouldn't it be able to undergo E1 as well? Why only SN1?
At9:40minutes, what's the reason to favor E1 over SN1? is it the heat? Can the reaction proceed with SN1?(1 vote)
- 1) You're asking the chloride ion to act as a base and reform its conjugate acid, hydrochloric acid, a strong acid. Strong acids like hydrochloric acid dissociate almost entirely and like to remain almost exclusively as their conjugate bases. So you won't see the chloride ion act as a base at all really, only as a nucleophile in a substitution reaction.
2) Yeah it's the heat that causes more E1 than SN1. Elimination reaction produce more products than reactants compared to substitution reactions. Reactions that produce more products than reactants generally increase the entropy of the system. By raising the temperature the (TΔS) term is increased in the Gibbs free energy equation, ΔG = ΔH - TΔS, contributing to a more favorable negative value for ΔG. So an increase in temperature generally favors eliminations while a decrease in temperature favors substitution.
It's important to keep in mind while these conditions favor E1 or SN1, that doesn't mean SN1 won't happen at all. You still get SN1 products, just less so than the E1 products.
Hope that helps.(1 vote)
- [Interviewer] Let's look at elimination versus substitution for a tertiary substrate. For this reaction, we have a tertiary alkyl halide, and we know that a tertiary alkyl halide will form a tertiary carbocation, which is a stable carbocation, and therefore an SN1 reaction is possible. An SN2 reaction is not possible because this tertiary alkyl halide has too much steric hindrance to undergo an SN2 mechanism. An E1 mechanism is also possible because an E1 mechanism requires a stable carbocation. And an E2 reaction is also possible. So we have three choices for a tertiary substrate. Next you want to look at your reagent, and here we have sodium chloride, and we know that the chloride ion is a very weak base, and it's gonna function only as a nucleophile. So if we have a nucleophile here, we're gonna go with substitution reaction, so an SN1 reaction must be the case for this example. And we know, in an SN1 mechanism, first step should be loss of our leaving group for this tertiary alkyl halide, so we form the iodide anion, and we take a bond away from this carbon in red. We form a tertiary carbocation. Let me draw this in. We have our six-membered ring, we have our methyl group, and the carbon in red is this one. It gets a plus one formal charge, so a positive charge on this carbon. Once we've formed our carbocation, in the next step of our mechanism, our nucleophile attacks, and our nucleophile is the chloride ion. Lone pair of electrons on the chlorine are gonna form a bond with that carbon in red, and we end up with our product. Let me draw in our ring. We have a methyl group and we have a bond to a chlorine now, so let me highlight a lone pair of electrons in magenta, form this bond between the chlorine and the carbon in red. We don't have any stereochemistry to worry about here because we don't have any chiral centers. This was an SN1 mechanism. First we look at our substrate, and we can see that this is a tertiary alkyl halide. The carbon that is directly bonded to our halogen is bonded to three other carbons. So a tertiary alkyl halide is too sterically hindered to undergo an SN2 reaction, so immediately we know that that is out, and when we look at our reagent, this is potassium tert-butoxide. Let me draw out the structure for that. We would have a negative one formal charge on this oxygen, and this potassium would be a plus one charge. Potassium tert-butoxide is too sterically hindered to function as a nucleophile, so instantly we know that SN1 is out, and potassium tert-butoxide is a strong base, so we know that E1 is out. And that leaves an E2 mechanism for this reaction. Let me draw the alkyl halide over here again. Let me put these in here. And we know that the carbon that's bonded to our halogen is our alpha carbon, and the beta carbons are bonded to that. So this would be a beta carbon, this would be a beta carbon, and this would be a beta carbon. Let's say our sterically hindered strong base comes along, and takes a proton from this beta carbon. So our base takes this proton, these electrons move into here. At the same time, these electrons come off onto the bromine to form the bromide ion. And we will get one product for this reaction, so let me draw it in here. We would now have a double bond, and let me show those electrons. The electrons in magenta moved into here to form our double bond. Let's say we took a proton from a different beta carbon. Let me redraw, let me just redraw this. I'll draw the alkyl halide in here like this, and again, we know that this is the alpha carbon, and let's take a proton from this beta carbon now. So I'll draw in a proton, and just think about our base coming along, so our base coming along with a negative charge, taking this proton. These electrons move into here. At the same time, these electrons come off onto the bromine, so that's our E2 mechanism. I'll draw the resulting alkene. There's a double bond in here this time. The electrons in magenta moved into here to form our double bond. When we look at our products, we know that the one on the right is actually more stable because the one on the right is a trisubstituted alkene, whereas the one on the left is only a disubstituted alkene. But when you're dealing with a sterically hindered base like potassium tert-butoxide, it turns out that the disubstituted, the less substituted, alkene is the major product. And that's just because our base is so sterically hindered. Also notice, if our base came along and took a proton from this beta carbon, we would get the same product as right here. Here's another tertiary substrate. So we know an SN2 mechanism is out, and when we look at our reagent, which is sodium methoxide, sodium has a plus one formal charge, and the oxygen has a negative one formal charge. The methoxide ion, we saw in an earlier video, is a strong nucleophile and a strong base. And whenever you think strong base, think E2 reaction. The E2 reaction is going to dominate here, and an E2 reaction means the methoxide ion is gonna function as a base and take a proton from our alkyl halide. Next we analyze our alkyl halide, the carbon bonded to our halogen is our alpha carbon, and the carbons directly bonded to the alpha carbon are the beta carbons. Here the beta carbons are identical, so I'm gonna say that's our beta carbon, and let's draw in one beta hydrogen. The methoxide ion comes along, so I'll draw that in, so we have a negative one formal charge on our oxygen. This is going to take this proton. At the same time, these electrons move in here to form our double bond, and these electrons come off to form our iodide ion as our leaving group. For our final product, we're gonna have a double bond here, and let's change colors again. The electrons in magenta moved into here to form our double bond. Since our beta carbons are all the same, this is the only product for this reaction. Here's the same tertiary alkyl halide we saw in the previous problem, so an SN2 reaction is out, and when we analyze our reagent, we know that water is a weak nucleophile and a weak base. And since water is a weak base, the E2 reaction is out. And that leaves the E1 reaction and the SN1 reaction, which both proceed via a carbocation. Let's draw the carbocation that we would form. These electrons come off to form the iodide ion, and we take a bond away from the carbon in red to form our tertiary carbocation. Our tertiary carbocation has a plus one formal charge on our central carbon, so this carbon in red. If water acts as a weak base, water can take a proton from one of the carbons next door to the carbon in red. I will draw in a hydrogen on this carbon, and if water takes this proton, these electrons would move into here to form a double bond. So that's one of the possible products for this reaction. We form an alkene, and our electrons in magenta moved into here to form our double bond. That's when water acts as a weak base, and that would be an E1 mechanism, but we also have the possibility of an SN1 mechanism, so let's draw our carbocation again. Here's our carbocation. Our carbon in red is this one, with a plus one formal charge, and this time, we're gonna show water acting as a nucleophile. Let's draw in our water molecule with two lone pairs of electrons on the oxygen. The nucleophile attacks our electrophile, and we form a bond between the oxygen and the carbon in red. Let's draw that in here. Now we have a bond between our oxygen and our carbon in red, which is this one, so a lone pair of electrons on our oxygen formed that bond in here, and our oxygen is still bonded to two other hydrogens. It still has one lone pair of electrons on the oxygen, which gives the oxygen a plus one formal charge. And we would also need, in our next step, to have an acid-base reaction. Another molecule of water could come along and take one of these protons. In the interest of time and space, I'm just gonna write minus H plus here. We lose one of our protons and we form our product, which would be this tertiary alcohol. So we have two possibilities for this reaction. We could have either an E1 reaction, which would give us this alkene, or an SN1 reaction, which would give us this alcohol. For our last example, let's look at this tertiary alcohol. We know SN2 is out and we have a strong acid present, not a strong base, so an E2 reaction is out. And whenever you see a tertiary alcohol with something like sulfuric acid or phosphoric acid and heat, think E1 reaction. So we need to form a carbocation here, but before we do that, let's analyze the structure of our alcohol. The carbon bonded to the OH is our alpha carbon, and the carbons directly bonded to that are our beta carbons. I'll call this one beta one, this one beta two, and this one beta three. We can't have loss of a leaving group right away because the hydroxide ion is a poor leaving group, but our sulfuric acid is a source of protons, so our first step is to protonate the alcohol. The lone pair of electrons on the oxygen pick up this proton, and let's draw what we would form here. The oxygen would now be bonded to two hydrogens. Still has a lone pair of electrons and a plus one formal charge on the oxygen. These electrons here in magenta pick up this proton to form this bond. And now we are ready for loss of a leaving group. When these electrons come off onto the oxygen, they would make water, which we know is a stable molecule. It's a good leaving group. And now we're taking a bond away from this carbon in red. So now we're gonna form a carbocation. Let's draw in our carbocation. So actually, let me make this a better methyl group here, and our carbon in red is this one, so that gets a plus one formal charge. In our E1 mechanism we know we're gonna have a weak base come along and take a proton from one of the beta carbons. I'll go with beta three for this carbocation here. I'll draw in a hydrogen coming off of this carbon, let me mark it in blue here. This carbon in blue was this carbon on our starting alcohol. A weak base would have to be water. We just formed water in our previous step, so water comes along and takes that proton. Let's draw in our lone pairs of electrons. We're gonna take this proton, and these electrons are gonna move in here to form our double bond, and that would, of course, get rid of the formal charge on the carbon in red, and give us our product, which is an alkene. Let's draw that in here. Our electrons in magenta moved into here to form our double bond, and that's a trisubstituted alkene. What about if we took a proton from one of the other beta carbons? Going back over here to our picture of the tertiary alcohol, beta one and beta two would actually both give us the same product. So we'll just take one of them, and first I'd have to redraw my tertiary carbocation. Here it is. Plus one formal charge on the carbon in red. And I'll draw in a hydrogen on this carbon. So water is going to function as a base. Let me draw in our water molecule here. And it's gonna take this proton, and these electrons are gonna move in to form our double bond. So when we draw in our product, we're gonna have a double bond here, and our electrons in magenta would move into here. This is only a disubstituted alkene, so this would be the minor product. And the major product would be the trisubstituted alkene, which is the more stable one. So this is the major product for our E1 reaction.