Elimination vs substitution: primary substrate
Figuring out if a primary substrate will undergo substitution or elimination.
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- At5:43, if H2O is a weak nucleophile AND a weak base, why can we rule out E2 before Sn2? Is it that primary tends to favor Sn2? Why?(17 votes)
- At1:40, (the second example) a rearrangement can make a tertiary carbocation,
and if we follow E1 mechanism, we get a more substituted product. So, shouldn't that be favored.(10 votes)
- A primary carbocation would have to form first before it could undergo a shift.
In almost all cases a primary carbocation can not form (there are a few exceptions, but this is not one of them).(3 votes)
- At3:20, why does Primary Halide favor Sn2? What makes it want Sn2 more than E2?(9 votes)
- For the second example, at1:40, why did we rule out Sn1 and E1? A tertiary carbonation can be formed by a hydride shift, can't it?(3 votes)
- It would only work if it was a secondary carbocation. Primary ones aren't observed(3 votes)
- at2:56you said that SN1 and E1 are out, why are they not being considered?(3 votes)
- 1) The SN1 and E1 mechanisms require the formation of a carbocation.
2) For the molecule shown (1-bromopentane), you can only get a primary carbocation.
3) Primary carbocation are almost always so unstable that they can't form (explained in earlier videos).
Jay mentions this for SN1 starting at0:12.(3 votes)
- Would someone please describe why (the underlying mechanism) it is that the Sn2 reaction is favored over the E2 reaction when you are dealing with a strong NUC and Base with a primary substrate.(3 votes)
- I think it is because the nucleophilic character outweighs the basic part. So SN2 is favored over E2(2 votes)
- Still cannot understand why hydride ion is not a nucleophile. In the rearrangement of carbocation, could the hydride ion shift (e.g. form tertiary carbocation) be viewed as a necleophilic reaction?(2 votes)
- Why does Sn2 require a strong nucleophile while E2 require a strong base?(1 vote)
- For SN2 to take place, we need a nucleophile to attack the alkyl halide at the same time when halide leaves the alkyl part. Only a strong nucleophile can attack the group simulataneouly when halid is leaving.
For E2 to take place we need base to attack the parent group when the leaving group is leaving. Only a strong base can be that reactive to undergo such a fast process.(3 votes)
- 5:50lol..from what i learnt here you should have ruled out SN2 and E2 because you need a strong nucleophile and a strong base..what?Please,can sb help me?(2 votes)
- At0:53, why will SH act only as a nucleophile and not as a base?(1 vote)
- He says SH is a weak base since its conjugate acid is pretty strong.(1 vote)
- [Lecturer] In this video we're going to look at primary substrates and figure out if the reaction is a substitution or an elimination. So for this primary alkele halide, we know an SN1 reaction is out because that would require a stable carbocation and we can't make one from this primary alkele halide. An SN2 reaction is possible because of the decreased steric hindrance of our primary alkele halides. An E1 reaction is out, again for the same reason as SN1, we can't form a stable carbocation. And an E2 mechanism is possible. So now the next step is to look at our reagent and figure out what the reagent is going to do. So for this reaction we have a sulfur nucleophile which we know is gonna act only as a nucleophile and not as a base. And since it's going to act as a nucleophile, that means that E2 is out. We need a strong base for an E2 reaction. So this must be an SN2 mechanism, which we know is a concerted mechanism. Our nucleophile attacks our electrophile at the same time that we get loss of a leaving group. So we're gonna form a bond between the sulfur and this carbon here in red. At the same time these electrons come off to form the iodide leaving group. So let's draw in our final product. So we have four carbons and then we have a bond to our sulfur which is bonded to a hydrogen. So the carbon in red is this one. And so this is our product and we don't have to worry about any stereochemistry here since we don't have any chiral centers. So again we have a primary alkele halide. And we've seen that SN1 and E1 are out. So we only have to decide between SN2 and E2. So we look at our reagent and we know that DBN is a strong base. So therefore this is gonna be an E2 reaction and not an SN2 reaction. So DBN has to function as a strong base. And we know that the halogen is directly connected to this carbon so that must be our alpha carbon. In an E2 mechanism we're gonna take a proton from a beta carbon. So this is a beta carbon right here and there's one, there's one hydrogen on this beta carbon. So DBN we know is a neutral base. So I'll draw in here just a generic base with a lone pair of electrons. This base is gonna take this proton and these electrons are gonna move into here. At the same time these electrons come off to form the iodide anion. So let's draw the final product. We have this long carbon chain here. So a double bond in this position. And then we're also gonna form a double bond in this position. So let's look at those electrons. So these electrons here in magenta, would move in here to form our double bond. For this primary alkele halide, again we've seen SN1 and E1 are out. So we're deciding between SN2 and E2. So next we look at our reagent. And we have potassium hydroxide. And we know that the hydroxide ion is a strong nucleophile and a strong base. So that's the category and we saw in a earlier video. And for a primary alkele halide that is unhindered, the SN2 reaction is going to win out. So a small nucleophile can attack this carbon, at the same time these electrons come off onto the bromine to form our bromide anion. So let's see our carbon in red is this one. So let's draw out our carbon chain. So we have one, two, three, four, five carbons. And this carbon is the one in red. And attached to the carbon in red is gonna be our OH. So we're gonna form an alcohol from this SN2 reaction. For this alkele halide, so again we're deciding between SN2 and E2. Now we have potassium tert butoxide as our reagent. And this looks similar to the previous problem where we had potassium hydroxide. Right we have a negative charge on this oxygen. So you would think this could act as a strong nucleophile or a strong base. But potassium tert butoxide is sterically hindered because of this large group over here. And because it's sterically hindered, it can't get close enough to act as a nucleophile. So an SN2 reaction is out which means this must be an E2 reaction. So let me redraw our alkele halide here just so we can see things a little better with the bromine going down in this direction. And we know that the carbon directly connected to the bromine, here is our alpha carbon so I'll mark that. And the carbon bonded to the alpha carbon is our beta carbon. So that's our beta carbon. And we have two beta hydrogens. I'll just draw one in here. And we know in our E2 mechanism, our strong base is going to take that beta proton. So our base takes this proton, at the same time these electrons move into here. And these electrons come off to form our bromide ion. So our final product, let's draw in our carbons here. So we should have five carbons and then a double bond between these two carbons. So our electrons in magenta moved in here to form our double bond. For our last example let's look at ethyl bromide reacting with water. We know it's not SN1, we know it's not E1. So we're deciding between SN2 and E2. We know that water is a weak nucleophile. And it's also a weak base. And since water is a weak base an E2 reaction is out. So this must be an SN2 reaction. So our nucleophile attacks our alkele halide and a bond forms between the oxygen and this carbon here, which I'll mark in red. So let's draw the results of that. This happens at the same time these electrons come off onto the bromine to form the bromide anion. So let's draw what we have in here. So this oxygen forms a bond with the carbon in red. And a lone pair of electrons on the oxygen, which I'll make magenta, would form this bond. The oxygen's still bonded to two hydrogens, so I'll draw in these two hydrogens here. And we still have a lone pair of electrons on that oxygen so that gives the oxygen a plus one formal charge. To get to a neutral product, we need to deprotonate this. And so another molecule of water comes along and acts as a base to take, let's say this proton here, leaving these electrons behind on the oxygen to give us our final product. So I'll just draw an OH in here. So we form ethanol. Now this reaction would need a lot of heat, it would need a lot of time. So it's not exactly the most practical reaction. Because an SN2 reaction needs a strong nucleophile and water is not that great of a nucleophile. Because this reaction is so slow, you might see some textbooks say it won't happen, because it isn't very practical.