- Identifying nucleophilic and electrophilic centers
- Curly arrow conventions in organic chemistry
- Intro to organic mechanisms
- Alkyl halide nomenclature and classification
- Sn1 mechanism: kinetics and substrate
- Sn1 mechanism: stereochemistry
- Carbocation stability and rearrangement introduction
- Carbocation rearrangement practice
- Sn1 mechanism: carbocation rearrangement
- Sn1 carbocation rearrangement (advanced)
- Sn2 mechanism: kinetics and substrate
- Sn2 mechanism: stereospecificity
- Sn1 and Sn2: leaving group
- Sn1 vs Sn2: Solvent effects
- Sn1 vs Sn2: Summary
Sn1 mechanism: carbocation rearrangement
Examples of Sn1 reactions involving carbocation rearrangements.
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- How do you determine what good and bad leaving groups are? Stability, yes, but how do you know what leaves something more\less ststabl? the carbon with the OH was secondary before and after the hydroxyl (OH) bound to a H so that doesn't explain it.(10 votes)
- Good leaving groups are weak bases. Weak bases are happy and stable on their own. They don't need the help of other groups.
Ex- weak bases- halide ions, water etc
Bad leaving groups are strong bases. They need the help of somebody else to be stable.
Ex- Strong bases - OH(14 votes)
- Is carbocation rearrangement necessary? Could the nucleophyle attach the secondary carbon before the rearrangement or is the rearrangement so fast that it always happens before the nucleophylic attack? Thanks(6 votes)
- Nucleophile should attach to the tertiary carbon.
This is Sn1 reaction. So nucleophile should be substituted with a leaving group .
In the given compound, we are given with hydroxide which is not a good leaving group. In order to continue the reaction, we must convert the hydroxide which is a bad leaving group into the good leaving group. Converting hydroxide into the water make it a good leaving group. Once the water is removed, we are left with a secondary carbocation which is least stable. So the compound will make a tertiary carbon by undergoing hydride shift. Since tertiary carbocation is an electrophile and we are provided with a nucleophile( methanol) , nucleophile will attack the carbocation which is on tertiary carbon.
Hence it is clear that, for the reaction to proceed and to get the required product, reactions should take place in an order.(5 votes)
- Are these shifts in any way possible via SN2?(2 votes)
- SN2 does not generate a carbocation so no chance for rearrangement.
The nucleophile attacking and the leaving group leaving happen at the same time.(11 votes)
- Yes, "EtOH" it's the abbreviation of "ethanol". The "Et" on "EtOH" stands for "Ethyl group" and the "OH" for the hydroxyl group.
Same thing with the "MeOH" on3:19. Now instead of an ethyl group you have a methyl group. Therefore "MeOH" is methanol.(6 votes)
- at4:12, I don't quite understand why the oxygen has a plus 1 formal charge?(1 vote)
- Formal charge = valence electrons - lone pair electrons - bonds
From this formula we get:
An oxygen with 2 bonds and 2 lone pairs will have a formal charge of 0
An oxygen with 3 bonds and 1 lone pair will have a formal charge of +1
An oxygen with 1 bond and 3 lone pairs will have a formal charge of -1(4 votes)
- I know this is somewhat stupid, but how do you know when to perform a hydride shift and a methyl shift when you're doing an Sn1 reaction. For instance if my teacher gave me a molecule and said predict the Sn1 reaction, is there any hints in the molecule that would tell me that I should let the molecule undergo a hydroxide or methyl shift?(2 votes)
- Check what type of carbocation would be formed.
In terms of stability primary < secondary < tertiary
If you can methyl or hydride shift to a more stable carbocation it’s probably going to happen.(1 vote)
- Why do we de-protonate on the last step? Is it to further stabilize the molecule? Do we de-protonate at the end of every SN1 reaction?(1 vote)
- Before deprotonating to yield to the final product for both of these reactions, we notice that there is a formal charge of +1 on the oxygens. In general formal charges create instability compared to neutral atoms/molecules. This is why certain molecules use resonance to spread out formal charges and reduce their instability. Furthermore having a positive formal charge on a highly electronegative atom like oxygen adds even more instability. It's essentially a molecule just begging to act as an acid where a base would extract the hydrogen to relieve the instability. This is similar reasoning as to why hydronium is so acidic.
Not every SN1 reaction will conclude with a deprotonation, and there are plenty of non-SN1 reactions which also conclude with a deprotonation step. In general having an electronegative atom like oxygen bearing a positive formal charge with a hydrogen will result in deprotonation. An SN1 reaction where the nucleophile is something like a halogen would not conclude with deprotonation simply because there are no hydrogens to extract from the new halogen group.
Hope that helps.(2 votes)
- Why tertiary carbocation is more stable than secondary and primary?(1 vote)
- Since a tertiary carbocation is bonded with 3 other alkyl groups, the electrons on the hydrogens of the alkyl groups stabilize the positive charge. However, a secondary carbonation only has 2 alkyl groups and thus, less electron density to stabilize the positive charge. Only alkyl groups can share their electron density to stabilize carbocations: single hydrogens cannot.(2 votes)
- At3:45, why is hydroxide a bad leaving group. Is it because you're in a polar aprotic solvent and hydroxide is a very strong base?(1 vote)
- Hydroxide is a strong base and a bad leaving group.
Once we protonate it, it will become water which is a weaker base and hence good leaving group(1 vote)
- @4:50, does the MeOH react with the Hydronium ion through an acid base mechanism? And what impact would this have on the overall target product, if any? Thanks(1 vote)
- [Lecturer] Since the SN1 mechanism involves the formation of a carbocation a rearrangement is possible. So let's look at this SN1 reaction. On the left is our alkyl halide, ethanol is our solvent and on the right is our product. The first step should be loss of a leaving group. So these electrons come off on to the iodine to form the iodide ion. We are taking a bond away from this carbon in red so the carbon in red is gonna get a plus one formal charge. So let's draw this in. We are gonna form a carbocation and the carbon in red is this carbon so this carbon gets a plus one formal charge. If we look at this carbocation it is secondary. The carbon in red is directly bonded to two other carbons so this is a secondary carbocation. If you think about the video on carbocations and rearrangements. We can have a rearrangement here to form a more stable carbocation. A methyl shift will make sense. So this methyl group is gonna shift over to this carbon and let's draw what we would make now. So we would have a methyl group now that's moved over so the carbon in red is this carbon. And we are taking a bond away from the carbon that I just circled in green so the carbon I circled in green is this carbon and that carbon gets a plus one formal charge now. So this is our carbocation. What kind of carbocation is it? Well if you look at the carbon that I circled in green that's directly bonded to one, two, three other carbons so it is a tertiary carbocation, which we know is more stable than a secondary carbocation so a methyl shift increases the stability going from a secondary to a tertiary carbocation. This carbocation is our electrophile. Now it is time for our nucleophile to attack our electrophile. Our nucleophile is our solvent ethanol. So this is a solvolysis reaction. So if I draw in an ethanol molecule here and put in two lone pairs of electrons on the oxygen, so one of those lone pairs is gonna form a bond with the carbon that I circled in green. So the nucleophile attacks the electrophile and we form a bond between the oxygen and the carbon. So if I sketch this in here, now we would have a bond to this carbon. This oxygen is still bonded to a hydrogen, still bonded to the ethyl and lets say that the lone pair of electrons form the bond. Let's make them these electrons in magenta. So those electrons in magenta on the oxygen form this bond. We still have a lone pair of electrons left on the oxygen so here they are. Here they are right here. And that's a plus one formal charge on the oxygen. If we compare this to our final product, notice all we have to do is deprotonate. So the last step of the mechanism is just a proton transfer and acid base reaction. Another molecule of ethanol could come along and serve as the base. So if I draw on my lone pairs of electrons on the oxygen one of those lone pairs could pick up this proton, leave these electrons behind on this oxygen and finally give us our product so you could draw your product a few different ways. But that is our SN1 mechanism with the carbocation rearrangement. Let's do another carbocation rearrangement in an SN1 mechanisms but this time we are starting with this alcohol. In the previous example, the first step was loss of a leaving group. But if we show the electrons going on to the oxygen now to form the hydroxide ion, that doesn't work because the hydroxide ion is a bad leaving group. So we actually have to protonate the alcohol first to form a good leaving group. So the first step of this mechanism is a proton transfer. The alcohol is gonna function as a base and HC3O plus is gonna donate a proton. So the hydronium ion is gonna act as our asset so that's HC3O plus and our alcohol acts as our base. Lone pair of electrons on the oxygen pick up a proton from hydronium. So our first step is a proton transfer. And that gives us this oxygen here with a plus one formal charge. So still one lone pair of electrons, a plus one formal charge on that oxygen and let's say that this lone pair of electrons picked up this proton to form this bond. Now we are ready for loss of a leaving group because when these electrons come off on to this oxygen now that gives us water and water is a good leaving group. So first step is proton transfer. Second step is loss of a leaving group and we are taking a bond away from this carbon in red. So that carbon in red gets a plus one formal charge. So let me draw this in here. So the carbon in red is this carbon. So let me write a plus one formal charge on this carbon. What kind of carbocation is that? The carbon in red is directly bonded to two other carbons. So this is a secondary carbocation and we think about the possibility of a rearrangement. So in the carbocation and rearrangements video, another shift that we did was a hydride shift. So on this carbon in magenta there's still a hydrogen and a hydride shift would give us a more stable carbocation. Remember a hydride shift think about this hydrogen and these two electrons here. Shifting over to this carbon in red. So let's draw in what we would have now. So now if you think about that carbon in red, we already had a hydrogen bonded to it. So let me go ahead and highlight this carbon as being the carbon in red. We already had a hydrogen bonded to it in the example, in the carbocation to the left I should say. So let me draw in that original hydrogen. With the hydride shift, we are adding in another hydrogen here to this carbon in red. So there is no more formal charge on the carbon in red. The formal charge moves to this carbon, the one I circled in green. It is losing a bond so that carbon that I circled in green now gets a plus one formal charge. What kind of carbocation did we form. Well the carbon in green is directly bonded to one, two, three, other carbons. So this is a tertiary carbocation. A hydride shift gives us a more stable carbocation. Now we formed our electrophile and our nucleophile will come along in the next step. And our nucleophile should be methanol. So let me draw in a molecule of methanol here. So put in a hydrogen and two lone pairs of electrons on the oxygen. So our nucleophile will attack our electrophile so lone pair of electrons on this oxygen are gonna form a bond with this carbon that I circled in green. So let's draw the result of our nucleophilic attack. So now we are forming a bond between the carbon in green and this oxygen here. So let me sketch in the rest of this. And then we still have a methyl on this oxygen. We still have a hydrogen on this oxygen and let me highlight those electrons. So our electrons, let's make them blue. So these electrons in blue here on this oxygen form a bond between the oxygen and that carbon. We still have a lone pair of electrons on this oxygen which gives the oxygen a plus one formal charge. And also notice what I did with this group right here. So these two carbons I just drew them going down this way. This free rotation around this bond right here so you can draw it however you want. I just drew it to look a little bit more like our product here. When we compare our two, the last step must be a proton transfer. All we have to do is take a proton away and we have our final product. I have a couple of choices here as my base. I could use either the water molecule that we lost back in this step or I could use another molecule of methanol as our base. It doesn't really matter what we use here. I'll just use the water molecule. Either water or methanol acts as a base in the last step of our mechanism. And takes this proton leaving these electrons behind and forming our final product.