- 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: kinetics and substrate
How an Sn1 reaction takes place. Created by Jay.
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- How come the mechanism started at7:35is in the SN1 reaction video? Isn't it considered an SN2 reaction since the reaction is involving 2 reactants in the first step instead of only 1?(13 votes)
- The reaction depending only upon the Chlorine leaving the t-butyl chloride molecule. I'm assuming this is the slow step of the overall reaction. The nucleophilic oxygen in the water only reacted with the carbonation because the Chlorine left, thus meaning that only one reactant was the cause of the overall "chain" reaction.(1 vote)
- Why are not use primary and secondery in Sn1 reaction(4 votes)
- That would be confusing, because 1° halides give SN2 reactions, 2° halides usually give SN2 reactions, and 3° halides give SN1 reactions.
The numbers in SN reactions come from the rate laws for the reactions. In an SN1 reaction, the rate law is 1stt order. That is, the reaction rate depends on the concentration of only one component, the alkyl halide. Hence the term Substitution Nucleophilic 1st order.
In an SN2 reaction, the rate law is 2nd order. That is, the reaction rate depends on the concentrations of two components, the alkyl halide and the nucleophile. Hence the term Substitution Nucleophilic 2nd order.
The 1 in SN1 and the 2 in SN2 come from the kinetics of the reactions, not from 1° and 2°.
Hope this helps.(11 votes)
- in the final part you showed, what happens to the H2OCH3 since the Oxygen will now have a + formal charge? Does it react with the Cl Leaving group?(6 votes)
- Yes, the oxygen will have a +1 formal charge. However, it is irrelevant for this problem because if the H2OCH3 reacted with the Cl- leaving group, the product of that reaction wouldn't be a product of the initial reaction with the t-butyl chloride if it is even possible to react.(1 vote)
- at8:41, he mentions that another molecule of water may come along. why cant the chloride anion generated in the rate determining step pluck a proton and stabilize the positive charge on oxygen? even then we will get stable molecules, the alcohol and HCl.(3 votes)
- HCl is a strong acid in water which means that Cl- is a very poor base, so that reaction (Cl- picking up the H) is very unlikely.
Something can be stable and charged.(6 votes)
- Why -O-H is not a good leaving group? Also, why CH3 groups in tert-butyl are electron donating groups?(1 vote)
- First, the C-O bond is relatively strong and difficult to break. Also the resulting products would be an organic cation R+ and a hydroxide ion OH-. You are separating an anion (OH-) from the attraction of the cation (R+), and this also takes energy.
If you were attempting to substitute the -OH with another nucleophile, (e.g., Br-), you would not get an alkyl bromide, because the C-O bond is stronger than the C-Br bond. In other words, hydroxide OH- is a stronger nucleophile than Br- and therefore a poorer leaving group.
To convert the -OH into a better leaving group, it is common to use a strong acid such as HCl or HBr. The acid protonates the OH group, forming C-OH2+ (a protonated OH group). The protonation greatly weakens the C-O bond and the bond easily breaks forming an organic cation R+ and a neutral water molecule. The leaving group (H2O) is now neutral and is not electrostatically attracted to the R+, so it is easier to remove the H2O from the R+.
The most common explanation for the electron donating ability of methyl groups is that C is slightly more electronegative than H, This puts a slight negative charge on the methyl carbon.(6 votes)
- At5:40it is said that tertiary alkyl halides undergo faster Sn1 than primary....but wont the approach-ability of the nucleophile difficult due to steric hindrance?(2 votes)
- Steric hindrance will play a very small role in depleting the approachability of the nucleophile. However, this is undermined by the fact that more groups are attached to the reacting area. This means that the carbocation (+ charged carbon) produced will have more possibilities for sharing it's positive charge and thus, be more stable, making the reaction more likely.(4 votes)
- At5:35What is hyper-conjugation?(1 vote)
- hyperconjugation is the interaction of the electrons in a sigma bond (usually C–H or C–C) with an adjacent empty (or partially filled) non-bonding p-orbital, antibonding σ or π orbital, or filled π orbital, to give an extended molecular orbital that increases the stability of the system.(4 votes)
- At5:32, which video was the one where he talked about hyperconjugation? I remember him talking about classification of carbons, but not the resulting stability.
- I believe this is the video in which he talked about hyperconjugation:
I do not remember him talking about hyperconjugation in too much detail before this video, so this is the best resource I could find right now on this topic.(2 votes)
- At7:14, can acetate attack a beta hydrogen as a nucleophile following E1 reaction?(2 votes)
- It could, but because the acetate anion is a relatively good nucleophile, the SN1 reaction will dominate. If you wanted to drive the E1 reaction mechanism over the SN1 mechanism, you could use a weaker nucleophile (for example, bisulfate, HSO4-) or heat the reaction.(2 votes)
- I was taught in class that SN1 reactions MUST have a neutral base to react with the alkyl halide. Is this not true? I noticed that in the reaction in the video, the ion HS(-) was used. We were told a charged nucleophile would result in an E2 reaction.(2 votes)
- [Instructor] In this video, we're gonna look at the SN1 mechanism, and we'll start with our alkyl halide. In the first step of our SN1 mechanism, we get loss of a leaving group, so the electrons in this bond come off onto the bromine to form the bromide ion. When that happens, we take a bond away from this carbon in red, so the carbon in red gets a plus one formal charge. Let me draw that in here. So the carbon in red is this one. It now has a plus one formal charge, and we have a carbocation. The carbon in red went from being sp3 hybridized in our alkyl halide to being sp2 hybridized in our carbocation, which means the geometry directly around the carbon in red is planar. We also have our bromine, so let me draw that in here. It has four lone pairs of electrons around it now, which gives it a negative one formal charge. That's the bromide ion. The electrons in this bond, these electrons in here come off onto the bromine to form our bromide ion, which is a good leaving group. In the first step of our SN1 mechanism, we get loss of a leaving group. And when that happens, we form our carbocation, and our carbocation has a plus one formal charge on this carbon. This is gonna function as our electrophile. And our nucleophile will be the hydrosulfide ion with a negative charge. Opposite charges attract, and in the second step of our mechanism, our nucleophile attacks our electrophile. On our second step we get nucleophilic attack, so nucleophilic attack, and a lone pair of electrons on our sulfur form a bond with our carbon in red. In our final product, this is our carbon in red, and I'll highlight a lone pair of electrons on this sulfur, and that lone pair forms this bond to give us our product. Let's go to the video so we can see this mechanism using a model set. Here's our alkyl halide, and I'm saying that the green is our bromine. In the first step of our mechanism, we get loss of a leaving group, so these electrons come off onto our bromine to form the bromide ion, and we form our carbocation. But using this model set, it doesn't look like we have a planar carbocation. Those carbons are not in the same plane, so let me grab another model set here. You can see that, actually, those carbons are in the same plane, and we have an sp2 hybridized carbon in the center. If we look at the carbocation from a top view, we can see the drawing, right? That's how we draw it in our mechanism. Next, our nucleophile comes along, which is our hydrosulfide ion, and our nucleophile could attack from either above or below, since we have a planar carbocation that is flat. Either way, we get the same product, so let's show the final product here. We're back to an sp3 hybridized carbon, so we have tetrahedral geometry in our final product. The first step of our mechanism, loss of a leaving group, turns out to be the slow step, and the second step, nucleophilic attack, turns out to be the fast step. And this mechanism is consistent with the experimentally determined rate law for this reaction. The rate of the reaction is equal to the rate constant, k, times the concentration of our alkyl halide. So experiments have determined this to be our rate law, and this is the concentration of our alkyl halide to the first power. The rate of the reaction depends on the concentration of the alkyl halide, but not on the concentration of the nucleophile. And that's because our first step is our slow step, and this is our rate determining step, if you remember this stuff from general chemistry. And that means if you increase the concentration of our alkyl halide, if you increase the concentration of our alkyl halide by a factor of two, you increase the rate by a factor of two, since it's first order with respect to our alkyl halides. But, if you try to increase the concentration of your nucleophile, so let's say you increase the concentration of your nucleophile by a factor of two, there's no effect on the rate. This is zero order with respect to our nucleophile. Our nucleophile can't attack until our carbocation is formed, and that's dependent only on the concentration of our alkyl halide. So that's why this reaction is first order with respect to our alkyl halides. We call this an SN1 reaction, so the S stands for Substitution, the N stands for Nucleophilic, and the one refers to the fact that this is a unimolecular, this is a unimolecular reaction, which means that the rate of the reaction depends on the concentration of only one thing, which is our substrate, our alkyl halide. So it's first order with respect to our alkyl halide, and the nucleophilic substitution means that our nucleophile has substituted for our leaving group in our product. So that's an SN1 mechanism. The structure of the substrate also affects the rate of the reaction. If we start with a tertiary alkyl halide, like we did in the example above, we're gonna get a tertiary carbocation. So if these electrons come off onto the bromine, we're left with a tertiary carbocation, a plus one formal charge on this carbon. And we know that tertiary carbocations are the most stable. We saw this in an earlier video. The more alkyl groups you have, the more electron density you can donate to help stabilize this positive charge. If we started with a secondary alkyl halide, when these electrons came off onto the bromine, we'd be left with a secondary carbocation. So let me draw this in here. And a secondary carbocation is only stabilized by two alkyl groups, so in this case, it'd be these two methyl groups here. And since a secondary carbocation is not as stable as a tertiary, a tertiary carbocation would form a lot faster. So tertiary alkyl halides are the most reactive in an SN1 mechanism. A primary alkyl halide or a methyl halide, these wouldn't have a very stable carbocation, so the carbocation is too unstable to exist, so generally, a primary alkyl halide does not react via an SN1 mechanism. And same with a methyl halide. In the previous example, we had a nucleophile with a negative one formal charge on it. What do you do if your nucleophile is neutral? In this case, the water molecule has no charge on the oxygen, but it could still act as a nucleophile in our SN1 mechanism. Our first step is loss of a leaving group, so these electrons come off onto chlorine to form the chloride anion, and we're taking a bond away from this carbon in red. So that carbon in red is gonna get a plus one formal charge. We draw our carbocation to try to show the planar geometry around that central carbon here, which has a plus one formal charge. And then the next step, we know our nucleophile attacks our electrophile. So our nucleophile has this oxygen with a partial negative charge, right? The oxygen's more electronegative than these hydrogens, so it withdraws some electron density. And a lone pair of electrons here on this oxygen can form a bond with this carbon. So our nucleophile attacks our electrophile, and let's draw what we would form. Let me draw in these groups here, and then we'd have our oxygen, which is now bonded to our carbon, and our oxygen is still bonded to two hydrogens. Let me show the electrons in magenta here, forming a bond between this oxygen and our carbons. So the electrons in magenta would be these electrons. We still have a lone pair of electrons on the oxygen that's left over, so that's this lone pair. Let me draw those in, and that would give this oxygen a plus one formal charge. If we compare this with our final product, notice we only need to do a proton transfer, right? An acid-base reaction. Another molecule of water could come along and act as a base. Let me draw that in here. And it could take one of these protons, so it could take this, and leave these electrons behind on the oxygen, so let's highlight those electrons in red. So these electrons in red here end up on that oxygen, and our final product is neutral. We have no charge, and we form tert-Butanol. This is called a solvolysis reaction. Let me write that in here. This is a solvolysis reaction, which just means that our solvent is the nucleophile. Our solvent is water, which also functions as the nucleophile in this mechanism.