So I have an organic chemistry final tomorrow and I am honestly worried. I’m probably going to end up with a C+ in this class... I need a 96 to pass with a B, and I don’t feel like I can do that. I was wondering if anyone knew anything helpful for Sn1/Sn2/E1/E2 reactions. I have the hardest time trying to figure out the product because my professor never really gave us any rules (I can tell if the reaction is going to be Sn1/Sn2/E1/E2, I just can’t get to the product). Can someone help?
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- Thread starter Ash2021
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I wish I remembered most of the helpful rules, for me now it's just a matter of seeing the problem and I can tell you the answer, but it's harder for me to explain it lol. Really though the SN1 and SN2 are just combining two molecules, as long as you know the binding site through the nucleophile/electrophile it's really pretty simple. I forget which it is, either E1 or E2 is, according to my professor, almost always a dehydration of an alcohol with a strong acid and then balancing the charges properly. H3C-CH-OH + H2SO4 ----> H2C=CH2 The H+ ion combines with the OH, and leaves as H2O leaving H3C-CH +, The hydride shift moves the hydrogen over for H2C-CH2 +, and then the double bond forms to create a neutral charge.
The others are just simple addition really...there are a few funky rules such as hydride shifts to form tertiary carbocations but mostly you'll have something like R-Br + Na-R ---> R-R + NaBr. The professor might try and make some large molecules just to confuse you, but if like you said you know how to identify which reaction is happening then the products should be fairly easy. Sorry if this isn't helpful, good luck to you pal.
The others are just simple addition really...there are a few funky rules such as hydride shifts to form tertiary carbocations but mostly you'll have something like R-Br + Na-R ---> R-R + NaBr. The professor might try and make some large molecules just to confuse you, but if like you said you know how to identify which reaction is happening then the products should be fairly easy. Sorry if this isn't helpful, good luck to you pal.
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Hi @Ash2021 - remembering all the details about SN1/SN2/E1/E2 seems like a daunting dask (which one has backside attack? which prefers a strong nucleophile?), but a lot of the details you might think you'd have to "memorize" you actually arrive at by using a series of logical steps. Here's an SN1 reaction:
First of all, an SN1 reaction has "1" in its name because it’s unimolecular, meaning that only one thing happens at a time. So first, the substrate loses a leaving group of its own volition. This is the SLOWEST step of the reaction, and when a reaction has multiple steps, the slowest step is going to determine the rate of the reaction. So the leaving group leaves, and then in step 2, the nucleophile attacks and replaces the leaving group. Since the leaving group was already gone, the nucleophile has its pick of where it wants to attack, from the front or the back? It turns out that the nucleophile will do half and half, so we get a racemic mixture of half with frontside attack and half with backside attack. So in summary: How many steps did we say there were? Two steps. And what does the rate law depend on? It depends on the first step, and the first step depended on the substrate. In case we forget, SN1 means it’s a unimolecular reaction, so the rate law must be rate = k[substrate]. What kind of substrate do you think is preferred here? Once that leaving group leaves, we want a substrate that’s going to be able to handle that positive charge in the transition state. The more substituted it is, the bigger it is, the more it can soak up that positive charge, mostly because those other groups spread out their electrons to balance it out. So with SN1 we prefer highly substituted substrates. The nucleophile is weak – it’s sort of like saying, the leaving group was already gone, so the nucleophile doesn’t have to be too pushy to jump right it. In terms of the preferred solvent, you’re going to be given two options: something polar protic, which means something with protons that is POLAR, or something polar aprotic, meaning a solvent that is polar but without protons. For SN1, we like polar protic solvents because the dipole of the solvent helps stabilize our positively charged substrate. You can remember that the transition state in SN1 is going to have a positive charge, and so do protons, to help you remember this. We already said that the nucleophile can attack from either side, so we are going to get a racemic mixture of products.
First of all, an SN1 reaction has "1" in its name because it’s unimolecular, meaning that only one thing happens at a time. So first, the substrate loses a leaving group of its own volition. This is the SLOWEST step of the reaction, and when a reaction has multiple steps, the slowest step is going to determine the rate of the reaction. So the leaving group leaves, and then in step 2, the nucleophile attacks and replaces the leaving group. Since the leaving group was already gone, the nucleophile has its pick of where it wants to attack, from the front or the back? It turns out that the nucleophile will do half and half, so we get a racemic mixture of half with frontside attack and half with backside attack. So in summary: How many steps did we say there were? Two steps. And what does the rate law depend on? It depends on the first step, and the first step depended on the substrate. In case we forget, SN1 means it’s a unimolecular reaction, so the rate law must be rate = k[substrate]. What kind of substrate do you think is preferred here? Once that leaving group leaves, we want a substrate that’s going to be able to handle that positive charge in the transition state. The more substituted it is, the bigger it is, the more it can soak up that positive charge, mostly because those other groups spread out their electrons to balance it out. So with SN1 we prefer highly substituted substrates. The nucleophile is weak – it’s sort of like saying, the leaving group was already gone, so the nucleophile doesn’t have to be too pushy to jump right it. In terms of the preferred solvent, you’re going to be given two options: something polar protic, which means something with protons that is POLAR, or something polar aprotic, meaning a solvent that is polar but without protons. For SN1, we like polar protic solvents because the dipole of the solvent helps stabilize our positively charged substrate. You can remember that the transition state in SN1 is going to have a positive charge, and so do protons, to help you remember this. We already said that the nucleophile can attack from either side, so we are going to get a racemic mixture of products.
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Here's SN2:
SN2 is a BIMOLECULAR reaction. With SN2, the nucleophile attacks right away, and we end up with this awkward pentavalent intermediate (5 attached groups) while the leaving group departs at the same time, like it’s being pushed off. If the leaving group is on one side, the nucleophile must attack from the other, so we get backside attack only with SN2. Then the product is going to have what we call inversion of configuration because the nucleophile had to attack from the opposite side. So to summarize: How many steps? Technically one step since nucleophile attack and loss of the leaving group occurred simultaneously. What’s the rate law? In bimolecular substitution, both the nucleophile and substrate are part of the main step, so rate = k[substrate][nucleophile]. What substrate is preferred? Remember that cluttered pentavalent intermediate we have? In contrast to SN1, SN2 is going to prefer LESS substituted substrates so there is less clutter and the nucleophile has more room to attack. I always just remember that SN1 (a low number) prefers a HIGHLY substituted substrate, and SN2 (the higher number) prefers a LESS substituted one, so the opposite of what you’d expect. SN2 will require a strong nucleophile because it has to be really aggressive to get in there are force the leaving group to leave. The solvent that’s preferred is polar aprotic because the lack of the hydrogen makes the nucleophile stronger. If the nucleophile is surrounded by hydrogens, forming a solvation shell, the nucleophile won’t get anywhere near the substrate. They are sort of like the security team for the substrate, and we need a STRONG nucleophile, so polar APROTIC is preferred here. We also said that inversion of configuration occurs because of backside attack from the nucleophile, which differs from SN1 reactions.
SN2 is a BIMOLECULAR reaction. With SN2, the nucleophile attacks right away, and we end up with this awkward pentavalent intermediate (5 attached groups) while the leaving group departs at the same time, like it’s being pushed off. If the leaving group is on one side, the nucleophile must attack from the other, so we get backside attack only with SN2. Then the product is going to have what we call inversion of configuration because the nucleophile had to attack from the opposite side. So to summarize: How many steps? Technically one step since nucleophile attack and loss of the leaving group occurred simultaneously. What’s the rate law? In bimolecular substitution, both the nucleophile and substrate are part of the main step, so rate = k[substrate][nucleophile]. What substrate is preferred? Remember that cluttered pentavalent intermediate we have? In contrast to SN1, SN2 is going to prefer LESS substituted substrates so there is less clutter and the nucleophile has more room to attack. I always just remember that SN1 (a low number) prefers a HIGHLY substituted substrate, and SN2 (the higher number) prefers a LESS substituted one, so the opposite of what you’d expect. SN2 will require a strong nucleophile because it has to be really aggressive to get in there are force the leaving group to leave. The solvent that’s preferred is polar aprotic because the lack of the hydrogen makes the nucleophile stronger. If the nucleophile is surrounded by hydrogens, forming a solvation shell, the nucleophile won’t get anywhere near the substrate. They are sort of like the security team for the substrate, and we need a STRONG nucleophile, so polar APROTIC is preferred here. We also said that inversion of configuration occurs because of backside attack from the nucleophile, which differs from SN1 reactions.
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With E1 or unimolecular elimination, the leaving group leaves first, just like in SN1. Then it has to decide whether it’s going to undergo an SN1 reaction or an E1 reaction. If it decides to undergo elimination, a base will abstract a hydrogen, and the electrons from that bond will go towards making a double bond. Elimination reactions always end in a double bond. Like SN1, there were two discrete steps here, and the slowest step involved the substrate only, so that’s the only species that will go into the rate law (rate = k[substrate]). Just like SN1, highly substituted substrates are preferred to stabilize the positive charge in the transition state. Here’s where you can decide between elimination and substitution. Elimination doesn’t just want any old nucleophile – it specifically wants a base to pair up with that lone hydrogen. You can recognize bases as often containing an oxygen or hydroxide in them. Like SN1, E1 also prefers polar protic for a similar reason. You might have asked yourself: how do you know where that double bound is going to form? The major product of elimination reactions will form the double bond at the MORE substituted position, which is more stable. This is called the Zaitsev rule.
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The E2 or bimolecular elimination reaction occurs when the base attacks at the SAME time that the leaving group departs. In other words, there is only 1 step, and both players take part in the rate law (rate = k[substrate][base]). Like SN2, less substituted substrates are preferred so the base can jump right in. To distinguish between E2 and SN2, a strong BASE is preferred, something like NaOH or a bulky base like tert butoxide even. Like SN2, E2 also prefers a polar aprotic solvent, and in almost all elimination reactions, the double bond will form at the more substituted position, forming the Zaitsev product. For example, the PCAT might show you an elimination reaction and ask you to identify the major product. Look for the one where the double bond falls at the more substituted position.
I recommend you trying drawing these out and then creating a table that summarizes all this information. Then recreate that table and try to fill it in from memory. Hopefully you can try to reason through the logic instead of just memorizing all these facts because that'll help you remember it better for your exam. Good luck!!
