Cumulative Substitution/Elimination - Online Tutor, Practice Problems & Exam Prep

In this section, students will apply their understanding of various mechanisms, including nucleophilic acyl substitution and electrophilic additions, to solve cumulative problems. Utilizing a flow chart, they will identify mechanisms and predict products, reinforcing concepts like Markovnikov's rule and anti addition. Mastery of these topics, such as the role of leaving groups and the significance of stereochemistry, is crucial for success in organic synthesis and reaction mechanisms. This practice will solidify knowledge and enhance problem-solving skills in organic chemistry.

Intro to Substitution/Elimination Problems

All right, guys. So now we're literally going to put everything together and do some cumulative problems based on everything we've learned about these four mechanisms and the big daddy flow chart. Okay? So what that means is that for these questions, I'm not going to tell you what the mechanism is. You're expected to use the flow chart to figure that out. And then on top of that, you're expected to actually know what the mechanisms do based on my descriptions of those mechanisms. So if you're weak on that, you might want to brush up on it before you start. And then you have to predict all the products as well. So this is literally a huge amount of practice, but this is going to help you guys solidify this chapter so well. So let's go ahead and get started with problem number 1.

Predict the mechanism for the following reactions. Provide the full mechanism and draw the final product.

• A.
• B.
• C.
• D.

Predict the mechanism for the following reactions. Provide the full mechanism and draw the final product.

• A.
• B.
• C.
• D. No reaction.

Predict the mechanism for the following reactions. Provide the full mechanism and draw the final product.

• A.
• B.
• C.
• D. No reaction.

What is the major product for reaction d ?

• A.
• B.
• C. A and B
• D. None of these

Predict the mechanism for the following reactions. Provide the full mechanism and draw the final product.

It's flowchart time. So what's the first question? And what is the first answer? Is this negatively charged or neutral? Okay? And it has a weird-looking nucleophile. Alright? I don't blame you if you have no clue what that is, but it is negatively charged because there's a k, right? So this would have a negative charge right there. And the k would dissociate, so I'm going to say that it's negative. I'm going to go down the left hand side. Okay? The second question is, is this one of my bulky bases? K? Actually, no. My 3 bulky bases are t-butoxide, LDA, lithium, and potassium. This is not any of those. So I'm going to say no and I'm going to move down to the third. Okay. The third question is what's my leaving group? Okay. So my leaving group, this time it's easy. It's just an alkyl halide. What type of alkyl halide? That would be tertiary. Okay? So there's a tertiary alkyl halide. By the way, this ring is drawn in 3D, but you don't need to draw your answer in 3D. That's just a weird way I'm trying to throw you guys off. Alright? So tertiary, do I know the mechanism already? No. I need to go to my last step, which is, okay, is this one of my good bases? Now, this is the part that a bunch of you guys probably got messed up on. Okay? And in advance, I'm sorry. This is just a tricky problem, but this is the way you learn. A lot of you guys thought that this is actually going to be OR negative, which is an oxide. Okay? That's actually the first strong base that I list because you see O and you see that this is an R. It turns out this is not OR negative. Okay? OR negative would be something like OCH3. See how it's just a single O, it's not 2 oxygens? What this actually is, is something that would be OOR negative. Okay? Because notice that there are 2 oxygens, so this is actually a different type of reagent altogether. Instead of being called an oxide, this is actually called an acetate. It's something complet

Predict the mechanism for the following reactions. Provide the full mechanism and draw the final product.

All right, guys. So what mechanism did you guys get? It should have been E2. Okay? Why is that? I'll tell you why. We're going to use the flowchart. So the first step was is this negative or neutral? It was negative because I had a sodium. So I had a negative charge right there. Okay? So now my second step was, is this one of my bulky bases? No. It's not. So then my third step was what kind of alkyl halide do I have or a leaving group? Look what I got again. One of those pesky sulfonate esters. All right? And that's going to be secondary. So I have a secondary leaving group. So that means that I'm going to have to make one more decision, which is the fourth one. Is my nucleophile a better base or is it a better nucleophile? So you have to think, is this molecule here on my good bases list? Yes, it is. This was actually the second base on the list. It's an alkynide. Remember that I said that alkynides are strong bases? So what that means is that this is going to be better at doing an elimination than a substitution. So since it's a good base, we're going to consider this E2. Is that okay so far? Cool. So now that this is E2, we have to remember everything about the mechanism. Where do we start with E2? Do you guys remember? We have to find beta protons, just like always. Okay? So where is my alpha? My alpha is here. So I have 2 betas. Right? I've got a beta at the top. I'll call it on the green one. I got a beta at the bottom. I'll call it on the blue one. Are they equivalent? Are they the same thing? No. The top one doesn't have a methyl, the bottom one does. So they're obviously different. Okay? Do they both have at least one hydrogen? Yeah, they do. The beta at the top, notice that the beta at the top actually has 2. It has a hydrogen in the front and a hydrogen in the back. And the blue one just has one, but that's fine. It's a hydrogen in the back. So really, we actually have 3 total hydrogens, but 2 different types of hydrogens. Are you guys following me so far? So here's the big question, how many final products did you guys draw? Did you draw 2? That would be wrong. If you drew 2, that's actually the wrong answer and you got 0 credit on this question and your professor tricked you. Okay? Why is that? Can anyone help me answer that question? I know someone's dying too. It's because of anticoplanar. Remember that I told you guys that every time you hear E2, from before when I told you about E2, you have to remember that anticoplanar is required. That means that your leaving group and your beta hydrogen have to be anti to each other. So how many of these hydrogens, out of the 3 hydrogens I have possible, how many of them are actually anti? This one's on the dash, which means that the little top. So now let's go ahead and draw our mechanism. And at this point, if you got this far, then you're just a rock star and you got the question right. It's just going to be in you grab this H, then you make a double bond between the alpha and the beta, and you kick out your leaving group, and your final answer should look like this with a double bond here. That's where the distinction is going to come, the double bond there and your methyl group there and obviously plus your leaving group. Okay? So anyway, like I told you guys earlier when we were discussing E2, this is one of the hardest types of problems that you guys can get. And that's why you just need to practice it. This is nothing but practice. It's about recognizing this over and over so that by the time you get to your test, it's second nature to look for anticoplanar. Because I know that 90% of you guys forgot to think about that and why you need to practice it. That's why this chapter is hard because you have to practice. Now, one more note I just wanted to make. Notice that in all 6 of these mechanisms, I completely ignored the solvent if there is one. If there was one. Acetone was here. DMF was here. DMSO was here. I completely ignored them. I didn't have them in my decision-making process. That's the way I want you guys to think when you're on the test. It's good to know what the solvents are for conceptual purposes, but for these questions, I don't want you thinking about them. Alright? It's just a distraction. It's not going to actually help you get the mechanism in the end. Alright? So I hope that made sense to you guys. Let me know if this video helped you or if these questions helped you. But also, let me know if there's anything I could do to explain it better. All right? So let's move on.

Here’s what students ask on this topic:

What is the role of a leaving group in nucleophilic substitution reactions?

The leaving group in nucleophilic substitution reactions is crucial because it must be able to stabilize the negative charge it acquires after departure. A good leaving group is typically a weak base, as weak bases are more stable with a negative charge. Common leaving groups include halides (Cl⁻, Br⁻, I⁻) and sulfonate esters (e.g., tosylate, TsO⁻). The ability of the leaving group to depart efficiently affects the reaction rate and mechanism (S_N1 or S_N2). In S_N1 reactions, the leaving group departure is the rate-determining step, while in S_N2 reactions, the leaving group leaves simultaneously as the nucleophile attacks.

How does Markovnikov's rule apply to electrophilic addition reactions?

Markovnikov's rule states that in the addition of HX (where X is a halogen) to an alkene, the hydrogen atom will attach to the carbon with the greater number of hydrogen atoms (the less substituted carbon), while the halogen will attach to the carbon with fewer hydrogen atoms (the more substituted carbon). This occurs because the more stable carbocation intermediate forms during the reaction. For example, in the addition of HBr to propene, the hydrogen attaches to the CH₂ group, and the bromine attaches to the CH group, resulting in 2-bromopropane as the major product.

What is the significance of stereochemistry in elimination reactions?

Stereochemistry is significant in elimination reactions, particularly in E2 mechanisms, where the anti-periplanar geometry is required. This means that the hydrogen atom being removed and the leaving group must be on opposite sides of the molecule. This arrangement allows for the proper orbital alignment necessary for the reaction to proceed. In cyclic systems, this requirement can lead to specific stereoisomers as products. For example, in the elimination of HBr from 2-bromobutane, the anti-periplanar requirement can lead to the formation of trans-2-butene as the major product.

How do you use a flow chart to determine the mechanism of a reaction?

A flow chart helps determine the mechanism of a reaction by guiding you through a series of questions based on the reactants and conditions. Typically, you start by identifying the type of substrate (primary, secondary, tertiary) and the nature of the nucleophile or base (strong/weak, bulky/non-bulky). The flow chart will then direct you to consider factors such as the solvent (protic or aprotic) and temperature. By following these steps, you can narrow down whether the reaction proceeds via S_N1, S_N2, E1, or E2 mechanisms. For example, a tertiary substrate with a strong base in a protic solvent likely undergoes an E2 elimination.

What are the key differences between S_N1 and S_N2 mechanisms?

The key differences between S_N1 and S_N2 mechanisms lie in their kinetics, stereochemistry, and conditions. S_N1 reactions are unimolecular and involve a two-step process: formation of a carbocation intermediate followed by nucleophilic attack. They exhibit first-order kinetics and often lead to racemization due to the planar nature of the carbocation. S_N2 reactions are bimolecular and occur in a single step where the nucleophile attacks the substrate simultaneously as the leaving group departs. They exhibit second-order kinetics and result in inversion of configuration at the carbon center. S_N1 is favored by tertiary substrates and polar protic solvents, while S_N2 is favored by primary substrates and polar aprotic solvents.