Reactions at Benzylic Positions - Online Tutor, Practice Problems & Exam Prep

Benzylic positions are highly reactive due to the stability of carbocation intermediates formed during reactions. In SN1 and E1 mechanisms, benzylic carbocations are stabilized by resonance, making them form faster than non-benzylic counterparts. E2 reactions favor elimination products due to the higher acidity of benzylic beta hydrogens, leading to conjugated alkene products. Additionally, weak oxidation of benzylic alcohols using manganese(IV) oxide selectively converts them to carbonyl compounds, highlighting their reactivity compared to non-benzylic alcohols.

Reactions at Benzylic Positions Concept 1

Hey, everyone. So in this series of videos, we're going to take a look at reactions that occur at the benzylic position. Now, here we're going to say that benzylic positions are reactive due to stable reaction intermediates. Now if we take a look at our structure to the right, we have our benzene ring, and branching off of this benzene is our carbon. This carbon represents the benzylic position.

We're going to learn that because of its special function and special properties, they'll be able to talk about different types of substitution and elimination reactions. So we'll talk about SN1 and E1, where we have carbocation formation. We'll talk about SN2 and E2, where we have benzylic halides. And then finally, we'll talk about the oxidation of benzylic alcohols. So just remember that this benzylic position is going to be reactive, and part of that is due to the resonance that it's able to do.

Alright. So click on the next video and let's take a look at SN1 and E1.

Reactions at Benzylic Positions Concept 2

Hey, everyone. So in this video, let's take a look at SN1 E1 reactions of benzylic compounds. Here, we're going to say that Carbocation Intermediate Formation is the rate-determining step. And we're going to say that benzylic carbocations form faster than their non-benzylic counterparts. So, if we take a look here, we'll talk about the alkyl carbocation versus our benzylic carbocation.

Now, remember that alkyl carbocations are stabilized by hyperconjugation, but benzylic carbocations are stabilized by resonance. If we take a look here, we have in the very first part, a Tertiary Alkyl Halide. Remember, tertiary alkyl halides can undergo SN1 more readily. What's going to happen here is that chlorine is more electronegative, so when it leaves, it takes the electrons and the bonds with it, leaving behind a positively charged carbon. So here we have a tertiary carbocation.

We could talk about its relative rate as being equal to 1. Here, we not only have a tertiary alkyl halide but more specifically a tertiary benzylic carbocation. It too can leave with the electrons and give us a carbocation. But here, it's a tertiary benzylic carbocation. Here, its relative rate would be approximately 620.

We can see that it's many more magnitudes faster than a typical tertiary alkyl halide. So this shows that benzylic carbocations are more easily made because of resonance. Now we're going to say ortho and para substituents that act as electron-donating groups. They're going to speed up SN1 rates at benzylic positions. Remember, our electron-donating groups, these are the groups that are attached to benzene rings.

And through either inductive effect or resonance effect, they can give electron energy to the benzene ring further activating them. Here, this has a speeding up effect on SN1 rates. Now, here are electron-donating groups. We have a nitrogen with a lone pair or oxygen with lone pairs. Here, we have a nitrogen with a lone pair, but because it's next to a carbonyl, it's not quite as activating.

And then finally, we have alkyl groups or carbon groups. They are donating through inductive effect instead of resonance effect like the ones above it. But all of them can help to speed up our SN1 rates when it comes to these benzylic positions. Alright. So just keep this in mind.

Our benzylic carbocations can more easily be formed which helps to speed up the process.

Reactions at Benzylic Positions Example 1

Hey, everyone. So in this example, it says, which of the following alkyl halides will form a carbocation the fastest? So if we take a look at the first one, we have here this carbon connected to the Halogen. That carbon is connected to two other carbons, so it's just a regular secondary alkyl halide. For b, this is secondary as well, but it is benzylic.

So it's a secondary benzylic. And here we have an R group that is para to it, which could help to speed up our SN1 reaction. If we look over here, this is just a primary alkyl halide, and then this one here is also secondary benzylic. So, we know that benzylic carbocations are faster in forming than just regular carbocations. So a and c are out.

Now, b versus d. Again, we have this alkyl group that is para towards this group here. Remember, the presence of an electron-donating group helps to speed up SN1 rates in the benzylic position. So because we have an ortho-para director here, electron-donating group, it's going to make that carbocation more easily to be formed. So b would be a better answer than d.

D doesn't have the presence of an electron-donating group anywhere else on the benzene ring, so it's not going to be further enhanced. Yes, the carbocation can form fast, but not as fast as option b. Alright. So, again, our final answer here would be option b.

Rank the following alkyl halides in increasing order of reactivity in S_N1 reaction.

• A. I < II < III < IV
• B. IV < I < II < III
• C. IV < II < I < III
• D. IV < I = II < III

Which of the following alcohols will undergo acid-catalyzed dehydration the fastest and which one the slowest?

• A. 1 and 2
• B. 3 and 2
• C. 2 and 3
• D. 3 and 4

Reactions at Benzylic Positions Concept 3

Hey, everyone. So in this video, let's take a look at E2 reactions of benzylic hydrogens. Now, we're going to say here, due to the higher acidity of benzylic beta hydrogens, we're going to say that elimination products predominate. And we're going to say alkene products of such eliminations are conjugated and are stabilized by resonance. So, if we take a look here, we're talking about the E2 reactions of benzylic halides.

So we're going to say here that we have this structure here, and we're going to say that this is alpha and this is beta. These beta benzylic hydrogens are acidic because of resonance. Here, we're using RO^- which represents the generic form of a strong base, and it's dissolved in an alcohol solvent. What's going to happen here following E2, this strong base would come in, remove one of these beta hydrogens, the bond would break, fall here, and kick out the chlorine. This would create our alkene product.

So we'd form this double bond here as our major product. Another thing that could happen, but it would be the minor, is a substitution because this is a primary alkyl halide. And remember, primary alkyl halides are susceptible to SN2 reactions when we use a negative nucleophile. So, another thing that could happen here instead of this alkene product forming, this could come in, hit this carbon, and kick out the Cl. This would be our minor product but it's still a possibility.

So we'd have an OR group here. So this is the structure we'd make as our minor. But again, remember due to the higher acidity of beta, benzylic beta hydrogens, elimination products are more likely to occur because what we make is something that's more conjugated.

We have more double bond, single bond, double bond. Conjugation leads to greater stability. All of this is helped by the idea of resonance. Alright? So just keep that in mind when you're looking at questions similar to this one.

Reactions at Benzylic Positions Example 2

Hey, everyone. So here it says in this example question, select a compound that will have the fastest rate of E2 reaction with a small strong base. Alright. So here, they're talking about us having a small strong base. And remember, E2 happens more readily when we have benzylic beta hydrogens.

So if we take a look, here is alpha, here is alpha, and here is alpha. Alpha carbon being the carbon with the halogen, the leaving group. And then here we have beta and beta, beta and then beta and beta. If we look, here's a beta hydrogen that's benzylic in nature, and here's one too. C does have benzylic hydrogens in the form of this one, but it's not in the beta position.

So it can't do elimination. It can't do E2. Alright. So C would not work the fastest. Here, for both A and B, we have our good leaving group in the form of bromine, and then we have benzylic hydrogens that are in the beta position.

We could deprotonate them, so we would remove this H, fall here to make a double bond and kick this out, giving us our alkene product. And the same thing here, we remove this H here, falls here to make a double bond and kick out this Br. So here we'd say that both A and B, because of the presence of their benzylic beta hydrogens, because we're using a small strong base, which is able to get into these areas to remove that beta hydrogen, they can readily undergo E2 to make our alkene products. So again, A and B are answers. C doesn't work because, yes, we have beta hydrogens, but they're not benzylic in nature.

So you're not going to form our alkene product as quickly, and that product that you make is not going to be conjugated as well. That's also another determining factor in why C doesn't work quite as fast as A and B.

Provide mechanism and structure of major elimination product of the following reaction.

Hey, everyone. So here it says to provide the mechanism and structure of the major elimination product of the following reaction. Here we have our alkyl halide, and we're using tert-butoxide. Remember, tert-butoxide represents a bulky base. And because it represents a bulky base, we're going to undergo Hofmann's rule of elimination and not Zaitsev's rule. So if we take a look here, our alpha carbon is the carbon with the halogen. It is connected to our beta carbons. Because tert-butoxide is a bulky base, it can't gain access as readily or as easily to this hydrogen here. So what it does instead, it's going to come in and remove one of these hydrogens, come here, remove this hydrogen, form a bond here to make a double bond, and kicks out this chlorine. This will give us our alkene product at the end. This will be our major product at the end. Now, let's say that we did the Zaitsev's rule instead. Again, it's less likely for this to occur because tert-butoxide is too bulky to reach that hydrogen. But if it did, what would the product look like? Here's tert-butoxide. It comes and removes this hydrogen here. Bond breaks, falls here, and kicks out the chlorine. So then we get here as our product. This would be our minor product. Now, yes, we've just made a more conjugated product at the end, and that's great, but we're still not going to be able to make it as easily because, again, tert-butoxide is a bulky base. It's hard for it to reach this beta hydrogen here, hard to access it. If we'd used a smaller base, like, let's say, an ethoxide ion or something, then we could more easily grab this hydrogen here by Zaitsev's rule and make this our major product. But again, we're not dealing with ethoxide ion, we're dealing with tert-butoxide. So while this is great, it's not going to be our major product. This is our major product following Hofmann's rule.

Reactions at Benzylic Positions Concept 4

Hey, everyone. So in this video, we're going to talk about Benzilic Alcohol Oxidation. Now here we're going to discuss a weak oxidation where Manganese(IV) oxide within a dichloromethane solution selectively oxidizes the benzylic OH group. Now here, benzylic alcohols are much more reactive than non-benzylic alcohols. If we take a look here, we have our benzylic alcohol.

We're using manganese(IV) oxide within our dichloromethane solution again. All it does is it oxidizes this alcohol into a carbonyl. So in this case, we've gone from our benzylic alcohol to this aldehyde group. We've just made Benzaldehyde as our product here. So just remember, this is selective in nature.

It's only attacking or oxidizing our benzylic OH groups. Okay. So that's what it's programmed to do in this case.

Reactions at Benzylic Positions Example 3

In this example question, we need to get the product from the following oxidation reaction. So in this question, we have Benzene, and directly connected to it is an OH group. We also have another substituent coming off of Benzene. Here is our benzylic carbon, and on it, our OH group. Here we're doing selective oxidation by using manganese 4 oxide within a dichloromethane solvent.

Remember, this is a very weak oxidation, a selective oxidation that only targets benzylic OH groups. Here is our benzylic carbon with its OH group. So, it's what's going to be oxidized. The other OH group, being directly connected to Benzene, is not benzylic in nature. So it stays as OH.

This alcohol gets oxidized, so we go from an OH group to a carbonyl group. So we're gonna get Cdouble bond O here. So we've gone from a secondary benzylic alcohol into a ketone. This would be our final product. Remember, this type of oxidation is only targeting our benzylic OH group.

The other OH group is unaffected.

Supply the reagents necessary to accomplish the following transformation.

• A. 1. CH₃CH₂Br/AlBr₃          2. NBS/hv                3. NaOEt/EtOH            4. OsO₄/NaHSO₃, H₂O       5. MnO₂/CH₂Cl₂            6. HNO₃/H₂SO₄   
• B. 1. HNO₃/H₂SO₄    2. CH₃CH₂Br/AlBr₃               3. NBS/hv             4. NaOEt/EtOH      5. OsO₄/NaHSO₃, H₂O           6. MnO₂/CH₂Cl₂  
• C. 1. CH₃CH₂Br/AlBr₃       2. NBS/hv               3. NaOEt/EtOH           4. BH₃ .THF/H₂O₂, NaOH    5. MnO₂/CH₂Cl₂           6. Br₂/FeBr₃   
• D. 1. HNO₃/H₂SO₄2. NaOEt/EtOH3. CH₃CH₂Br/AlBr₃4. NaOEt/EtOH5. H₃O⁺6. PCC/CH₂Cl₂

Provide the chemical steps necessary for the following synthesis.

Here in this practice question, it says, provide the chemical steps necessary for the following synthesis. If we take a look here, we're starting out with a benzylic alcohol. And at the end, what we've done is we've created a ketone. In addition to this, what else do we notice? Well, here's my benzylic carbon which must be the same carbon here, and somehow I added these 4 carbons. Now a key giveaway here is we're adding carbons, and at the end, we have a carbonyl or an aldehyde or an alcohol. Like, if you have these 3 types of functional groups involved, you have alcohol, ketone, aldehyde, additional carbons being added. That's usually indicative of a Grignard reagent being used in some way. So what we're going to do here is we know we've seen benzylic alcohols before. And in this lesson, we've dealt with benzylic alcohols when we're talking about the selective oxidation of them. Remember, that happens through the use of manganese (IV) oxide with our dichloromethane solvent. This would oxidize this benzylic alcohol into an aldehyde. So now we have an aldehyde, but we still need to get to this ketone and add these 4 additional carbons. At this point, we know that if we have an aldehyde and we need to add additional carbons, we can utilize a Grignard reagent. So, we need to add 4 carbons. 1, 2, 3, 4. This would break, hit here and kick this bond up. And we'd have it over H₃O⁺ to help protonate that O⁻. Or that O negative to make it into an OH. Okay, so here go the 4 carbons we've added: 1, 2, 3, 4. And we've protonated this OH. So we've just increased our carbon chain. All we have to do now is find a way to change this OH group into a carbonyl group. Again, what do we have? Yes. This is not a benzylic alcohol like before, but it's still the benzylic position here, and it still has an OH group there. So we can still do selective oxidation. So, again, we can use manganese (IV) oxide with our dichloromethane solvent. That would oxidize this into our ketone. And in that way, we've gone from our initial benzylic alcohol to this final ketone product. Right? So these are the steps that we need to undergo in order to go from our reactant to our final answer with our product.

Beginning with benzene, provide a method to prepare the following compound. 

Hey, everyone. So here it says, beginning with Benzene, provide a method to repair the following compound. Now here, going on with the themes of oxidation and going on with the ideas that we've covered within this section, one thing we could do is start with Benzene. We see that we started with Benzene, and we've added 2 carbons, and the end carbon happens to be an aldehyde. Alright. So if I want to add 2 carbons to benzene, I can do this through Friedel-Crafts alkylation. The 2 carbons I want to add, and then here we'd have Br, and then over AlBr3. Here goes the 2 carbon chain I've added. Next, I could do NBS over HV. Remember, this will replace 1 benzylic hydrogen with a bromine. Next, I can do an elimination reaction. So here's alpha, and this is beta. We can remove one of these betas with this strong base. Bond breaks and falls here to make a double bond and kicks out the Br. So, we've just created an alkene. Now let's think about this. I've made an alkene. The end product is an aldehyde. Alright. So how do I get an aldehyde to the end? Well, I can get an aldehyde through oxidation of an alcohol. So now I say to myself, okay, let's say that this, this carbonyl group was an OH group. How in the world can I go from an alkene to an alcohol? Well, this requires us going a little back in time. We need to go to hydroboration oxidation here, or we can do there's a couple of different ways. We could do hydroboration oxidation. Let's actually, we could do that. BH3 THF H2O2, NaOH. So this would add OH there. And then we can just oxidize it with PCC over dichloromethane. So this is yet one way we can make this. This is the list of steps that we can utilize in order to get to our final answer. Another method that we could have done is we could have gone this route, where we do HBr over a peroxide, which would add H and Br by anti-Markovnikov. Then we could have done NaOH and done SN2 to kick this out. And then PCC again to give us this final answer. Right? So 2 different methods, either one could be utilized to get to this final answer. The beauty of synthesis is that sometimes, there's more than one way to get there. All you have to do is give a reasonable pathway starting from our reactant, getting to our final product. Right? So either one of these options is an acceptable answer. Right? So, hopefully, you're able to follow along and not get lost. But remember, some of this does require you to remember certain reactions that you learned quite a bit ago in order to get from our starting benzene to this aldehyde product at the end.

Here’s what students ask on this topic:

Why are benzylic positions highly reactive in organic chemistry?

Benzylic positions are highly reactive due to the stability of the intermediates formed during reactions. The key factor is resonance stabilization. When a benzylic carbocation is formed, the positive charge can be delocalized over the aromatic ring, making the carbocation more stable compared to non-benzylic carbocations. This increased stability facilitates various reactions, such as SN1 and E1 mechanisms, where the rate-determining step involves carbocation formation. Additionally, the benzylic position's reactivity is enhanced in elimination reactions (E2) due to the higher acidity of benzylic beta hydrogens, leading to conjugated alkene products.

What is the role of resonance in stabilizing benzylic carbocations?

Resonance plays a crucial role in stabilizing benzylic carbocations. When a benzylic carbocation is formed, the positive charge on the carbon adjacent to the benzene ring can be delocalized over the aromatic system. This delocalization distributes the positive charge over multiple atoms, reducing the energy of the carbocation and making it more stable. The resonance structures allow the positive charge to be shared by the carbon atoms in the benzene ring, which significantly lowers the energy of the intermediate and increases the rate of reactions like SN1 and E1.

How do electron-donating groups affect SN1 reactions at benzylic positions?

Electron-donating groups (EDGs) enhance SN1 reactions at benzylic positions by stabilizing the carbocation intermediate through resonance or inductive effects. EDGs, such as -OH, -OR, and -NH2, donate electron density to the benzene ring, which in turn stabilizes the positive charge on the benzylic carbocation. This stabilization lowers the activation energy for the formation of the carbocation, thereby increasing the rate of the SN1 reaction. Ortho and para substituents are particularly effective in this role due to their ability to participate in resonance with the benzylic position.

Why do E2 reactions favor elimination products at benzylic positions?

E2 reactions favor elimination products at benzylic positions due to the higher acidity of benzylic beta hydrogens. These hydrogens are more acidic because the resulting conjugated alkene product is stabilized by resonance. When a strong base removes a beta hydrogen, the electrons from the C-H bond form a double bond with the adjacent carbon, and the leaving group is expelled. The resulting alkene is conjugated with the benzene ring, which provides additional stability through resonance. This makes elimination products more favorable compared to substitution products in E2 reactions at benzylic positions.

How does manganese(IV) oxide selectively oxidize benzylic alcohols?

Manganese(IV) oxide (MnO2) selectively oxidizes benzylic alcohols due to the unique reactivity of the benzylic position. In a dichloromethane solution, MnO2 specifically targets the benzylic -OH group, converting it into a carbonyl group (aldehyde or ketone). This selectivity arises because benzylic alcohols are more reactive than non-benzylic alcohols, likely due to the resonance stabilization of the intermediate formed during the oxidation process. For example, benzylic alcohols can be oxidized to benzaldehyde, highlighting the selective nature of MnO2 in oxidizing benzylic positions.