Carbocation Stability

Now we’re going to discuss the most important intermediate of all organic chemistry, the carbocation.

Carbocation Stability

Concept: Determining Carbocation Stability

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Video Transcript

Let's talk about intermediates. Remember that intermediates are high-energy compounds that can be isolated but exist at a higher energy state than the products or the reactants. That's the definition of an intermediate. It turns out that the most popular, the most widely used type of intermediate in this course is going to be the carbocation. What that means is I want to go more into depth on exactly what are the properties of carbocations because those are the most popular intermediates.
Remember that carbocations are just carbons that have three bonds instead of four, so they're always going to have a empty orbital. That empty orbital we give it a positive charge because it's missing a bond. And carbocations it turns out are going to be stabilized by something. There is something that can make that positive charge a little bit better. That's something called hyperconjugation.
What hyperconjugation says is that the electron density from neighboring sigma bonds, so adjacent sigma bonds, can share electron density with the carbocation, with that empty orbital. So if I have – the more sigma bonds I have close to it that are able to donate, the more stable my positive is going to be. Because of the fact that my positive is missing a ton of electrons, if I can share a little bit of these electrons into that positive charge, it's going to make it a lot more stable.
Basically, hyperconjugation is defined by the delocalization of charge of the interaction of an empty p orbital with an adjacent eclipsed sigma bond. It turns out that there's only one group that's really good at having eclipsed sigma bonds to p orbitals and that's R groups because R groups remember they always have hydrogens coming off of them. So those hydrogens are going to be able to eclipse with the p orbital. What that means is that the more substituted my carbocation, or the more R groups I have attached to it, the more stable.
So now I want to show you guys a chart of carbocation stability and show you guys the differences between what's a really bad carbocation and what's a good one.
A bad one would be all the way down here at primary. This would be called a primary carbocation because I only have one R group that's going to have H's that can donate electrons. Because remember an R group just means it's a carbon with some H's on it. That would be an example of a really, really bad carbocation because it only has one side donating electrons.
Now on the opposite end of the spectrum, a tertiary is awesome because a tertiary is going to have hydrogens here, here, and here. And they're all going to be able to donate electron density to that positive charge stabilizing it. So what that means is tertiary carbocations are always going to be your most stable no matter what.
Then in between, we have other ones like secondary and then these guys here. What are those? These are actually going to be ones that can resonate. This is called allylic. And this is benzylic. All these words mean is that – allylic just means next to a double bond so that's what this is. Benzylic means it's next to a benzene.
Remember when we talked about resonance structures way back that a positive charge next to a double bond could resonate and it would form another resonance structure that looked like this. That is a form of delocalization where now my positive charge isn't just stuck in one place, it can be distributed over several atoms so that's going to be more stable.
So allylic carbocations are more stable than secondary because secondary, it's good, it's better than primary because it has the ability to donate with hyperconjugation, but it doesn't have the resonance capability, so it's not as good as allylic. Same with benzylic. Benzylic is also better than secondary.
But the only thing is that tertiary is actually always going to be the best even if it can't resonate, that's fine. Because the tertiary one has so much hyperconjugation that that one's actually going to be more stable overall than even the benzylic or the allylic ones that can resonate. So you might just want to put here that these can resonate. But then this one just has – the tertiary's just the most stable overall.
Now I want to do this example. And I want you guys to think about this for a second and say which of the following alkyl halides would generate the most stable carbocation.
Now I am kind of asking you guys to think about an extra step, so I'm going to tell you what that extra step is. Just so you guys know. Maybe you've learned this already or not or maybe you remember it or you don't, but that's fine. Alkyl halide, which we're going to talk about more in a little bit, have the ability to leave. And if alkyl halides have the ability to leave the molecule completely that means that they're going to leave a carbocation behind.
What you need to do is think after this F leaves and becomes an F negative and I get a positive charge here, what kind of carbocation would that be. Do the same thing with all four of these and then figure out which of them is going to be the most stable carbocation. So go ahead and try to solve this question. Then I'll answer it. 

Carbocations are stabilized by a phenomenon called hyperconjugation.

Hyperconjugation is the delocalization of charge by the interaction of an empty p-orbital with an adjacent, eclipsed σ-bond. Basically it’s like resonance for single bonds (more complicated than that but don’t worry about it). 

Since this is only possible with -R groups, the more substituted the carbocation, the more stable

Which of the following leaving groups would generate the most stable carbocation?

Example: Which of the following leaving groups would generate the most stable carbocation?

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Carbocation Stability Additional Practice Problems

Which of the following carbocations is the most stable?  

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Which carbocation is the least stable? 

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This carbocation rearranges. In the box to the right, draw the carbocation that is the result of the rearrangement. Include any nonzero formal charges in your drawing.


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Rank the substrates below in terms of relative SN1 reaction rate. (1 = fastest, 4 = slowest)

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For the following carbocations, rank them from 1-3 according to overall carbocation stability, with a 1 under the LEAST STABLE CARBOCATION and a 3 under the MOST STABLE CARBOCATION.

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When 2-methylbutene is reacted with hydrochloric acid, where does the carbocation form and why?

1. The carbocation forms at the tertiary position because of steric hindrance.

2. The carbocation forms at the secondary position because of hyperconjugation from neighboring carbons.

3. The carbocation forms at the primary position because nature says so.

4. The carbocation forms at the tertiary position because of hyperconjugation from neighboring carbons.

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Rank the following molecules in order of increasing relative rate of SN1 solvolysis with methanol and heat (slowest to fastest reacting).

A) 5 < 4 < 3 < 2 < 1

B) 1 < 2 < 5 < 4 < 3

C) 2 < 3 < 4 < 1 < 5

D) 3 < 2 < 4 < 5 < 1

E) 2 < 3 < 4 < 5 < 1

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Circle the correct answer

 

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The following statements are true. Choose from among the following three possibilities and in the space provided, write the letter of the one or more phenomena that best explain the true statement.

A. The inductive effect     B. Hyperconjugation     C. Resonance delocalization of a charged species

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The driving force for carbocation rearrangements is that the less stable 2° carbocation is converted into a more stable 3° carbocation ( True or False )

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Identify which of the following substrates will undergo an SN1 reaction more rapidly. Explain your choice

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The overall equation for the addition of HCl to alkenes is:  

 If the transition state for proton transfer from HCl to the alkene (arrow  5 ) resembles a carbocation and this step is rate-determining, what should be the effect of alkene structure on the rate of the overall reaction?

 Fastest rate                                    Slowest rate

A. H2C=CH2    CH3CH=CHCH3    (CH3)2C=C(CH3)2

B. CH3CH=CHCH3    (CH3)2C=C(CH3)2    H2C=CH2

C. CH3CH=CHCH3    H2C=CH2    (CH3)2C=C(CH3)2

D. (CH3)2C=C(CH3)2    CH3CH=CHCH3    H2C=CH2

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When tert-butyl bromide undergoes SN1 hydrolysis, adding a “common ion” (e.g., NaBr) to the aqueous solution has no effect on the rate. On the other hand, when (C6H5)2CHBr undergoes SN1 hydrolysis, adding NaBr retards the reaction. Given that the (C6H5)2CH+ cation is known to be much more stable than the (CH3)3C+ cation (and we shall see why in Section 15.12A), provide an explanation for the different behavior of the two compounds.
 

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Explain the following observations: When tert-butyl bromide is treated with sodium methoxide in a mixture of methanol and water, the rate of formation of tert-butyl alcohol and tert-butyl methyl ether does not change appreciably as the concentration of sodium methoxide is increased. However, increasing the concentration of sodium methoxide causes a marked increase in the rate at which tert-butyl bromide disappears from the mixture.
 

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Which SN1 reaction of each pair would you expect to take place more rapidly? Explain your answer.
 

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Two stereoisomers of 1-bromo-4-methylcyclohexane are formed when trans-4-methylcyclohexanol reacts with hydrogen bromide. Write structural formulas of: 

(b) The carbocation intermediate in this reaction

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Select the compound in each of the following pairs that will be converted to the corresponding alkyl bromide more rapidly on being treated with hydrogen bromide. Explain the reason for your choice.

 (g) 1-Cyclopentylethanol or 1-ethylcyclopentanol 

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Select the compound in each of the following pairs that will be converted to the corresponding alkyl bromide more rapidly on being treated with hydrogen bromide. Explain the reason for your choice.

(f) 1-Methylcyclopentanol or trans -2-methylcyclopentanol 

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Select the compound in each of the following pairs that will be converted to the corresponding alkyl bromide more rapidly on being treated with hydrogen bromide. Explain the reason for your choice.

 (d) 2-Methylbutane or 2-butanol

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Select the compound in each of the following pairs that will be converted to the corresponding alkyl bromide more rapidly on being treated with hydrogen bromide. Explain the reason for your choice. 

(c) 2-Methyl-2-butanol or 2-butanol 

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Select the compound in each of the following pairs that will be converted to the corresponding alkyl bromide more rapidly on being treated with hydrogen bromide. Explain the reason for your choice. 

(b) 2-Methyl-1-butanol or 2-butanol 

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Select the compound in each of the following pairs that will be converted to the corresponding alkyl bromide more rapidly on being treated with hydrogen bromide. Explain the reason for your choice. 

(a) 1-Butanol or 2-butanol 

 

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Predict whether each of the following carbocations will rearrange. If so, draw the expected rearrangement using curved arrows.

 

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Predict whether each of the following carbocations will rearrange. If so, draw the expected rearrangement using curved arrows.

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Predict whether each of the following carbocations will rearrange. If so, draw the expected rearrangement using curved arrows.

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Predict whether each of the following carbocations will rearrange. If so, draw the expected rearrangement using curved arrows.

 

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Rank the three carbocations shown in terms of increasing stability:

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Rank the three carbocations shown in terms of increasing stability:

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For each of the following reactions identify the arrow-pushing pattern that is being utilized:

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For each of the following reactions identify the arrow-pushing pattern that is being utilized:

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Which two compounds ionize with loss of bromide ion to form the same carbocation?

a. 1 and 2

b. 2 and 3

c. 1 and 4

d. 3 and 4

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Which alcohol is dehydrated fastest in concentrated H2SO4?

 

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Which carbocation is the most stable?

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Rank the following carbocations in order of stability: 

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