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Short link: mendelset.com/chapters/269
Let's go over how a carbocation can form from an alcohol.
Write in the curved arrows to show the formation of the protonated alcohol, and water acting as a leaving group to form a carbocation.
Carbocations aren't very stable and so don't last very long after they are formed.
Use curved arrows to show:
a) how a carbocation reacts with a halide ions to form an alkyl halide.
b) how a carbocation reacts with water to form an alcohol.
c) how a carbocation reacts with a base to form an alkene.
For a), the product is neutral and so you are done after the nucleophile (Cl-) attacks the carbocation.
For b), the the intermediate is a protonated alcohol, and so you must do a proton exchange step (also called a hydrogen exchange or deprotonation) to get the final alcohol product, which is neutral.
For c), there are two different types of beta hydrogens, and so two different alkenes are possible.
The more stable alkene is the one that will form, and this will always be the most highly substituted alkene. This is Zaitsev's rule. The rationale for this is hyperconjugation: neighboring carbon atoms stabilize an alkene.
For the reaction below, draw the structures of the carbocation intermediate and the final product.
The delta (Δ) in the reaciton arrows means that heat is being added to this reaction, which tends to favor elimination over substitution. Also, the reaction is using a non-nucleophilic acid (H2SO4), which tends to favor elimination reactions (H3PO4 is another common reagent for E1 reactions, while HCl or HBr tend to go SN1).
Because this reaction is taking place in acid, a carbocation is likely to form, so this is an E1 reaction. Since water is lost over the course of the reaction, this is a dehydration, which is a type of elimination reaction.
The alcohol below is protonated and contains an oxygen with a positive charge. Using curved arrows, show the two "legal moves" that result in a neutral oxygen.
Protonation is a common first step in mechanisms. When you have a protonated alcohol, it's likely that either:
Either one is possible. If you're not sure which one happens, remember that when writing out mechanisms, never go backwards- don't draw structures you have drawn before.
Let's work through an elimination reaction. Draw the structures for each of the species in the three boxes below (protonated thiol, carbocation, and alkene). Also draw curved arrows to show electron movement.
The first step in this reaciton, like many reactions, is an acid-base reaction- when you see an H+ (acid), the first step is usually something getting protonated. This starting material is a thiol (sulfur), but the same thing would happen with an alcohol or ether (oxygen).
When you have a protonated thiol (or alcohol, ether, etc.), you know the reaction isn't finished yet- products are rarely charged. There are two legal moves to get that positive charge off of the protonated thiol. Either the thiol deprotonates, or it takes off as a neutral leaving group. It can't deprontate because that would be going backwards, so your only "legal move" is for the HSCH2CH3 to act as a leaving group (See problem 518).
But then you're left with a carbocation, which is definitely not the final product. How can we get rid of it? (See problem 335 for general ways carbocations react). If the HSCH2CH3 attacked in an SN1 type reaction that would be OK, but it would be going backwards! So the only way to get rid of the carbocation is to do a beta-elminiation (E1).
For a molecule to undergo an E2 reaction, the leaving group and the beta-proton must be in an anti-coplanar conformation (one atom straight up, the other straight down). Based on this, which compound undergoes E2 reaction with KOtBu faster? Why?
The methyl group is bulky and so is most stable in the equatorial position.
The 1,4 cis compound's most stable chair form has its leaving group (Br) and beta-hydrogen in an anti-coplanar (anti-periplanar in some textbooks) conformation.
So most of the time, 1,4 cis is in a chair conformation that is able to undergo an E2 reaction.
The opposite is true for the 1,4 trans compound. Most of the time it is in a chair conformation that is unable to undergo an E2 reaction.
Therefore, the 1,4 cis compound will undergo an E2 elimination reaction faster than the 1,4 trans compound.
Predict the product(s) of the reaction below, and used curved arrows to show a mechanism.
You know this is an elimination reaction because no nucleophile is present; H3PO4 and H2SO4 are non-nucleophilic acids. IF the reagent were HCl or HBr on the other hand, this would be a substitution reaction.
Since acid is present, a carbocation will probably form, so we know that this is an E1 mechanism.
The 2º carbocation that is initially formed will undergo a 1,2-hydride shift to become a 3° carbocation, but that doesn't affect the final product of the reaction; with either the 2° or 3° carbocation, the most stable alkene product is 2-methyl-2-butene.
To predict the stereochemistry of the alkene product from an E2 reaction, we have to rotate the leaving group (Br) and the beta-proton into anti-coplanar geometry. It's easiest to put the H and Br in the plane of the page.
After we do this, we see that both the ethyl and phenyl groups are wedges, and the two methyls are dashes. So after the E2 reaction, the resulting alkene will have its two methyl groups cis to each other.
Textbook and Chapters: Carey and Giuliano 8th Ed. (2010), Chapters 4, 5
Keywords: carbocation rearrangement, free radical halogenation, mechanism
Description: Practice mechanisms from the first half of Fall semester orgo. Dehydration (E1), substitution with carbocation rearrangement (SN1), and free-radical bromination.
Total Problems: 3
Textbook and Chapters: Carey and Giuliano 8th Ed. (2010), Chapters 4, 5, 6
Keywords: alkene addition, carbocation
Description: Identify the intermediates (carbocation, radical, borane intermediate, etc.) and products for important reactions dealing with alkenes. Good review for an orgo1 midterm.
Total Problems: 7
Keywords: carbocation, carbocation formation, carbocation rearrangement
Description: This mendel set guides you through everything you have to know about carbocations:
Also includes some practice problems: addition to an alkene, dehydration (E1), and substitution (SN1).
Total Problems: 8
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