Each of the carbocations below will spontaneously rearrange. Draw the structure of the expected rearrangement product.
Be on the lookout for a 1,2-shift when you have a carbocation adjacent to a carbon atom that is more substituted (ex: a 2° carbocation next to a 3° or 4° carbon).
MendelSet practice problem # 332 submitted by Matt on June 7, 2011.
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.
MendelSet practice problem # 335 submitted by Matt on June 7, 2011.
For the reaction below, draw the structures of the carbocation intermediate and the final product.
This is a substitution reaction. Because the reaction takes place in acid and the leaving group is on a 2º carbon, it will probably form a carbocation (SN1 mechanism).
-OH is a poor leaving group so the alcohol will protonate first, so it can leave as H2O, which is neutral.
MendelSet practice problem # 347 submitted by Matt on June 7, 2011.
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:
The alcohol will deprotonate to form a neutral alcohol (right), or
The alcohol will leave as a neutral molecule to form a carbocation.
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.
MendelSet practice problem # 518 submitted by Matt on June 30, 2011.
Predict the product(s) of the reaction below, and used curved arrows to show a mechanism.
The 2° carbocation formed immediately undergoes a 1,2-methyl shift (a rearrangement) to form the more stable 3° carbocation, so the product is the 3° alkyl chloride instead of the 2º alkyl chloride, which would have formed in the absence of rearrangement.
MendelSet practice problem # 337 submitted by Matt on June 7, 2011.