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.
For the reaction below, draw the structures of the borane intermediate and the final product.
This is a hydroboration reaction (also called hydroboration-oxidation in some textbooks). The product is an alcohol.
Boron bonds to the less substituted carbon, and is replaced with an oxygen after the borane intermediate is treated with peroxide, resulting in an anti-Markovnikov addition.
Note that one equivalent of BH3 reacts with three equivalents of alkene; the borane intermediate is not BH2(alkyl) but rather B(alkyl)3.
MendelSet practice problem # 344 submitted by Matt on June 7, 2011.
For the reaction below, draw the structures of the radical intermediate and the final product.
When HBr reacts with in an alkene in the presence of peroxides (which are free radical initiators), the reaction goes through a free radical mechanism and adds in an anti-Markovnikov manner. The product is an alkyl bromide.
The reaction goes anti-Markovnikov because the more substituted radical intermediate is only formed if the bromine adds to the less substituted carbon.
MendelSet practice problem # 345 submitted by Matt on June 7, 2011.
Let's work through an alkene addition reaction. Draw the structures for each of the species in the three boxes below (3º carbocation, protonated thiol, and thiol). Also draw curved arrows to show electron movement. Note: thiol = RSH, like an alcohol, but with sulfur instead of oxygen.
Note that this is the reverse of problem 519. Instead of going from thiol (alcohol) to carbocation to alkene, we're going from alkene to carbocation to thiol. In each step ask yourself "what arrows can I draw?" and choose the step that doesn't go backwards.
The first step is the alkene picks up a proton to form the more stable carbocation. (See problem 334 if you don't understand why only the 3º carbocation is formed). To get rid of the carbocation, we can either do a beta-elimination (E1) to form an alkene, or an addition reaction (SN1) to form a protonated thiol. Since forming the alkene would be going backwards, the only choice is addition.
Then you're left with a protonated thiol. How do we get rid of the positive charge on the sulfur? There are two legal moves- either the RSCH2CH3 acts as a leaving group, or the sulfur deprotonates. Since the RSCH2CH3 leaving would form a carbocation and take us backwards, the only option is for the sulfur to deprotonate.
Because the sulfur added to the more substituted carbon, this was a Markovnikov addition.
MendelSet practice problem # 520 submitted by Matt on June 30, 2011.
Rank the carbocations below in order of decreasing stability. (1 = most stable)
The more substituted the carbocation, the most stable it is. This is because of the inductive effect- adjacent carbon atoms donate some of their electron density to neighboring carbocations (which are electron deficient), making them closer to neutral and more stable.
So the 3° carbocation is the most stable.
MendelSet practice problem # 331 submitted by Matt on June 7, 2011.
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.
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.
Let's work through a halogenation reaction. Draw the structures for each of the species in the four boxes below (3º carbocation, halonium ion, protonated thiol, and thiol). Also draw curved arrows to show electron movement. Note: thiol = RSH, like an alcohol, but with sulfur instead of oxygen.
Because the chlorine is more electronegative than iodine, the iodine will have a partial positive charge and will be attacked by the alkene. It forms the more stable carbocation as normal (as in problem 335). The nucleophile (HSCH2CH3) will eventually attack the carbocation, but the iodine does something special first- it forms a cyclic intermediate.
The formation of the cyclic halonium ion is probably the most important concept you will learn in first semester organic chemistry. With iodine, it's called an iodonium ion. (with chlorine or bromine, it's called a chloronium or bromonium ion).
The HSCH2CH3 will attack the carbon that was the carbocation, but it's forced to attack from the side opposite to that of the iodine. So the product will always have anti stereochemistry (you get the trans product).
MendelSet practice problem # 521 submitted by Matt on July 1, 2011.
Indicate the major organic product of the reaction below. Include stereochemistry.
This is a hydrogenation reaction. The product is an alkane. But this product has two stereocenters.
Normally, the product would be a racemic mixuture of enantiomers (equal amounts of each stereoisomer). But in this case, the starting material is optically active (chiral but not racemic), so the products won't be equal and opposite as usual.
Since there is a substitutent blocking the top face of the molecule (isopropyl is a wedge), the hydrogens will preferentially add to the opposite face of the molecule (dashes).
MendelSet practice problem # 529 submitted by Matt on July 2, 2011.
Let's work through anti and syn additions to alkenes.
Show the product for each reaction below, and indicate whether the product will be a racemic mixture of enantiomers, or a meso compound (which is achiral).
The reaciton of an alkene with Br2 is an anti-addition, and hydrogenation (H2 or D2) is a syn addition. From this we can figure our the relative stereochemistry of each product.
Each starting material is achiral, and therefore not optically active, so the products cannot be optically active.
There are three ways for products not to be optically active:
products can be achiral
products can be a racemic mixture
products can be meso
In both a) and b) each product has a sterocenter, so the products can't be achiral.
In a), one stereocenter is S and the other stereocenter is R, and both with the same substituents, so this is a mirror image relationship, and the products are meso.
In b), the products have no internal mirror planes, so the products are chiral, and must be a racemic mixture.
MendelSet practice problem # 530 submitted by Matt on July 2, 2011.