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.
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.
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.
MendelSet practice problem # 348 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.
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).
MendelSet practice problem # 519 submitted by Matt on June 30, 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.