Organic Chemistry

It's easiest to assign R or S configuration when the lowest priority substituent is "in the back" or "behind" the molecule, that is, a dash. Most of the time the lowest priority substituent will be a hydrogen atom.

So a) is straightforward. Because hydrogen is already a dash, we can ignore it and see in which direction the other three substituents decrease in priority (according to Cahn–Ingold–Prelog priority rules). Since they decrease in a counter-clockwise way, a) has S absolute configuration.

b) is a little harder to assign. If hydrogen were a wedge we would be able to just take the opposite of whatever answer we get. But in this case, H is neither a wedge nor a dash. So we use a trick: if we swtich any two pairs of substituents, the absolute configuration of the molecule remains the same. So we switch two pairs so that the hydrogen becomes a dash. On the resulting molecule, the priorities 1-3 decrease in a clockwise manner, so this molecule has an R absolute configuration.

To be a stereocenter you need to:

• be a carbon
• be sp³ hybridized (all single bonds)
• have four different substituents

Are there other types of stereocenters? Of course! ( sp³ sulfur atoms, for example). But in undergraduate organic chemistry, 99% of all sterocenters you come across will be sp³ hybridized carbon atoms.

So in the molecule below, there are four stereocenters.

Note that the sp² (alkene) carbons can't be sterocenters.

This is an S_N1 reaction, but with a twist.

The top reaction is a regular S_N1 reaction; the leaving group (Br^-) leaves and the nucleophile (CH₃OH) attacks the carbocation.

The twist is that the carbocation is allylic, and has resonance, so there is another carbon that the nucleophile can attack. 

So instead of just one S_N1 product, two products are formed. The product that doesn't involve resonance is called the direct addition or 1,2 product. The product that involves allylic resonance is called the conjugate addition or 1,4 product.

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).

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 (S_N1) 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.

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 S_N1 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).

Protonation is a common first step in mechanisms. When you have a protonated alcohol, it's likely that either:

1. The alcohol will deprotonate to form a neutral alcohol (right), or
2. 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.

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 S_N1).

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.

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 (S_N1 mechanism).

^-OH is a poor leaving group so the alcohol will protonate first, so it can leave as H₂O, which is neutral.

This is a free radical halogenation. The hint is that you see light (hν) in the reaction arrow, indicating radicals are involved.

The more substituted the carbon atom, the more stable the corresponding radical, so the major product of this reaction is the 3º alkyl halide.