Chapter 6 Reactions of Alkyl Halides

Chapter 6 – Reactions of Alkyl Halides

Nucleophiles
Nucleophiles are Lewis Bases, so they must have a lone pair.
Negatively charged nucleophiles are stronger than their neutral counterparts.
Nucleophile strength decreases from left to right on periodic table.
Nucleophile strength increases going down the periodic table.
Bulky groups decrease nucleophile strength.
Ex. Hydroxide is a better leaving group than t-butoxide because the t-butoxide is so large that it has difficulty getting to the site of reactivity.
Leaving groups – halide ions in this chapter
Must be electron-withdrawing to create a partial positive on the carbon
Take their electrons with them
They’re that kid on the playground who takes the ball when he doesn’t get his way.
The weaker the base, the better the leaving group

Substitutions vs. Elimination
These are words in the English language and they don’t have new definitions for chemistry.
With substitution, your nucleophile comes in and takes the place of the leaving group.
If you had a substitute teacher in high school, a new teacher came in and replaced your regular teacher.

With elimination you are getting rid of the leaving group and the nothing takes its place.
If the teacher had been eliminated, you would have come into class and not had a teacher at all.

SN2 – Second Order Nucleophilic Substitution
Nucleophile just knocks off the halogen and takes its place

One-step
No intermediate, so no rearrangement
rate=k[RX][nuc]
Both the alkyl halide and nucleophile are involved in the rate-limiting step (the only step in this reaction) so they both affect the rate.
Backside attack leads to inversion of configuration
Most of the time your question won’t look like the mechanism drawn above.
When you are given a substrate in line-angle form and asked to draw the product, draw the product with the nucleophile where the halogen was, but with a dash where a wedge once was or vice versa.

Needs strong nucleophile
At this point in the course the strong nucleophiles given will all be negatively-charged, though ammonia and amines are also strong nucleophiles which can undergo SN2 reactions.
Polar, aprotic solvents can help this along
Aprotic means that the solvent is not a good source of protons.
No hydrogen bonding.
Mr. Baker requires you to know four of these polar, aprotic solvents.
Acetone

DMSO (dimethylsulfoxide)

Acetonitrile

CH3CN

DMF (dimethylformamide)

A polar, aprotic solvent is not required for this reaction.
So if he gives you a reaction and doesn’t give a solvent, still do the reaction.

Suitability of alkyl halides as substrates
Methyl>1°>2°>3° (wont’ happen)
This should make sense as adding carbon groups will increase the steric hindrance.
The rate will be faster with more suitable substrates (so an SN2 on a methyl will go faster than on a 2°).

SN1
First, the halogen falls off (slow step)

Sometimes you’ll see AgNO3 here.
The silver complexes with the halide and precipitates out of solution.
It’s just a hint that you have SN1/E1 conditions.
Do not write Ag or NO3 on your product (unless for some reason you’re asked for the inorganic byproducts.

Now you have a carbocation, which can rearrange

Rearrangements will only occur when the resulting carbocation is more stable than the initial carbocation.
Ex. If a secondary carbocation becomes tertiary after shift, tertiary carbocation becomes tertiary allylic, etc.
There are two types of rearrangements:
Hydride shifts
In a hydride shift, a hydrogen next door moves over taking its electrons with it.
Alkyl shifts
In an alkyl shift, an alkyl group next door moves over taking its electrons with it.
You only do alkyl shifts when there’s a quaternary position next door to the initial carbocation.

The nucleophile (often a polar, protic solvent) attaches to the positively-charged carbon

Deprotonation finishes it off

Major substitution product always comes from the rearranged carbocation
Happens with weak nucleophiles
For this test the weak nucleophile will be your polar, protic solvent
Water and alcohols
Order of reactivity
3°>2°>1°>methyl (won’t happen)
This should make sense, as the slow step is forming the carbocation.
The more stable the resulting carbocation, the more likely it is that the first step will happen.
Partial racemization
Means you’ll get both R and S where you added the nucleophile.
You’ll also get both R and S where the methyl moved if you have a methyl shift.
You will see throughout the course that if a reaction goes through a trigonal planar intermediate, then there is no way to select for R or S.
This time the trigonal planar intermediate is a carbocation.
rate=k[RX]
The first step is the slow step and only the alkyl halide is involved with that.

Same first step as SN1, so they’ll both happen at the same time
Rearrangement still possible

A hydrogen next-door to the + falls off as a proton, leaving it’s electrons behind to form a double bond

Get all possible stereoisomers

Choosing the major elimination product
Look at the products that come from the rearranged carbocation
Zaitsev’s rule
The more carbons there are coming off a double bond, the more stable the double bond is.
More stable products are more likely to form.
If there’s a tie, look to see if one of the more substituted products was reached from two different paths
If so, then that’s the major product.
rate=k[RX]

Drawing all the products when you have SN1/E1 conditions
You should expect to see a problem like this at some point.

I think this is easiest when you make a template for yourself.
Here’s my template:

So now you go through and fill out the templateTo get the initial carbocation, take off the halogen and put a positive charge there.

Then to get the SN1product(s) just let your solvent plop on there and if the position is chiral remember you get both R and S.

To get the E1 product(s) make a double bond in any possible direction from the carbocation and then consider whether you have E/Z double bonds. If you do, you’ll get both E and Z.

Now rearrange the carbocation.

Just move the positive charge to the more substituted spot next door.
Notice that you lose the wedge to that methyl and now it’s flat.
That is because carbocations are trigonal planar.

And to get the new products just do what you did with the unrearranged products.

Notice that you get one of the E1 products from both the rearranged carbocation and the unrearranged carbocation.

Picking the major products.
Substitution
This one’s easy.
It’s always the rearranged product.
Elimination
There are two factors which can compete.
The most substituted double bond is the most stable.
More stable products are more likely to form.
There is always an E1 product that you get from both the unrearranged and the rearranged carbocation.
If you get to it through multiple paths then it is more likely to form.
If the factors compete, he will not ask you for the major E1 product.
In this situation he would ask you for the most stable E1 product; go with Zaitsev!

E2
Happens with secondary and tertiary alkyl halides combined with strong bases.
At this point in the course, the only strong bases you are responsible for knowing are hydroxide and alkoxides.
-OH and -OR
One step

Strong base abstracts proton next-door to halogen
The electrons that were attached to the hydrogen kick off the halogen

The hydrogen and halogen must be anti-coplanar to each other

rate=k[RX][base]

Zaitsev’s Rule
Applies to elimination reactions
Whenever possible, the more substituted double bond will form
When your base is particularly bulky (think t-butoxide or similar), you will get the less substituted product.
Which reaction should I do?

Alkyl Halide / Strong Nucleophile
Weak Base Strong Base
(RCOO-, -CN, -SH (-OR, -OH)
-SR, N3-, X-, NH3
or amines) / Weak Nucleophile
(H2O, ROH)
Methyl / SN2 SN2 / NR
1° / SN2 SN2 / NR
2° / SN2 E2 / SN1/E1
3° / NR E2 / SN1/E1

When you have things in solution that tell you to do different reactions, the strong thing wins!

The strong nucleophile wins and we do an SN2.