Chem 343 PSet L The SN1 Reaction

The SN1 Reaction is a substitution reaction in which one group is exchanged for another. The group swapping in is called the nucleophile. The group swapping out is called the leaving group. The carbon where everything is occurring is the alpha carbon (it is the carbon attached to the leaving group). The molecule containing the alpha carbon will be dubbed the electrophile. Draw in the lone pairs in the substitution reaction below. Substitution Reaction: leaving group Br I αC I Br

nucleophile electrophile

There are three ways this can happen: 1 The leaving group leaves, then the nucleophile adds to the alpha carbon. 2 The nucleophile adds as the leaving group leaves (simultaneous process). 3 The nucleophile adds, then the leaving group leaves. Option 3 is the one we can rule out if the alpha carbon is sp3 hybridized, because that would mean five bonds to a carbon. Option 1 is the SN1 reaction and option 2 is the SN2 reaction. The example above is the SN1 reaction.

Mechanism of the SN1: Arrows show the movement of electrons. First step: leaving group leaves. The electrons from the alpha carbon leaving group bond become a lone pair on the leaving group. The arrow starts at the C-Br bond and points to the Br. This creates an intermediate. Draw in the lone pairs and the arrows below. What is the formal charge of the alpha carbon in the intermediate? Draw the formal charge as well. Answer is on next page.

Br Br

intermediate

Br Br

intermediate

The intermediate is a carbocation. As a strong Lewis acid, it is a very high energy intermediate that has a short lifetime. What makes it a strong Lewis acid?

The second step is just a Lewis acid/base reaction. The nucleophile adds to the alpha carbon (the carbocation). A nucleophile is a Lewis base that likes to form bonds with another atom (typically carbon). The arrow starts at the nucleophilic atom and goes to electrophilic carbon (Lewis acidic carbon). This creates the bond between the alpha carbon and the nucleophile. Draw the curved arrows and lone pairs below for the entire sequence. Answer is on next page.

I Br Br I Br

intermediate The SN1 reaction is a two step reaction. Which reaction step is the slowest step? (Which one looks the least favorable?)

I have some blank space so… Notes on arrows: It is all too tempting to start them at the negative formal charge. Resist this urge. Arrows can only start at bonds or atoms with lone pairs. That’s it.

What are the conjugate pKa’s of the following molecules? O

O O Cl O S H H H O

I Br Br I Br

The slow step is the carbocation formation. After all, you are creating a molecule with an incomplete octet. That means that this is the rate determining step. Notice, the nucleophile is not involved in this step only the electrophile. The rate of an SN1 reaction is dependent on the concentration of the electrophile and not on the nucleophile. Double the concentration of the electrophile, you double the rate of the reaction. Double the concentration of the nucleophile, nothing happens to the rate of the reaction. (This is the origin of the 1 in SN1. It is a unimolecular rate reaction only dependent on the concentration of one molecule).

The reaction can also be described by an energy diagram. An energy diagram described the free energy of reaction as it proceeds. The y axis is energy, and the x-axis is the reaction coordinate which you can think of as time.

I I Br Br I Br

The rate of is dependent on the formation of the carbocation. The lower the energy of the transition state (barrier) to the carbocation, the faster the reaction. The energy of transition states are hard to model, but they can be approximated by the intermediates on the path closest in energy to it. This is the Hammond postulate. Simply put, the intermediate is closer in energy to the starting material, so the transition state’s energy can be approximated by the structure of the intermediates. More simply put: the lower the energy of the intermediates, the lower the energy of the transition state, and the faster the reaction.

First Requirement: A good leaving group attached to a sp3 carbon

The SN1 reaction is dependent on the stability of the carbocation and the leaving group on its own. Let’s talk about the leaving group first. The more stable the leaving group after it has left, the faster the reaction. There is actually a straightforward way of recognizing good leaving groups. Good leaving groups are conjugate bases of strong acids (pKa ≤ 0).

Circle the good leaving groups:

O

F O I OH O OH

O Br Cl O S OH C O O N (Tosylate TsO )

H2O

3 SN1 reactions require the alpha carbon to be sp hybridized. Put a rectangle around the molecules that have a good leaving group attached to an sp3 hybridized carbon. In those molecules, circle the entire leaving group.

O F Cl

OH

O Br O S OH C O O N

H2O

O H

O OTs I O I want to point out that with limited exception (OTs), molecules that just have neutral oxygens do not have good leaving groups. Molecules with positive oxygens contain good leaving groups. In the case of protonated ethers, there are two leaving groups because there are two alpha carbons.

If you do not have a good leaving group, sometimes it is possible to make one during the course of the reaction if a strong acid is present.

Sometimes, you have to do an extra acid base step to create that good leaving group. In the following example, draw in the curved arrows, lone pairs, and formal charges for the following reaction.

H H H H Cl H H O O Cl O Cl Cl

In other situations, you have to do an extra acid base step at the end of a reaction as organic molecules with pKa’s less than 5 are rarely isolated as they are frequently unstable. Another way to put this: molecules with oxygens with a positive formal charge are rarely isolated. In the following example, draw in the curved arrows, lone pairs, and formal charges for the following reaction.

H H H Cl O H Cl Cl Cl O O H H

The product of a predict the product in an organic reaction should not have a positive oxygen. Think of it as a convention or a pet peeve of mine.

Second Requirement: Carbocation stability. The more stable the carbocation the faster the reaction. Carbocations are unstable due to possessing an atom with an incomplete octet. By sharing electron density with this atom, the energy of the carbocation is diminished and thus its stability is increased.

Two main ways to stabilize a carbocation: First way resonance: If the carbocation is attached to an oxygen or nitrogen with a lone pair, then it is a resonance stabilized carbocation. Alkenes, alkynes, and benzene rings can also stabilize the carbocation. Draw resonance structures of the following carbocations.

O

N

Carbonyls do not stabilize a carbocation. They actually destabilize a carbocation. What is wrong with the following curved arrow?

O O

Second way is hyperconjugation: This is sharing of electrons between a filled and unfilled orbital. It is not as efficient as resonance. Hyperconjugation allows the empty p orbital of the carbocation to syphon electron density from the bonding orbitals of adjacent C-C or C-H bonds. Hydrogen possessing just an s orbital cannot supply electron density to a p orbital as an s orbital does not have the correct symmetry.

H H + + 9 + + C = C C C H = C H

Hyperconjugation Good overlap is cancelled Only good overlap by bad overlap

Easier way to remember it: Tertiary carbocations (carbocations attached to 3 nonhydrogens) are more stable than secondary carbocations which are more stable than primary carbocations.

Tertiary carbocations form the fastest. Secondaries can form but are slow to do so. Primary carbocations generally do not without special circumstances. Methyl carbocations do not.

Looking at the stability of carbocations: a tertiary carbocation, resonance stabilized carbocation, or a secondary carbocations are the allowable carbocations that we will encounter.

Third requirement: Nucleophile must be a weak base. Conjugate pKa less than 5. Strong acids and strong bases cannot coexist. Carbocations are strong acids, so a strong base cannot be present. If one were present, a side reaction (E2) would take place. Nucleophiles that are unlikely to do an E2 reaction are weak bases with conjugate pKa’s less than 5. Azide is a bit of an exception. It has a conjugate pKa close to 5 and it typically prefers SN1 over E2. (Nucleophiles must also be able to form a bond. Nitrogen can form up to four bonds, oxygen and sulfur can form up to three bonds, and the halogens typically stick to one bond). Put a rectangle around the molecules that can be used as SN1 nucleophiles and circle the atom that the new bond will go to.

Br I Cl N N N N azide SH S H2O O HO H H N N

Stereochemistry:

The SN1 reaction goes through a carbocation intermediate. The alpha carbon goes from sp3 hybridization to sp2 hybridization and then back to sp3 hybridization. This has implications, because the resulting nucleophile can attack either face of the carbocation. As a result, two products can occur. This is especially true if the starting material is optically active and the resulting product is chiral. It is important to determine what the relationship is between the two molecules. If they are the same molecule, then you draw the molecule once and write optically active. If they are enantiomers, then you draw one of the two molecules and write racemic. If they are diastereomers, then you draw them both and write optically active under each one.

Problems: 1.) For the following reactions, draw out the products if the nucleophile attacks from the above (becomes a wedge) or below (becomes a dash) in the two rectangles. Determine what the relationship is between the two molecules. Write out the final answer in the rectangle that says final answer. (The final answer according to the bold statement up above.) D is the abbreviation for deuterium (the heavy isotope of hydrogen). It has the same reactivity of hydrogen but is considered to be different atom for chiral center purposes. It has atomic number of 1 but is considered higher priority than H.

D A.

optically Br H O + active 2 CD3

from above from below D

optically Br H O active 2 CD3

FINAL ANSWER (what you will write on exams)

D B.

optically Br H O + active 2

from above from below D

optically Br H O active 2

FINAL ANSWER (what you will write on exams)

C.

optically Br H O + active 2 CD3

from above from below

optically Br H O active 2 CD3

FINAL ANSWER (what you will write on exams)

2.) Predict the products of the following reactions. Draw the mechanism for the following reactions. a. NaI Br

b. OH HBr

c. Br Cl O H H

3.) Explain why the reaction of benzyl bromide with water gives only one product, but the reaction of 1-bromo-3-methyl-2-butene gives two products. Draw the mechanisms for the two reactions. a.

Br O OH H H

b. OH Br O H H

OH

4.) Here is the question from the Quiz from last semester. The following molecule has resonance structures. Draw a different resonance structure in each rectangle. If a molecule does not have enough resonance structures to fill the rectangles, then leave some blank. If there are only 3 possible resonance structures, one box will be blank.

O

O

5.) Here are some student answers. What is incorrect about some of these answers and why? a.)

b.)

c.)

d.)

e.)

f.)

6.) Draw all the resonance structures of the following molecule:

O

O

7.) Draw all the resonance structures of the following molecule:

HN

8.) Circle the most acidic H on each molecule. (It may be necessary to draw it in). Write the its pKa in the box below it.

NH2 O

O Cl H OH O O

N

9.) The Newman projection of the following molecule is given. On the flat cyclohexane next to it, draw in the missing chlorines and bromines:

Br Br H H

Cl Me H Cl Me

10.) Draw one resonance structure of the following molecule:

O N H

11.) How many types of carbon does the following molecule have? (Assume room temperature)