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Worksheet 14

CH 235-2F

November 5, 2014

Short Answer:

1. Define elimination. What are the two elimination reactions you have learned?

Elimination reactions involve the elimination of a hydrogen to form an alkene. The two elimination reactions you have learned are E1 and E2.

2. Draw a general E1 mechanism.

3. Draw the reaction energy diagram for an E1 mechanism. What is the rate law?

Rate=k[RX]

4. What is the stereochemical outcome of an E1 reaction? How do you determine the major/minor product?

Due to the planar cation intermediate, a hydrogen can be removed from either the same face as the cation or the opposite face. Because of this, both the cis and trans product are possible. The trans alkene product is more stable than the cis. Also, remember that the cation can rearrange if it will become more stable and this will lead to multiple products. Another way you can get multiple products is if there is more than one site that the hydrogen can be taken off of. The more stable alkene is the one that is most internal (most substituted).

5. Draw a general E2 mechanism.

6. Draw the reaction energy diagram for an E2 reaction. What is the rate law?

7. Knowing that SN2 competes with E2, what are some ways that you can favor one reaction over the other?

8. What is the required geometry for an E2 reaction? Why?

In order to undergo an E2 reaction, the alkyl halide must be able to adopt anticoplanar geometry where the H being removed and the leaving group and 180 degrees from each other. This geometry is necessary because it allows for maximal orbital overlap. This lowers the energy of the transition state and allows for easier bond breaking/bond formation.

On a Newman Projection, anticoplanar geometry is when the H and Br and anti to each other. On a chair conformation, anticoplanar geometry occurs when both the H and the leaving group and on side by side carbons and in axial positions.

9. For the molecules shown below: 1) Draw the two chair conformations, 2) Give the alkene product(s) you would expect to see if it undergoes an E2 reaction (If no reaction occurs, explain why) 3) Decide which will undergo E2 faster and explain why.

B A

See attached photo on my website. PLEASE NOTE THIS IS A LITLE DIFFERENT THAN THE ONE I DID IN MY SESSION.

Molecule A:

1. The chair conformations are shown in the picture on my website. Only the chair conformation with the leaving group in the axial position (1A) can do E2.

2. Shown on the picture.

3. Molecule B will undergo E2 faster because it’s more stable chair conformation can undergo E2 whereas with molecule A it’s less stable chair conformation does E2.

Molecule B:

1. The chair conformations are shown in the picture on my website. Only the chair conformation with the leaving group in the axial position (2B) can do E2.

2. Shown on the picture.

3. Molecule B will undergo E2 faster because it’s more stable chair conformation can undergo E2 whereas with molecule A it’s less stable chair conformation does E2.

Multiple Choice:

1. Given the reaction shown below, which is/are the possible products?

Cl2, Δ

A. 1-chloro-2-methylethane

B. 2-chloro-2-methylethane

C. 2-chloro-3-methylethane

D. 1-chloro-3-methylethane

E. All are possible products.

2. Which of the following statements are false about the following reaction?

CH3OH

I. The reaction will undergo an E1 reaction.

II. The reaction will undergo an SN1 reaction.

III. There are two possible major products (counting a racemic mixture as 1).

IV. The rate law for this reaction is Rate=k[RX][Nuc]

A. I and II only.

B. III only.

C. I, II, III, and IV

D. I, II, III

E. IV only.

3. At which of the indicated carbons will the alkyl halide undergo SN2 and E1/SN1, respectively?

1 2

A. 1, 2

B. 2, 1

C. Neither will undergo SN2 or SN1/E1

D. Both will undergo SN2 and SN1/E1 equally

E. 1, neither will undergo E1/SN1

Explanation to Multiple Choice:

1. This is a radical chlorination reaction. Radical chlorination adds a chlorine at every carbon that can form a radical.

2. The alkyl halide shown has a good leaving group. It is a tetrahedral carbon so it can do E1/SN1 and POSSIBLY E2. To decide if it can do E2, look at the nucleophile. In this case, the nucleophile is an alcohol. This is a weak base and a good nucleophile to be used for Sn1 and E1. It is not a good nucleophile for E2. Remember, for E2 you want a strong base. Both I and II are true because E1 and Sn1 compete with each other. III is correct because there is one product from the SN1 reaction and a set of enantiomers for the major product of E1. IV is false because it gives the rate law for E2, which we determined canot happen under these conditions.

3. Both carbon 1 and carbon 2 are primary. The number 1 will undergo SN2 faster because that end of the molecule is less cluttered (the other end has a methyl group nearby). Neither will do E1/SN1 because they are primary carbons.