# Organic Chemistry I & II

Organic Chemistry I & II is designed for instructors who want an active, dynamic, and understandable approach to support their own efforts in the classroom. This ever-evolving textbook includes auto-graded questions, videos and approachable language in order to make difficult concepts easier to understand and implement.

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## Comparison of Organic Chemistry Textbooks

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### Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

### McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

### Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

### Pricing

Average price of textbook across most common format

#### $219 Hardcover print text only ####$301

Hardcover print text only

#### $301 Hardcover print text only ### Always up-to-date content, constantly revised by community of professors Content meets standard for Introduction to Organic Chemistry course, and is updated with the latest content ### In-Book Interactivity Includes embedded multi-media files and integrated software to enhance visual presentation of concepts directly in textbook Only available with supplementary resources at additional cost Only available with supplementary resources at additional cost Only available with supplementary resources at additional cost ### Customizable Ability to revise, adjust and adapt content to meet needs of course and instructor ### All-in-one Platform Access to additional questions, test banks, and slides available within one platform ## Pricing Average price of textbook across most common format ### Top Hat Steven Forsey, “Organic Chemistry”, Only one edition needed #### Up to40-60%more affordable Lifetime access on any device ### McGraw-Hill Carey & Giuliano, “Organic Chemistry”, 10th Edition ####$219

Hardcover print text only

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

#### $301 Hardcover print text only ### Wiley David R. Klein, “Organic Chemistry”, 3rd Edition ####$301

Hardcover print text only

## Always up-to-date content, constantly revised by community of professors

Constantly revised and updated by a community of professors with the latest content

### Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

### McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

### Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

## In-book Interactivity

Includes embedded multi-media files and integrated software to enhance visual presentation of concepts directly in textbook

### Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

### McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

### Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

## Customizable

Ability to revise, adjust and adapt content to meet needs of course and instructor

### Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

### McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

### Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

## All-in-one Platform

### Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

### McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

### Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

#### Dr. Steven Forsey, Ph.D.University of Waterloo

Steven Forsey is currently a Professor at University of Waterloo, teaching a variety of organic chemistry courses to Chemistry, Science, Chemical Engineering, Nanotechnology and distance education students. He received his Ph.D. (2004) for Synthetic Organic Chemistry from University of Waterloo, Ontario. He is a recipient of the Excellence of Science Teaching Award and has acted as the Teaching Fellow for the Department of Chemistry since 2016.

## Explore this textbook

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# Chapter 12: Alkenes

## Learning Objectives

• Evaluate a series of carbocations and rank their relative stability from most to least stable.
• Understand carbocation stability and how carbocation stability relates to Markovnikov’s rule.
• Recognize when a carbocation rearrangement will occur and determine the products formed in hydrohalogenation reactions.
• Recognize different reaction conditions needed for the hydration of alkenes and the dehydration of alcohols.
• Understand the mechanism and predict the products formed for addition reactions of alkenes, using reagents: HX, H2SO4/H2O, BH3⋅THF, Hg(OAc)2, X2 and X2/H2O.
• Understand how electronic and steric effects influence the addition of reagents to double bonds and predict the regiochemistry and stereochemistry of the products produced.
• Calculate oxidation states of carbon and predict the products produced in the oxidation and reduction of alkenes.
• Identify reagents needed to change alkenes into carboxylic acids, ketones, aldehydes and alcohols. (OsO4, KMnO4, H2/Pd, O3).
• Understand the mechanism of the addition of HBr with peroxides to an alkene and predict the products formed.
• Predict the intermediates and the final product in multi-stepped synthesis questions.

## 12.1 Definitions

Hofmann Rule: This rule refers to a special β-elimination reaction in which the alkene having the smallest number of alkyl groups attached to the double bonded carbon atoms will be the predominant product (less stable alkene forms as the major product). The Hofmann rule is observed in elimination reactions with leaving groups like quaternary ammonium salts, tertiary sulfonium salts, and with bulky strong bases (NaOC(CH3)3) reacting with tertiary alkyl halides.

Markovnikov Rule: In the ionic addition of a polar reagent to an unsymmetrical alkene, the positive portion of the reagent being added attaches itself to a carbon atom of the double bond forming the more stable carbocation as an intermediate. Older definition: the hydrogen becomes attached to the carbon atom of the double bond with the greater number of hydrogens (for addition of HX). It can also be stated; the hydrogen adds to the less substituted carbon.

Regiochemistry: A reaction that can occur with different regions on a molecule to produce different products. A region is defined as a site on a molecule where a reaction can occur. Regiochemistry is the result of the differences in the reactivity of various sites in the reactants.

Regioselective reaction: When there is more than one reactive site a regioselective reaction can occur. A regioselective reaction is one that produces one major product when many possible products are possible. For example, the molecule 2-bromo-2-methylbutane has eight hydrogens on β carbons that can undergo elimination reaction in the presence of a base such as sodium ethoxide. Removal of a Ha hydrogen produces the minor product 2-methyl-but-1-ene while removal of a Hb hydrogen produces the major product.

Stereoselective reaction: A reaction in which a single reactant can produce two or more stereoisomeric products and one of these products is preferred over another. For example, the reaction of either enantiomer of 2-bromobutane will stereoselectively produce (E)-but-2-ene as the major product. Notice that it does not matter which enantiomer the starting material is; the product ratio is the same.

Stereospecific reaction: A reaction in which a single reactant can produce two or more stereoisomeric products and one of these products is exclusively formed over the other(s). For example; the elimination reaction of (1R,2R)-1-bromo-1,2-diphenylpropane with a strong base produces the (Z) stereoisomer whereas the elimination of the (1R,2S) diastereomer produces the (E) stereoisomer.

Similarly, a SN2 reaction with 2-bromobutane and sodium methanethiolate (CH3SNa) will produce either (S)-sec-butyl(methyl)sulfane or (R)-sec-butyl(methyl)sulfane depending on which enantiomer you start with. This is another example of a stereospecific reaction

Zaitsev’s Rule: An empirical rule that states; when two or more alkenes can be produced in an elimination reaction, the thermodynamically most stable alkene will predominate. The most thermodynamically stable alkene will be the alkene that has the most alkyl groups attached to the alkene carbons.

## 12.2 Carbocation Stability

Before examining reactions with alkenes, we need to review carbocation stability because carbocation intermediates play an important role in determining the regiochemistry of the products.

Neighboring functional groups can stabilize carbocations through hyperconjugation, adjacent lone pairs, and adjacent π bonds. Carbocations can also be destabilized by nearby partial positive charges.

## 12.2.1 Hyperconjugation

The stability of a carbocation increases as the number of alkyl groups attached to the electron-deficient carbon increases. Review: Chapter 3.4.1: Effect of structure on the rate of reaction.

This can be explained by the electron donation of the adjacent C–C or C–H σ bonds into the empty carbocation’s p orbital, which stabilizes the electron-deficient carbon. Alkyl groups donate electron density inductively through σ bond conjugation or hyperconjugation.

Atoms adjacent to the carbonium ion that have lone pairs of electron (N, O, S, and halogens) will stabilize carbocations through resonance.

You may be thinking that electronegative atoms, such as oxygen and chlorine, would destabilize the carbocation due to their electronegativity; however, in most cases, resonance with the heteroatom has a greater influence than inductive effects.

Carbocations that are adjacent to π bonds (allyl, benzyl, and cyano) are stabilized through resonance, which delocalizes the charge onto different atoms.

## 12.2.4 Inductive Effects

Carbocations are destabilized by electronegative atoms that inductively create a partial positive charge adjacent to the carbocation.

Q12.1 - Level 1

Order the following carbocations from most stable to least stable.

A

1 > 2 > 3

B

2 > 3 > 1

C

3 > 1 > 2

D

2 > 1 > 3

E

1 > 3 > 2

Watch the following video to see the explanation to Q12.1.

## 12.3.1 General Reaction and Mechanism

Electrophilic addition of hydrogen halides involves the addition of a hydrogen halide across a double bond.

The standard mechanism showing electron pushing is given below.

Alkenes are weak nucleophiles; however, the weakly bound π electrons will react with strong electrophiles and an electrophilic addition will occur.

The first step in the mechanism is the electrophilic addition of a proton to the π bond to form a carbocation. This is a slow endothermic step and is also the rate-determining step.

The carbocation intermediate formed is an electrophile; in the second step, the halogen acts as a nucleophile and adds to the strongly electrophilic carbocation. The second step is a fast exothermic step and forms the product. The overall reaction is exothermic.

Notice, from the intermediate onward, the mechanism is the same as an SN1 reaction. Thus, the knowledge gained concerning carbocation stability, stereochemistry, and regiochemistry for SN1 reactions can be applied to the second half of this mechanism

## 12.3.2 Hydrohalogenation: Regiochemistry and Markovnikov Rule

When determining the regiochemistry of the products being formed it is very important to think about the transition state of the forming carbocation intermediate. The transition state will have a substantial positive charge developing on the carbon adjacent to the forming H–C σ bond.

Stability of transition state: The formation of the carbocation intermediate is an endothermic process and the transition state resembles the intermediate. Thus, when thinking about the stability of the transition state of an unsymmetrical alkene, a more stable transition state will have a positive charge forming on the carbon that stabilizes the partial positive charge to the greatest extent.

Q12.2 - Level 1

Which of the following transition states would be the most stable?

A

Transition State 1

B

Transition Sate 2

For help with these concepts: Review Hammond–Leffler postulate and transition states, Chapter 3.8, Chapter 8.4 and Chapter 10.4.

Electrophilic addition of hydrogen halides to differentially substituted alkenes is a regioselective process because of the difference in stability between the two transition states.

Let us explore why.

When 2-methylprop-1-ene is treated with HBr, two carbocation intermediates are possible.

Which intermediate will form more readily?

Remember in our discussion of kinetics vs. thermodynamics (Chapter 10.4), for most reactions we can say that the rate of the formation of the most stable product will be faster than the formation of the less stable product. This is because the most stable product (or intermediate in this reaction), follows a pathway that has a more stable, lower energy transition state. A lower energy transition state means a smaller energy barrier and a faster rate of reaction.

Thus, for the example presented, the rate of formation of the more stable tertiary carbocation will be faster than the rate of formation of the less stable primary carbocation. This leads to the formation of only one product: 2-bromo-2-methylpropane.

This phenomenon is known as Markovnikov’s Rule: the hydrogen becomes attached to the carbon atom of the double bond with the greater number of hydrogens (for addition of HX). It can also be stated that the hydrogen adds to the less substituted carbon.

The modern definition of Markovnikov’s Rule: In the ionic addition of a polar reagent to an unsymmetrical alkene, the positive portion of the added reagent attaches itself to a carbon atom of the double bond forming the more stable carbocation as an intermediate (you already know this concept from SN1 reactions; the most stable carbocation intermediate forms more readily).

Q12.3 - Level 2

What is the product produced in the following reaction?

A

1)

B

2)

Watch the following video to see the explanation to Q12.3.

As you can see from the above question, the modern definition of Markovnikov’s Rule is needed. However, these definitions do not need to be memorized as long as you understand the mechanism. When doing a question, draw the two carbocations that can form. Think about the stability of the intermediates. The most stable intermediate will go on to form the major product. Look in Section 2.0 to review carbocation stability and answer the following question.

Q12.4 - Level 1

What is the major product formed in the following reaction?

A

1)

B

2)

Watch the following video to see the explanation to Q12.4.

Q12.5 - Level 1

Click on the more substituted sp$^2$ carbons in each of the given alkenes. Click inside the circle.

Q12.6 - Level 2

What is the major product formed in the following reaction?

A

1)

B

2)

C

3)

D

4)

Q12.7 - Level 2

What is the major product formed in the following reaction?

A

1)

B

2)

C

3)

D

4)

## 12.3.3 Hydrohalogenation: Regiochemistry, Stereochemistry and Markovnikov Rule

The carbocation intermediate formed in the hydrohalogenation reaction is sp2 hybridized and planar; just as in an SN1 reaction, the nucleophile can attack from either face of the carbocation.

Q12.8 - Level 2

What is/are the major product(s) formed in the following reaction?

A

1)

B

2)

C

3)

D

4)

E

A racemic mixture of 1 and 2

F

A racemic mixture of 3 and 4

Watch the following video to see the explanation to Q12.8.

Now try the following questions.

Q12.9 - Level 1

Click on the sp$^2$ carbon(s) in the given molecules that will produce a chiral center in the major product when treated with HBr. (Think about the mechanism.) Click in the circle.

Q12.10 - Level 2

What is the stereochemistry of the major product(s) formed? Are they enantiomers, diastereomers, or is the major product achiral and only one product is produced?

Premise
Response
1

1)

A

Achiral only one product produced

2

2)

B

Diastereomers

3

3)

C

Enantiomers

## 12.3.4 Hydrohalogenation: Carbocation Rearrangement

Any time a 2˚carbocation intermediate is produced in a reaction, rearrangement of the intermediate may occur. This only occurs in 2° carbocations, since 1˚ carbocations won’t even form and 3˚ carbocations won’t rearrange (Chapter 8.3.5).

Q12.11 - Level 2

What is/are the major product(s) formed in the following reaction?

A

1)

B

2)

C

3)

Watch the following video to see the explanation to Q12.11.

## 12.4 Hydration: Acid-catalyzed

Hydration is the addition of the elements of water to a π bond.

Water is a weak acid and is not strong enough to react with an alkene. An acid, such as sulfuric (H2SO4) or nitric acid (HNO3), must be added as a catalyst. Shown in the mechanism below, sulfuric acid is added to the reaction mixture to generate hydronium ions. The hydronium ion is a strong electrophile and adds a proton to the alkene to generate a carbocation intermediate. In the next step, water attacks the electrophilic carbocation to form an oxonium ion (R–OH2+), which upon deprotonation regenerates the catalyst and produces an alcohol.

You may be asking, why doesn’t the hydrogen sulfate ion (HSO3) act as a nucleophile? The reason for this is that the hydrogen sulfate ion (HSO3-) is less nucleophilic than water and has a greater affinity for a proton than the carbon atom. Also, notice that all steps are reversible; thus, conditions must be used to favor the formation of the alcohol. Dilute acid solutions (containing water) are used to push the equilibrium to formation of the alcohol.

Notice that the hydration mechanism is the same as the hydrohalogenation reaction except the conjugate base of the acid used for the hydrohalogenation reaction (X-) is a good nucleophile, whereas in the hydration reaction the conjugate base of the acid (HSO4-) is less nucleophilic than water and water becomes the nucleophile.

Thus, the concepts learned in the hydrohalogenation reaction, such as stability of carbocations, carbocation rearrangements, and regio- and stereochemistry, can be applied to predict the products of the hydration reaction.

Q12.12 - Level 2

What is the major product formed in the following reaction?

A

1)

B

2)

C

3)

D

4)

Q12.13 - Level 2

What is the major product formed in the following reaction?

A

1)

B

2)

C

3)

D

4)

Q12.14 - Level 2

What are the major products formed in the following reaction?

A

B

3) and 4) are the major products

C

3) and 2) are the major products

D

1) and 2) are the major products

E

1) and 4) are the major products

Q12.15 - Level 2

Would the following reactions produce a pair of enantiomeric alcohols or an achiral product?

Premise
Response
1

1)

A

Enantiomeric alcohols produced

2

2)

B

Achiral alcohol produced

3

3)

C

Enantiomeric alcohols produced

D

Enantiomeric alcohols produced

E

Achiral alcohol produced

F

Achiral alcohol produced

## 12.5 Hydroboration-oxidation of Alkenes to Produce Alcohols: anti-Markovnikov Addition

3D Molecule*: 3-methylbut-1-ene

3D Molecule*: 3-methylbutan-1-ol

### 12.5.1 Reaction Overview: Regio- and Stereospecific Anti-Markovnikov Addition

The hydroboration–oxidation reaction occurs in two steps.

Step 1: The borane reagent is added to an alkene to form an alkylborane. This reaction is regioselective in that the boron atom almost exclusively adds to the least substituted carbon and the hydrogen atom to the most substituted carbon. Since this is the opposite of a Markovnikov addition, this is called an anti-Markovnikov addition. Boron then reacts two more times with two additional alkenes to form trialkylborane.

Step 2: Oxidation of the alkylborane with basic aqueous hydrogen peroxide (NaOH, H2O2, H2O) replaces the boron atom with a hydroxyl group.

Hydroboration–oxidation is an important two-step reaction because it adds the hydroxyl group to the least substituted carbon (anti-Markovnikov). It is also important because a carbocation is not formed, thus, rearrangements do not occur. Below is a comparison between a hydroboration–oxidation reaction and an acid-catalyzed hydration.

Below is a comparison between a hydroboration–oxidation reaction that does not undergo carbocation rearrangement and an acid-catalyzed hydration reaction that undergoes carbocation rearrangement.

The hydroboration reaction is also a stereospecific reaction. The borane reagent adds the hydrogen and the boron to the same side of the alkene in a syn addition reaction. Since the reagent can add to either face of the alkene, a pair of enantiomers is formed.

### Nobel Prize in Chemistry

Alkene functionalization reactions have led to several Nobel Prizes in Chemistry. One winner was H. C. Brown, who received the Prize in 1979 for pioneering hydroboration chemistry.

One notable example of Brown’s chemistry is in the synthesis of artemisinin developed by Hoffmann-La Roche.

Artemisinin is one of the most potent anti-malarial drugs available on the market and is currently accessed by isolation from plants and “semisynthesis”. Malaria is very prominent in poor countries and 1–3 million people die from malaria each year.

Click here for a review of the importance and synthesis of Artemisinin.

## 12.5.2 Hydroboration Mechanism

Boron has three valence electrons and can form three σ bonds to three other atoms. When bonded to three other atoms, boron only has six electrons in its outer shell. Thus, boron is very electrophilic because it wants to obtain a full octet. It is similar to a carbocation.

Because borane is highly electrophilic, free BH3 does not exist and spontaneously dimerizes to diborane B2H6.

Borane is commercially available in diethyl ether or tetrahydrofuran (THF). Diborane, when dissolved in THF, forms a borane-THF complex. For simplicity, when discussing the mechanism, the borane-THF complex is not shown.

As stated in Section 5.1, boron adds to the least substituted carbon and the hydrogen atom to the most substituted carbon in an anti-Markovnikov fashion. Does this go against everything we have learned so far? Thankfully, everything you have learned about electronic and steric interactions can be applied here. And, as always, by examining the transition states we will have a better understanding of why the reaction proceeds along the anti-Markovnikov pathway.

The mechanism is believed to be a one-step concerted reaction in which two new bonds are forming at the same time in the transition state. This is supported by the fact that only syn addition occurs because if it were a two-step mechanism in which a carbocation formed by a combination of syn and anti additions, then this would be expected (like acid-catalyzed hydration).

Borane is electron-poor and the alkene is electron-rich. It is believed that a Lewis acid–base interaction initially occurs to form a π complex, which then proceeds to form the transition state and then the products.

In the transition state, it is conceivable that the carbon–boron σ bond will be forming to a greater extent than the carbon–hydrogen σ bond (π bond electrons donating to empty p orbital). This interaction would create a partial negative charge on the boron atom and a partial positive charge on the carbon adjacent to the forming boron-carbon σ bond. Below are given the mechanisms that produce the Markovnikov and anti-Markovnikov products. Answer the following question.

Q12.16 - Level 1

Which transition state is more stable and why?

A

The transition state that produces the Markovnikov product is more stable because a partial positive charge is forming on tertiary carbon

B

The transition state that produces the Markovnikov product is more stable because a partial positive charge is forming on primary carbon

C

The transition state that produces the anti-Markovnikov product is more stable because a partial positive charge is forming on tertiary carbon

D

The transition state that produces the anti-Markovnikov product is more stable because a partial positive charge is forming on primary carbon

Another factor is steric interaction. There is less steric interaction in the transition state with the addition of boron to the primary carbon.

The combination of steric and electronic factors makes the activation energy to form the anti-Markovnikov product far smaller than the activation energy to form the Markovnikov product. Thus, by understanding the energies of the transition state we can see why the reaction follows an anti-Markovnikov pathway.

The combination of steric and electronic factors makes the activation energy to form the anti-Markovnikov product far smaller than the activation energy to form the Markovnikov product. Thus, by understanding the energies of the transition state we can see why the reaction follows an anti-Markovnikov pathway.

## 12.5.3 Oxidation and Hydrolysis of Alkylboranes: Mechanism

Oxidation and hydrolysis of the alkylborane to the alcohol takes place with retention of stereochemistry at the carbon bonded to boron (i.e., converts boron to a hydroxyl group, -OH). The mechanism of the alkylborane oxidation involves the nucleophilic attack of hydroperoxide ion on the electron-poor boron atom. The intermediate that forms is unstable and the alkyl group migrates with retention of configuration. This occurs two more times to form a trialkyl borate ester. The alkyl borate ester is then hydrolyzed by base to form the alcohol and sodium borate.

Mechanistically, the most important step is the alkyl group migration from the unstable intermediate while retaining its stereochemistry. This is an important step because when you are solving hydroboration–oxidation problems you need to only think about the syn anti-Markovnikov addition of BH3 to the alkene to form the trialkylborane. In the next step, you replace the boron atom with a hydroxyl group without worrying about changing the stereochemistry or regiochemistry.

Q12.17 - Level 1

The hydroboration mechanism involves the syn addition of borane across a double bond. Both electronic and steric factors play a role in where the boron atom bonds. Which drawing represents the most stable transition state?

A

1)

B

2)

C

3)

D

4)

Q12.18 - Level 1

Hydroboration of an alkene follows an anti-Markovnikov mechanism. Click on the carbon that would form the carbon-boron bond in the creation of the alkylborane intermediate, for each alkene.

Q12.19 - Level 2

What is the major product formed in the following reaction?

A

1)

B

2)

C

3)

D

4)

Q12.20 - Level 2

The hydroboration-oxidation of the given alkene produces:

A

Diastereomeric alcohols

B

Enantiomeric alcohols

C

A meso alcohol

D

An achiral molecule because the alcohol is bonded to an achiral carbon

Q12.21 - Level 2

What are the major products produced in the following reaction?

A

1) and 2)

B

3) and 4)

C

1) and 4

D

2) and 3)

E

All are produced

## 12.6 Oxymercuration–demercuration of Alkenes to Produce Alcohols: Markovnikov Addition

3D Molecule*: 3-methylbut-1-ene

3D Molecule*: 3-methylbutan-2-ol

## 12.6.1 Overview of Reaction: Regioselective and Stereospecific

An oxymercuration–demercuration reaction occurs in two steps.

Step 1 Oxymercuration: Mercury acetate is added to an alkene to form an alkylmercuric acetate. This reaction is regioselective in that the HgOAc group adds to the least substituted carbon and the hydroxyl group to the most substituted carbon. This reaction is stereospecific in that the HgOAc and hydroxyl groups are trans to each other. This is called an anti-addition.

Step 2 Demercuration: Reduction of alkylmercuric acetate with sodium borohydride replaces the mercury acetate group with a hydrogen atom. In Figure 12.38 the reduction of the alkylmercuric acetate enantiomers leads to only one product because no chirality centers are present.