Adding H and OH, Markovnikov

Careful comparison of the starting material and prodnct will reveal that we have performed a hydration, with Markovnikov regiochemistry. Notice the reagent that we used (H3O+). Essentially, this is water (H2O) and an acid sonrce (such as sulfuric acid). There are many ways to show this reagent. Sometimes, it is written as H3O+ (as above), while at other times, it might be written like this H2O, H+. You might even see it like this, with brackets around the acid  

These brackets indicate that H+ is not consumed in the reaction. In other words, H+ is a catalyst, and therefore, we call this reaction an acid-catalyzed hydration. In order to understand why this reaction proceeds via a Markovnikov addition, we turn our attention to the mechanism. The proposed mechanism of an acid-catalyzed hydration 

In each mechanism above, the first step involves protonation of the alkene to form a carbocation. Then, in both cases, a nucleophile (either X or H2O) attacks the car-bocation to give a product. The difference between these two reactions is in the nature of the product. The first reaction above (hydrohalogenation) gives a product that is neutral (no charge). However, the second reaction above (hydration) produced a charged species. Therefore, one more step is necessary at the end of the hydration reaction— we must get rid of the positive charge. To do this, we simply deprotonate  

Notice what reagent we use to pull off the proton. We use H2O, rather than using a hydroxide ion (HO ). To understand why, remember that we are in acidic conditions there really aren t many hydroxide ions floating around. But there is plenty of water, and a mechanism must always be consistent with the conditions that are present. 

Now that we have seen all of the individual steps, let s look at the entire mechanism  

Over the next two sections, we will learn how to add H and OH across a double bond. The process of adding water across an alkene is called hydration, and it can be done through a Markovnikov addition or through an anti-Markovnikov addition. We just need to carefully choose our reagents. In this section, we will explore the reagents that give a Markovnikov addition of water. Then, in the next section, we will explore how to do an anti-Markovnikov addition of water. 

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Answer This reagent (H3O+) suggests that we have an acid-catalyzed hydration. Therefore, we are adding H and OH, and the regiochemistry will follow a Markovnikov addition. The stereochemistry of an acid-catalyzed hydration is only complex when two new stereocenters are formed. In this case, we are not forming two new stereocenters. In fact, we are not even forming one new stereocenter. Without any stereocenters, we expect only one product ... 

A quick glance at the products indicates that we are adding H and OH across the alkene. Let s take a closer look and carefully analyze the regiochemistry and stereochemistry of this reaction. The OH is ending up on the less substituted carbon, and therefore, the regiochemistry represents an anti-Markovnikov addition. But what about the stereochemistry Are we seeing a syn addition here, or is this anti addition ... 

For the regiochemistry, we notice that the boron ends up on the less substituted carbon (and that is where the OH group will ultimately end up). Now we can understand one of the sources of this regiochemical preference. We are adding H and BH2 across the double bond. BH2 is bigger and bulkier than H, so it will have an easier time fitting over the less substituted carbon (the less sterically hindered position). Therefore, we get an antf-Markovnikov addition. 

ANSWER In this case, H and OH are added across the pi bond in an anti-Markovnikov addition. No stereocenters were formed, so the stereochemical outcome is not relevant. We have only seen one way to achieve an anti-Markovnikov addition of water across a pi bond hydroboration-oxidation. Therefore, our answer is ... 

The hydroboration-oxidation procedure is a valuable method to hydrate an alkene with anti-Markovnikov orientation and with syn addition of the H and OH groups. ° Addition of BH3 (which may be added to the reaction mixture as diborane, B2Hg) to an alkene occurs readily in diethyl ether, THF, or similar solvent. The hydroboration is strongly exothermic, with a AH of -33kcal/mol per B-H bond that reacts. If stoichiometry and the steric requirements of the alkyl substituents on the boron atom permit, the reaction proceeds until three alkyl groups are attached to each boron atom. The trialkylborane can then be oxidized with hydrogen peroxide in aqueous base to produce the alcohol. 

The next few sections discuss three methods for adding water (H and OH) across a double bond, a process called hydration. The first two methods proceed through Markovnikov additions, while the third method proceeds through an < r/-Markovnikov addition. In this section, we will explore the first of these three reactions. 

The problem statement indicates a hydroboration-oxidation. The net result of this two-step process is the anfi -Markovnikov addition of H and OH across the tc bond. That is, the OH group is placed at the less-substituted position, while the H is placed at the more substituted position. In this case, two chirality centers are created. Therefore, the stereochemical requirement for syn addition determines that the H and OH are added on the same face of the alkene, giving the following products ... 

In the absence of a catalyst, water does not react with alkenes. But, if an acid catalyst such as sulfuric acid is added, water adds to carbon-carbon double bonds to give alcohols. In this reaction, which is called hydration, a water molecule is split in such a way that —H attaches to one carbon of the double bond, and —OH attaches to the other carbon. In Reactions 2.9-2.11, HjO is written H—OH to emphasize the portions that add to the double bond. Notice that the addition follows Markovnikov s rule ... 

Addition of water (H—OH) to alkenes also occurs according to Markovnikov s rule. Hydrogen (— H) of water is added to the carbon atom of the double bond that already has the greater number of hydrogen atoms. —OH is added to the carbon atom of the double bond that has the smaller number of hydrogen atoms. In this reaction, H2SO4 is used as catalyst, and a monoalcohol is produced. 

There are three possible products when NOCl is added to alkenes, a (3-halo nitro-so compound, an oxime, or a (3-halo nitro compound.The initial product is always the (3-halo nitroso compound, but these are stable only if the carbon bearing the nitrogen has no hydrogen. If it has, the nitroso compound tautomerizes to the oxime, H—C-N=0 C=N-OH. With some alkenes, the initial (3-halo nitroso compound is oxidized by the NOCl to a (3-halo nitro compound. Many functional groups can be present without interference (e.g., COOH, COOR, CN, OR). The mechanism in most cases is probably simple electrophilic addition, and the addition is usually anti, although syn addition has been reported in some cases.Markovnikov s mle is followed, the positive NO going to the carbon that has more hydrogens. 

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