Adding H and

In this type of reaction, called hydrogenation, the regiochemistry will always be irrelevant, regardless of what alkene we use (we are adding two of the same group). However, we do need to explore the stereochemistry of hydrogenation reactions. In order to do this, let s take a close look at how the reaction takes place. 

Notice the reagents that we use to accomplish a hydrogenation reaction (H2 and a metal catalyst). A variety of metal catalysts can be used, such as Pt, Pd, or Ni. The hydrogen molecules (H2) interact with the surface of the metal catalyst, effectively breaking the H—H bonds  

This forms individual hydrogen atoms adsorbed to the surface of the metal. These hydrogen atoms are now available for addition across the alkene. The addition reaction begins when the alkene coordinates with the metal surface  

Surface chemistry then allows for the following two steps, effectively adding H and H across the alkene  

Notice that the alkene grabs both hydrogen atoms on the same side of the alkene. Therefore, we get a syn addition. 

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Draw the products that you would expect when adding H and OH in the following way ... 

Answer HBr indicates that we will be adding H and Br across the double bond. The presence of peroxides indicates that the regiochemistry will be anti-Markovnikov. To determine whether stereochemistry is relevant in this particular case, we need to look at whether we are creating two new stereocenters. When we place the Br on the less substituted carbon (and the H on the more substituted carbon), we will only be creating one new stereocenter. With only one stereocenter, there are not four possible stereoisomers but just two possible products (a pair of enantiomers). And we will get this pair of enantiomers regardless of whether the reaction was syn or anti ... 

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. 

The hydrocarbalkoxylation of olefins with palladium catalysts has been known for a long time and has been reviewed adequately (14, 15). Adding H- and -COOR to the olefin occurs with a cis-stereochemistry (16) the most likely mechanism proposed up to now is reported in Scheme III (17). A complete discussion of the mechanism will be published elsewhere. 

Other illustrative examples of carbanionic ion-pair dissociation/aggregation are lithium triphenyhnethide, which exists as a tight ion pair in diethyl ether and as a solvent-separated ion pair in tetrahydrofuran, as shown by UV/Vis spectrophotometric measurements [287], and lithium 10-phenylnonafulvene-10-oxide, which exists as a tight ion pair (2b) in tetrahydrofuran solution and as a solvent-separated ion pair (3b) when hexamethylphosphoric triamide or dimethyl sulfoxide are added ( H and NMR measurements) [288]. 

A certain buffer is made by dissolving NaH2P04 and Na2HP04 in water. Write equations to show how this buffer neutralizes added H+ and OH-. 

The theory of ideal thermochromatography suggests that, at any point, there exists thermodynamical (adsorption) equilibrium between the gas-phase and surface concentrations of the adsorbable molecules. The ideal elution TC, like the ideal IC of nonradioactive species, would yield a narrow band at some 7 d - a function of A ads H and Aac s.S, as well as of the processing time and temperature profile. Figure 4.1 illustrates different situations. 

Using a CNDO/2 model the energies for adding H+ and H io la have been calculated No geometry optimization of these reaction products was made. Therefore these results have also a more or less qualitative character. It is found that the 3 position can be seen to be the preferred position for electrophilic attack. It is predicted that it wiU take place as facile as with benzene. 

Cation exchangers, which are likely to improve tartrate stability by removing K+ and Ca +, acidify wine by adding H" and, possibly, prevent ferric casse by reducing Fe +. 

Hemoglobin (HHb) is a weak acid that forms the Hb anion. Write equations to show how the HHb/Hb buffer would react to added H and added OH . 

Complete the following table by adding -H and — signs to the AH° and columns so as to correspond to the effect of temperature on a reversible reaction. 

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