Equilibrium between Cyclic and Linear Sugars

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1 Equilibrium between Cyclic and Linear Sugars Outline Nucleophiles and Electrophiles Cyclic Forms of Glucose Thermodynamics Homework Linear glucose is in equilibrium with four cyclic forms. All five of these isomers are present in any solution of this sugar. Nucleophiles and Electrophiles Electrophiles and Lewis Acids An electrophile is a molecule that forms a bond to its reaction partner (the nucleophile) by accepting both bonding electrons from that reaction partner. Electrophilic reagents are Lewis acids. You've already seen that carbon dioxide is an electrophile. Below are some examples of electrophiles. Chemistry 104 Prof. Shapley page 1

Nucleophiles and Lewis Bases A nucleophile is a molecule that forms a bond with its reaction partner (the electrophile) by donating both electrons for that bond. Nucleophiles are Lewis bases.

2 Nucleophiles and Lewis Bases A nucleophile is a molecule that forms a bond with its reaction partner (the electrophile) by donating both electrons for that bond. Nucleophiles are Lewis bases. As you've seen, hydroxide is an example of nucleophile that adds to carbon dioxide. Below are some examples of nucleophiles. Addition of Nucleophiles to the Carbonyl Group Hydroxide, a nucleophile, adds to the electrophilic carbon of CO2. The purple arrows indicate the migration of a pair of electrons. First a pair of electrons moves from the nucleophile (hydroxide) to the electrophile (carbonyl carbon). Because carbon can't have more than 8 total electrons around it, a pair of electrons associated with the C=O unit must move to the oxygen atom. Similarly, alcohols are nucleophiles that can add to the electrophilic carbon of an aldehyde. Chemistry 104 Prof. Shapley page 2

After the addition of the nucleophile to the electrophile, there is an acid/base reaction. The acidic proton on the positively charged oxygen transfers to the basic, negatively charged oxygen.

3 After the addition of the nucleophile to the electrophile, there is an acid/base reaction. The acidic proton on the positively charged oxygen transfers to the basic, negatively charged oxygen. Cyclic Forms of Glucose Internal Nucleophilic and Electrophilic Sites Glucose includes an aldehyde group with an electrophilic carbon. Call this carbon C-1. It is also a polyalcohol. Each of the other carbon atoms is an alcohol carbon and bears an OH group. These are nucleophiles. When a nucleophile and an electrophile are in the same molecule, they react together much more quickly than separate nucleophilic and electrophilic molecules do. Take a look at the linear form of glucose below. If the OH group on C-4 adds to the aldehyde carbon it will make a 5-membered ring, and if the OH group on C-5 adds to C-1 it will make a 6-membered ring. Rings smaller than 5 atoms or larger than 6 atoms are not very stable so the OH groups on C-2, C-3, and C-6 don't react with the aldehyde carbon. Glucose is a pretty complicated molecule with all of its functional groups. To see the reaction more clearly, The scheme below shows only the reactive OH group on C-5 and the aldehyde group. Chemistry 104 Prof. Shapley page 3

4 All the carbon atoms are free to rotate around the C-C bonds in the linear molecule and they do this rapidly. In one of the conformations that places the oxygen near the C-1 carbon, a O-C bond can form. Both of the electrons are donated from the nucleophilic oxygen atom. Because carbon can have a maximum of 8 electrons, when the OH group donates a pair of electrons, electron density must shift to the carbonyl oxygen. This oxygen is then very basic and it can abstract the acidic proton from the former alcohol oxygen atom. Adding to the Two Sides of the Carbonyl Group In the scheme above, the new OH group formed from the aldehyde is on the same side as the R group (CH2OH). Another isomer results from addition of the C-5 OH group to the other side of the flat CHO group. Forming the 5-Membered Ring Isomers To form the 5-membered ring isomer of glucose, the nucleophilic OH group on C-4 is the one that adds to the electrophilic carbon at C-1. After the oxygen donates its electron pair to form the new bond, there is a proton transfer between an acidic site and a basic site. Chemistry 104 Prof. Shapley page 4

5 When the OH group attached to C-4 adds to the other side of the carbonyl (CO) group, another isomer is formed. Thermodynamics Enthalpy Changes: Analysis Based on Bond Energy The linear form of glucose is in rapid equilibrium with 4, different cyclic forms in solution. Anytime chemical species are in rapid equilibrium, the concentration of each of these species depends on its relative energy. Let's look at one of those reactions. Chemistry 104 Prof. Shapley page 5

On the left, you can see the linear form of glucose. This molecule has an aldehyde C=O on the first carbon (C1) of the chain and an alcohol OH group on all of the other carbons.

6 On the left, you can see the linear form of glucose. This molecule has an aldehyde C=O on the first carbon (C1) of the chain and an alcohol OH group on all of the other carbons. The oxygen bonded to C4 can use one of its lone pairs of electrons to form a bond with the electron-poor carbon of C1. This gives the intermediate. Finally, a proton is transferred from the oxygen on C4 to the oxygen on C1 to give the product. What is the energy difference between the linear reactant and the cyclic product? The first things to consider are the number and type of bonds that are broken (requires energy, positive ΔH) and the number and types of bonds that are formed (releases energy, negative ΔH). 1 O-H bond broken 1 O-H bond made 1 C=O bond broken 2 C-O bonds made The O-H bond make and breaking cancel each other out. The 2 C-O bonds made release a little more energy than breaking the C=O requires. Based on bond energies, we would expect that the energy difference between the linear and cyclic forms should be small but should favor the cyclic form. Enthalpy Changes: Steric Strain Steric interactions, or repulsive interactions between electron clouds that are too close together, increase the energy of molecules. Consider the simple molecule ethane, CH3CH3. The molecule is free to rotate around the C-C bond but it spends most of it time in the lower energy staggered conformation. This form minimizes the repulsive interactions between the hydrogen atoms on the adjacent carbon atoms. The staggered form is more stable than the eclipsed form by 12.5 kj/mol. Chemistry 104 Prof. Shapley page 6

7 If we look down any of the C-C bonds in the cyclic sugar with a 5-membered ring, we can see that the substituents above and below the ring line up. This eclipsing strain raises the overall energy of this form of the sugar. The cyclic isomers with 6-membered rings don't have this eclipsing interaction and so should be more stable (lower energy) than the isomers with 5-membered rings. So based on bond energies and steric interactions, we would expect that the 6-membered ring form of the cyclic sugar should be favored. Entropy Changes Remember that the tendency for a reaction to proceed as written depends on the change in Gibbs free energy. ΔG = ΔH - TΔS What about entropy? Do you think a cyclic structure or a linear form has greater entropy? Think about the ability of the molecule to rotate freely around its bonds. Observations At room temperature in water, all five forms of glucose are rapidly interconverting. The two 6-membered ring forms have the greatest concentration and the two 5-membered ring forms have the lowest concentration. Chemistry 104 Prof. Shapley page 7

Calculating Gibbs Free Energy We can use the equilibrium concentrations of the five isomers at room temperature to calculate the ΔGrxn. Remember that ΔGrxn depends on the equilibrium constant, Keq.

8 Calculating Gibbs Free Energy We can use the equilibrium concentrations of the five isomers at room temperature to calculate the ΔGrxn. Remember that ΔGrxn depends on the equilibrium constant, Keq. ΔGrxn = - R T ln(keq) The equilibrium constant for each of these reactions is the ratio of product concentration to reactant concentration. Temperature in Kelvin is 300 and the gas constant, R, is 8.31 J K -1 mol -1. Chemistry 104 Prof. Shapley page 8

9 Chemistry 104 Prof. Shapley page 9

Chapter 6 An Overview of Organic Reactions

John E. McMurry www.cengage.com/chemistry/mcmurry Chapter 6 An Overview of Organic Reactions Why this chapter? To understand organic and/or biochemistry, it is necessary to know: -What occurs -Why and