Model 2: Aldohexoses, aldopentoses, ketohexoses and ketopentoses

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1 Model 1: D and L Sugars CHEM1405 Worksheet 13: Sugars and Amino Acids The simplest sugar is glyceraldehyde. This is a chiral molecule and the two enantiomers are shown opposite with wedges and dashes and as Fischer projections. To distinguish these isomers, chemists call them L-glyceraldehye and D-glyceraldehyde. By convention: D sugars have the OH group on the bottom chiral carbon on the right in the Fischer projection and L sugars have the OH group on the bottom chiral carbon on the left in the Fischer projection. Nearly all naturally occurring sugars have the D configuration. 1. Label the stereogenic carbon atoms in the structures above with an asterisk. 2. By using the convention, label the two forms above as either D-glyceraldehyde or L-glyceraldehyde. Five sugars are shown below. 3. Using the convention, add the appropriate D or L label to the name of each of these sugars. 4. Label all of the stereogenic carbon atoms in the structures with an asterisk. 5. Two forms are fructose are shown. What is the relationship between the stereogenic centres in the L and D isomers of a simple sugar? Model 2: Aldohexoses, aldopentoses, ketohexoses and ketopentoses Aldoses contain an aldehyde group. Ketoses contain a ketone group. Most common sugars have 5 or 6 carbons and so are called pentoses or hexoses. These terms are mixed to produce the names aldohexoses, aldopentoses, ketohexoses and ketopentoses. 1. Label each sugar above as an aldohexose, aldopentose, ketohexose or ketopentose.

2 Model 3: Cyclic structures and Haworth Projections As you saw in Worksheet 9, alcohols can react with carbonyls to form hemiacetals. The sugars in Model 1 contain both of these groups and can react with themselves to form 5 and 6 membered hemiacetal rings. These are the forms in which sugars usually exist. The cyclic structures are represented using Haworth projections. The formation of the cyclic structures from D-glucose is shown below with the Haworth projections on the right. In D sugars, the CH 2 OH group points upwards. In L sugars, the CH 2 OH group points downwards. 1. Starting at the carbonyl carbon, label the carbon atoms in the Fischer projection as C1, C2, C3, C4, C5 and C6. 2. Add these labels to the equivalent carbon atoms in the bent structure and to the Haworth projections. 3. Circle the OH group in the Fischer projection which ends up as the O in the ring structure. 4. By following what happens to the position of the other OH groups, complete the following: an OH group on the right in the Fischer projection is up / down in the Haworth structure an OH group on the left in the Fischer projection is up / down in the Haworth structure 5. By using your answers to the Q1 Q4, draw the Haworth structures * for the two sugars below. * 5 and 6 membered rings are possible for ribose.

3 Model 4: Amino Acids There are 20 different amino acids in nature. These all contain a carboxylic acid group (α-cooh) and a basic amino group (α-nh 2 ). They differ in the nature of the side chain, labelled R in the scheme below. As shown in the scheme, the form in which they exist depends on the ph of the solution they are in. In solutions with a ph near neutral, the zwitterionic form in the middle dominates. When the solution ph equals the pk a of a functional group, 50% of the molecules are protonated and 50% of the molecules are deprotonated. The deprotonated species predominates above the pk a value, whilst below the pk a value, the protonated form is the dominant species. 1. Write the overall charge of the ion underneath the 3 structures in the scheme above. 2. Draw the structure of the following amino acids as they would exist in the conditions stated. (a) Valine (R = -CH(CH 3 ) 2 in the stomach at ph 1.5. Data: pk a (α-cooh) = 2.29 and pk a (α-nh 3 + ) = (b) Serine (R = -CH 2 OH) in the small intestine at ph 8. Data: pk a (α-cooh) = 2.17 and pk a (α-nh 3 + ) = The zwitterion is the predominant species at or near the isoelectric point, pi. For amino acids such as those in Q2 which have neutral side chains, pi is equal to the average of pk a (α-cooh) and pk a (α-nh 3 + ). 3. At what ph is the zwitterionic form of serine the predominant species? Model 5: Amino Acids with Acidic and Basic Side Chains Seven amino acids contain side-chains that can also be protonated or deprotonated depending on the ph of the solution. Three have side-chains that contain basic groups, two have side-chains that contain carboxylic acid groups and two (tyrosine and cysteine) contain functional groups that can be deprotonated at high ph. The structure of histidine in the stomach is shown to the right. The R group contains a basic imidazole ring and pk a (α-cooh) = 1.77, pk a (α-nh 3 + ) = 9.18 and pk a (side-chain) = As stomach ph is lower than each of these values, each of these groups is protonated.

4 1. Draw the structure of histidine that predominates at a ph of Is it possible to protonate the side chain and not protonate the α-nh 2 group? If your answer is yes, what ph value would achieve this? If your answer is no, briefly explain why not. The structure of glutamic acid in the stomach is shown to the right. The R group contains a carboxylic acid group and pk a (α-cooh) = 2.10, pk a (α-nh 3 + ) = 9.47 and pk a (side-chain) = As stomach ph is lower than each of these values, each of these groups is protonated. 3. Draw the structure of glutamic acid that predominates at a ph of Is it possible to deprotonate the side chain and not deprotonate the α-cooh group? If your answer is yes, what ph value would achieve this? If your answer is no, briefly explain why not. The isoelectric point for amino acids with basic side chains, such as histidine, is the average of the two highest pk a values. The isoelectric point for amino acids with acidic side chains, such as glutamic acid, is the average of the two lowest pk a values. 5. What are the values of the isoelectric point for each of the two amino acids above? (a) histidine; pi = (b) glutamic acid; pi = 6. Calculate the pi value for the two amino acids below and draw the structure of their zwitterions. (a) Aspartic acid (R = -CH 2 COOH) Data: pk a (α-cooh) = 2.10, pk a (α-nh + 3 ) = 9.82 and pk a (side chain) = 3.86

5 (b) Lysine (R = -CH 2 CH 2 CH 2 CH 2 NH 2 ). Data: pk a (α-cooh) = 2.18, pk a (α-nh 3 + ) = 8.95 and pk a (side chain) = Model 6: Formation of Peptides The amino group of one amino acid and the acid group of another can undergo a condensation reaction to form a dipeptide containing the amide functional group. Further condensations can lead to tripeptides, tetrapeptides, etc and ultimately to polypeptides. The new bond that is formed is called a peptide bond or amide bond. The scheme below shows the condensation of glycine and alanine to make the dipeptide glycylalanine. 1. Draw a box around the part of glycylalanine that originated in glycine. 2. Draw a box around the part of glycylalanine that originated in alanine. 3. Draw an arrow to the peptide bond and label it. 4. Would the dipeptide alanylglycine (ala-gly) be same molecule? If not, draw its structure. 5. A tripeptide is shown below. Label the two peptide bonds and draw a circle around the side chains (R groups). What three amino acids condensed to make this tripeptide? Model 7: Reactions of Amino Acids and Peptides All amino acids and peptides contain a weakly basic amine group and a weakly acidic carboxylic acid group. These groups are protonated or deprotonated according to the ph as discussed in Model 1. The

6 peptide bond is extremely stable and can only be broken by using hot, concentrated acid or hot, concentrated base. Under these conditions, the condensation reaction is reversed and the peptide is hydrolysed back to the amino acids from which it was made. The scheme below shows this for gylcylalanine. 1. Draw an arrow to the peptide bond in glycylalanine and label it. 2. Why are the amine groups in the products not protonated? 3. Draw the products of the hydrolysis reactions below. (a) (b)