CHEM1239 Organic Chemistry Exam (PDF)

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Summary

This is an RMİT exam paper for CHEM1239, focusing on organic chemistry topics including carbohydrates and proteins. The exam has various questions requiring explanations, drawings, and calculations related to specific examples.

Full Transcript

CHEM 1239 Chemistry for Food and Life Science CHEM1239: Exemplar assessment (Organic Chemistry) SCHOOL OF SCIENCE You must submit this assessment within the allocated...

CHEM 1239 Chemistry for Food and Life Science CHEM1239: Exemplar assessment (Organic Chemistry) SCHOOL OF SCIENCE You must submit this assessment within the allocated period. Please read this paper carefully and ensure that you leave sufficient time for checking your answers. To save time, where possible use the structures already provided and then just write your answers onto this copy of the assessment paper. PLEASE READ THE QUESTIONS CAREFULLY AS YOU ARE OFTEN ONLY ASKED TO ANSWER SOME PARTS. DO NOT ANSWER MORE PARTS THAN NECESSARY. YOU MUST SUBMIT THIS ASSESSMENT PAPER AT THE COMPLETION OF THE ASSESSMENT AND SIGN THE COVER SHEET Attempt all questions and answer these in the script book provided. Name: Student Number: CHEM 1239 Chemistry for Food and Life Science ORGANIC CHEMISTRY Please attempt ALL questions. 1. CARBOHYDRATES (a) D-Ribose is a typical monosaccharide. The Fischer CHO structure for the open chain form is given to the right: H OH H OH D-ribose H OH CH2OH (i) Explain why it is a D sugar (ii) Draw the Fischer structure of: (iii) Draw the Fischer structure of: N-acetyl-2-amino-2-deoxyribose altrose, the C-3 epimer of D-ribose (iv) Give the products of the reaction between (v) Give the organic product of the reaction D-ribose and Ag(NH3)2+ (Tollens between D-ribose and HNO3. reagent). This reaction with Tollens reagent is used as a test for reducing sugars; what do you observe in a positive test? This is a redox reaction; what is being reduced, and what is being oxidised? CHEM 1239 Chemistry for Food and Life Science 1. CARBOHYDRATES, CONTINUED… (b) D-Mannose is a typical monosaccharide. The Fischer structure for the open chain form is given to the right: CHO HO H HO H H OH H OH CH2OH (i) Draw the Haworth (ring) structure of α-D-mannose (α-D-mannopyranose). (ii) Go onto give the chair structure of α-D-mannose (α-D-mannopyranose). (iii) Give the structure of the disaccharide formed between two molecules of D-mannopyranose, joined by a β-1,4-glycoside bond. ([4 + 4 + 4 + 8 + 4] + [4 + 6 + 6] = 40 marks) CHEM 1239 Chemistry for iFood and Life Science QUESTION 2. PROTEINS Below are the structures of glycine and lysine, and their pKa data: NH2 CH2 COOH NH2 CH COOH (CH2)4NH2 Glycine (gly) Lysine (lys) The pKa and pI values for the amino acids above are: pKa1 pKa2 pKa3 pI acid amine side-chain glycine (gly) 2.3 9.2 – 6.7 lysine (lys) 2.2 8.9 10.5 9.7 (a) Draw the structure for the major form of lysine that would be found: (i) at very low pH (less than pH = 1); (ii) at very high pH (more than pH = 12). (b) Using the structures of the amino acids above, draw the structure of the tripeptide glycylglycyllysine acid (gly-gly-lys), showing the appropriate charges assuming the compound is in solution at pH 7. Identify the N-terminal and C-terminal amino acids, and the peptide bonds. CHEM 1239 Chemistry for Food and Life Science QUESTION 2. PROTEINS, CONTINUED… (c) (i) Define "secondary structure" as it applies to proteins. (ii) Define "prosthetic group" as it applies to proteins. (d) Describe briefly how amino acids with polar side chains, such as serine (right), contribute to the overall shape of a protein. NH2 CH COOH CH2OH Serine ([2.5 + 2.5] + 10 + [5 + 5] + 5 = 30 marks) CHEM 1239 Chemistry for Food and Life Science QUESTION 3. LIPIDS (a) Consider the following data for selected fatty acids: Common name Carbons Double Structure mp bonds (Note: all double bonds are cis) (°C) myristic acid 14 0 COOH 54 palmitic acid 16 0 COOH 64 palmitoleic 16 1 COOH 0 acid (i) Explain the difference in melting point between myristic acid and palmitic acid. (ii) Explain the difference in melting point between palmitic acid and palmitoleic acid. (iii) Draw the structure of glyceryl trimyristate, the triester of glycerol with myristic acid. (iv) Give the reaction for the saponification of glyceryl trimyristate with NaOH. (v) Describe the nature of the solution which results when sodium myristate is dissolved in water. Further, describe how a solution of a soap (such as sodium myristate) cleans greasy or oily items. Use diagrams to assist your discussion wherever possible. CHEM 1239 Chemistry for Food and Life Science QUESTION 3. LIPIDS, CONTINUED… (b) (i) Lipids are a diverse groups of compounds. What common property do they have which allows them to be classed together? (ii) Emulsifying agents are used to stabilise emulsions - mixtures of liquids that are normally immiscible (that is, don't normally mix), for example, oil and water. Explain why a compound such as a lecithin, below, acts as an emulsifying agent CHEM 1239 Chemistry for Food and Life Science QUESTION 3. LIPIDS, CONTINUED… (c) Myristoleic acid (C14H26O2) is an uncommon mono-unsaturated fatty acid. Its melting point is –4°C and it occurs in marine animal fats like whale blubber, shark liver, eel and turtle or, in low quantities, in terrestrial animal fats like cow milk and butter fats. COOH (i) Give the structure of glyceryl trimyristoleate (the triester of glycerol and myristoleic acid). (ii) Calculate the iodine number of glyceryl trimyristoleate. Use the following molecular weights (in g mol-1) for your calculations: C = 12, O = 16, H = 1, I = 126.9. ([1 + 1 + 2 + 3 + 3] + [5 + 5] + [5 + 5] = 30 marks) Total marks for this paper: 100 marks

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