1 / 176

Chemistry and Food

Chapter 47. Chemistry and Food. 47.1 Principal Components of Food 47.2 Food Preservation 47.3 Food Additives. 47.1 Principal Components of Food (SB p.279). 6 basic ingredients supply from a nutritional diet: Proteins Carbohydrates Fats Minerals Vitamins Water.

zoe-stark
Download Presentation

Chemistry and Food

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 47 Chemistry and Food 47.1Principal Components of Food 47.2Food Preservation 47.3Food Additives

  2. 47.1 Principal Components of Food (SB p.279) • 6 basic ingredients supply from a nutritional diet: • Proteins • Carbohydrates • Fats • Minerals • Vitamins • Water

  3. 47.1 Principal Components of Food (SB p.279) Essential food components serve 3 main functions: 1. Provide energy for the essential biological processes to take place 2. Provide materials for building and repairing body tissues 3. Provide materials need for regulation of life processes such as the transport of oxygen and digestion

  4. Food provides energy for the athletes to run 47.1 Principal Components of Food (SB p.279) • Fats andcarbohydrates supply most of the energy needs • Proteinssupply most of the building blocks for tissue • Trace amount of vitamins andminerals acts asregulators of vital body processes

  5. 47.1 Principal Components of Food (SB p.280) Proteins Importance of Proteins in Our Body

  6. 47.1 Principal Components of Food (SB p.280) Some functions of proteins in our body (cont’d)

  7. Proteins are needed for building muscles Food rich in proteins 47.1 Principal Components of Food (SB p.280)

  8. 47.1 Principal Components of Food (SB p.281) Structure of Proteins • Proteins are organic compounds of large molecular masses • In addition to C, H, O, they always contain N, usually S and sometimes P • The basic structural units of proteins is amino acids • Amino acid is a group of chemical and around twenty of them occurred in proteins

  9. 47.1 Principal Components of Food (SB p.281) • Amino acid contains a basic amino group (–NH2) and an acidic carboxyl group (–COOH) • Variation in the structure of R gives rise to the 20 amino acids which are the building blocks of different proteins

  10. 47.1 Principal Components of Food (SB p.281) • Common proteins are considered as macromolecules of amino acids • Different proteins are formed when the 20 different amino acids are linked up inspecific sequences

  11. 47.1 Principal Components of Food (SB p.281) • The first step in the synthesis of a protein molecule is the condensation reaction between two amino acidswith a water molecule being eliminated

  12. The molecule formed from two amino acids is called dipeptide • The group is called amide group or the peptide linkage 47.1 Principal Components of Food (SB p.282) • Either ends of the peptide can undergo condensation reaction with another amino acid to form a tripeptide • Further combinations with amino acids extend the length of the chain to form a polypeptide

  13. 47.1 Principal Components of Food (SB p.282) Hydrolysis of Proteins • The peptide linkages in the protein molecules are broken when it hydrolyzed by mineral acids or enzymes • On complete hydrolysis, the protein molecule is broken down into its constituent amino acids

  14. 47.1 Principal Components of Food (SB p.282) Separation of Amino Acids by Paper Chromatography As each amino acid has its own adsorption ability on a piece of paper, they can be separated and identified by means of paper chromatography 1. In paper chromatography, a piece of filter paper is used as the adsorbent 2. A capillary tube is used to spot the paper with a solution of the mixture of amino acids (U) to be analyzed 3. By using different capillary tubes, a number of solutions of known amino acids (A to F) are also spotted for comparison

  15. 47.1 Principal Components of Food (SB p.282) 4. In order to achieve a good separation, the spots should be placed along the pencil line near the bottom of the paper 5. When the paper become dry, the paper is then rolled into a cylindrical tube with the ends stapled together

  16. 47.1 Principal Components of Food (SB p.283) • 6. The paper is then placed into a beaker containing a small amount of a suitable solvent such as a mixture of NH3 and propan-2-ol • The spot must be above the initial level of the solvent, otherwise they will be removed completely from the paper • 8. The beaker is covered tightly with a piece of aluminium foil, so that the air inside is saturated with the solvent vapour

  17. 47.1 Principal Components of Food (SB p.283) • 9. The solvent moved up the filter paper by capillary action. As solvent moved across spots, partition of the spots between the stationary phase of trapped water in the paper and the mobile phase of the solvent occurs • The states of partition of the spots depend on: • 1. The tendency of the amino acids to attach to the adsorbed water, • 2. The tendency of the amino acids to dissolve in the solvent

  18. 47.1 Principal Components of Food (SB p.283) • As different amino acids have different states of partition between the mobile and stationary phase, they would be moved up in different extent • Amino acids which are more soluble in the solvent than in water are moved up by the mobile phase at a faster rate • Amino acids which are more soluble in water than in the solvent are moved up by the mobile phase at a slower rate • As fresh solvent is moving up continuously, solvent extraction occurs continuously along the filter paper.

  19. 47.1 Principal Components of Food (SB p.283) • Sufficient time is allowed for the solvent to rise up the paper so that the amino acids are separated far enough for easy identification • The paper is then removed from the solvent, and a pencil is used to mark the solvent front • The paper is allowed to dry and a developing agent (e.g. ninhydrin) is then sprayed on the paper to reveal the positions of the colourless amino acids

  20. A pencil is then used to mark the positions of the spots. For each amino acid, a Rf value (retardation factor) is calculated by using the following formula • Rf value must be smaller than 1 • At the same temperature and with the same solvent, an amino acids always has the same Rf value • When the Rf value of various amino acids under a particular condition are known, it may be possible to identify an unknown amino acid by paper chromatography 47.1 Principal Components of Food (SB p.283)

  21. 47.1 Principal Components of Food (SB p.284) • The Rf value of some amino acids can be very similar in particular solvent •  amino acids would not appear as separate spots on the developed chromatogram • 2-D chromatography is applied for further separation •  a chromatogram is produced in the usual way and dried •  It is then turned 90° and placed in the second solvent with polarity different from that of the first one •  The second solvent completes the separation

  22. 47.1 Principal Components of Food (SB p.284) 2-D paper chromatography

  23. 47.1 Principal Components of Food (SB p.284) Example 47-1 The figure below shows the chromatogram of an unknown sample U. The distances travelled by the three amino acids A1, A2 and A3 are 5.2 cm, 3.4 cm and 7.5 cm respectively. The distances travelled by the spots above sample U are 5.3 cm and 7.6 cm and the distance travelled by the solvent front is 10 cm. Calculate the Rf values of the three amino acids and determine the amino acids in the unknown sample U. Answer

  24. Solution: Rf of A1 = Rf of A2 = Rf of A3 = The Rf values of the spots above U are i.e. 0.53 and 0.76. By comparing these values with the Rf values of A1, A2 and A3, we can deduce that the unknown sample contains A1 and A3. 47.1 Principal Components of Food (SB p.285)

  25. A peptide linkage is a bond formed by the condensation reaction between two amino acids with a water molecule being eliminated. 47.1 Principal Components of Food (SB p.285) Check Point 47-1 What is a peptide linkage? Illustrate your answer with 2-aminopropanoic acid Answer

  26. Simple carbohydrates are synthesized by chlorophyll-containing plant through photosynthesis • 6CO2(g) + 6H2O(l)  C6H12O6(s) + 6O2(g) Light chlorophyll 47.1 Principal Components of Food (SB p.285) Carbohydrates • Carbohydrates are the important source of energy in the diet • They comprise a large group of organic compounds which C, H and O. The general formula is CxH2yOy

  27. 47.1 Principal Components of Food (SB p.285) Carbohydrates provide a significant amount of the energy for our daily activities

  28. 47.1 Principal Components of Food (SB p.286) Classification of Carbohydrates • Carbohydrates are divided into 3 groups: • monosaccharides • disaccharides • polysaccharides

  29. 47.1 Principal Components of Food (SB p.286) • Monosaccharides • A group of sweet, soluble crystalline molecules of relatively low molecular masses which cannot be further hydrolyzed into simpler compounds • General formula: C6H12O6 • Most important example: glucose and fructose (found in many fruits, honey; glucose is found in the blood of animals)

  30. 47.1 Principal Components of Food (SB p.286) • Disaccharides • Sweet, soluble and crystalline • General formula: C11H22O11 • May be formed from the condensation reaction of two monosaccharide molecules • Under suitable conditions, they may be hydrolyzed by water to yield its constiuent monosaccharides

  31. 47.1 Principal Components of Food (SB p.286) • Common dissarcharides:1. Sucrose (found in sugar cane), consists of a glucose unit and a fructose unit • 2. Maltose (found in malt), consists of 2 glucose units • 3. Lactose (found in milk), consists of a glucose unit and a galactose unit

  32. 47.1 Principal Components of Food (SB p.286) • Polysaccharides • Condensation polymers of monosaccharides • General formula: (C6H10O5)n (where n is a large number) • Examples of polysaccharides: • 1. Starch; it is commonly found in rice, bread and potatoes • 2. Cellulose; it is found in fruits, vegetables, cotton and wood

  33. 47.1 Principal Components of Food (SB p.286) Open Chain and Ring Structures of Glucose and Fructose • Glucose can exist in acyclic (also describe as open-chain) and cyclic forms • In acyclic structure, it contains an aldehyde group known as aldohexose

  34. 47.1 Principal Components of Food (SB p.287) • In solid state, glucose does not exist as open-chain structure • It exists in one of the two cyclic structures which differ in the carbon C1 only -glucose and -glucose

  35. 47.1 Principal Components of Food (SB p.287) • Two cyclic forms of glucose can be converted to the other form when dissolved in water • In equilibrium mixture, 2 forms will exist together with trace amount of acyclic form • Although the acyclic form only exist in trace amount, most of the reaction of glucose in aqueous solution are due to the presence of free aldehyde group in open-chain form

  36. 47.1 Principal Components of Food (SB p.287) • Fructose can exist as an acyclic form, 6-membered cyclic form and 5-membered cyclic form • It contains a keto group in open-chain form known as ketohexose

  37. 47.1 Principal Components of Food (SB p.288) • Similar to glucose, most of the reactions of fructose in aqueous solution are due to the presence of free keto group in the open-chain form

  38. 47.1 Principal Components of Food (SB p.288) Glycosidic Linkage in Carbohydrates • The bond formed between two monosaccharides is called glycosidic linkage •  formed from the condensation reaction between 2 –OH groups of two monosaccharides with the elimination of H2O molecule

  39. 47.1 Principal Components of Food (SB p.288) • Sucrose is made up of a glucose unit and fructose unit with the elimination of H2O molecule

  40. 47.1 Principal Components of Food (SB p.288) • Maltose is made up of 2 glucose units with the elimination of H2O molecule

  41. 47.1 Principal Components of Food (SB p.288) Food containing sucrose and maltose

  42. 47.1 Principal Components of Food (SB p.289) • The condensation process can be repeated to build up a giant molecule of polysaccharides • e.g. Starch and Cellulose • Starch:

  43. Potatoes contain starch, and cabbage contains cellulose 47.1 Principal Components of Food (SB p.289) Cellulose:

  44. Sucroseis hydrolyzed by dilute mineral acids (or enzymes) to give glucose and fructose • C12H22O11 + H2O  C6H12O6 + C6H12O6 H+ glucose sucrose fructose • The solution of starch is hydrolyzed into maltose in the presence of suitable enzyme • 2(C6H10O5)n + nH2O  nC12H22O11 amylase maltose starch 47.1 Principal Components of Food (SB p.289) Hydrolysis of Sucrose and Starch

  45. When maltose is further treated with a dilute acid or enzyme maltase, 2 glucose units will be formed • C12H22O11 + H2O  2C6H12O6 H+ or maltase glucose maltose • When starch is boiled with dilute sulphuric(VI) acid, it is hydrolyzed to give glucose • (C6H10O5)n + nH2O  nC6H12O6 H+ starch glucose 47.1 Principal Components of Food (SB p.289)

  46. 47.1 Principal Components of Food (SB p.290) Fehling’s test • A reducing sugar can reduce a basic solution of Cu2+(aq) (Fehling’s solution) or Ag+(aq) (Tollen’s solution) • Reducing sugars reduce Fehling’s solution •  The Fehling’s solution changes from blue to green and a yellow-orange ppt. of Cu2O(s) is produced • Glucose, fructose, maltose are reducing sugars • Sucrose, starch and cellulose are non-reducing sugars

  47. 47.1 Principal Components of Food (SB p.290) Results of Fehling’s test on glucose, maltose and sucrose (from left to right)

  48. 47.1 Principal Components of Food (SB p.290) • Fehling’s solution is used to distinguish reducing and non-reducing sugars • Glucose is a reducing sugar∵ the presence of the aldehyde group in the acyclic form in aqueous solution in equilibrium • Fructose is a reducing sugar∵ the presence of the keto group in the acyclic form in aqueous solution in equilibrium • Disarccharides also need to have free carbonyl groups for them to have reducing action on Fehling’s solution

  49. 47.1 Principal Components of Food (SB p.290) • Maltose is formed by the condensation of 2 glucose units which are joined up by a glycosidic linkage • The glycosidic linkage is formed by the condensation of the –OH group of C1 of one glucose ring with the –OH group on C4 of another glucose ring • The ring can open up to form acyclic structure with a free aldehyde group when dissolved in water

  50. 47.1 Principal Components of Food (SB p.291) • In sucrose, the glycosidic linkage is formed between C1 of glucose and C2 of fructose • Both constituent rings in sucrose cannot equilibrate with acyclic form containing a free carbonyl group which possess the reducing properties sucrose is a non-reducing sugar

More Related