1. 1 Oxidation of Monosaccharides Monosaccharides are reducing sugars if their carbonyl groups oxidize to give carboxylic acids. In the Benedict’s text, D-glucose is oxidized to D-gluconic acid. Glucose is a reducing sugar.

2. 2 The chemistry of Benedict’s test Reducing sugars, those that contain a free aldehyde group, are oxidized by Cu 2+ which is reduced to Cu 1+ and precipitates in a form of a red Cu 2 O. Benedict’s reagent is an alkaline solution of CuSO 4, Na 2 CO 3, and sodium citrate (Na 3 C 6 H 5 O 7 ∙2H 2 O). Sodium citrate forms a soluble complex with Cu 2+ and prevents it from precipitating out as blue Cu(OH) 2 or black CuO.

3. 3 Fructose is a ketose that changes to aldose in a basic solution.

4. 4 Oxidation – Reduction Oxidation (LEO) Loss of electrons Gain of O Loss of H (H + & e) Increase of the oxidation number (state) Reduction Gain of electrons Loss of O Gain of H (H + & e) Decrease of the oxidation number (state)

5. 5 Three views: 1.Oxidation = addition of oxygen Reduction = removal of oxygen 2 Mg + O 2 → 2 MgO (oxidation of magnesium) Mg has no oxygen → Mg gained oxygen Mg is oxidized O 2 is reduced

6. 6 2.Oxidation = loss of hydrogen atoms Reduction = gain of hydrogen atoms CH 4 + 2 O 2 → CO 2 + 2 H 2 O (combustion of methane) Carbon lost hydrogen atoms – oxidized Oxygen gained hydrogen atoms – reduced C and H in CH 4 had no oxygen atoms, but after the reaction C gained two oxygen atoms (CO 2 ) and H gained one oxygen atom (H 2 O)

7. 7 3.Oxidation is the loss of electrons Reduction is the gain of electrons 2 Mg 0 + O 2 0 → 2 MgO (2 Mg 2+ + 2 O 2- ) Mg looses electrons while O gains electrons LEOGER Mg is oxidized (reducing agent) O is reduced (oxidizing agent) Half reactions: Mg 0 → Mg 2+ + 2 e│2 O 2 0 + 4 e → 2 O 2-

8. 8 Oxidation numbers: Oxidation number = number of electrons in uncombined atom – number of electrons in atom in compound Oxidation number = number of valence electrons of the free atom – number of valence electrons of the atom in the compound Aox. no. for mono-atomic ions Box. no. for covalent compounds and poly- atomic ions.

9. 9 Aoxidation number = ionic charge NaCl Na atom11 eNa + ion10 e ox. no.11 – 10 = 1(I A group) valence e = 1valence e = 0 ox. no. = 1 – 0 = 1 Cl atom17 eCl - ion18 e ox. no.17 – 18 = -1(VII A group – 8 = -1) valence e = 7valence e = 8 ox. no. = 7 – 8 = -1

10. 10 B oxidation numbers of covalently bonded atoms SO 3 S (electropositive) 6 – 0 = 6 O (electronegative) 6 – 8 = -2 x + 3 (-2) = 0 x – 6 = 0 x = 6

11. 11 SO 3 2- (as in Na 2 SO 3 ) x + 3 (-2) = -2 x – 6 = -2 x = 4

12. 12 H 2 SO 4 H is assigned the ox. no. = 1 O is assigned the ox. no. = -2 Finding the oxidation number of sulfur by algebra: 2 (1) + x + 4 (-2) = 0 2 + x – 8 = 0 x = 6

13. 13 ClO 3 1- & Cl 2 O 7 x + 3 (-2) = -1 x – 6 = -1 x = 5 2x + 7 (-2) = 0 2x – 14 = 0 2x = 14 x = 7 Max ox. no. = group number Min ox. no. = group number – 8 Min ox. no. for Cl = VII -8 = -1 (= ionic charge).

14. 14 Oxidation – Reduction (oxidation state or oxidation numbers) In simple ionic compounds, the chemical bond is formed by a complete transfer from the more electropositive to the more electronegative element – the actual charge of the ion is equal to its oxidation number (e.g. NaCl – the oxidation number of Na + is +1 and that of Cl - is -1. In covalent compounds (such as CH 4 ) and in polyatomic ions with covalent bonds (such as SO 4 2- ), electrons are shared between bonded atoms. For calculations of oxidation numbers, the electrons are assigned completely to the more electronegative atom (electron hog). In CH 4 the oxidation number of C is -4 and that of H is +1. In SO 4 2- the oxidation number of S is +6 and that of O is -2. The sum of the oxidation numbers in the polyatomic ion is equal to the charge of the ion. The sum of the oxidation numbers in the compound is equal to zero. (In most compounds hydrogen is assigned the oxidation number of +1 and oxygen is -2.) For an element (Ag, Na, Cl 2, O 2, etc.), the oxidation number of each atom is equal to zero.

15. 15 Half – reactions (Reduction) Gain of Electrons Ag + + e → Ag (+1) (0) Loss of Oxygen NO 3 - + 2H + + 2e → NO 2 - + H 2 O (+5)(+3) Gain of Hydrogen 2 CO 2 + 2H + + 2e → H 2 C 2 O 4 (+4) (+3)

16. 16 Oxidation – Reduction in organic chemistry Reduction of an organic molecule usually corresponds to increasing its hydrogen content or to decreasing its oxygen content. Oxidation is the opposite of reduction, thus increasing the oxygen content of organic molecule or decreasing its hydrogen content is an oxidation.

17. 17 Method of assigning an oxidation state to a carbon atom of an organic compound: Base the assignment on the groups attached to carbon. A bond to hydrogen (or anything less electronegative than carbon) makes it -1. A bond to oxygen (or anything more electronegative than carbon, like nitrogen or halogen) makes it +1. A bond to another carbon makes it 0. H | Methane, CH 4 (H – C – H) – oxidation state of carbon is -4. | H Carbon dioxide, CO 2 (O=C=O) – oxidation state of carbon is +4. H | Methanol, CH 3 OH (H – C – OH) – oxidation state of carbon is | H 3(-1)+1=-2.

18. 18 Oxidation of aldehyde to carboxylic acid R – C = O → R – C = O (oxidation) | | H OH (+1) (+3) Cu 2+ + 1e → Cu 1+ (reduction) (+2)(+1)

19. 19 Reduction of Monosaccharides The reduction of the carbonyl group produces sugar alcohols, or alditols. D-Glucose is reduced to D- glucitol also called sorbitol.