Oxidation of monosaccharides
Download
1 / 19

Support Current user: unauthenticateduser UM.SiteMaker Message - PowerPoint PPT Presentation


  • 64 Views
  • Uploaded on

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. The chemistry of Benedict’s test.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about ' Support Current user: unauthenticateduser UM.SiteMaker Message' - lin


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.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
Oxidation of monosaccharides
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.


The chemistry of benedict s test
The chemistry of Benedict’s test

Reducing sugars, those that contain a free aldehyde group, are oxidized by Cu2+which is reduced to Cu1+ and precipitates in a form of a red Cu2O.

Benedict’s reagent is an alkaline solution of CuSO4, Na2CO3, and sodium citrate (Na3C6H5O7∙2H2O). Sodium citrate forms a soluble complex with Cu2+ and prevents it from precipitating out as blue Cu(OH)2 or black CuO.



Oxidation reduction

Oxidation (LEO) solution.

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)

Oxidation – Reduction


Three views 1 oxidation addition of oxygen reduction removal of oxygen
Three views: solution.1. Oxidation = addition of oxygen Reduction = removal of oxygen

  • 2 Mg + O2→ 2 MgO

  • (oxidation of magnesium)

  • Mg has no oxygen → Mg gained oxygen

  • Mg is oxidized

  • O2 is reduced


Oxidation loss of hydrogen atoms reduction gain of hydrogen atoms
Oxidation = loss of hydrogen atoms solution.Reduction = gain of hydrogen atoms

  • CH4 + 2 O2→ CO2 + 2 H2O

  • (combustion of methane)

  • Carbon lost hydrogen atoms – oxidized

  • Oxygen gained hydrogen atoms – reduced

  • C and H in CH4 had no oxygen atoms, but after the reaction C gained two oxygen atoms (CO2) and H gained one oxygen atom (H2O)


Oxidation is the loss of electrons reduction is the gain of electrons
Oxidation is the loss of electrons solution.Reduction is the gain of electrons

  • 2 Mg0 + O20→ 2 MgO

    • (2 Mg2+ + 2 O2-)

    • Mg looses electrons while O gains electrons

    • LEO GER

    • Mg is oxidized (reducing agent)

    • O is reduced (oxidizing agent)

    • Half reactions:

      Mg0 → Mg2+ + 2 e │2

      O20 + 4 e → 2 O2-


Oxidation numbers
Oxidation numbers: solution.

  • 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

  • A ox. no. for mono-atomic ions

  • B ox. no. for covalent compounds and poly- atomic ions.


A oxidation number ionic charge
A oxidation number = ionic charge solution.

  • NaCl

  • Na atom 11 e Na+ ion 10 e

  • ox. no. 11 – 10 = 1 (I A group)

  • valence e = 1 valence e = 0

  • ox. no. = 1 – 0 = 1

  • Cl atom 17 e Cl- ion 18 e

  • ox. no. 17 – 18 = -1 (VII A group – 8 = -1)

  • valence e = 7 valence e = 8

  • ox. no. = 7 – 8 = -1


B oxidation numbers of covalently bonded atoms
B oxidation numbers of covalently bonded atoms solution.

  • SO3

  • S (electropositive)

  • 6 – 0 = 6

  • O (electronegative)

  • 6 – 8 = -2

  • x + 3 (-2) = 0

  • x – 6 = 0

  • x = 6


So 3 2 as in na 2 so 3
SO solution.32- (as in Na2SO3)

  • x + 3 (-2) = -2

  • x – 6 = -2

  • x = 4


H 2 so 4
H solution.2SO4

  • 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


Clo 3 1 cl 2 o 7
ClO solution.31- & Cl2O7

  • 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).


Oxidation reduction oxidation state or oxidation numbers
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 CH4) and in polyatomic ions with covalent bonds (such as SO42-), electrons are shared between bonded atoms. For calculations of oxidation numbers, the electrons are assigned completely to the more electronegative atom (electron hog). In CH4 the oxidation number of C is -4 and that of H is +1. In SO42- 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, Cl2, O2, etc.), the oxidation number of each atom is equal to zero.


Half reactions reduction
Half – reactions (Reduction) numbers)

  • Gain of Electrons

    Ag+ + e → Ag

    (+1) (0)

  • Loss of Oxygen

    NO3- + 2H+ + 2e → NO2- + H2O

    (+5) (+3)

  • Gain of Hydrogen

    2 CO2 + 2H+ + 2e → H2C2O4

    (+4) (+3)


Oxidation reduction in organic chemistry
Oxidation – Reduction in organic chemistry numbers)

  • 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.


Method of assigning an oxidation state to a carbon atom of an organic compound
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, CH4 (H – C – H) – oxidation state of carbon is -4.

      |

      H

    • Carbon dioxide, CO2 (O=C=O) – oxidation state of carbon is +4.

      H

      |

    • Methanol, CH3OH (H – C – OH) – oxidation state of carbon is

      |

      H

      3(-1)+1=-2.


Oxidation of aldehyde to carboxylic acid
Oxidation of aldehyde to carboxylic acid an organic compound:

  • R – C = O → R – C = O (oxidation)

    | |

    H OH

    (+1) (+3)

  • Cu2+ + 1e → Cu1+ (reduction)

    (+2) (+1)


Reduction of monosaccharides
Reduction of Monosaccharides an organic compound:

  • The reduction of the carbonyl group produces sugar alcohols, or alditols.

  • D-Glucose is reduced to D-glucitol also called sorbitol.


ad