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reducing and non-reducing sugars (1 Viewer)

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if the question was "describe the chemical difference between reducing and non-reducing" how would you answer it. and also the prac that you did.
 

thejosiekiller

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think carbohydrates with an hydroxide group off an aromatic carbon ;) also allows for H-bonding

non reducing sugars im sure have an hydroxide group, but while linked to other carbohydrate groups it cannot be a reduxing agent

my memory is fuzzy on this
 

currysauce

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Reducing sugars contain either a free aldehyde group ( O=C-H) or a free ketose group ( O=C) :)
 

currysauce

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;) god i hate chemistry :(

Oh and the practical is testing various saccharides (spelling?) with Benedicts solution, if the sugar is reducing, it will cause the Copper ions to reduce in the solution causing the colour to go browny - hope that helps

(you dont really need to know the equation for that reaction, but really the most simplest is the reduction of Copper 2+ to Copper)
 
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Dreamerish*~

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azza_3761 said:
To determine whether a sugar is a reducing or non-reducing, we use a reagent that can oxidise a -CHO group (also known as an aldehyde group) or a –CO-CH2OH group but will oxidise the many ordinary alcohol groups present in all sugars. Three regents are available:
1. Tollens’ reagent, a colourless solution of silver nitrate in aqueous ammonia
2. Benedict’s solution, a deep blue solution of copper sulfate in alkaline citrate
3. Fehling’s solution, a deep blue solution of copper sulfate in alkaline tartrate

Since the cations in the three solutions usually precipitate out of alkaline solutions the Ag+ and the Cu+2 ions are presented as complex ions. When a colourless solution of a reducing sugar is warmed with colourless Tollens’ reagent, metallic silver is formed, depending on the conditions, it may appear as a black precipitate of finely divided metal or as a shiny mirror on the walls of a very clean test tube.
E.g. R-CHO + 2Ag+ + 3OH- --> 2 Ag(s) + R-COO- + 2H2O
R-CO-CH2OH + 2Ag+ + 3OH- --> 2 Ag(s) + R-CHOH-COO- + 2H2O

When a solution of a reducing sugar is warmed with either deep blue Benedict’s or Fehling’s solutions, a reddish-brown precipitate of copper (I) oxide is formed. Cu+2 is reduced to Cu+.
Eg. R-CHO + 2Cu+2 + 5OH- --> Cu2O(s) + R-COO- + 3H2O
R-CO-CH2OH + 2Cu+2 + 5OH- --> Cu2O(s) + R-CHOH-COO- + 3H2O
Go the Conquering Chem ay? ;)

Know about Benedict's solution, know of the others, and don't worry about the equations.
 

azza_3761

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To determine whether a sugar is a reducing or non-reducing, we use a reagent that can oxidise a -CHO group (also known as an aldehyde group) or a –CO-CH2OH group but will oxidise the many ordinary alcohol groups present in all sugars. Three regents are available:
1. Tollens’ reagent, a colourless solution of silver nitrate in aqueous ammonia
2. Benedict’s solution, a deep blue solution of copper sulfate in alkaline citrate
3. Fehling’s solution, a deep blue solution of copper sulfate in alkaline tartrate

Since the cations in the three solutions usually precipitate out of alkaline solutions the Ag+ and the Cu+2 ions are presented as complex ions. When a colourless solution of a reducing sugar is warmed with colourless Tollens’ reagent, metallic silver is formed, depending on the conditions, it may appear as a black precipitate of finely divided metal or as a shiny mirror on the walls of a very clean test tube.
E.g. R-CHO + 2Ag+ + 3OH- --> 2 Ag(s) + R-COO- + 2H2O
R-CO-CH2OH + 2Ag+ + 3OH- --> 2 Ag(s) + R-CHOH-COO- + 2H2O

When a solution of a reducing sugar is warmed with either deep blue Benedict’s or Fehling’s solutions, a reddish-brown precipitate of copper (I) oxide is formed. Cu+2 is reduced to Cu+.
Eg. R-CHO + 2Cu+2 + 5OH- --> Cu2O(s) + R-COO- + 3H2O
R-CO-CH2OH + 2Cu+2 + 5OH- --> Cu2O(s) + R-CHOH-COO- + 3H2O
 

karatekid72

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OMG!!!!! i love everyone who replied above!!! thanks =]
 

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