labo 3

Question 1

what are glucose, fructose, sucrose and xylose

Answer 1

Glucose is a hexoaldose reducing monosaccharide sugar.
Fructose is a hexoketose reducing monosaccharide sugar.
Sucrose is a di-hexose and aldoketose non-reducing disaccharide sugar.
Xylose is a pentoaldose reducing monosaccharide sugar.

Question 2

Benedecit’s test: composition, color of the reaction, react with what

Answer 2

• composed of sodium citrate, sodium carbonate and cupper sulphate which only react with reducing end sugars.
• When heated in an alkali solution (provides by the sodium carbonate), reducing sugars tautomerize and form enediols which will reduce cupric ions (Cu2+) from the cupper sulphate to cuprous oxide (Cu+). This ion precipitates and is responsible for the red brick color of the solution.
• Glucose, fructose and xylose react positively to the Benedict’s test since they are three REDUCING SUGAR which is not the case of the sucrose.
  The solution with the sucrose stayed blue, the color of a solution when it contains non-reducing sugar.
• Sodium citrate forms complexes with Cu2+ to make sure they do not deteriorate to Cu+ during storage.
• FORMATION OF OF A PRECIPITATE IN A BRICK RED SOLUTION

Question 3

Bial’s test

Answer 3

• composed of HCL, FeCl_3 and orcinol
• reacts with pentose sugar such as xylose where HCl hydrolyses the sugar to form furfural derivative. It will then react with orcinol to form a blue color (without precipitate) in the presence of ferric ions.
• Since glucose and fructose are hexose sugar and sucrose is made of two hexose, they did not react positively to the test.
  COLOR: FORMATION OF BLUE COLOUR

Question 4

Seilwanoff’s test

Answer 4

Seilwanoff’s reagent is composed of HCl and resorcinol.

• distinguish ketone from aldose since its reaction with ketone is much faster than with aldose.
• HCl reacts with the ketone group to form furfural derivative which then reacts with resorcinol in a condensation reaction to form a red wine colour.
• Fructose and sucrose are the ones that positively reacts with this reagent since fructose is a ketone and one of the two sugar in sucrose is fructose (the other one is glucose).
• Since glucose and xylose are aldose, they will also react to the Seilwanoff’s reagent but at a much lower rate (which had not been observed during the lab).

Question 5

Molish’s test

Answer 5

• used to identify sugar, which explain why glucose, fructose, sucrose and xylose all positively react with it.
• sugar is hydrolysed with the sulphuric acid that we add in the solution to form furfural in case of pentose sugar or hydroxymethyl furfural in case of hexose. These products react with α-naphthol presents in the reagent to form a purple color (a layer between the acid and the solution where the acid is at the bottom is at the bottom since it is denser).