Carbohydrates

molisch’s test

This is a common or general test for all

carbohydrates larger than tetroses.

- Monosaccharide gives a rapid positive test.

- Disaccharides and polysaccharides react slower.moli

monosaccharides, furfural, hydroxymethylfurfural,purple

Principles:

- The test is on the basis that pentoses

With concentrated sulfuric acid, H2SO4, the

polysaccharides and dissacharides are first

hydrolyzed to ______ before they are

dehydrated.

- Pentoses yield ____ whereas hexoses yield

____ which is further

degraded into levulinic acid.

 A positive reaction for Molisch's test is given by

almost all carbohydrates, except tetroses and

trioses. Some glycoproteins and nucleic acids also

give positive results for this test. They tend to

undergo hydrolysis when exposed to strong mineral

acids and form monosaccharides.and

hexoses are dehydrated by the test reagent or

the conc. sulfuric acid (H2SO4) to form furfural or

hydroxyl methyl furfural, respectively.

- These products condense with α-naphthol to

form ____ condensation product or a violet

ring between the junction of two layers, which is

the basis of Molisch’s test.

Reducing sugars

act as reducing agents in chemical

reactions (i.e. donates electrons to other molecules).

include all monosaccharides (e.g.

glucose and fructose) and some disaccharides (e.g.

lactose and maltose).

Non-Reducing sugars

______ do not act as reducing agents

in chemical reactions. They include most disaccharides

(e.g. sucrose) and simple polysaccharides (starch).

FEHLING’S TEST

 To determine the presence of reducing sugars and

aldehydes.

reddish brown

precipitate of cuprous oxide

The test used the mixture of Fehling’s A or

aqueous copper (II) sulfate, and Fehlin’s B or

alkaline potassium sodium tartrate (Rochelle

salt) that act as the chelating agent.

- On heating the aldehyde or reducing sugar with

Fehling’s solution gives ______ that denotes the

presence of reducing sugar.

benedict’s test

To detect the presence of reducing sugars.

- All monosaccharides are reducing sugars; they

have a free reactive carbonyl group.

- Some disaccharides have exposed carbonyl

group and are also reducing sugars. Other

disaccharides such as sucrose are non-reducing

sugars and will not react with the Benedict’s

solution.

- Large polymers of glucose, such as starch, are

not reducing sugars, since the concetration of

hemiacetal groups are very low

copper sulfate, aldehyde or ketone group, reddish precipitate

Principles:

- The _____ (CuSO4) present in

Benedict's solution reacts with electrons from

the _______ of the reducing

sugar in alkaline medium.

- Reducing sugars are oxidized by the copper ion

in solution to form a carboxylic acid and a

_______ of copper (I) oxide.

BARFOED’S TEST

 To distinguish between mon-, di-, and

polysaccharides.

copper ion, reddish precipiratate,disaccharides

Principles:

- Barfoed’s test used copper (II) ions in a slightly

acidic medium.

- Reducing monosaccharides are oxidized by the

_______ to form a carboxylic acid

and a _____ of copper (I) oxide

within three minutes.

- Reducing ______ undergo the same

reaction but do so at a slower rate.

- The nonreducing sugars give negative result.

SELIWANOFF’S TEST

 Used to distinguish between aldoses (like

glucose) and ketoses (like fructose)

ketohexoses to

form 5-hydroxymethylfurfural, cherry red,Aldohexoses, yellow to faint pink, Ketoses

color,

Principles:

- Seliwanoff’s test used a seliwanoff’s reagent

(mixture of 6M HCl as dehydrating agent and

resorcinol as condensation reagent)

The test reagent dehydrates _________.5-

hydroxymethylfurfural further condenses with

resorcinol present in the test reagent to produce

a _____ product within two minutes.

- ______ react to form the same product but

do so more slowly giving _____

-_______ give a cherry red color within a few

minutes (approximately 2 minutes) while

aldoses require a longer time.

- This occurs because ketoses existing in the

furanose or pyranose form can readily attain,

under acidic conditions, the 2- ketofuranose

structure which favors the formation of furfural

by dehydration.

BIAL’S ORCINOL TEST

 To distinguish between pentose monosaccharide

and hexose monosaccharide.

pentoses, furfural, bluish or green

Principles:

- Bial’s test uses Bial’s reagent (mixture of

concentratred HCl as a dehydrating acid and

orcinol + traces of ferric chloride [FeCl3] as a

condensation reagent).

- The test reagent dehydrates ____ to form

_______.

- Furfural further reacts with orcinol and the iron

ion present in the test reagent to produce a

_____ product, while the hexoses yield

muddy-brown to grey condensation products.

IODINE TEST, helical structure of amylose, blue black color

 To differentiate a starch from mono-, di-, and

polysaccharides through the reaction of starch and

iodine interaction.

 Principles:

- The iodine test used iodine in water or the

mixture of aqueous solution of molecular iodine

(I) and potassium iodide (Ki) commonly referred

to as the Lugol’s Iodine.

- This involbed the _______

interact with starch that forms a charge transfer

complex with iodine, whein the iodine is present

inside the spiral or helical structure of the

amylose—resulting in a _______

PICRIC ACID TEST, red coloured Picramic Acid

 To detect reducing sugars

Principles:

- It uses picric acid (toxic yellow crystalline solid)

also chemically known as 2,4,6-trinitrophenol

(TNP) to form a ______

represented as a mahogany red color.

- All monosaccharides and disaccharides

containing the potentially free aldehyde or

ketone group possess reducing properties.

Reducing sugars contains a free aldehyde or

ketone group possess reducing property.

TOLLEN’S TEST, aldehyde, carboxylate salt, metallic silver

 To distinguish between an aldehyde and a ketone.

- The _____ is oxidized by the Tollens’ reagent

and forms a _____

- The silver ions present in the Tollens’ reagent are

reduced into _____.

 Principles:

- The Tollens’ reagent is the solution of alkaline

silver nitrate (AgNo3) combined with aqueous

ammonia solution (NH4OH), leading to the

development of a complex.

- Silver nitrate in water produces a silver-aqua

complex, in which water acts as a ligand.

- Hydroxide ions convert the aqua complexes into

silver oxides (Ag2O).

- The [Ag(NH3)2]+ complex is formed when silver

oxide develops a brown precipitate, which

would be subsequently dissolved by aqueous

ammonia.

- The complex subsequently forms a carboxylic

acid by oxidizing the aldehyde group in a given

compound.

- The silver ions in the reagent will be converted

to metallic silver.

- A silver mirror forms on the bottom side of a test

tube as silver ions are reduced to metallic silver.

- A terminal hydroxy ketone produces a positive

Tollens test because this Tollen's reagent will

convert the ketone into an aldehyde

MUCIC ACID TEST, galactose

 To identify the presence of the sugar galactode.

 Principles:

- When the _______ solution is heated with

concentrated nitric acid, a dicarboxylic acid

called mucic acid or galactaric acid formed as

white precipitate, which counts as a positive

result

OSAZONES, crystalline

yellow colored oazones

 To identify the presence of the sugar galactode.

 Principles:

- Reducing sugar solution when heated with

phenyl hydrazine, characteristic __________ are formed.

- Needle shaped crystals by glucose, sunflower

shaped crystals by maltose and powder puff

shaped crystals by lactose are formed.

NYLANDER’S TEST

 Nylander’s Test is a type of chemical test used to

determine the presence of reducing sugars. Sugars

like fructose and glucose can reduce bismuth

oxynitrate down to bismuth when under alkaline

conditions. If Nylander’s test reagent which is

comprised of bismuth Nitrate along with potassium

sodium tartrate as well as potassium hydroxide

mixed into a solution containing reduction sugars, a

dark bismuth precipitate has created.

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