ACTIVITY 2 Flashcards
(30 cards)
carbohydrates, also know as saccharides, include a large group of ____________.
polyhydroxyaldehydes or polyhydroxyketones and their derivatives
are widely distributed in nature and make up most of the organic structures of all plants as well as being present to some extent in all animals
carbohydrates
the presence of carbohydrates can be tested based on:
A. production of furfural and its derivatives
B. their reducing property
when a monosacchaaride is treated with a strong mineral acid, ______ occurs
dehydration
the dehydrated product is ______ if the monosaccahride is a pentose and __________ if the monosachharide is a hexose
furfural, hydroxymethylfurfural (5-hydroxymethylfurfural)
various phenolic compounds like α-naphtol, orcinol, and resorcinol will condense with the furfural or hydroxymethyl furfural to form
COLORED DYES.
The formation of these colored condensation compounds is a positive test for the _________.
presence of carbohydrates
Trioses and tetroses don’t undergo this reaction since _____________
they don’t possess the requisite minimum of five carbon atoms.
uses a-naphthol is the most general test for the presence of
carbohydrates because it gives a positive test with all carbohydrates larger than tetroses.
The Molisch Test
In this test, the furfural and hydroxymethyl furfural derivatives react with a-naphthol to form purple colored condensation products.
The Molisch Test
is useful in differentiating between hexoses and pentoses because orcinol condenses with furfural from pentoses to form a blue-green compound and with hydroxymethyl furfural from hexoses to form a yellow-brown product.
The Orcinol or Bial’s Test
uses resorcinol distinguishes aldohexoses from ketohexoses based on their differential rates of reaction with hot HCI.
The Sellwanoffs Test
(Seliwanoff’s test)
Ketoses are dehydrated more rapidly than aldoses to give hydroxymethyl furfural compounds which then condense with resorcinol (m-dihydroxybenzene) to form a
cherry red complex
(Seliwanoff’s test)
Aldoses give ______ color that takes a longer time to develop. This test is most sensitive for fructose which is a ketose.
a light pink color
is one which has a free hydroxyl group at the anomeric C.
reducing sugar
As the cyclic structure reverts to its open-chain form, the free hydroxyl group is converted back to the aldehyde. The aldehyde group can be oxidized by reagents such as ________ reagent. The oxidizing agent is ______ that can be reduced to Cu+.
Benedict’s reagent; Cu2+
all the sugars that react with Benedict’s reagent are
reducing sugars.
All the monosaccharides and some disaccharides have
the ability to reduce an alkaline solution of Cu2+ ion. The Cu2+ ion which is less soluble precipitates out of the alkaline solution as a __________.
brick-red precipitate of cuprous oxide (+ = present)
distinguishes between reducing monosaccharides and reducing disaccharides by a difference in rate of reaction.
Barfoed’s test
the ______ reagent consists of cupric ions, like in
Benedict’s reagent,
reacts with reducing monosaccharides, to produce cuprous oxide, faster than with reducing disaccharides,
Barfoed reagent
Molisch Test:
- sucrose
- starch
- dextrin
- glucose
- arabinose
- distilled water
purple, with carbohydrates (+)
- +
- +
- +
- +
- +
- -
Orcinol or Bials’ test result
glucose, fructose, galactose, arabinose, distilled water
yellow brown precipitate = hexose: glucose, galactose and fructose
blue green precipitate = pentose: arabinose
Seliwanoff’s test
fructose, glucose, galactose
cherry red = ketohexose : fructose
light pink = aldohexose : glucose and galactose
Benedict’s and Barfoed’s test
glucose, fructose, lactose, sucrose, maltose
reducing monosacch =brick red ppt, faster: glucose, fructose
recuding disacch = brick red ppt, slower: lactose, maltose
nonreducing = no ppt: sucrose
