How to chemically identify glucose, sucrose, starch and phenol solutions

Glucose (chemical formula C6H12O6) is also known as corn glucose, maize sugar, or glucose. Synonyms: dextrose, cornsugar, grapesugar, bloodsugar.

It is the most widely distributed and important monosaccharide in nature, and it is a polyhydroxyaldehyde. Pure glucose is a colorless crystal, with a sweet taste but not as sweet as sucrose (ordinary people can't taste sweetness), soluble in water, slightly soluble in ethanol, insoluble in ether. The aqueous solution is rotated to the right, so it belongs to "dextrose".

Glucose has an important position in the field of biology and is the energy of living cells** and the metabolic intermediate, that is, the main energy supplier of living organisms. Plants can produce glucose through photosynthesis. It has a wide range of applications in the field of confectionery manufacturing and medicine.

Sucrose has been around for thousands of years. It is the main product of photosynthesis and is widely distributed in plants, especially in sugar beets, sugarcane and fruits. Sugar with sucrose as the main ingredient is divided into the following according to the purity from high to low

Rock sugar, white sugar, white sugar and red sugar (also known as brown sugar or brown sugar), sucrose is the most abundant in sugar beet and sugar cane, and the white sugar and brown sugar used are sucrose.

Chemical sucrose, organic compounds, molecular weight. Colorless crystals with optical rotation but no variable rotation. Molecular formula of sucrose:

c12h22o11。Sucrose is susceptible to acid hydrolysis, which produces equal amounts of D-glucose and D-fructose. Non-reducible.

The caramel formed by fermentation can be used as a color enhancer for soy sauce.

The identification process is as follows:

Four test tubes were added to the four solutions to be identified, and FeCl3 was added to each of them, and phenol was formed into a purple solution.

The remaining 3 sticks are added with silver ammonia solution, and the silver mirror is formed by glucose.

The last two were iodized, and the starch that turned blue was the starch.

The last one is sucrose.

2.First of all, starch can be identified by the reaction of iodine solution with starch to turn blue. When heated, a brick-red precipitate appears as glucose. Phenol was identified by the reaction of ferric chloride with phenol in purple, and glucose is a reducing sugar.

3.Because sucrose is a non-reducing sugar, it can be identified with Filin reagent1

It is phenol that forms the purple solution.

The last two are iodized.

The remaining 3 sticks were added with silver ammonia solution.

The last one is sucrose, the silver mirror is glucose, and the starch is blue, and the four test tubes are added to the four solutions to be identified, and FeCl3 is added to each

1.First of all, starch can be identified by the reaction of iodine solution with starch to turn blue.

2.Phenol was identified by reacting ferric chloride with phenol in purple color.

3.Because sucrose is a non-reducing sugar, and glucose is a reducing sugar, it can be identified by Filin reagent. When heated, a brick-red precipitate appears as glucose.

Samples were taken separately. Bromine water was added to the four test tubes, and it was phenol that produced a white precipitate.

Iodine water was added to the remaining three tubes, and the solution turned blue with starch.

The remaining two tubes were filled with silver ammonia reagent, heated, and the silver mirror was produced.

Glucose is a reducing sugar that can be tested with Filin reagent.

The starch turns blue when exposed to iodine.

Phenol turns purple when exposed to ferric ions.

The rest is sucrose.

First starch was identified, then glucose was identified, and sucrose was not visible.

1. Use iodine solution to identify starch.

Take an appropriate amount of the solution of the three sugars, add them to the three test tubes, and add the same amount of iodine solution to the three test tubes. Only starch can discolor the solution. Therefore, the blue color is the starch.

2. Use silver ammonia solution to identify glucose.

Prepare the silver ammonia solution first, and then drop the silver ammonia solution into two test tubes containing glucose and sucrose solutions respectively, shake and warm the test tubes in hot water. Soon it can be seen that a layer of bright metallic silver, which is glucose, is attached to the wall of one of the test tubes in the heated area.

Identification of glucose.

The easiest way to use fructose, sucrose, and starch is chromogenic methods.

First of all, an appropriate amount of iodine solution is added to each of these four sugars, only the starch turns blue, and the other three sugars do not change color.

Then in glucose, fructose, sucrose respectively in the overtime reagent, glucose, fructose is a simple sugar.

It is reducible and can be oxidized by Ban's reagent, whereas sucrose is a non-reducing disaccharide.

No response. Then add bromine water to glucose and fructose respectively, because glucose has reducing property and make bromine water fade, fructose has no reducing property, can not make bromine water fade, so that the purpose of distinguishing these two kinds of sugar can be achieved.

Glucose is an aldose bai, which can be oxygenated by bromide

Change. Sucrose does not contain aldehyde groups, and contains zhi

It is a ketone group and cannot make silver ammonia solution or newly made hydroxide inner copper produce silver mirror or brick red precipitate.

Silver mirror reaction of glucose: first prepare a silver ammonia solution, then drop into a dropper of glucose solution, shake and put the test tube in hot water to warm up. Soon it will be seen that a layer of mirror-like metallic silver is attached to the wall of the heated area of the test tube.

The silver mirror reaction is a chemical reaction in which the solution of silver (Ag) compounds is reduced to metallic silver, and it is called the silver mirror reaction because the metallic silver generated is attached to the inner wall of the container and is as bright as a mirror.

Experimental Procedure:1Take a small amount to dissolve separately.

liquid with 2 tubes.

2.Add an equal amount of silver ammonia solution (or a new inner copper hydroxide solution) to each of the 2 test tubes

Water bath with capacitive heat.

Observe the inner wall of the test tube.

experimental phenomena and results; If there is a silver mirror (or brick-red precipitate) on the inner wall of the test tube, it is a glucose solution.

On the contrary, if there is no obvious phenomenon, it is sucrose solution.

Explanation: Glucose contains aldehyde groups, so it can react with silver ammonia solution and freshly made copper hydroxide to produce silver mirror and brick-red precipitate.

Sucrose does not contain aldehyde groups, but contains ketone groups, which cannot make silver ammonia solution or newly made copper hydroxide produce silver mirror or brick red precipitate.

The reaction to identify the aldehyde group can be used, glucose has an aldehyde group, and sucrose does not.

If you use hydrogen to test the copper oxide solution of Chaohu Copper Oxide, glucose will react with it and there will be brick-red precipitate, but sucrose does not. There is also the silver mirror reaction.

Samples were taken separately.

Silver ammonia reagent was added to the two test tubes, heated, and the silver mirror reaction occurred. The other is sucrose.

Since glucose (CH2OH-(CHOH)4-CHO) contains an aldehyde group (-CHO), while sucrose (C12H22O11, which contains hydroxyl (-OH) and ketone groups) is absent.

The aldehyde group has unique and special properties: it can be a silver mirror reaction with silver ammonia solution (CH2OH-(CHOH)4-CHO+2[AG(NH3)2](+2OH(-)CHH2OH-(CHOH)4-COOH+2AG +H2O+4NH3), and the reaction with copper hydroxide solution to form a brick-red precipitate (CH2OH-(CHOH)4-CHO+2Cu(OH)2 CH2OH-(CHOH)4-COOH+CU2O+2h2o）。 Therefore, it can be identified.

The first step is Portugal.

Copy the identification method of four substances: glucose, maltose, fructose, and lactose.

Zhi: The aqueous solution of these four substances is added to the clarified lime water respectively, and the white precipitate is fructose, because fructose will combine with Ca(OH)2 to form insoluble substances, while the other three sugars do not have this phenomenon.

The second step, a certain amount of glucose, maltose, lactose are mixed with water respectively, because lactose in cold water (room temperature) solubility is very small, can not be completely dissolved after adding water and stirring, so after mixing turbid is lactose, can be completely dissolved to form a transparent solution for the other two.

In the third step, glucose and maltose react with phenylhydrazine to form glycostringene, which is glucose faster and maltose is slower.

Silver mirror reaction: glucose is produced with bright silver.

Starch is identified with iodine water.

For the remaining three gafilin reagents, there is a brick red precipitate for fructose

The remaining two are heated and hydrolyzed by adding water and dilute sulfuric acid, and after hydrolysis, the reagent of Gafilin has precipitation to produce sucrose, otherwise it is maltose.

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1. Use iodine solution to identify starch: starch is detected with iodine solution, and the starch solution is blue, and the others do not change color.

2. Use silver ammonia solution: the silver mirror reaction is maltose, fructose, and glucose. Maltose reacts with silver ammonia solution in silver mirror, and can also react with newly made alkaline copper hydroxide to form a brick-red precipitate.

3. Add Filin reagent to maltose glucose: Brick red precipitate is glucose when heated in a water bath, and maltose is maltose if there is no reaction, and maltose plus Filin reagent has no reaction, and sulfuric acid hydrolysis can be added to react with Filin reagent. In the remaining two lactose and sucrose, sodium hydroxide test solution is added, slightly hot, the solution is yellow at first, and then it will turn brown-red, and then copper sulfate test solution is added, and red precipitates as lactose.

Starch can be identified by adding iodine solution. With the addition of bromine water, glucose can fade the bromine water. Then add film reagent and dilute sulfuric acid, which can make the film reagent turn brick red and precipitate sucrose, and the rest is fructose.

1. First of all, iodine water is added, and the blue one is starch.

2. Silver mirror experiment: glucose and fructose, and sucrose without silver mirror.

3. Glucose and fructose are mixed with lime water respectively, and there is a precipitate of fructose.