The following can react with Filin reagent to form a brick red precipitate

Answer] A Answer Analysis] Test Question Analysis: Soluble reducing sugar contains aldehyde group, which is heated with Feilin reagent water bath to produce brick-red precipitate, common soluble reducing sugars are glucose, fructose and maltose, starch, sucrose and cellulose are not reducible, and cannot react with Feilin reagent to produce brick-red precipitate.

Test point: The identification of reducing sugars is intended to test the candidate's ability to use the knowledge and opinions learned to explain and reason about certain biological problems through comparison, analysis and synthesis, and make reasonable judgments or draw correct conclusions.

Answer] ABC

Answer analysis] Test question analysis: The reagent used to identify reducing sugars in cells is Feilin reagent, which can produce brick-red precipitates. Cellulose is a polysaccharide, which is a non-reducing sugar, and glucose, fructose, and maltose are all reducing sugars, so they can react with Feilin reagent to form a brick-red precipitate.

Test Topic: Identification of organic matter in cells.

Answer] D Answer Analysis] Test Question Analysis: Feilin reagent reacts with reducing sugars to form brick-red precipitates, and common reducing sugars include glucose, fructose, and maltose; Therefore, choose D.

Test point: This question examines the types and identification of reducing sugars, this question is relatively simple, and can be trained to improve ability.

Answer] A Answer Analysis] Test Question Analysis: Under the condition of heating the water bath, the reducing sugar and the Feilin reagent appear brick-red precipitate, so A is correct.

Test Center: This question mainly examines the identification of organic matter, which is intended to test the candidate's ability to understand the key points of the knowledge learned and grasp the internal connection between knowledge.

d Because there is an aldehyde group in 14, the carbon chain of 2 can be broken to produce an aldehyde group (this content is not required in high school chemistry).

Four options, A is a ketone, which cannot react with the film reagent, and B is acetal, which is stable in nature, and it is impossible to react with the film reagent. D is an aromatic aldehyde and does not react with the film reagent. Only c is a hemiacetal.

The structure, unstable, can be hydrolyzed into aldehydes, hemiacetals and aldehydes coexist, so it can react with film reagents.

c, this compound, its structure is hydroxyl with an ether bond. is a semiacetal. Hemiacetal can also react with film reagents. So C should be the answer, and the other three compounds do not have a suitable aldehyde group.

Feilin reagent detects reducing substances, and can detect highly reducing organic substances such as aldehyde groups.

a.Benzoic acid.

b.Acetone. c.Glycerol.

d.Formic acid. Correct answer to the case of the empty state: d

CA, is maltose has an aldehyde group, reducibility, and can react with Feilin reagent to produce brick-red precipitate.

b. Glucose has an aldehyde group and is reducible, and can react with Feilin reagent to produce brick-red precipitate.

c. Sucrose has no aldehyde group, no reducibility, and will not react with Feilin reagent to produce brick-red precipitate.

d. The urine of diabetic patients has glucose, aldehyde group, and reducibility, which can react with Feilin reagent to produce brick-red precipitate.

Reaction equation structure of glucose and film reagent Simplified equation:

2 CH3(CH2OH)4CHo+ 2Cu(OH)2 --CH3(CH2OH)4COO)2Cu + Cu2O(red precipitate) +2H2O

In short, glucose is a polyhydroxyaldehyde, and the Filin reagent is essentially a new copper hydroxide. The Filin reagent is made by mixing CuSO4 and NaOH and its useful part is the brick-red precipitate generated by the reaction of newly prepared Cu(OH)2 with reducing sugar, which is Cu2O For fructose, molecular formula: C6H12O6 Structure: Simplified formula:

ch2oh(choh)3coch2oh。Fructose can undergo ketone-enol interchange in dilute alkali solution. Under alkaline conditions, fructose can be converted into glucose and mannose by epimerization, which means that the ketone group of fructose is converted into aldehyde group through enolation.

Therefore, the fructose solution under alkaline conditions is actually a mixture of fructose &glucose &mannose, which is of course reducible. Organic chemistry believes that the reducing property is not a property of fructose itself, especially under acidic conditions, the reducing property of fructose almost disappears. The essence is still glucose and film reagents to produce brick-red precipitates.

The above is the equation of the reaction between glucose and film reagent expressed in a simple formula, you can change glucose to fratome, if you make it easier to write the molecular formula, you can make it yourself.

The slag in the following groups of substances cannot be identified by Filin reagent () aBenzene modified formaldehyde and nuclear dispersed formaldehyde.

b.Formaldehyde and acetaldehyde.

c.Acetone and propionaldehyde.

d.Acetaldehyde and propionaldehyde.

Correct answer: d

Which of the following compounds can be oxidized by the Filin reagent (Hall Wheel) aAcetaldehyde. b.Benzene ambent with formaldehyde.

c.Acetone. d.Pentanedione.

Correct Answer: a

Aldehydes can be mixed with Filin reagent (copper sulfate and potassium sodium tartrate in equal volumes to obtain a blue solution, in which the role of potassium sodium tartrate is to make copper ions form a complex without forming copper hydroxide precipitates in the alkaline solution, and the oxidizing effect is divalent copper ions), aldehyde molecules are oxidized into carboxylic acids (carboxylates are obtained in alkaline solution), and divalent copper is reduced to red cuprous oxide precipitates.

r-cho+2cu(oh)2+naoh=r-ccona+cu2o↓+3h2o

1) Due to glucose.

It belongs to reducing sugar, and sucrose per acre.

It is a non-reducing sugar, so the same amount of glucose solution and sucrose solution are divided into coarse acacia and the same amount of Feilin reagent.

After heating over low heat for 2 3 min, the color reaction of the two was different, indicating that the color reaction of Feilin's reagent could be used to identify reducing sugars such as glucose

2) Since the sucrose solution reacts orange with Feilin reagent after being treated with dilute hydrochloric acid, it indicates that sucrose can be decomposed under the action of dilute hydrochloric acid to produce reducing sugar

3) Reducing sugars include all monosaccharides and maltose in disaccharides.

Sugars that can react with the brick-red color of Filin reagent like lactose, etc., like glucose, are maltose and fructose

So the answer is: 1) reducing sugars.

2) Catalyzed by dilute hydrochloric acid, sucrose was decomposed into reducing sugar rocks.

3) Maltose, lactose, etc.