What are the skills for doing organic inference questions in the second year of high school? In part

For organic inference, the change of color is the key, and then it is to memorize the equation, the organic equation is a little easier to remember than the inorganic, remember the properties of some functional groups, because the structure determines the properties of this point to keep in mind, there is any structure has the properties of this structure, for example, there is -COOH can be retroesterified, it can react with Na and so on. Finally, the focus of organic inference is to introduce a substance and ask you to write about its isomers and its reaction with a substance; The equation for the esterification reaction must be written.

When doing inference questions, you must calm down and recall the properties of some substances, don't give up because a substance can't be pushed, even if you can't push any of them, you have to look at the topic, sometimes even if you can't push one out, you can fill in one or two blanks, and after looking at the problem, it may also bring you ideas for solving the problem.

Key points (1) Origin comparison: Pay attention to the changes in groups and possible reactions, such as the formation of carbon-carbon double bonds, and think about elimination.

2) Retrospective traceability of the nascent group -Cooh may be CH3 => -CH2BR => -CH2OH => -CHo => -COOH

3) The group that does not break or stand must be destroyed before it is generated.

CH3-CH=CH-CH3 CH2=CH-CH=CH2 was added to BR2 and then eliminated by co-heating with NaOH alcohol solution.

4) Familiarize yourself with the model of the book (which is the most important thing) and the so-called "ever-changing" is just that.

Inference is mainly about the information given to you in the question, and you have to be able to apply it, and when you are organic, you usually give it something that you haven't seen in the textbook, so you have to memorize all the basic knowledge of the textbook. After that, you have to use the basic knowledge you have learned to apply it to it, and the chemical equations are not difficult because they are all applied, mainly to see if you will draw inferences. There are many types of organic reactions, just keep in mind.

The skill is to review it well after every time, and my chemistry grades back then were raised by doing inference questions! After a year and a half of sleep, it can exhaust me to death! It is to try first, if you can't do it, just look at the answer, and then memorize the title of the question, such as some special reaction phenomena learned in high school, the characteristics of substances, and the conditions required for reactions!

The most important thing to do organic inference is to guess! Have faith!!

Master the rules and use the skills.

The first step is to be fully familiar with the properties of various hydrocarbons.

Organic chemistry mainly examines functional groups, followed by reaction conditions, such as sodium hydroxide alcohol solution represents the elimination reaction, and an indispensable relationship in organic is esterification reaction or hydrolysis of esters, which generally occurs.

Analysis: From the mass ratio of CO2 to H2O is 44:9, it can be obtained that the number of atoms C and H in its molecular formula is equal.

The organic matter cannot make the FeCl3 solution colored, and the silver mirror reaction cannot occur, indicating that there is no CHO and no phenol OH.

1molA can react with 1mol acetic acid, then this reaction can only be esterification, so it should have 1 alcohol OH.

After oxidizing A under certain conditions, 1molA can consume 2molNaOH, which oxidizes alcohol OH to COOH. The instructions should also have a COOH in the first place.

Therefore, the substance is Ho CH2 Cooh. where , denotes the benzene ring. Its molecular formula is C8H8O3, which is in line with the title.

I don't know exactly what it's called. It is a benzene ring with two groups, one is -Cooh, the other is -CH2OH, because it cannot make the FeCl3 solution colored, so it is not a phenol, and the silver mirror reaction cannot occur, so there is no aldehyde group, what can react with NaOH is phenol and acid, excluding phenol, it can only be acid, and because alcohol is oxidized into aldehyde, aldehyde is oxidized into acid, so A is alcohol. Alcohol and acid esterification reaction, 1mola can react with 1mol acetic acid, the introduction contains only 1 OH, and the mass ratio of CO2 to H2O is 44:

9. It means that the number of c and h atoms in a is equal, so it can be obtained.

There are three types of critical pairs in the simple structure.

FeCl3 can not fade, indicating that there is no phenol hydroxyl group, can not occur silver mirror reaction, indicating that there is no aldehyde group, can react with acetic acid, indicating that there is an alcohol hydroxyl group, the mass ratio of carbon dioxide and water to the mass ratio of carbon and hydrogen substances is 1:1, after oxidation, the alcohol hydroxyl group becomes a carboxyl group, phenyl C6H5 + hydroxyl OH so the branch chain should be CoH = CH2, after oxidation, the branch chain is broken, CO2 is produced, and 2molNaOH is consumed, so C8H8

A can be esterified with ethanol and acetic acid, so that A contains two groups: OH and COOH.

If A contains only x COOH, and one COOH reacts with one OH to form an H2O, so the relative molecular mass of the product when A reacts completely with ethanol is .

m = 134 + 46x - 18x =190 The solution gives x = 2, the molecular weight of two COOH and one OH is 45 * 2 + 17 =107, and the molecular weight of the remaining group is 134 - 107 = 27, so it is C2H3, so the molecular formula is written:

Hoocchohch2COOH (malic acid).

22. Solution: (1) The density of organic matter is 45 times that of H2 under the same conditions, so the relative molecular mass of organic matter = 45 * 2 = 90;

2) The amount of the substance =

Concentrated sulfuric acid increases, then the mass of the generated water is, the generation of n(H2O) =, the amount contained in N(H2O) =, the weight of soda lime, the generation of m(CO2) is, N(CO2) =, so N(C) is, indicating that the organic matter contains oxygen elements, then the amount of oxygen-containing atoms in the organic molecule = (

n(a):n(c):n(h):n(o)=:::3:6:3, then the molecular formula of organic matter is:

c3h6o3；

3) The amount of the substance =

A and sodium bicarbonate reaction shows that A contains a carboxyl group, forming a standard condition), N(CO2)=, so it indicates that it contains a carboxyl group;

Both the alcohol hydroxyl group and the carboxyl group can react with sodium metal to form hydrogen, and the reaction with sufficient sodium metal will form a standard state), n(h2)=, when the carboxyl group or hydroxyl group reacts with sodium to form hydrogen, the ratio of the amount of carboxyl or hydroxyl group to the amount of hydrogen is 2:1, and when a reacts with sodium, the ratio of the amount of a substance to the amount of hydrogen is 1:1, which means that a contains a hydroxyl group in addition to a carboxyl group;

4) According to the NMR hydrogen spectrum, there are 4 peaks in the organic matter, which contains 4 types of equivalent hydrogen atoms;

5) According to the NMR hydrogen spectrum, it can be seen that there are 4 peaks in the organic matter, then there are 4 types of equivalent hydrogen atoms, and the ratio of the number of hydrogen atoms is 3:1:1:1, so the structure is simply CH3CHCOOH

23、（1）c8h8o3 ho-c6h4-cooh

2) 2HCO+O2 catalyst 2HCOoh

3）ho-c6h4-cooch3+2naoh→nao-c6h4-coona+ch3oh+h2o

Aromatic compound A is formed into B and F under dilute sulfuric acid and heating conditions, B can catalyze oxidation to C, and C can catalyze oxidation to D, indicating that B is alcohol, C is aldehyde, D is carboxylic acid, and F is carboxylic acid. B and D are used as catalysts and heating conditions in concentrated sulfuric acid to generate E, E is an ester, and the ester can also undergo a silver reaction, indicating that E is methyl formate, B is methanol, C is formaldehyde, and D is formic acid; F reacts with NaHCO3 to produce gas, and there are two monochlorine substituents on the benzene ring of F, then there should be two substituents on the para-position of the benzene ring, if these two substituents are carboxyl groups with more than 180 relative molecules, which does not meet the topic, so the substituents on the para-position of F are -OH and -COOH respectively.