Sophomore chemistry organic solution, sophomore organic chemistry problem solving

Hello, glad to answer for you.

DA answer Ethanol can react with acidic potassium permanganate, oxidize to acetaldehyde and then become acetic acid, and can not undergo addition reaction, but can undergo substitution (reaction with Na to generate sodium acetate).

B Answer: Acetaldehyde can be added with hydrogen to form ethanol, and it can also be substituted and reacted with acidic potassium permanganate to form acetic acid.

C answer ethylene can be added with hydrogen to form ethane (or react with bromine water to form 1,2-dibromoethane, this reaction is the best reaction for removing ethylene in methane mixed with ethylene) (there are many addition reactions, not to list them one by one) It is difficult to replace the reaction, but it can be replaced, and the reaction with acidic potassium permanganate to form carbon dioxide (can not be analogized, not all olefins react with potassium permanganate to form CO2, the reaction mechanism is that the double bond of ethylene is broken, and each oxygen atom becomes formaldehyde, and formaldehyde is further oxidized, The position of the H atom becomes OH, which is equivalent to carbonic acid, and carbonic acid decomposes to CO2).

D Answer: Benzene reacts with H2 to obtain cyclohexane, reacts with liquid bromine (not bromine water) to form bromobenzene, reacts with concentrated nitric acid (catalyzed by concentrated sulfuric acid) to form nitrobenzene, and reacts with concentrated sulfuric acid sulfonation to form benzene sulfonic acid, but benzene cannot fade the acidic potassium permanganate (proving that the benzene ring is not an alternating structure of single and double bonds) Note: Most of the homologues of benzene can react with acidic potassium permanganate (this is a method to identify benzene and benzene homologues).

If you have any questions, you can ask them.

a No addition reaction can occur; b can make acidic KMNO4 solution fade; c. No substitution reaction can occur; So choose D.

Benzene can be added with hydrogen, can be substituted with liquid bromine, etc., and cannot be oxidized by acidic KMno4 solution.

Answer: The DA option does not have a double bond and cannot be added, but OH- is reducible and can fade the acidic KMno4 solution.

b Aldehyde CHO- will discolor potassium permanganate.

c Double bonds fade potassium permanganate.

d Benzene can undergo substitution reaction and addition reaction; Does not fade acidic KMno4 solution.

Aso's Ga will be restored to health, Ga Ga No.

(1) M= VM at the same temperature and pressure can be deduced from =m v, n=v vm, n=m m, so the relative molecular mass of the substance is 8; The amount of the burning sample substance is that the complete combustion product passes through the concentrated H2SO4, and the mass increased by the concentrated H2SO4 is water [in fact, looking at the second question, it can be seen that the substance contains N elements, but N is completely burned to form N2], and the amount of the substance is , which contains H; What is absorbed through soda lime is CO2, the amount of the substance, which contains c; Also, the same amount of sample produces the amount of N2 substance, which contains n is.

Therefore, from the conservation of atoms, we can know that the molecular formula A is C9H9NOX; and from the relative molecular mass of x is 3

Hence the molecular formula A: C9H9NO3 (or C9H9O3N).

2) Here is a biochemical common sense, that is, glycine: in fact, glycine, or 2-aminoacetic acid, is the simplest amino acid, with the simple structure formula H2NCH2COOH and the molecular formula C2H5NO2; Glycine and B are dehydrated and condensed to form A, and water participates in the process, so H and O have changes, but C atoms can be directly added or subtracted, so the number of C atoms in B is 9-2=7; It is also composed of "B substances, which are composed of only three elements: C, H, and O. The ratio of the quantities of its matter is 7:

6:2 "It can be seen that the molecular formula B is C7H6O2, which is 10 H less than the saturated paraffin C7H16, and the unsaturation is 5, which is very likely to contain a benzene ring (in fact, it is not necessarily, it may also contain double, triple bonds, etc., but in that case, there are too many possibilities, it is too troublesome to list them one by one, and generally in this case, the topics to be examined are all benzene rings), so B is likely to be benzoic acid PH-COOH (PH- stands for benzene ring).

B is dehydrated and condensed with C2H5NO2 to form A, so A is PH-Co-NH-CH2-COOH [PH- stands for benzene ring].

First of all, you have to know the structure of phosphoric acid.

Then it is necessary to know that esterification is acid dehydroxyl group, alcohol dehydrogenation.

Compare it and you'll understand.

If you have any questions, please ask.

-ch2oh ho

ch2oh + ho p=o-ch2oh ho

The rest of the side remains unchanged, and then according to the acid dehydroxyl group, alcohol dehydrogenation, I hope it will help you (the three hydroxyl groups of phosphoric acid have three single bonds connected to p, and they can't be drawn.)

First of all, you have to be clear about the form of ester bond: -coor, which is an organic acid, and inorganic oxygenated acids, such as nitric acid, sulfuric acid, phosphoric acid, etc. are all OK. This question is in the form of an inorganic oxygenate formed by phosphoric acid, phosphoric acid is a ternary acid, which can provide three -OH, and the quaterniol given by the question undergoes an esterification reaction to remove three molecules of water to obtain the ester in the question, so it belongs to the ester substance.

The so-called ester is CoO-R, where this R group cannot be -H, if it is -H, then it becomes a carboxyl group!

There is a chemical saying, acid dehydroxyl alcohol dehydrogenation, from this structural formula should be n-pentanol and phosphoric acid esterification reaction!

Just think of that p as a c atom.

CH3CH2CH2CH2BR + NAOH --CH3CH2CH=CH2 + NABR + H2O (alcohol solution, heated).

ch3ch2ch=ch2 + hbr --ch3ch2chbrch3

CH3)2CHch = CH2 + HBR --CH3)2CH2CH2BR (under peroxide conditions).

CH3) 2CH2CH2BR + NAOH --CH3) 2CH2CH2CH2OH + NABR (aqueous solution, heated).

When asymmetric olefins are added to HBR (HX), hydrogen atoms are generally added to the C-C bond carbon containing more H, which is called Markov's empirical law, referred to as "Mahalanobis rule", and when peroxide is present, it is added to the C-C bond carbon with less H, which is called the "anti-Marb rule"; The halogenated hydrocarbons undergo substitution reaction in the aqueous solution of alkene NaOH to form alcohol, and the elimination reaction occurs in the concentrated NaOH alcohol solution to form alkene. According to the information given in the question, to convert from 1-bromobutane to 2-bromobutane, you can first convert 1-bromobutane into 1-butene, and then add HBR according to the "Marhalobis rule"; In order to convert 3-methyl-1-butene into 3-methyl-1-butanol, according to the "anti-horse rule", under the condition of peroxide, the olefin should be converted into 3-methyl-1-bromobutane, and then hydrolyzed to obtain 3-methyl-1-butanol.

Answer: (1) ch3ch2ch2ch2br thick naohEthanol. △ ch3ch2ch＝ch2

ch3ch2ch＝ch2 hbr ch3ch2chbr-ch3

2)(ch3)2chch＝ch2 hbr.Peroxide (CH3) 2CH2CH2BR

CH3)2CH2CH2BR dilute NAOHAqueous solution (CH3) 2CHch2CH2OH

In the middle are the reaction conditions.

1.Monochlorinates are compounds in which one hydrogen atom of an organic substance, usually a hydrocarbon, is replaced by a halogen atom.

There are two main ways to look at monochlorides: there are several different types of hydrogen through the symmetry of the molecular structure, such as CH3-CH3 There is only one; The second is the number of the main chain, which can be equivalent to the same carbon atom, and the hydrogen on it is the same.

2.Single bonds are saturated hydrocarbons. Saturated hydrocarbons require each of their carbon atoms to be attached to four other atoms. That is, all carbon atoms are single bonds (including naphthenes), so benzene is unsaturated.

3.Bromine is elemental, and the last two are solutions, but the solvents are different. The carbon tetrachloride solvent of bromine is an organic carbon tetrachloride, a non-polar solvent that is miscible with benzene and cannot be extracted; Bromine is directly soluble in benzene; Bromine water can be extracted due to the incompatibility between solvents.

Toluene is the same. Polyethylene is added first, and then analyzed to resemble benzene.

4.Liquids of different polarities are incompatible with each other (similar to each other), and organic matter is usually nonpolar. However, the low-grade alcohols, aldehydes and carboxylic acids are more polar and miscible with water.

It cannot be rotated. c=c, one is head-to-head, and the other is shoulder-to-shoulder, and it cannot be rotated under any circumstances.

It is a covalent bond, because the common electrons are paired into bonds, the polarity is not strong, so it cannot be ionized.

ch3-ch2-ch2-ch-ch3

CH3 2-methylpentane.

ch3-ch2-ch-ch2-ch3

CH3 3-methylpentane.

The arrow is attached to the carbon that has only one hydrogen atom.

Below is the systematic nomenclature of alkanes.

1。The longest C chain in the selected molecule (the one with the most C) is the main chain, and there are several C atoms on the main chain, which is a few anenes. For example, c-c-c-c-c-c is hexane (A, B, C, D, E, H, G, and octane).

2。Take the closest end of the main chain from the branch as the starting point, and use the number to mark the order of the c atom of the main chain. Such as:

c-c-c-c

3。Use the branch chain as the substituent, write the name of the substituent (such as methyl-ch4, ethyl-ch2-ch4) before "halane", and write the number corresponding to the c atom of the main chain connected to the substituent at the front, and separate it from the text with **"-". Such as:

c-c-c-c

2-methylbutane when there is a methyl group on the 2nd position;

When c-c-c-c

2-dimethylbutane when there are two methyl groups on the 2nd position;

When c-c-c-c

The 2,3 bits have methyl groups when they are 2,3-dimethylbutane.

2 kinds of CH3CH2CH2CH(CH3)2 Dimethylpentane CH3CH(CH2CH3)2 trimethylpentane.

The structure of this hydrocarbon is CH3CH==C(CH3)CH(CH3)2

The name is 3,4-dimethyl-2-pentene

The structure of hydrocarbon is simply is.

ch3 ch3

ch3—ch— c =ch—ch3

The name is dimethyl2pentene.

I figured out that it was C, and I couldn't find the error myself.

C4H8 (G) + 6O2 (G) = 4CO2 (G) + 4H20 (G) CH4 (G) + 2O2 (G) = CO2 (G) + 2H2O (G) The methane combustion volume remains unchanged, let the original volume of butene be X, and the amount of gas increased is also X, then X=(1+4)*(

The volume fraction is:

50%C4H8(G)+6O2(G)=4CO2(G)+4H20(G)CH4(G)+2O2(G)=CO2(G)+2H2O(G) Let butene be B CH4 is A

a+b=13a+8b+(4-2a-6b)=(4+1) 2a+6bLess than or equal to 4

a=b=In fact, this question is very loose, full combustion means that the least amount of material just reacts completely, and no further reaction, it is entirely possible that the amount of oxygen is not enough, resulting in carbon monoxide production, the amount of O2 is not enough, and all of it can not be converted into carbon dioxide, this question has been discussed.