Chemistry questions, organic matter, chemistry questions about organic matter

You can draw a carbon chain.

For a cc—c—c—c

c The name of the organism should be 2,2-dimethylbutane, because it is counted from the lowest side of the number.

For c c—c

c—c—c—c

The name of this organic substance should be 3-methylpentane, because C is the main chain with the most.

For dc c

c—c—c—c

c The name of the organic substance is 2,2,3-trimethylbutane, because the sum of the numbers should be the smallest.

A: When naming, you must make the previous number the smallest (minimum number principle), 3,3-dimethylbutane, if it is written out is the same as 2,2-dimethylbutane, then according to the principle of minimum number, it can be concluded that 2,2-dimethylbutane is correct, that is, A is wrong, B is right.

C: The longest carbon chain must be used as the main chain (longest chain principle) when naming, 2-ethylbutane is the same as 3-methylpentane if written as a chemical formula, but the main carbon chain of the latter is obviously longer than the former, so according to the longest chain principle, it can be seen that the substance should be named 3-methylpentane, that is, C is wrong.

d: Same as a, d should be called 2,2,3-trimethylbutane when naming, so as to conform to the principle of minimum number, d is also wrong

b correct. A should be 2,2-dimethylbutane.

c should be 3 methylpentane.

d should be 2,2,3-methylbutane.

The nomenclature is available in chemistry books.

(1) Let the molecular formula be CNHM, so 58=12N+M

When n=4, m=10; When n=3, m=22, it is unreasonable; When n=5, m=-2, it is unreasonable, so n=4, m=10, the molecular formula is C4H10, and the simple structural formula may be n-butane, or 2-methylpropane.

2) The molecule contains methyl groups, the molecular weight is 15, so the remaining molecular weight is only 43, due to the oxygen content, when containing one oxygen atom, the molecular mass of the remaining atom is 27, 12n+m=27, when n=1, m=15; n=2, m=3 is reasonable, so the molecular weight of an oxygen atom is c3H6O, which may be acetone or propionaldehyde, because it can undergo silver mirror reaction, so it should be propionaldehyde.

If it contains two oxygen atoms, the molecular mass of the remaining atoms is 11, 12n+m=11, n=0, m=11, which is unreasonable.

3) First of all, the silver mirror reaction can occur, there must be an aldehyde group, if there is only one aldehyde group, there must be a methyl group, methylene, if there are two aldehyde groups, the molecular weight is just right. So it is glyoxal.

(1) Divide by 12 to get 4. If C is 4 and the molecular formula is C4H10, it is an alkane. It may be butane or 2-methylpropane.

If the carbon content is 3 and the molecular formula is C3H22, it is impossible, and of course less than 3 is even more unlikely.

2) Oxygen-containing derivatives and can undergo silver mirror reaction, then containing aldehyde group-CHO, remove 29, contain a methyl group, remove 15, and leave 14, it is impossible to contain oxygen, and it cannot be 14 h, so the rest is CH2

Possible structure: CH3CH2CHO

3) The silver mirror reaction can occur, containing CHO, leaving 29, no methyl group and no OH, and just one aldehyde group is 29, so it may be a combination of 2 aldehyde groups together, that is, glyoxal. ohc-cho

Hydroxyl CH3-CH2-CH2-OH (propanol).

Addition reaction. 1) CH3-CH2-CH2-OH = ===CH3-CH=CH2 + H2O (propanol intramolecular dehydration becomes propylene).

2) CH3-CH=CH2+BR2====CH3-CHBR-CH2BR (addition reaction of propylene to bromine).

3) O2+2ch3-ch2-ch2-oh===2ch3-ch2-cho+2H2O (that is, catalytic oxidation of propanol).

Reaction conditions are added by looking at the topic by yourself.

Hello The answer is d

First of all, methane and chlorine produce methyl chloride and HCl under light conditionsThen, methylene chloride and the chlorine gas produced by HCL form methylene chloride and HCLThen, dichloromethane and HCl produce chlorine gas to form chloroform (also known as chloroform).

Finally, chloroform (also known as chloroform) and HCL produce chlorine gas to form tetrachloroform (also known as carbon tetrachloride).

The reaction process is sequential. The main reaction formula CH3Cl side reaction is generated 2 CH2CI2 3 CHCi3 4 CCI4

Regardless of the ratio of methane to chlorine, their reaction products are 5.

CH4 + Cl2 --CH3Cl+ HCLCH3Cl + Cl2 --CH2Cl2 + HCLch2Cl2 + Cl2 --CHCl3 + HCLCl3 + Cl2 --CCL4 + HCl, because every reaction has reactants and conditions that are present, that is, even the exact amount of the product cannot be determined.

But the product is there.

D should be chosen, and all four products are available.

The substitution reaction of alkanes, unlike other organic reactions, is a chain reaction. When one chlorine atom replaces one hydrogen on methane, the second chlorine atom immediately continues the reaction ,......Until the reaction produces CCL4, which is a complete chain initiating chain termination, in which CH3Cl, CH2Cl2, CHCL3, CCL4 are produced in turn

Due to the randomness of the chlorine atomic reaction, all the reactions occur at the same time, and the condition given in the title is the amount of the same substance, it is obvious that chlorine is not enough to completely replace methane with Ccl4, so that all the products will exist at the moment when the reaction is terminated. So choose D. If there is an absolute excess of chlorine, it can be considered that the product consists of only Ccl4.

<> accumulation of double dilution affirmative vertical look at the figure specifically.

S with double bond is SP2, triple bond SP, single bond SP3.

The c with the double and triple bonds cannot be twisted, and must be in the same plane.

Sulfuric acid is dehydrated and condensed to form double bonds. Sulfuric acid has an up-oxidizing dehydration line, where double bonds can be produced.

The resulting pathway is shown in Fig. Carbocation rearrangement, methyl group migration, and removal of H+ occurs:

88*88* So there are 5 c, 12 h, and the rest is 1 och3

ch3-c-o-ch3ch3