Detailed explanation of the steps of organic chemistry problems in the second year of high school

Because the mass ratio of vapor to the same volume of hydrogen in the same condition is 30, the relative molecular mass of organic matter is 30, and the number of 2=60c is (

Similarly, the number of h is (

The number of o is ((1

So the molecular formula of this organic substance is C2H4O2

The possible structural formula is CH3COOH and HCOOCH3 differentiation: CH3COOH turns red with litmus.

Because CH3COOH reacts with Na2CO3 solution to produce gas.

So the organic matter is CH3COOH

1) Molecular formula: C2H4O2, may be CH3COOH (acetic acid) or Hoch2CHO (hydroxyacetaldehyde), the aldehyde group can be identified by detecting the aldehyde group with silver ammonia solution (generated by silver mirror) or copper hydroxide suspension (brick-red precipitate heated in a water bath).

2) Acetic acid.

If the mass ratio to the same volume of hydrogen is 30, it can be set to one mo, so its relative molecular mass is 60, and the number of carbons is x, hydrogen y, and oxygen z.

12x+y+16z=60

60*, it can be known that the number of carbon atoms is two, hydrogen = 60* hydrogen atoms, so its chemical formula is C2H4O2, it can be known as CH3COOH, HCOOCH3, with litmus, acetic acid is red, or smell, one is acidic, the other is lipids, and lipids are fragrant.

CH3COOH+NAC3====CH3COONA+CO2, which is acetic acid.

Under the same condition, the density of the gas is proportional to the molecular weight, so the molecular weight of the organic matter is 2*30=60, the number of C is 60*, H is 60*, and the oxygen is (60-2*12-4) 16=2, so the molecular formula of the organic matter is C2H4O2, and the possible structural formula is CH3COOH and HCOOCH3, CH3COOH can make the litmus solution red, and HCOOCH3 can undergo silver mirror reaction; If the organic matter reacts with Na2Co3 solution to produce gas, it is indicated as an acid, so the structure is simplified to CH3CoOH

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.

The answer is a. From the reaction of "1mol C and 2mol B residue of socks, a chlorine-containing ester (C6H8O4Cl2) is obtained. It can be seen that C contains 2 hydroxyl groups (-OH), and the hydrolysis of Chasa into C, combined with the oxidation of C to B (Clch2CoOH), can lead to the answer.

A B is a carboxylic acid Clch2Cooh containing chlorine in Sakura, and a chlorine-containing ester (C6H8O4Cl2) is obtained. Knowing that propylene is diol hoch2ch2oh, so, methyl contains alcohol hydroxynavir,

Let the molecular weights of ab c be 8x 22x 29x, respectively;

8x + 22x*2 + 29x) 4 = * 4x=4 molecular weights are 32 88 116, respectively

Because it is a monoalcohol, there is only one OH, subtract 17 respectively, and the rest are the masses of C and H;

Get 15 71 99;

CH3 C5H11 C7H15, respectively

So the answer is ch3oh c5h11oh c7h15oh

1mol of aldehyde group can react with 2mol Cu(OH)2 to form 1mol Cu2OCu2O's relative molecular mass = 64*2 + 16 = 144 Now Cu2O is generated, that is, Cu2O is generated

So the amount of aldehyde-based species = =

1mol aldehyde group can react with 2mol silver ammonia solution to generate 2mol ag, so the aldehyde group can generate ag, that is, 108* = ag, so a maximum of = = 20 b can be made

It does not reflect with Br2 in carbon tetrachloride solution, indicating that the hydrocarbon is a saturated hydrocarbon, that is, an alkane.

12n + 2n +2 =156, n = 11 , 235-156 + 1 = 80 (substitution reaction is taken as generation one).

There are only two kinds of products with a relative molecular mass of 235, that is, there are only two monobromine substitutes for the alkane, and the product with a relative molecular weight of 314 can be obtained as a dibromine substitute. 314-156+2=2*80

Combined with the title, the hydrocarbon structure formula can be obtained as.

CH3)3CC(CH3)2C(CH3)3, i.e., 2,2,3,3,4,4-hexamethylpentane.

Structural symmetry is preferred when there are one or two x substituents for multi-carbon alkanes).

1: Isopropanol or 2-propanol.

2.Pick B. Analysis: Cuo and ethanol will form aldehydes and Cu, dilute nitric acid will dissolve Co and Cu, hydrochloric acid will dissolve Cuo, only lime water does not react and only Cuo.

1: ch3ch(oh)ch3 how to name ch3?

2-Propanol. 2: b lime water.

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.