High School Organic Chemistry Questions Close at 11 a.m

CH3)2C=CHch2CH3, CH3CH=CHCH(CH3)2, CH2=CHC(CH3)3, CH2=C(CH3)CH(CH3)2, CH3CH=C(CH3)CH2CH3, so there are 5 kinds, CXH2X+BR2=CXH2XBR2

So n(br)=, so n(olefin)=,m=

So the olefin is C5H10

If you have any questions, we will serve you wholeheartedly.

c=c-c(ch3)3,c=c(ch3)-c(ch3)2,ch3-c=c(ch3)-c-ch3, ch3-c=c-c(ch3)2,ch3-c(ch3)=c-c-ch3

2 d c5h10 product weight gain = the mass of the reacted bromine = grams = 8 grams, then the number of moles of bromine in the reaction = 8 160 = the number of olefin moles, the amount of olefin formula =, so it is c5h10

Write out three methyl groups and four methylhexane, and then add five double bonds (not counting cis-trans isomerism) It's cumbersome and ugly to represent in letters, don't think about it.

m(br)=8g

n(br)=

The options are all monoolefins, so olefins are.

m=70g select d

Oh, no, how is it different from him.

1---3 2-methyl-2-pentene, 3-methyl-2-pentene, 4-methyl-2-pentene.

Benzyl chloride requires one mole of sodium hydroxide to give the benzyl alcohol hydroxyl group.

The chlorine on the benzene ring requires one mole of sodium hydroxide to get the phenolic hydroxyl group, and the phenolic hydroxyl group needs one mole of sodium hydroxide to get the sodium phenol.

So it's 1+1+1=3

The products obtained by coheating and acidifying the four products known as D or E and alkaline KMNO4 solution are Hooc-(CH2)10-COOH and CH3-(CH2)4-COOH or CH3-(CH2)7-COOH and Hooc-(CH2)7-COOH, (the isomers have the same number of carbon atoms) can be up to the molecular formula CH3-(CH2)10-CH=CH-(CH2)4-COOH, CH3-(CH2)7- ch=ch-(ch2)7-cooh, and the structure of b can be obtained by combining the two formulas: ch3-(ch2)7-ch=ch-ch2-ch=ch-(ch2)4-cooh

Since D and E are isomers of each other, we must first ensure that the number of carbon atoms is equal, then we can consider Hooc-(CH2)10-COOH and CH3-(CH2)4-COOH to be the products of D.

CH3-(CH2)7-COOH and HOOC-(CH2)7-COOH are E products according to the reaction mechanism.

d is ch3-(ch2)4-ch=ch-(ch2)10-coohe is ch3-(ch2)7-ch=ch-(ch2)7-coohb and hydrogen are addition reactions, d and e are different from the place where the addition should be generated, so according to the title, b can be obtained with two double bonds ch3-(ch2)4-ch=ch-ch2-ch=ch-(ch2)7-cooh

According to the conditions, the average molar mass of the mixed gas is 26g mol, that is, the molar mass of a gas is greater than 26, and the mass of a gas is less than 26, so one of the gases is CH4 (whether in alkanes or alkenes, only CH4 is less than 26).

Olefins, alkynes, and benzene can fade bromine water.

Because they have carbon-carbon double bonds.

There are only alkenes here so alkynes and benzene are not discussed.

Let the amount of methane be x mol and the amount of unknown olefin be y mol, and x+y= can be obtained from the title

Let the molecular formula of the unknown olefin be his general formula.

The molar mass of the olefin is 14n g mol

According to the relation: molar mass of an unknown olefin * amount of an unknown olefin substance = mass of bromine water weight gain gram can be obtained: the product of y and n is.

At this point, we're going to return to the first condition.

It is to substitute the three olefins in the answer.

Calculate which olefin's n value meets the condition that the molar mass of the gas mixture is 26, and the result is n=4, so d is selected

The relative density of option d to hydrogen is 13, that is, the relative molecular weight = 13 * 2 = 26, and only the relative molecular weight of methane is less than 26 among all hydrocarbons, and all must have methane.

In the standard case) i.e., the mass = where the methane mass = , i.e., so the unsaturated hydrocarbon has, its relative molecular mass = , is C4H8.

dMean molar mass = 13 * 2 = 26 g mol

Of all hydrocarbons, only methane has a relative molecular weight of less than 26, and all must have methane.

Let the amount of methane be x mol and the amount of unknown olefin be y mol and the molar mass m

From the proposition, we can get x+y= y*m= m*(y(y+x))+16*(x(x+y))=26

If m=56 is obtained, it is c4h8

A is correct, there is hydrogen generation, the original hydrogen element valency is not 0, and the h valency of the generated hydrogen is 0, so a redox reaction occurs.

b is wrong, it can be calculated by averaging the valency, h in b5h9 is +1 valence, b is -9 5 valence, b in b2o3 is +3 VALENCE, SO THE NUMBER OF TRANSFERRED ELECTRONS IS (3+9 5)*10=48mol. It can also be calculated from the element oxygen. It turns out that o in oxygen is 0 valence, and o in the product is -2 valence, so the number of transfer electrons is 12*2*2=48mol.

C is wrong, the general formula is bnhn+4, b8h12

d correct. It's just that according to the general formula of B10H14 and B5H9, there is no diagram.

b5h9 is unfamiliar, it doesn't matter. Calculate with o2:

O2 is 0 valence before the reaction and -2 valence after the reaction.

A total of 24 O had a change, so 24 electrons were transferred.

i.e. burn 2 mol of B5H9 to transfer 24 mol of electrons. Burn 1 mol B5H9 to transfer 12 mol electrons.

So b is wrong.

Option A, the so-called nitrification reaction is essentially a substitution reaction, the benzene ring reacts with nitric acid, and the -NO2 (nitro group) in nitric acid replaces the position of hydrogen on the benzene ring, and each nitro group is ortho-substituted. Generally speaking, the substitution reaction can occur with a benzene ring (nitrification is included in the substitution reaction), so it can also be directly said that nitrification can occur with a benzene ring.

In option a, both can nitrify on the benzene ring with concentrated nitric acid, so a is correct;

In option B, the functional groups contained in the fluorescein and oxidized fluorescein molecules are different, and they are not homologues, so B is wrong;

In option C, only -COOH can react with NaHCO3, and the phenolic hydroxyl group (the hydroxyl group is directly attached to the benzene ring) cannot react with NaHCO3, so C is wrong;

In option d, all atoms directly connected to the benzene ring must be in the same plane, so there are at least 7 carbon atoms in the fluorescein and oxidized fluorescein molecules, so d is wrong.

How to judge: Generally speaking, nitrification can occur if there is a benzene ring! Nitrification is the process of introducing a nitro group (NO2) into the molecule of an organic compound, which is a monovalent group formed by the loss of a hydroxyl group in nitric acid.

Benzene, concentrated nitric acid: C6H6 + HNO3 (Ho-NO2)--C6H5NO2 + H2O

The figure above shows the nitrification reaction of naphthalene and concentrated nitric acid.

Like this one, it should be done using the method of elimination:

a, first of all, it is not sure whether nitrification can occur.

b, it's certainly not a homologue. One has -COOH and the other has C=OC, and the product cannot react with NaHCO3

D, it is a wrong drop, and the product has taken off a C, which should be less than 7, so choose A at this time

Good luck with your studies

The nitrification reaction is a reaction that replaces the nitro group, and the phenols in the above picture can be nitrated.

Definitely choose A, it will not be a homologue, the standard of homologues is more or less a number of hydrocarbons2, silver oxide can not react with sodium bicarbonate, because the acidity of phenol is too small, there can be up to 8 or more carbon atoms on the fair side.

OK. Nitrification refers to the formation of nitro substitution products by attacking the benzene ring after nitric acid removes the hydroxyl group to become a nitrate cation. It is enough to have an active h.

Nitrification is a reaction process in which nitro (-NO2) is introduced into organic molecules.

.The molecular weight of NH4 is 18, the molecular weight of CH4 is 16, and the molecular weight of H2 is 2, so the volume ratio is x:y:z.

18x+16y+2z=10*(x+y+z) simplification yields 4x+3y=4z, takes x=1, y=1, and solves z=7 4, so x:y:z =4:4:7

2.(1) Because of the excess oxygen, methane is completely burned to form CO2 and H2O equations CH4+2O2=CO2+2H2O

16 64 44 36 Soda lime only absorbs CO2 and H2O produced by the complete combustion of methane, so it is the total mass of both. According to the equation.

The mass of ch4 is 16*v=

The remaining gases are oxygen, v=840-168=672ml(2), and the volume ratio of methane to oxygen can be obtained from the above question: 168:672=1:4

The molecular mass of ammonia is 17, methane 16, hydrogen 2, so the ratio of ammonia to hydrogen is 2:17 Methane, hydrogen 2:16, the quantity ratio of substances between ammonia and methane has no connection, it is arbitrary ratio 2The amount of weight gain of soda lime is the amount of CO2.

As soon as the amount of CO2 is calculated, you can know the amount of CH4 and the amount of O2 participating in the reaction, and by subtracting the volume of CH4 from the total volume and the volume of O2 participating in the reaction, you can deduce the volume of O2 that does not participate in the reaction, so that the volume ratio of methane to oxygen in the original gas mixture can be calculated.

1.The molecular mass of ammonia is 17, methane 16, hydrogen 2, so the ratio of ammonia to hydrogen is 2:17 Methane, hydrogen 2:16, the quantity ratio of substances between ammonia and methane has no connection, it is arbitrary ratio 2The amount of weight gain of soda lime is the amount of CO2.

As soon as the amount of CO2 is calculated, you can know the amount of CH4 and the amount of O2 participating in the reaction, and by subtracting the volume of CH4 from the total volume and the volume of O2 participating in the reaction, you can deduce the volume of O2 that does not participate in the reaction, so that the volume ratio of methane to oxygen in the original gas mixture can be calculated.

Gain weight for the mass of CO2 absorbed.

v=2)ch4+2o2==co2+2h2o

x yx=y=

When there is an excess of methane, the amount of excess is.

Methane is 305 + 230 = 535 ml with a volume ratio of 535:305 When there is an excess of oxygen, oxygen is 305 + 230 = 535ml with a volume ratio of 305:535

1) Let the hydrocarbon be cxhy, y=12, x>=5

2）c5h12

C(CH3)4 neopentane or 2,2-dimethylethane.

If the structure is simplified, it is a C with methyl CH3 on the top and bottom and left and right - solution steps: 1) CXHY+(X+Y4)O2=X CO2+ Y2 H2O

The increase in volume y 4 in the equation 1 a is equal to the actual volume increase (3+a)-(1+a).

Available; y=12。Whereas, x>=5 is derived from the saturation of hydrocarbons.

2) Substitute y=12, a=9=x+3, x=6.

That's when you find out that the answer is wrong, but it says that there is an excess of oxygen, and the 6 we get is obtained with the right amount of oxygen. That is to say, only x<6 is the correct solution, so 5<=x<6, i.e., x=5

Because water is gaseous at 400 K (i.e., about 127 degrees Celsius), the equation CxHY+(X+Y 4)O2 Xco2+Y 2H2O shows that Y=12 is solved by Y2+X-(X+Y4)-1=2, so the number of hydrogen atoms in the molecule (1) is 12(2) Because x+y 4 9, get x 6Therefore, it may be C6H12 or C5H12 (no matter how little carbon is possible).

3) is monodilute, so C5H12, containing four methyl groups, is (CH3)2C=C(CH3)2,2,3-dimethyl-2-butene.