Senior 1 Chemistry Questions Organic Matter, High School Organic Chemistry Questions

The mixture in the other test tube is shaken under the light and then stands, the solution is stratified, and the upper and lower layers of liquid are almost colorless", which is due to the substitution reaction of n-ethane and bromide water under the light to generate ethyl bromide and hydrogen bromide, ethyl bromide is heavier and is in the lower layer, and hydrogen bromide is volatile in the upper layer (like hydrochloric acid), and the volatilized hydrogen bromide encounters the same volatile concentrated ammonia to produce ammonium bromide, ammonium bromide is a salt and a solid particle, so "there is white smoke".

Add 10 ml of bromine water and 5 ml of n-hexane to each of the two tubes. -- What happens at this point is extraction, where n-hexane extracts BR2 from bromine water.

The mixture in a test tube was shaken in a dark place and allowed to stand, and the liquid was divided into two layers, the upper layer was an orange-red bromine hexane solution, and the lower layer was dilute bromine water that had almost faded into colorlessness. The mixture in the other test tube is shaken under light and then standed, and the solution is also stratified, and the upper and lower layers of liquid are almost colorless,-- at this time, pay attention to the difference in whether there is light or not, that is, under light conditions, hexane and Br2 have a substitution reaction, similar to the substitution of methane to Cl2.

Dip a glass rod into concentrated ammonia and stick it above the liquid level of the test tube, and white smoke appears. - At this time, it should be noted that white smoke is a characteristic phenomenon of the combination of NH3 and HCl, so it is considered that the HCl generated by the substitution reaction in the previous step is dissolved in water to obtain hydrochloric acid, but it has a certain volatility, and when it encounters a glass rod dipped in concentrated ammonia, NH3 volatilized by ammonia meets HCl volatilized by hydrochloric acid, and NH4Cl white smoke is formed.

What conclusion do you draw from the phenomenon of "white smoke appears when concentrated ammonia water is dipped in a glass rod and stretched above the liquid level in the test tube"? --Hydrochloric acid was formed in the front.

Is it related to the properties of hexane? --Related to the nature of hexane, i.e., a substitution reaction has occurred earlier.

Under light conditions, bromine water and n-ethane can undergo substitution reactions to generate hydrogen bromide combined with ammonia to form ammonium bromide as white smoke.

The density of the vapors of this substance to the air is.

So the molecular weight of the gas is.

Generate water and. It is explained that 1 mol of organic matter produces 4 mol of water and 3 mol of carbon dioxide.

So the molecular formula is C3H8O

ch3ch2ch2oh

It should be n-hexane. Ethane is a gas.

Under light conditions, bromine water and n-hexane can undergo substitution reactions to form hydrogen bromide combined with ammonia to form ammonium bromide as white smoke.

Obvious mistakes ......

Ethane is in gaseous state at room temperature, and free radical substitution requires anhydrous.

Avogadro's law corollaries that at the same temperature and pressure, the gas density ratio is equal to the molar mass ratio, i.e., 1 2

m1 m2 is known at the same temperature and pressure.

The relative density of the gas mixture to hydrogen is 13

Based on the above inference, we can see that m(mixture) m(h2)=13, that is, the molar mass of the mixed gas is 26g mol

One gaseous alkane and one gaseous olefin.

The molar mass of the gas mixture is somewhere in between.

Alkanes are methane (ethane molar mass is 30g mol, large) and mixed gas is 26g mol

The mass is 65g, and the weight of bromine water increases.

This is the quality of the olefin.

There is methane, and the amount of olefin material is.

The quantity ratio of matter is::1

The molar mass of the olefin is 35, divided by 14 (CH2), and 4 is obtained, which means that butene has three isomers.

Two backbones of 4 carbon (1-butene, 2-butene) and one backbone of 3 carbons (2-methyl-1-propylene).

The relative density of organic vapor to H2 is 23, and the formula of organic matter is carbon dioxide and water respectively, and the oxygen mass is 2:6:1, and then the organic matter is C2H6O according to the formula

m=23×2=46

n=n(c)=

n(h) = so 1mol of organic matter contains.

n(c)=n(h)=

n(o)=(46-12 2-6) 16=1mol, so this organic matter is c2H6O

The relative molecular mass is 46, the organic matter is, where carbon, hydrogen, and this organic matter is C2H6O

In fact, this problem must first know that the weight it adds is the amount of the initial reactant, then it means that there is no gas and water generated, and then start to write the reaction equation, note that it is the total reaction equation, for example: 2 H2+ O2=2 H2O(1),2 Na2O2 +2 H2O= 4 Na2O2+ O2(2), substitute (1) into (2) to get 2 H2+ O2+2 Na2O2=4 NaOH+O2, cancel both sides of this total equation, and become 2 H2+2 Na2O2=4 Naoh, obviously there is no generation of gas and water, then it is the correct solution, in fact, I am very bad, I want to tell you the correct answer, choose D, the other few are the same solution, practice it yourself, say a lot of words, Oxygen is generated, so no.

Combustion produces H2O and CO2

2 Na2O2+2 H2O=4 NaOH+ O2 Na2O2 weight gain mass is equivalent to H mass.

2 Na2O2+2 CO2=2 Na2CO3+ O2 Na2O2 weight gain mass is equivalent to CO mass.

According to the question, the mass of combustibles is the same as the mass of gain, so the amount of substance c in the original combustible = the amount of substance of o.

So choose D

B4 and 5 are fully burned in oxygen, the products are water and carbon dioxide, and the total mass increases the mass of oxygen.

Water + carbon dioxide + sodium peroxide = sodium carbonate + oxygen, oxygen is released, and the solid mass increases wg

The molar mass of organic matter = 2g mol 43 = the amount of substance of 86g mol organic matter =

The amount of CO2 = 66g 44g mol=, the mass of carbon in organic matter = the amount of H2O substance=, the mass of hydrogen in organic matter =

As it can be seen, it does not contain oxygen.

The molecular formula of 1mol of organic matter is CXHY x= =6 y= so the molecular formula is C6H14

There are four methyl groups, which means that there are 2 carbons left in the molecule, so there are only two possibilities, 1) there are 2 methyl groups on each carbon, obtain: (CH3)2CH-CH(CH3)2, dimethylbutane.

2) One carbon and one methyl, and the other carbon has three methyl groups to obtain: CH3-CH2-C(CH3)3,2,2-dimethylbutane.

Solution: (1) Anaerobic elements.

2) The possible structure of the hydrocarbon is shortened and named:

CH3C(CH3)2CH2CH3 2,2-DimethylbutaneCH3CH(CH3)CH(CH3) CH3 2,3-Dimethylbutane.

c3h8+5o2===3co2+4h20

ch4+2o2===co2+2h20

From the two equations, it can be seen that when the gas appliance of liquefied petroleum gas (represented by C3H8) is switched to natural gas, the amount of oxygen consumed by the same mentioned gas is reduced, so the air inlet of the air should be reduced and the air inlet of the gas should be increased.

In this topic, pay attention to one place, "Removal of carbon dioxide and water vapor generated", indicating that the temperature is greater than 100 degrees Celsius. Alkanes below 8 carbons, at 100 degrees Celsius, are gases.

So here, it can be simply estimated that at this temperature, the alkane is a gas. According to the information in the topic, the amount of oxygen consumed by the combustion of 2 ml of alkanes is between 13 and 19 ml, which cannot be equal to 13 or 19. According to the combustion general formula of alkanes, the following formula can be obtained.

2 19<2 (3N+1)<2 13, and the molecular formula of 4 this alkane is C5H12. There are three isomers, which are relatively simple.

Isn't this topic finished?。。 I won't do it, just sit back and wait for the master.

Alkane combustion: CNH2N+2+(3N+1) 2O2=**

3n+1) 2 is greater than 13 and less than 19 solution n=5

According to the title, 1 mol of mixed gas is fully burned to produce 2 mol of CO2 and molH2O, so 1 mol of mixed gas contains 2 mol of carbon atoms and mol. of hydrogen atoms

The molecular formula of ethylene is C2H4, so there are also 2 carbon atoms in the molecule of another hydrocarbon A, and the hydrogen atom is greater than 4, and only C2H6 is eligible

Therefore, the molecular formula of a is C2H6

Suppose the amount of substance containing x mol of ethylene and ethane in 1 mol of mixed gas is (1-x) mol, then 4x + 6 (1-x) =

x = , i.e., 1 mol of mixed gas contains mol of ethylene and mol ethane.

The mass fraction of ethylene is, and the mass fraction of ethane is.

Through the ratio, it can be seen that mc:mh=, 1mol contains 2mol of carbon and hydrogen; Ethylene is C2H4, so it can be seen that the number of C atoms in the other hydrocarbon A is the same as that of ethylene, which is 2, and H atoms are more. Among hydrocarbons, there are more H atoms than alkene, that is, there are alkanes.

Therefore, it is known as ethane: C2H6. The ratio is obtained by the ratio of hydrogen atoms:

2x+3（1-x）=，x=；

The mol ratio is:; The ratio of mass fraction can be obtained by multiplying the mol ratio of the copper drum by the atomic weight, which will not be described here.

d Analysis:

There is -OH and -COOH in molecule a, so esterification can occur.

b (way of doing questions, calculating the molecular formula and putting it to the final exclusion, special for the college entrance examination) c molecule has no benzene ring.

d has a c=c unsaturated bond, so it can discolor the bromine water.

Your question has been answered (meow.

I'm very happy if I adopt it

A pair, containing carboxyl and hydroxyl groups can be esterified.

c is wrong, because that is a ring with a double bond, not a benzene ring.

It contains -OH and -COOH, which can be esterified.

b.The chemical formula is C7H10O5, the six-membered ring is C6H12, there is a double bond, subtract 2 is C6H10, and add the number of O atoms to get it.

c.Wrong, it's not the benzene ring, the benzene ring is three c=c, c-c alternately arranged.

d.False, shikimic acid contains c=c, which can discolor bromine water.