How to identify hydrogen, methane and ethylene

Example 1] How to identify methane and ethylene by experiment?

Analysis: The essence of this question is how to distinguish saturated hydrocarbons from unsaturated hydrocarbons. Saturated hydrocarbons are relatively stable in nature and are not easy to react under normal conditions. Unsaturated ethylene can undergo addition and oxidation reactions under normal conditions.

Bromine water or potassium permanganate acidic solution should be used as reagent for the reaction with obvious phenomena.

Answer: The two gases are respectively introduced into the bromine water, and the one that can fade the bromine water is ethylene, and the non-fading one is methane. (Or the two gases are respectively passed into the potassium permanganate acid solution, which can fade ethylene, and methane that does not fade.) ）

Dial: Bromine water and potassium permanganate acidic solutions are often used to identify substances. They are both oxidizing agents in inorganic reactions and can react with substances containing equal amounts.

In addition to reacting with unsaturated hydrocarbons in organic reactions, it can also interact with many substances, and I hope that students will pay attention to it in their future studies.

How to identify hydrogen, carbon monoxide, and methane?

Ignite each of the three gases, cover each of the three flames with a cold, dry beaker, and after a while, you will see that the inner walls of the two beakers are blurred and water vapor condenses, indicating that one of the two gases is hydrogen and the other is methane. The kind of gas that does not condense with water vapor on the inner wall of the beaker must be carbon monoxide.

Then inject a small amount of clear lime water into the other two beakers, shake it, and see that the lime water becomes muddy, it means that the gas is methane. If the lime water is not muddy, it means that the gas is hydrogen.

It is ethylene that fades bromine water.

Hydrogen and methane can be ignited and passed through the clarified lime water, and it is methane that becomes turbid, and you can also listen to whether there is a popping sound when ignition, and some are hydrogen.

It is ethylene that fades bromine water.

The products of the remaining two are then passed into the clarified lime water, and the faded methane is methane.

1. First identify alkanes: olefins and alkynes can have an addition reaction with bromine, which makes the carbon tetrachloride solution of bromine fade, while alkanes cannot have an addition reaction. Alkanes can therefore be identified by reacting with a carbon tetrachloride solution of bromine.

2. Secondly, there are two ways to identify olefins and alkynes.

First: Although both alkenes and alkynes can react with the carbon tetrachloride solution of bromine to cause them to fade, the speed of discoloration is different. Olefins can cause the carbon tetrachloride solution of bromine to fade immediately, while alkynes take a few minutes to fade.

Olefins and alkynes can be judged by the speed of fading of the carbon tetrachloride solution of bromine.

Second: alkynes are passed into silver ammonia solution or cuprous ammonia solution, and white and red-brown alkynes can be precipitated respectively, while olefins cannot. Therefore, the gas can be passed into the silver ammonia solution or cuprous ammonia solution, and the phenomenon can be observed by static closure and rest, if the white precipitate or reddish-brown precipitate is precipitated, it is an alkyne, otherwise it is an alkene.

Extended Materials. The general formula of single-stranded olefin molecule is CNH2N, and C2-C4 is gas at room temperature, which is a non-polar molecule, insoluble or slightly soluble in water. The double bond group is a functional group in the olefin molecule, which is reactive, and can undergo addition reactions such as hydrogenation, halogenation, hydration, halohydrogenation, hypohalogenation, sulfate, epoxidation, polymerization, etc., and can also oxidize the cleavage of double bonds to generate aldehydes, carboxylic acids, etc.

The physical properties of olefins can be compared to those of alkanes. The physical state is determined by the mass of the molecule. In standard or room temperature, ethylene, propylene, and butene are gases in simple olefins, linear olefins containing 5 to 18 carbon atoms are liquids, and higher olefins are waxy solids.

Under standard conditions or at room temperature, C2 and C4 olefins are gases; C5 and C18 are volatile liquids; C19 or above solids.

In n-olefins, the boiling point increases as the relative molecular weight increases. The boiling point of n-olefins with the same carbon number is higher than that of alkenes with branched chains. For the alkenes of the same carbon frame, the double bond moves from the end of the chain to the middle of the chain, and the boiling point and melting point are increased.

Olefins are chemically stable but more reactive than alkanes. Considering that the carbon-carbon double bond in alkenes is stronger than the carbon-carbon single bond in alkanes, most of the reactions in olefins have the double bond broken and two new single bonds formed.

ignite and place a dry beaker above the flame: if water droplets are formed on the walls of the beaker, the gas is hydrogen or methane; If there are no water beads on the walls of the beaker, it is carbon monoxide.

Then put it on the table, add an appropriate amount of clarified lime water, if the lime water becomes turbid, the gas is methane; If the lime water is not turbid, the gas is hydrogen.

Methane is relatively stable and does not react with strong oxidants such as potassium permanganate, nor with strong acids and bases. But under certain conditions, methane can also undergo certain reactions.

1. Through the acidic potassium permanganate, ethane cannot make it fade. (Reason: Double and triple bonds can fade potassium permanganate solution).

2. The remaining two gases are passed into the silver nitrate solution, and acetylene can produce a white precipitate with silver nitrate. (Reason: Silver nitrate can react with triple bonds to form a white precipitate).

It is ethylene that fades into the bromine water

CH2 = CH2 + BR2 = CH2BR2BR into acidic potassium permanganate solution:

5 ch2=ch2+12kmno4+18h2o==12mnso4+6k2so4+10co2+28h2o

Ethane is chemically identified, ethyl bromide, ethylene is chemically identified, and by acidic potassium permanganate, ethane cannot be discolored. The remaining two gases are passed into the silver nitrate solution, and acetylene and silver nitrate can form a white precipitate.

Methods to distinguish between olefins and alkanes can be used in this way. Because carbon-carbon double bonds are reductive. Acidic potassium permanganate has strong oxidizing properties. The redox reaction occurs in pairs. It has a color variation, so the identification of olefins and alkanes can be used this method.

Method 1: Unknown gas is introduced into the acidic potassium permanganate solution, and it is ethylene that can fade it, and ethane that cannot be discolored. (Based on the fact that ethylene double bonds can be oxidized by potassium permanganate).

Method 2: Unknown gases are introduced into bromine water, and ethylene is the one that can fade the bromine water, and ethane that cannot be discolored. (based on ethylene double bond energy addition).

Hello, ethylene is an unsaturated alkane, according to the difference between the properties of saturated alkanes and unsaturated alkanes, with bromine water or acidic potassium permanganate solution identification, can make it fade is ethylene, otherwise it is ethane, I hope it will help you, come on!

Ethylene can undergo an addition reaction, ethylene and bromine water undergo an addition reaction, and bromine water fades. Ethane and bromine water do not react.

Light it to see the black smoke emitted.

The darker it is, the greater the concentration of C, the more ethane CH4 and CH2=CH2, and of course the more ethylene C. The darker the smoke, the more vinyl.

It is identified according to the fact that ethylene is not a saturated hydrocarbon (containing double bonds).

A small amount of acidic potassium permanganate solution is respectively introduced into ethylene acetylene, which is not faded into ethane, and then ethylene acetylene is respectively passed into the ammonia aqueous solution of silver nitrate, and the white precipitate (generation of acetylene silver) is acetylene, and ethylene is not obvious.

The equations are c2h4 2br c2h4br2, c2h2 2br c2h2br2.

Potassium permanganate is an oxidizing agent, which is used for organic synthesis, disinfection, oxidation, etc. Contact with ether, ethanol, sulfuric acid, sulfur, hydrogen peroxide, etc.**; In case of glycerin, it immediately decomposes and burns strongly.

The three are respectively passed into the CCL4 solution (or potassium permanganate acidic solution) of BR2, and the discolored one is acetylene, and the non-fading one is ethane.

The remaining two are respectively passed into the ammonia solution of silver nitrate, and the white precipitate produces acetylene, and the non-phenomenon is ethylene.

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The specific operation steps are as follows: 1. Through the acidic potassium permanganate, ethane cannot make it fade. (Reason:.)

Double bonds and triple bonds can fade potassium permanganate solution) 2. The remaining two gases are passed into the silver nitrate solution, and acetylene can form a white precipitate with silver nitrate. (Reason: Silver nitrate can react with triple bonds to form white precipitate)

Ethane structure Ethylene structure Acetylene structure.

First of all, methane and ethylene are both gases, while the last two are liquids. Put two gases into bromine water, the discolored is ethylene, and the non-fading is methane Identify the two liquids with acidic potassium permanganate, the discolored is toluene, and the non-fading is benzene.

Ethylene: is a compound made up of two carbon atoms and four hydrogen atoms. The two carbon atoms are connected by a double bond with each other. Ethylene exists in some tissues and organs of plants, and is converted from methionine under the condition of sufficient oxygen supply.

Benzene: It is a colorless and dry transparent liquid with a sweet smell at room temperature, and has a strong aromatic odor. Benzene is flammable, highly toxic, and a carcinogen.

The identification of methane and ethylene can be based on two aspects: physical and chemical properties. Here are some of the ways to do this:

1.In terms of physical properties:

Methane and ethylene are both gaseous at room temperature, but methane is more stable than ethylene and has less density than ethylene. So you can distinguish between the two by the amount of density.

2.In terms of chemical properties:

1. Methane and ethylene are added to liquid nitrogen respectively, and oxygen is added to ignite the mixed gas. When ignited, methane burns to produce a blue flame, while ethylene burns to produce a yellow flame.

2. Tested with bromine water. Methane or ethylene gas is introduced into bromine water, which exhibits a color change, methane does not undergo the sinlingkai reaction, and ethylene fades the water.

3) Detected with potassium permanganate solution. Methane or ethylene gas is introduced into the potassium permanganate solution, and the potassium permanganate fruit does not change after the methane is introduced, but the ethylene water swim will make it lighter.

These are common methods for identifying methane and ethylene, but it is important to note that these methods are not always very accurate and more accurate detection methods may be relied upon for more complex cases.

Identification of methane and ethylene:

1. Bromine water identification.

Due to its unsaturation, ethylene can be additionally reacted with bromine water at room temperature, which makes the bromine water fade; Methane cannot, so methane and ethylene can be distinguished by bromine water.

2. Identification of potassium permanganate.

In methane and ethylene, a potassium permanganate acid solution is introduced respectively, and the solution is discolored by ethylene and methane. Because ethylene can be oxidized by acidic potassium permanganate solutions.

3. Ignition method.

Igniting two gases, ethylene has a bright flame and black smoke when burning, because ethylene has a higher carbon content than methane, but methane does not.

Extended Materials. The differences between methane and ethylene are:

1. Saturation.

Normally, methane is relatively stable due to its saturation, and ethylene has a carbon-carbon double bond and is unsaturated.

2. Stability.

Methane is stable in nature, does not react with strong oxidants such as potassium permanganate, and does not react with strong acids and alkalis. Ethylene can react with oxidants, strong acids, etc.

3. Addition reaction.

Methane does not undergo an addition reaction, whereas ethylene can undergo an addition reaction.