Is C2H5OH O2 CH3CHO H2O a dehydrogenation elimination reaction from adjacent C and O atoms

No, it is hydroxyl hydrogen and alpha hydrogen (that is, carbon linked to the carbon attached to the hydroxyl group, or hydrogen separated by 1 carbon-oxygen bond, 1 carbon-carbon bond and 1 carbon-hydrogen bond to the hydroxyl group) participate in the oxidation reaction. If there is no alpha hydrogen, this reaction cannot take place.

No, the definition of elimination reaction is.

The reaction in which an organic compound removes a small molecule (such as water, hydrogen halide, etc.) from a molecule under appropriate conditions to form an unsaturated (double or triple bond) compound is called a elimination reaction. Alcohols and halogenated hydrocarbons can be eliminated.

In this reaction, hydrogen is sent down from adjacent C and O atoms, but it is not a dissipation reaction.

No, the reaction in which an organic compound removes a small molecule (such as water, hydrogen halide, etc.) from a molecule under appropriate conditions to form an unsaturated (double or triple bond) compound, which is called an elimination reaction.

It is hydroxyl hydrogen and alpha hydrogen (that is, carbon connected to the carbon attached to the hydroxyl group, or hydrogen separated by 1 carbon-oxygen bond, 1 carbon-carbon bond and one carbon-hydrogen bond separated from the hydroxyl group) to participate in the oxidation reaction. If there is no alpha hydrogen, this reaction cannot take place.

Yes. It is a carbon-oxygen double bond formed by the simultaneous elimination of a hydrogen atom from the atom of carbon attached to oxygen.

You can write out his structure and look at it.

Answer]: Because of CH3

The O-H bond in OH is highly polar, hydrogen is easy to dissociate, and oxygen is also easy to accept negative charges.

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ch3ch2)2c=ch2 is oxidized after hydrogen boride.

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Hello dear, it is a pleasure to serve you. (ch3ch2)2c=ch2 after hydrogen boride and potato in oxidized mu, anti-martensis addition. The Marshalls Rule states:

In the electrophilic addition reaction of alkenes, the positive group (H atom) of the addition reagent (HBR) will be added to the carbon atom with fewer substituents (or more hydrogen) in the double bond (or triple bond) of the olefin, and in the presence of H2O2, it is an anti-Marhalobis addition CH3CH2CH(CH3)CH2BR

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1.Because of the silver mirror reaction, a has an aldehyde group, and the aldehyde group is hydrogenated and reduced to alcohol, and the alcohol and concentrated sulfuric acid are aluminized and eliminated to form an alkene, so A: ch3ch2cho b: ch3ch2ch2oh c: ch2=chch3

2.Because the silver mirror reaction can not occur, so A is aldehyde-free, there is a ketone group, and then it is also hydrogenated to the original alcohol, because C can not be bromine water fading, so C is an alcohol under the catalysis of concentrated sulfuric acid substitution reaction to form ether so A: CH3CoCH3 B:

ch3chohch3 c:ch(ch3)2-o-ch(ch3)2

3.According to the title, A is alcohol, and it can fade bromine water, so it is an enol, so B reduces the carbon-carbon double bond to form alcohol, and the alcohol is eliminated to form an ene. So A:

CH2 = CHCH2OH B: CH3CH2CH2OH C: CH2 = CHCH3,1,1, propionaldehyde, 1-propanol, propylene.

2, acetone 2-propanol diisopropyl ether.

3, propylene alcohol 1-propanol, propylene. ,0,The molecular formula of organic matter A is C3H6O, which is reduced to B, and B can be coheated with concentrated sulfuric acid to form C

1) If A can have a silver mirror reaction, C can make the bromine water fade;

2) If A cannot occur the silver mirror reaction, C can not fade the bromine water;

3) If A cannot have a silver mirror reaction, but can react with sodium metal to release H2, A and C can fade bromine water.

Write out the structure of each sub-question a, b, and c according to the requirements of the omission.

ch3ch2oh + ch3ch2oh→ ch3ch2-o-ch2ch3+h2o

The hydroxyl H in the first ethanol molecule is replaced by CH3CH2O- in the other ethanol molecule, so it is a substitution reaction.

The substitution reaction and the metathesis reaction are very similar.

ch3ch2oh + ch3ch2oh→ ch3ch2-o-ch2ch3+h2o

The hydroxyl H in the first ethanol molecule is replaced by CH3CH2O- in the other ethanol molecule, which is a substitution reaction.

The substitution reaction and the complex fraction reaction are very similar to the lack of macros.