There are 4 carbon atoms on the C6H12O backbone, which can undergo a silver mirror reaction, and its

Conditions for the substance in which the silver mirror reaction occurs:

1.Formaldehyde, acetaldehyde, glyoxal and other aldehydes contain aldehyde groups (such as various aldehydes, as well as a certain ester of formate, etc.).

2.Formic acid and its salts, such as HCOOH, HCOONA, etc.3Formate esters, such as ethyl formate HCOOC2H5, propyl formate HCOOC3H7 and so on4Sugars containing aldehyde groups in molecules such as glucose and maltose.

Because there is only one oxygen in the molecular formula, the condition is excluded.

Therefore, a substance with the formula C6H12O must contain an aldehyde group-cho listing 6 c atoms, and all possible structures of four c's on the backbone c c-c-c-c c-c

cAccording to the location of the aldehyde group, there are the following structures.

ch₃cho ch₃ch₃ cho ch₃ch₃-ch-ch-ch₃ ch₃-ch-ch-cho ch₃-ch₂-c-ch₃ ch₃-ch₂-c-cho

ch₃ ch₃ch₃

cho-ch₂-c-ch₃

ch Therefore, there are five types of structures that meet this question.

Hope it helps.

a. It can interact with phenylhydrazine but does not react with silver mirror.

Aldehydes can be excluded; d can occur silver mirror reaction, but not iodoform reaction, can exclude acetaldehyde.

and methyl ketones; E, on the other hand, can undergo iodoform reaction without silver mirror reaction, then E should be methyl ketone. From the above, it can be concluded that:

A: CH3CH2C = OCH (CH3)2,2-methyl-3 pentanone;

B: CH3CH2CH(OH)CH(CH3)2,2-methyl-3pentanol, C:CH3CH2CH=C(CH3)2, i.e., 2-methyl-2-pentene, D: CH3CH2CHo, propionaldehyde;

e: Acetone. 2.It has optical rotation.

That is, there must be chiral carbon. C can react with Na to release H2, which is formaldehyde or ethanol.

Extrapolating backwards, you can see:

A: H-C-O-CH(CH3)CH2OH,3-hydroxy-carboxylic acid isopropanol.

B: Hoch(CH3)CH2OH,1,2-propanediol.

c: Formaldehyde. <>

Summary. C10H12O3 is the molecular formula of phenylethyl alcohol. Phenylethanol is an isomer with two substituents, and the substituents are located on the benzene ring.

These two substituents are methyl and ethyl respectively. The two isomers are:1

Methylphenylethanol (O-methylphenylethanol), with the molecular formula C8H10O2, is capable of silver mirror reaction. 2.Ethylphenyl ethanol (P-ethylphenylethanol), with the molecular formula C8H10O2, can react with sodium bicarbonate to form acetic acid.

Both can only undergo a silver mirror reaction or react with sodium bicarbonate if the substituents on the benzene ring are methyl or ethyl.

Come on. C10H12O3 is the molecular formula of phenylethyl alcohol. Phenylethanol is an isomer with two substituents, and the substituents are located on the benzene ring.

These two mutation groups are methyl and ethyl respectively. The two isomers are:1

Methylphenyl ethanol (O-methylbenzene resistant ethanol), the molecular formula is C8H10O2, which can undergo silver mirror reaction. 2.Ethylphenyl ethanol (P-ethylphenylethanol), with the molecular formula C8H10O2, can react with sodium bicarbonate to form acetic acid.

Both can only undergo a silver mirror reaction or react with sodium bicarbonate if the substituents on the benzene ring are methyl or ethyl. Socks defeated.

How many kinds are there?

Both. How can it be.

Which two. C10H12O3 is the molecular formula of phenylethyl alcohol. Phenylethanol is an isomer with two substituents, and the substituents are located on the benzene ring.

These two mutation groups are methyl and ethyl respectively. The two isomers are:1

Methylphenyl ethanol (O-methylbenzene resistant ethanol), the molecular formula is C8H10O2, which can undergo silver mirror reaction. 2.Ethylphenyl ethanol (P-ethylphenylethanol), with the molecular formula C8H10O2, can react with sodium bicarbonate to form acetic acid.

Both can only undergo a silver mirror reaction or react with sodium bicarbonate if the substituents on the benzene ring are methyl or ethyl. Socks defeated.

In the isomer of C10H12O3, there are only two substituents on the benzene ring, and there are many kinds of benzene ring that can not only undergo silver mirror reaction, but also with bicarbonate.

That's sodium bicarbonate.

C10H12O3 is the molecular formula of phenylethyl alcohol. There are many isomers of phenyethanol, but there are only two substituents on the benzene ring, and only one can react with both silver mirror and sodium bicarbonate. Specifically:

Liling Jiajia rubberylphenyl ethanol (O-methylphenylethanol), with the molecular formula C8H10O2, can undergo silver mirror reaction and reaction with sodium bicarbonate. This is an isomer with only two spring substituents on the benzene ring in phenylethanol and can react with both silver mirror and sodium bicarbonate.

Unsaturation = ((2*4)+2)-8) 2=1

It can also be silver, so there is an aldehyde group, which can be hydrolyzed, so it is an ester, combined with two points, and it is a formate ester.

So HCOOCH2CH2CH3, propyl carbamate, or HCOOCH(CH3)2, isopropyl formate.

The molecular formula C6H12O can react with phenylhydrazine, but does not undergo silver mirror reaction. A is catalytically hydrogenated to obtain B with the molecular formula C6H14O, compounds A and B, the molecular formula is C3H6O, both can react with the bright cluster phenylhydrazine to form phenylhydrazone, A can Zheng crude silver mirror reaction, and B can't, B can undergo iodoform reaction, and A can't, try to speculate the macro structure and name of A and B.

It is aldehydes that occur in the silver mirror reaction, so it is ch3ch2chob that can be used to react iodoform methyl ketone is a group containing the dactyl group with -coch3, so a ch3ch2chob ch3coch3

According to my speculation, the auspicious nature of the substance is: CH3CH2COCH(CH3)2, and the reaction equation is as follows:

1. Oximme reaction balance: CH3CH2COCH(CH3)2+NH2OH-->CH3CH2C(NH)CH(CH3)2

2. Hydrogenation reaction: CH3CH2COCH(CH3)2+H2-->CH3CH2CH(OH)CH(CH3)2

3. Dehydration reaction: ch3ch2ch(oh)ch(ch3)2--> ch3ch2ch=c(ch3)2

4. Oxidation reaction: ch3ch2ch=c(ch3)2+o3-->ch3ch2cho+ch3coch3

5. Silver mirror reaction: ch3ch2cho+ag(nh3)2-->ch3ch2coonh4+ag+nh3+h2o

6. Iodoform reaction: CH3COCH3+NAIO-->CH3+CH3COONA

Answer] D Answer Analysis] Test Question Analysis: After deducting -CHO, C5H11, that is, the pentyl group, is left, and the position isomerization and carbon chain isomerization are carried out, and it can be seen that there are 8 isomers.

Test site: Isomers with aldehyde groups.

Comments: The difficulty of this question is moderate, you should be careful when doing the question, the key is to determine that there is an aldehyde group that can occur in the silver mirror reaction, and the remaining alkyl group is discussed with isomers. Find out all the isomers.

There are only four types of structures that are required for compounding.

For the top two structures, the NMR hydrogen spectrum has only four sets of peaks.

C5H10O, the unsaturation is 1°, judging from the silver mirror reaction and addition reaction that can occur in the Huai oak blind, there must be an aldehyde group-CHO. The aldehyde group is at the end of the carbon chain, and the lead is empty after addition with hydrogen, which is the primary alcohol with the hydroxyl group at the end.

Let's start by drawing five carbon chain isomers like burying a carbon atom: c c

On this basis, c-c-c, c-c-c-c, c-c-cc, hydroxyl group is added to the end of the carbon chain

ch3ch2ch2ch2ch2oh、(ch3)2chch2ch2oh、ch3ch2ch(ch3)ch2oh、(ch3)3cch2oh