A chemistry problem about the structure of a molecule

ABCE is either polyline, or V-shaped.

Among them, of2 and scl2 can be regarded as H2O and H2S in H2S being replaced by F and Cl.

O3 is the V-type which is to be remembered and is the isoelectron of SO2.

And SO2, NO2 - can be calculated using the theory of mutual exclusion of valence shell electron pairs.

Hello, why do I think there is more than one answer? According to the two points to determine a straight line, d is a diatomic molecule, so it must be a straight line. Personally, I think e is also the correct answer.

Since E and CO2 are isoelectrons, both should have the same configuration, CO2 is linear, then E is also linear. O3 is polygonal due to its unique delocalization bond, and AB has two options, and the number of lone pairs can be calculated, both of which are 2, so it is polyline·.

I remember that it seems to be, the outermost electrons of the A element, the two unpaired ones are on one side of the nucleus, so I don't think the one with O and the S in the problem will work. There are only two atoms of iodine chloride, and the two points must be collinear.

Don't spray if you say something wrong.

d is 2 atoms. 2 points determine a straight line. That's definitely straight. I don't know the rest except that O3 is V-shaped.

d is just two atoms, and it has to be linear even if the repulsive force between the atoms is ignored. As for the easiest to choose, D is because there are three covalent bonds and one coordination bond between N and O, which are of different lengths, so they are V-shaped.

Organic matter must have c so 74 12 = 6 surplus 2 but it is impossible to have 6 c, there must always be other elements, so the number of c should be reduced, consider 5 c, so that there is 14 left, if it is c5h14 then it will be supersaturated, saturated to c5h12 then subtract c 4 c to leave 26

Then it can be considered to add o 1 o relative atomic mass 16 and 10 left, so C4 H9 OH is butanol. If you are in the minus c then there is 38 left, consider 2 o and leave 6 to make up h, which can be an ester or an alcohol with a double bond, in short, the unsaturation is determined to be 1

Subtract another C to leave 50 o 3 and 2 to make up H can be written as anhydride.

I don't know if there is any more, after all, the questions are too wide to be comprehensive, so it's best to start with a simple one.

1) Carbon atoms cannot form 5 covalent bonds.

2) The chemical formula of the C10H16A molecule is C4H4 plus 4 CH2 for C10H16

3) B B molecule has 6 C atoms, and each C atom has an H atom, so its molecular formula is C6H6, an isomer of B is the simplest aromatic hydrocarbon is benzene, if the benzene chemical bond is alternating single and double bonds, then its ortho binary substitutes have two kinds.

4) n c6h5-ch=ch2 ch-ch2 under certain conditions ch-ch2]n-

The number of C atoms of C6H5C is 8, and there is an H atom on each C atom, so its molecular formula is C8H8, and an isomer of C belonging to aromatic hydrocarbons is a monomer for the production of a plastic, indicating that an addition reaction can occur in its isomer, and it can be determined that it is C6H5-CH=CH2 that is, styrene according to the unsaturation of aromatic hydrocarbons.

Analysis: (1) The carbon atom in d has 5 covalent bonds, which is not in accordance with the principle.

2) A is a tetrahedral with a total of 6 edges, the molecular formula is C4H4, and a -CH2- group is inserted at the midpoint of each edge, a total of 6 -CH2- groups, and the molecular formula is C10H16

3) The molecular formula of B is: C6H6, and the simplest aromatic hydrocarbon is benzene C6H6 single and double bonds alternately, and the bond length must be inconsistent, so the most telling fact is the C option.

4) The molecular formula of C is C8H8, because it belongs to aromatic hydrocarbons, it contains benzene ring -C6H5, and the remaining -C2H3 group must be styrene C6H5-CH=CH2

n c6h5-ch=ch2 high temperature, high pressure, catalyst) ch-ch2]n-

c6h5

Although this question is written, the molecular formula is randomly guessed, please guide the clear idea, thank you! MA MB = VPA VPB, i.e. MA = MB * (PA PB) = X * MB aromatic hydrocarbons with at least benzene rings; Fourfold.

ch3-ch(oh)-cooh

The 3 carbons are labeled 123 from left to right.

The key test point of this problem is the structure of the COOH carboxyl group, which has the structure of OH, and the other C=O is connected by double bonds, which is called the carbonyl group.

Carbon with double bonds (carbon No. 3) is sp2 hybridized, which means that the double bond of this carbon and the other two single bonds are in a plane, and the bond angle is about 120 degrees.

In this way, carbon 3 and oxygen on the carbonyl group, as well as oxygen attached to the carbonyl carbon and carbon 2 attached to the carbonyl group are all in one plane and are four atoms. The carboxyl group of OH and the last hydrogen can also be in a plane, plus up to five of them.

Then look at carbon 2, since all are single bonds, the bond angle is about 109 degrees and not in a plane, so either the methyl group of carbon 1 rotates to the plane discussed earlier, or the hydroxyl oh above rotates to the previous plane. In either case, there are up to two more atoms that can be in the same plane, so 7.

7 two cases.

1。Hydroxyl carbon carboxyl group.

2。Hydrogen (choose one of three) carbon carbon carboxyl group.

All but 3 Hs in CH3 on the left can be coplanar.

It should be saturated sodium bisulfite, right? Sodium bisulfite can undergo nucleophilic addition reaction with the carbonyl group of aldehydes or ketones to form white crystals of sodium hydroxysulfonate (see figure). Sodium hydroxysulfonate can react with acids or bases to regenerate sodium bisulfite and aldehydes or ketones.

This reaction is often used to purify aldehydes or ketones.

The chain is broken by the oxidation of acidic potassium permanganate, which is generally a carbon-carbon double bond or a carbon-carbon single bond.

Compound A (C8H14O) can rapidly fade the carbon tetrachloride solution of bromine, and can undergo addition reaction with saturated sodium sulfite solution to obtain white crystals. It shows that a has a carbon-carbon double bond, and the addition reaction with saturated sodium sulfite solution is that the double-bonded carbon atom is combined with the sodium atom and bisulfate respectively. A is oxidized by acidic potassium permanganate to two acidic compounds B(C3H6O2) and C, indicating that A double bond is broken, B is propionic acid, and the double bond is in the first.

Between three or four carbon atoms, and there is a carboxyl group in the C molecule, which contains 5 carbon atoms. C reacts with sodium hydroxide solution of iodine to form sodium succinate and iodoform, which is the reaction of iodoform to generate iodoform, which proves that there is CH3CO- in C, so C is CH3COCH2CH2COOH. Therefore, it is inversely deduced that the simple formula of a structure is ch3coch2ch2c=ch2ch2ch3.

Attached: Compounds that can cause haloform reactions:

1 Compounds with a structure of CH3CO- attached to H or C2 Compounds with a structure of CH3CHOH-R (can be oxidized by hypohalites to CH3CO-).

Hello subject, these two substances are not the same kind of substance, and you can't judge whether the substance is the same kind just by the number of atoms. Although the number of atoms is the same, the structure is different, the molecular structure is different, and the chemical properties are not the same, so they are not the same kind of substances. I don't know if you understand??

The first one, there are two kinds of them, 1The electrons of the carbon-oxygen double bond are transferred to the oxygen atom, so that the oxygen atom is negatively charged, the carbon atom connected to the oxygen atom is positively charged, and the negatively charged carbon atom remains unchanged. But this structural instability doesn't make much sense.

2.The electron transfer of the carbon-oxygen double bond is transferred to the oxygen atom, which makes the oxygen atom negatively charged, and then the negatively charged carbon atom and its connected carbon atom share electron pairs to form a double bond. This structure is stable.

Second, maintain the principle that electrons do not produce additional electropositive centers when they are transferred. Such also have two stable hybrids. 1.

The center of negative electricity is on carbon No. 3 (clockwise).2.The center of negative electricity is on carbon No. 5.

You should be able to draw the position of the two double buttons yourself. There is only one way.

The answer is as follows, and the last one forgot how to calculate.

In addition to the benzene ring, option b also has a double bond and a carbon-carbon triple bond.

The carbon atoms on the double bond are all on a plane, but they must no longer be in a straight line, so a is wrong;

The carbon atoms connected by the triple bond must be in a straight line, so c is wrong;

For the leftmost carbon atom is connected by a single bond, the single bond can be rotated, so it can be in a straight line, or it may not be in a straight line, so d is wrong.

Since this compound contains a carbon-carbon double bond, the leftmost carbon must not be in the same line as the carbon of the double bond, so the carbon atoms except the benzene ring cannot be in the same line.

There is a carbon-carbon double bond in the molecular structure, forming a planar structure, for example, the carbon-carbon double bond is ** in a rectangular plane, and the C, H, H, and benzene rings connected on the double bond are each on the four corners of the rectangle, so B is correct.