For example, let's take hydrocarbons of the saturated and unsaturated series.
Definition
First, let's find out what the phenomenon of isomerism is. Depending on the number of carbon atoms in the molecule, the formation of compounds that differ in structure, physical and chemical properties is possible. Isomerism is a phenomenon that explains the diversity of organic substances.
Isomerism of saturated hydrocarbons
How to compose isomers, name representatives of this class of organic compounds? In order to cope with the task, let us first highlight the distinctive characteristics of this class of substances. Saturated hydrocarbons have the general formula SpH2n+2; their molecules contain only simple (single) bonds. Isomerism for representatives of the methane series presupposes the existence of different organic substances that have the same qualitative and quantitative composition, but differ in the sequence of arrangement of atoms.
If saturated hydrocarbons contain four or more carbon atoms, isomerism of the carbon skeleton is observed for representatives of this class. For example, you can create a formula for substances of isomers of the composition C5H12 in the form of normal pentane, 2-methylbutane, 2,2-dimethylpropane.
Subsequence
Structural isomers characteristic of alkanes are composed using a specific algorithm of actions. In order to understand how to compose isomers of saturated hydrocarbons, let us dwell on this issue in more detail. First, a straight carbon chain with no additional branches is considered. For example, if there are six carbon atoms in a molecule, you can create the formula for hexane. Since all alkanes have single bonds, only structural isomers can be written for them.
Structural isomers
To compose the formulas of possible isomers, the carbon skeleton is shortened by one C atom, it turns into an active particle - a radical. The methyl group can be located at all atoms in the chain, excluding the outermost atoms, thereby forming various organic derivatives of alkanes.
For example, you can formulate the formula 2-methylpentane, 3-methylpentane. Then the number of carbon atoms in the main (main) chain is reduced by one more, resulting in two active methyl groups. They can be placed at the same or adjacent carbon atoms, resulting in various isomeric compounds.
For example, you can create formulas for two isomers: 2,2-dimethylbutane, 2,3-dimethylbutane, which differ in physical characteristics. With subsequent shortening of the main carbon skeleton, other structural isomers can be obtained. So, for hydrocarbons of the limiting series, the phenomenon of isomerism is explained by the presence of single (simple) bonds in their molecules.
Features of alkene isomerism
In order to understand how to compose isomers, it is necessary to note the specific features of this class of organic substances. We have the general formula SpN2n. In the molecules of these substances, in addition to a single bond, there is also a double bond, which affects the number of isomeric compounds. In addition to the structural isomerism characteristic of alkanes, for this class one can also distinguish isomerism of the position of a multiple bond, interclass isomerism.
For example, for a hydrocarbon with the composition C4H8, you can create formulas for two substances that will differ in the location of the double bond: butene-1 and butene-2.
To understand how to form isomers with the general formula C4H8, you need to understand that, in addition to alkenes, cyclic hydrocarbons also have the same general formula. Examples of isomers belonging to cyclic compounds include cyclobutane and methylcyclopropane.
In addition, for unsaturated compounds of the ethylene series, the formulas of geometric isomers can be written: cis and trans forms. Hydrocarbons that have a double bond between carbon atoms are characterized by several types of isomerism: structural, interclass, geometric.
Alkynes
Compounds that belong to this class of hydrocarbons have a general formula - SpN2n-2. Among the distinctive characteristics of this class is the presence of a triple bond in the molecule. One of them is simple, formed by hybrid clouds. Two bonds are formed when non-hybrid clouds overlap; they determine the features of isomerism of this class.
For example, for a hydrocarbon with the composition C5H8, you can create formulas for substances that have an unbranched carbon chain. Since there is a multiple bond in the parent compound, it can be positioned differently, forming pentine-1, pentine-2. For example, you can write an expanded and abbreviated formula for a compound with a given qualitative and quantitative composition, in which the carbon chain will be reduced by one atom, which will be represented in the compound as a radical. In addition, for alkynes there are also interclass isomers, which are diene hydrocarbons.
For hydrocarbons that have a triple bond, you can create isomers of the carbon skeleton, write formulas for dienes, and also consider compounds with different arrangements of the multiple bond.
Conclusion
When composing the structural formulas of organic substances, oxygen and carbon atoms can be arranged in different ways, obtaining substances called isomers. Depending on the specific class of organic compounds, the number of isomers may vary. For example, hydrocarbons of the limiting series, which include compounds of the methane series, are characterized only by structural isomerism.
For ethylene homologues, which are characterized by the presence of a multiple (double) bond, in addition to structural isomers, it is also possible to consider isomerism of the position of the multiple bond. In addition, other compounds that belong to the class of cycloalkanes have the same general formula, that is, interclass isomerism is possible.
For oxygen-containing substances, for example, for carboxylic acids, the formulas of optical isomers can also be written.
Let's look at the example of an alkane C 6 H 14.
1. First, we depict the linear isomer molecule (its carbon skeleton)
2. Then we shorten the chain by 1 carbon atom and attach this atom to any carbon atom of the chain as a branch from it, excluding extreme positions:
(2) or (3)
If you attach a carbon atom to one of the extreme positions, the chemical structure of the chain does not change:
In addition, you need to ensure that there are no repetitions. Yes, the structure
identical to structure (2).
3. When all positions of the main chain have been exhausted, we shorten the chain by another 1 carbon atom:
Now there will be 2 carbon atoms in the side branches. The following combinations of atoms are possible here:
A side substituent may consist of 2 or more carbon atoms connected in series, but for hexane there are no isomers with such side branches, and the structure
identical to structure (3).
A side substituent - C-C can only be placed in a chain containing at least 5 carbon atoms and can only be attached to the 3rd and further atom from the end of the chain.
4. After constructing the carbon skeleton of the isomer, it is necessary to supplement all carbon atoms in the molecule with hydrogen bonds, given that carbon is tetravalent.
So, the composition C 6 H 14 corresponds to 5 isomers:
2) 
3) 
4) 
5) 
Rotational isomerism of alkanes
A characteristic feature of s-bonds is that the electron density in them is distributed symmetrically relative to the axis connecting the nuclei of the bonded atoms (cylindrical or rotational symmetry). Therefore, the rotation of atoms around the s-bond will not lead to its breaking. As a result of intramolecular rotation along C–C s-bonds, alkane molecules, starting with ethane C 2 H 6, can take on different geometric shapes.
Various spatial forms of a molecule that transform into each other by rotating around C–C s-bonds are called conformations or rotary isomers(conformers).
Rotational isomers of a molecule are its energetically unequal states. Their interconversion occurs quickly and constantly as a result of thermal movement. Therefore, rotary isomers cannot be isolated in individual form, but their existence has been proven by physical methods. Some conformations are more stable (energetically favorable) and the molecule remains in such states for a longer time.
Let's consider rotary isomers using ethane H 3 C–CH 3 as an example:
When one CH 3 group rotates relative to another, many different forms of the molecule arise, among which two characteristic conformations are distinguished ( A And B), characterized by a rotation of 60°:
These rotary isomers of ethane differ in the distances between the hydrogen atoms connected to different carbon atoms.
In conformation A The hydrogen atoms are close together (obscure each other), their repulsion is great, the energy of the molecule is maximum. This conformation is called “eclipsed”, it is energetically unfavorable and the molecule goes into the conformation B, where the distances between H atoms of different carbon atoms are greatest and, accordingly, repulsion is minimal. This conformation is called “inhibited” because it is energetically more favorable and the molecule remains in this form for more time.
As the carbon chain lengthens, the number of distinguishable conformations increases. Thus, rotation along the central bond in n-butane
leads to four rotary isomers:
The most stable of them is conformer IV, in which the CH 3 groups are maximally distant from each other. Construct the dependence of the potential energy of n-butane on the angle of rotation with students on the board.
Optical isomerism
If a carbon atom in a molecule is bonded to four different atoms or atomic groups, for example:
then the existence of two compounds with the same structural formula, but differing in spatial structure, is possible. The molecules of such compounds relate to each other as an object and its mirror image and are spatial isomers.
This type of isomerism is called optical; isomers are called optical isomers or optical antipodes:
Molecules of optical isomers are incompatible in space (like left and right hands); they lack a plane of symmetry.
Thus, optical isomers are spatial isomers whose molecules are related to each other as an object and a mirror image incompatible with it.
Optical isomers have the same physical and chemical properties, but differ in their relationship to polarized light. Such isomers have optical activity (one of them rotates the plane of polarized light to the left, and the other by the same angle to the right). Differences in chemical properties are observed only in reactions with optically active reagents.
Optical isomerism manifests itself in organic substances of various classes and plays a very important role in the chemistry of natural compounds.
There are several types of structural isomerism:
carbon skeleton isomerism;
isomerism of the position of multiple bonds;
isomerism of the position of functional groups.
To derive formulas for isomers that differ in the sequence of bonds of carbon atoms in the molecule (carbon skeleton isomerism):
a) create a structural formula for a carbon skeleton of normal structure with a given number of carbon atoms;
b) gradually shorten the chain (each time by one carbon atom) and perform all possible rearrangements of one or more carbon atoms and thus derive the formulas of all possible isomers.
EXAMPLE: Draw up the structural formulas of all isomeric hydrocarbons of the composition C 5 H 12.
1. Let’s create formulas for a carbon skeleton with a normal chain of 5 carbon atoms.
S – S – S – S – S
2. Let's shorten the chain by one carbon atom and carry out all possible rearrangements.
S – S – S – S
4. Let us arrange the required number of hydrogen atoms.
1. CH 3 – CH 2 – CH 2 – CH 2 – CH 3
2. CH 3 – CH – CH 2 – CH 3
3. CH 3 – C – CH 3
To derive the structural formulas of all isomers due to different positions of multiple bonds, substituents (halogens) or functional groups (OH, - COOH, NO 2, NH 2), proceed as follows:
remove all structural isomers associated with isomerism of the carbon skeleton;
graphically move a multiple bond or functional group to those positions to which this is possible from the point of view of carbon tetravalency:
EXAMPLE: Write the structural formulas of all pentenes (C 5 H 10).
1. Let’s create the formulas of all isomers that differ in the structure of the carbon skeleton:
a) S – S - S – S – S b) S – S - S – S c) S – S - S
2. Let's move the multiple connection for cases a) and b)
C = C - C – C – C CH 2 = CH – CH 2 – CH 2 – CH 3
A) C – C - C – C – C C – C = C – C – C CH 3 – CH = CH – CH 2 – CH 3
B) C – C - C – C C = C - C – C CH 2 = C – CH 2 – CH 3
C - C = C – C CH 3 - C = CH – CH 3
C – C - C – C C CH 3
C C - C - C = C CH 2 - CH – CH = CH 2
Thus, for C 5 H 10, five isomers are possible.
EXAMPLE: Draw up the structural formulas of all aromatic hydrocarbons of the composition C 8 H 10.
In the case of aromatic compounds, isomerism of the side chain skeleton and isomerism of the position of substituents in the aromatic ring are possible.
1. Let’s create a structural formula with a normal side chain:
2. Let's shorten the side chain by one carbon atom and make possible rearrangements of CH 3 in the benzene ring.
There are 4 isomers of the composition C 8 H 10.
1. When performing exercises, you must pay special attention to the correct writing of the structural formulas of organic compounds. It is most convenient to use semi-expanded (simplified) structural formulas, in which the bonds between atoms are indicated with dashes, with the exception of bonds with hydrogen atoms. It is advantageous, if possible, to write formulas with a horizontally written carbon chain so that the functional groups located at the end of the chains are located on the right, and the substituents at non-terminal carbon atoms are located under or above the carbon chain:
CH 3 - CH – CH 2 - OH CH 3 – CH 2 – CH - CH 3
CH 2 – CH 2 – C CH 3 - CH – COOH
2. In initial exercises with formulas of aromatic compounds in benzene rings, it is better to write all the C and H atoms. In a simplified representation of benzene rings, substituent atoms and groups must be clearly connected to the atoms of the benzene ring by valence lines.
3. In most cases, it is advisable to write inorganic compounds in reaction equations using structural or semi-expanded structural formulas:
For example: HON instead of H 2 O,
HOSO 3 H instead of H 2 SO 4,
HONO 3 instead of HNO 3
This is not necessary if such compounds participate in ionic reactions, for example when acids react with amines to form salts.
4. Organic reactions can be expressed by equations in which coefficients are added and the number of atoms on the right and left sides is equalized. However, it is often not equations that are written, but reaction schemes. This is done in cases where the process occurs simultaneously in several directions or through a number of successive stages, for example:
Cl 2 CH 3 – CH 2 – CH 2 – Cl + HCl
CH 3 – CH 2 – CH 3
light CH 3 – CH – CH 3 + HCl
or NaOH, t 0 C Cu, t 0 C
CH 3 – CH 2 - Cl CH 3 – CH 2 – OH CH 3 – CH = O
As shown in the examples given in the diagrams, the active reagent is shown above the arrow. For simplicity, the coefficients on the right or left sides of the diagram are not equalized, and some substances, such as, for example, H 2, HCl, H 2 O, Na Cl, etc. those formed during reactions are either not shown in the diagrams at all, or are indicated under the arrow with a minus sign. The direction of transformation of substances in reactions is indicated by an arrow. The reagents and reaction conditions, catalyst, etc. are indicated above the arrow.
For example:
H 2 O, H 2 SO 4,130 0 C
CH 3 – CH 2 – CH 2 – NO 2 CH 3 – CH 2 – COOH + NH 2 OH* H 2 SO 4
Sometimes, under the arrow (preferably in parentheses), the starting substances are indicated, which, as a result of interaction with each other, form a reagent (indicated above the arrow). For example:
R – NH 2 R – OH + N 2 + H 2 O
Consequently, in this case, the reagent - nitrous acid - is formed from sodium nitrite and hydrochloric acid taken in the reaction. Naturally, sodium chloride NaCl is also obtained here, but this compound may not be designated in the diagram as not being directly related to the process. As a rule, redox transformations of organic substances are depicted with simplified reaction schemes in order to draw attention to the oxidation or reduction of an organic compound; the details of the transformations, respectively, of the oxidizing agent or the reducing agent, may not be reflected in the diagram.
For this purpose, the oxidizing agent is represented by the symbol [O], and the reducing agent by the symbol [H] above the arrow. If necessary, the active reagents can be indicated under the arrow (preferably in parentheses).
For example:
CH 3 OH CH 2 = O + H 2 O CH 3 OH CH 2 = O + H 2 O
(K 2 Cr 2 O 7 + H 2 SO 4)
C 6 H 5 NO 2 C 6 H 5 – NH 2 + 2 H 2 O
In recent years, in scientific as well as in educational chemical literature, the nomenclature of organic compounds developed by the International Union of Pure and Applied Chemistry, the IUPAC nomenclature, is predominantly used as a systematic one; it is usually called the “international systematic nomenclature”. In educational literature, rational nomenclature is also used.
1. When starting to perform exercises on nomenclature, you must first of all study this issue in the textbook, where the recommendations for this class of nomenclature systems are discussed in detail. Here only brief characteristics of the recommended nomenclatures are given and examples are given.
2. It is necessary to pay attention to the correct spelling of names. in names according to the international nomenclature, numbers should be separated from words by dashes, and numbers from numbers by commas: 1,4 dibromo – 2,3 – dimethylbutene – 2.
Although it is customary to write the component parts of names together, for didactic reasons, complex names can be separated by dashes.
For example: Name
Methylethylpropyl isobutylmethane can be written and is recommended to be written as follows: methyl - ethyl - propyl - isobutyl - methane.
When the name is broken down into its component parts, the structure of the compound and its formula are more clearly presented.
TASKS FOR CONTROL WORK
Okay, maybe not so much.
To go through everything and not miss a single one, you can come up with several approaches. I like this one: Take ethene (ethylene) CH2 = CH2. It differs from heptene by 5 carbon atoms (C5H10). To sort through all possible isomers, you need to take one hydrogen atom from ethene and give it to the C5H10 fragment. The result is an alkyl C5H11, and it must be added to the ethene residue (ethenyl CH2=CH-) in place of the removed hydrogen.
1) The C5H11 alkyl itself can have several isomers. The simplest one with a straight chain is CH2-CH2-CH2-CH2-CH3 (pentyl or amyl). From it and ethenyl, heptene-1 (or 1-heptene, or hept-1-ene) is formed, which is simply called heptene CH2=CH-CH2-CH2-CH2-CH2-CH3.
2a) If in a pentyl we move one hydrogen from the C2 atom to the C1 atom, we get pentyl-2 (or 2-pentyl, or pent-2-yl) CH3-CH(-)-CH2-CH2-CH3. The dash in parentheses means that the stick needs to be drawn up or down, and that there is an unpaired electron here, and this is where the pentyl-2 will attach to the ethenyl. The result is CH2=CH-CH(CH3)-CH2-CH2-CH3 3-methylhexene-1 or 3-methyl-1-hexene or 3-methylhex-1-ene. I hope you understand the principle of the formation of alternative names, so for the compounds mentioned below I will give only one name.
2b) If in a pentyl we move one hydrogen from the C3 atom to the C1 atom, we get pentyl-3 CH3-CH2-CH(-)-CH2-CH3. Combining it with ethenyl we get CH2=CH-CH(CH2-CH3)-CH2-CH3 3-ethylpentene-1
3a, b) Pentyl is isomerized into a chain of 4 carbon atoms (butyl) having one methyl group. This methyl group can be attached to the C2 or C3 atom of the butyl. We obtain, respectively, 2-methylbutyl -CH2-CH(CH3)-CH2-CH3 and 3-methylbutyl -CH2-CH2-CH(CH3)-CH3, and adding them to ethenyl we obtain two more isomers C7H14 CH2=CH-CH2-CH( CH3)-CH2-CH3 4-methylhexene-1 and CH2=CH-CH2-CH2-CH(CH3)-CH3 5-methylhexene-1.
4a, b) Now in butyl we move the line to the C2 atom, we get 2-butyl CH3-CH(-)-CH2-CH3. But we need to add one more carbon atom (replace H with CH3). If we add this methyl to one of the terminal atoms, we get the pentyl-3 and pentyl-2 already discussed. But the addition of methyl to one of the middle atoms will give two new alkyls CH3-C(CH3)(-)-CH2-CH3 2-methyl-2-butyl- and CH3-CH(-)-CH(CH3)-CH3 2 -methyl-2-butyl-.
By adding them to ethenyl we get two more isomers C7H14 CH2=CH-C(CH3)2-CH2-CH3 3,3-dimethylpentene-1 and CH2=CH-CH(CH3)-CH(CH3)-CH3 3.4 -dimethyl-pentene-1.
5) Now, when building an alkyl, we will leave a chain of 3 carbon atoms -CH2-CH2-CH3. The missing 2 carbon atoms can be added either as ethyl or as two methyls. In the case of addition in the form of ethyl, we obtain the already considered options. But two methyls can be attached either both to the first, or one to the first, one to the second carbon atoms, or both to the second. In the first and second cases we get the already considered options, and in the last we get a new alkyl -CH2-C(CH3)2-CH3 2,2-dimethylpropyl, and adding it to ethenyl we get CH2=CH-CH2-C(CH3)2- CH3 4,4-dimethylpentene-1.
Thus, 8 isomers have already been obtained. Note that in these isomers the double bond is at the end of the chain, i.e. binds atoms C1 and C2. Such olefins (with a double bond at the end are called terminal). Terminal olefins do not exhibit cis-trans isomerism.
Next, we divide the C5H10 fragment into two fragments. This can be done in two ways: CH2 + C4H8 and C2H4 + C3H6. From the CH2 and C2H4 fragments, only one variant of alkyls can be constructed (CH3 and CH2-CH3). From the C3H6 fragment, propyl -CH2-CH2-CH3 and isopropyl CH3-CH(-)-CH3 can be formed.
From the C4H8 fragment, the following alkyls can be constructed -CH2-CH2-CH2-CH3 - butyl-1, CH3-CH(-)-CH2-CH3 - butyl-2, -CH2-CH(CH3)-CH3 - isobutyl (2-methylpropyl ) and -C(CH3)2-CH3 - tert-butyl (2,2-dimethylethyl).
To add them to alkyls, we remove two hydrogen atoms from the ethene molecule. This can be done in three ways: by removing both hydrogen atoms from the same carbon atom (this will produce terminal olefins), or by removing one from each. In the second option, these two hydrogen atoms can be removed from the same side of the double bond (cis isomers are obtained), and from different sides (trans isomers are obtained).
CH2=C(CH3)-CH2-CH2-CH2-CH3 - 2-methylhexene-1;
CH2=C(CH3)-CH(CH3)-CH2-CH3 - 2,3-dimethylpentene-1;
CH2=C(CH3)-CH2-CH(CH3)-CH3 - 2,4-dimethylpentene-1;
CH2=C(CH3)-C(CH3)2-CH3 - 2,3,3-trimethyl butene-1.
CH2=C(CH2CH3)-CH2-CH2-CH3 - 2-ethylpentene-1 or 3-methylenehexane;
CH2=C(CH2CH3)-CH(CH3)-CH3 - 2-ethyl-3-methylbutene-1 or 2-methyl-3-methylenepentane.
CH3-CH=CH-CH2-CH2-CH2-CH3 - heptene-2 (cis and trans isomers);
CH3-CH=CH-CH(CH3)-CH2-CH3 - 4-methylhexene-2 (cis and trans isomers);
CH3-CH=CH-CH2-CH(CH3)-CH3 - 5-methylhexene-2 (cis and trans isomers);
CH3-CH=CH-C(CH3)2-CH3 - 4,4-dimethylpentene-2 (cis and trans isomers);
CH3-CH2-CH=CH-CH2-CH2-CH3 - heptene-3 (cis and trans isomers);
CH3-CH2-CH=CH-CH(CH3)-CH3 - 2-methylhexene-3 (cis and trans isomers).
Well, with olefins it seems like everything. What remains are the cycloalkanes.
In cycloalkanes, several carbon atoms form a ring. Conventionally, it can be considered as a flat cycle. Therefore, if two substituents are attached to the ring (at different carbon atoms), then they can be located on the same side (cis-isomers) or on opposite sides (trans-isomers) of the ring plane.
Draw a heptagon. Place CH2 at each vertex. The result was cycloheptane;
Now draw a hexagon. Write CH2 at five vertices, and CH-CH3 at one. The result was methylcyclohexane;
Draw a pentagon. Draw CH-CH2-CH3 at one vertex, and CH2 at the other vertices. ethylcyclopentane;
Draw a pentagon. Draw CH-CH3 at two vertices in a row, and CH2 at the remaining vertices. The result was 1,2-dimethylpentane (cis- and trans-isomers);
Draw a pentagon. At two vertices, draw CH-CH3 through one, and CH2 at the remaining vertices. The result was 1,3-dimethylpentane (cis- and trans-isomers);
Draw a quadrilateral. Draw CH2 at three vertices, and CH at one, and attach -CH2-CH2-CH3 to it. The result was propylcyclobutane;
Draw a quadrilateral. Draw CH2 at three vertices, and CH at one, and attach -CH(CH3)-CH3 to it. The result is isopropylcyclobutane;
Draw a quadrilateral. Draw CH2 at three vertices, and C at one, and attach the CH3 and CH2-CH3 groups to it. The result was 1-methyl-1-ethylcyclobutane;
Draw a quadrilateral. Draw CH2 at two vertices in a row, and CH at the other two. Add CH3 to one CH, and CH2-CH3 to the other. The result was 1-methyl-2-ethylcyclobutane (cis- and trans-isomers);
Draw a quadrilateral. At two vertices, draw CH2 through one, and at the other two, CH. Add CH3 to one CH, and CH2-CH3 to the other. The result was 1-methyl-3-ethylcyclobutane (cis and trans isomers);
Draw a quadrilateral. At two vertices in a row, draw CH2, at one CH, at one C. Draw CH3 to CH, and to C two groups of CH3. The result was 1,1,2-dimethylcyclobutane;
Organic chemistry is not that easy.
You can guess something using logical reasoning.
And somewhere logic will not help, you need to cram.
As, for example, in this question.
Here's a look at the formulas:
Hydrocarbons corresponding to the formula C17H14 belong to both alkenes and cycloalkanes. Therefore, as Rafail told you in the comment, there are a lot of them. In alkenes (intraclass isomerism) there are three types of isomerism: 1). isomerism of double bond position; 2). carbon skeleton isomerism; 3). and some alkenes have spatial cis- and trans-isomers. And cycloalkanes within this class have closed ring isomerism, and some cycloalkanes have cis and trans isomers. It is necessary to decide on the class of connections.
In fact, there are quite a lot of them, so I won’t list them all:
Here are some of their representatives:
But there are still many of them and, frankly speaking, it is very difficult to remember all the representatives of all isomers of this composition.
Not a very simple task, or rather not a very quick one. I can’t give you all, but more than 20 isomers for the indicated composition:
If your task is to compose drawings, then I sympathize with you, but I found several images with compiled isomer chains:
In general, be strong!
1. Structural isomerism.
2. Conformational isomerism.
3. Geometric isomerism.
4. Optical isomerism.
Isomers- these are substances that have the same composition and molecular weight, but different physical and chemical properties. Differences in the properties of isomers are due to differences in their chemical or spatial structure. In this regard, two types of isomerism are distinguished.
isomerism
structural
spatial
carbon skeleton
Configuration
Conformational
Functional position
Optical
Interclass
Geometric
1. Structural isomerism
Structural isomers differ in chemical structure, i.e. the nature and sequence of bonds between atoms in a molecule. Structural isomers are isolated in pure form. They exist as individual, stable substances; their mutual transformation requires high energy - about 350 - 400 kJ/mol. Only structural isomers - tautomers - are in dynamic equilibrium. Tautomerism is a common phenomenon in organic chemistry. It is possible through the transfer of a mobile hydrogen atom in a molecule (carbonyl compounds, amines, heterocycles, etc.), intramolecular interactions (carbohydrates).
All structural isomers are presented in the form of structural formulas and named according to the IUPAC nomenclature. For example, the composition C 4 H 8 O corresponds to structural isomers:
A)with different carbon skeleton
unbranched C-chain - CH 3 -CH 2 -CH 2 -CH=O (butanal, aldehyde) and
branched C-chain -
(2-methylpropanal, aldehyde) or
cycle - 
(cyclobutanol, cyclic alcohol);
b)with different position of the functional group
butanone-2, ketone;
V)with different composition of the functional group
3-butenol-2, unsaturated alcohol;
G)metamerism
A heteroatom functional group may be included in a carbon skeleton (cycle or chain). One of the possible isomers of this type of isomerism is CH 3 -O-CH 2 -CH=CH 2 (3-methoxypropene-1, ether);
d)tautomerism (keto-enol)
enol form keto form
Tautomers are in dynamic equilibrium, with the more stable form, the keto form, predominating in the mixture.
For aromatic compounds, structural isomerism is considered only for the side chain.
2. Spatial isomerism (stereoisomerism)
Spatial isomers have the same chemical structure and differ in the spatial arrangement of atoms in the molecule. This difference creates a difference in physical and chemical properties. Spatial isomers are depicted in the form of various projections or stereochemical formulas. The branch of chemistry that studies the spatial structure and its influence on the physical and chemical properties of compounds, on the direction and rate of their reactions, is called stereochemistry.
A)Conformational (rotational) isomerism
Without changing either bond angles or bond lengths, one can imagine many geometric shapes (conformations) of the molecule, differing from each other in the mutual rotation of carbon tetrahedra around the σ-C-C bond connecting them. As a result of this rotation, rotary isomers (conformers) arise. The energy of different conformers is not the same, but the energy barrier separating different conformational isomers is small for most organic compounds. Therefore, under ordinary conditions, as a rule, it is impossible to fix molecules in one strictly defined conformation. Typically, several conformational isomers easily transform into each other coexist in equilibrium.
The methods of depiction and the nomenclature of isomers can be considered using the example of the ethane molecule. For it, we can foresee the existence of two conformations that differ maximally in energy, which can be depicted in the form perspective projections(1) (“sawmill goats”) or projections Newman(2):
inhibited conformation eclipsed conformation
In perspective projection (1) the C-C connection must be imagined going into the distance; The carbon atom on the left is close to the observer, and the carbon atom on the right is farther away from him.
In the Newman projection (2), the molecule is viewed along the C-C bond. Three lines diverging at an angle of 120° from the center of the circle indicate the bonds of the carbon atom closest to the observer; the lines “poking out” from behind the circle are the bonds of the distant carbon atom.
The conformation shown on the right is called obscured . This name reminds us that the hydrogen atoms of both CH 3 groups are opposite each other. The eclipsed conformation has increased internal energy and is therefore unfavorable. The conformation shown on the left is called inhibited , implying that free rotation around the C-C bond is “inhibited” in this position, i.e. the molecule exists predominantly in this conformation.
The minimum energy required to completely rotate a molecule around a particular bond is called the rotation barrier for that bond. The rotation barrier in a molecule like ethane can be expressed in terms of the change in the potential energy of the molecule as a function of the change in the dihedral (torsion - τ) angle of the system. The energy profile of rotation around the C-C bond in ethane is shown in Figure 1. The rotation barrier separating the two forms of ethane is about 3 kcal/mol (12.6 kJ/mol). The minima of the potential energy curve correspond to inhibited conformations, and the maxima correspond to occluded conformations. Since at room temperature the energy of some molecular collisions can reach 20 kcal/mol (about 80 kJ/mol), this barrier of 12.6 kJ/mol is easily overcome and rotation in ethane is considered free. In a mixture of all possible conformations, inhibited conformations predominate.
Fig.1. Potential energy diagram of ethane conformations.
For more complex molecules, the number of possible conformations increases. Yes, for n-butane can already be depicted in six conformations that arise when rotating around the central C 2 - C 3 bond and differing in the mutual arrangement of CH 3 groups. The different eclipsed and inhibited conformations of butane differ in energy. Inhibited conformations are energetically more favorable.
The energy profile of rotation around the C 2 -C 3 bond in butane is shown in Figure 2.
Fig.2. Potential energy diagram of n-butane conformations.
For a molecule with a long carbon chain, the number of conformational forms increases.
The molecule of alicyclic compounds is characterized by different conformational forms of the cycle (for example, for cyclohexane armchair, bath, twist-forms).
So, conformations are different spatial forms of a molecule that has a certain configuration. Conformers are stereoisomeric structures that correspond to energy minima on the potential energy diagram, are in mobile equilibrium and are capable of interconversion by rotation around simple σ bonds.
If the barrier to such transformations becomes high enough, then stereoisomeric forms can be separated (for example, optically active biphenyls). In such cases, we no longer talk about conformers, but about actually existing stereoisomers.
b)Geometric isomerism
Geometric isomers arise as a result of the absence in the molecule of:
1. rotation of carbon atoms relative to each other is a consequence of the rigidity of the C=C double bond or cyclic structure;
2. two identical groups at one carbon atom of a double bond or ring.
Geometric isomers, unlike conformers, can be isolated in pure form and exist as individual, stable substances. For their mutual transformation, higher energy is required - about 125-170 kJ/mol (30-40 kcal/mol).
There are cis-trans-(Z,E) isomers; cis- forms are geometric isomers in which identical substituents lie on the same side of the plane of the π bond or ring, trance- forms are geometric isomers in which identical substituents lie on opposite sides of the plane of the π bond or ring.
The simplest example is the isomers of butene-2, which exists in the form of cis-, trans-geometric isomers:
cis-butene-2 trans-butene-2
melting temperature
138.9 0 C - 105.6 0 C
boiling temperature
3.72 0 С 1.00 0 С
density
1,2 – dichlorocyclopropane exists in the form of cis-, trans-isomers:
cis-1,2-dichlorocyclopropane trans-1,2-dichlorocyclopropane
In more complex cases it is used Z,E-nomenclature (Kanna, Ingold, Prelog nomenclature - KIP, nomenclature of seniority of deputies). In connection
1-bromo-2-methyl-1-chlorobutene-1 (Br)(CI)C=C(CH 3) - CH 2 -CH 3 all substituents on carbon atoms with a double bond are different; therefore, this compound exists in the form of Z-, E- geometric isomers:
E-1-bromo-2-methyl-1-chlorobutene-1 Z-1-bromo-2-methyl-1-chlorobutene-1.
To indicate the isomer configuration, indicate the arrangement of senior substituents at a double bond (or ring) is Z- (from the German Zusammen - together) or E- (from the German Entgegen - opposite).
In the Z,E system, substituents with a large atomic number are considered senior. If the atoms directly bonded to the unsaturated carbon atoms are the same, then move on to the “second layer”, if necessary - to the “third layer”, etc.
In the first projection, the senior groups are opposite each other relative to the double bond, so it is an E isomer. In the second projection, the senior groups are on the same side of the double bond (together), so it is a Z-isomer.
Geometric isomers are widespread in nature. For example, natural polymers rubber (cis-isomer) and gutta-percha (trans-isomer), natural fumaric (trans-butenedioic acid) and synthetic maleic (cis-butenedioic acid) acids, in the composition of fats - cis-oleic, linoleic, linolenic acids.
V)Optical isomerism
Molecules of organic compounds can be chiral and achiral. Chirality (from the Greek cheir - hand) is the incompatibility of a molecule with its mirror image.
Chiral substances are capable of rotating the plane of polarization of light. This phenomenon is called optical activity, and the corresponding substances are optically active. Optically active substances occur in pairs optical antipodes- isomers, the physical and chemical properties of which are the same under normal conditions, with the exception of one thing - the sign of rotation of the plane of polarization: one of the optical antipodes deflects the plane of polarization to the right (+, dextrorotatory isomer), the other - to the left (-, levorotatory). The configuration of optical antipodes can be determined experimentally using a device - a polarimeter.
Optical isomerism appears when the molecule contains asymmetric carbon atom(there are other reasons for the chirality of a molecule). This is the name given to the carbon atom in sp 3 - hybridization and associated with four different substituents. Two tetrahedral arrangements of substituents around an asymmetric atom are possible. In this case, two spatial forms cannot be combined by any rotation; one of them is a mirror image of the other:
Both mirror forms form a pair of optical antipodes or enantiomers .
Optical isomers are depicted in the form of projection formulas by E. Fischer. They are obtained by projecting a molecule with an asymmetric carbon atom. In this case, the asymmetric carbon atom itself on the plane is designated by a dot, and on the horizontal line are indicated by the symbols of the substituents protruding in front of the plane of the drawing. The vertical line (dashed or solid) indicates substituents that are removed beyond the plane of the drawing. Below are different ways to write the projection formula corresponding to the left model in the previous figure:
In projection, the main carbon chain is depicted vertically; the main function, if it is at the end of the chain, is indicated at the top of the projection. For example, the stereochemical and projection formulas of (+) and (-) alanine - CH 3 - * CH(NH 2)-COOH are presented as follows:
A mixture with the same content of enantiomers is called a racemate. The racemate does not have optical activity and is characterized by physical properties different from the enantiomers.
Rules for transforming projection formulas.
1. Formulas can be rotated 180° in the drawing plane without changing their stereochemical meaning:
2. Two (or any even number) rearrangements of substituents on one asymmetric atom do not change the stereochemical meaning of the formula:
3. One (or any odd number) rearrangement of substituents at the asymmetric center leads to the formula for the optical antipode:
4. A 90° rotation in the drawing plane turns the formula into an antipode.
5. Rotation of any three substituents clockwise or counterclockwise does not change the stereochemical meaning of the formula:
6. Projection formulas cannot be derived from the drawing plane.
Optical activity is possessed by organic compounds in whose molecules other atoms, such as silicon, phosphorus, nitrogen, and sulfur, are chiral centers.
Compounds with several asymmetric carbon atoms exist in the form diastereomers , i.e. spatial isomers that do not constitute optical antipodes with each other.
Diastereomers differ from each other not only in optical rotation, but also in all other physical constants: they have different melting and boiling points, different solubilities, etc.
The number of spatial isomers is determined by the Fischer formula N=2 n, where n is the number of asymmetric carbon atoms. The number of stereoisomers may decrease due to partial symmetry appearing in some structures. Optically inactive diastereomers are called meso-forms.
Nomenclature of optical isomers:
a) D-, L- nomenclature
To determine the D- or L-series of an isomer, the configuration (position of the OH group at the asymmetric carbon atom) is compared with the configurations of the enantiomers of glyceraldehyde (glycerol key):
L-glyceraldehyde D-glyceraldehyde
The use of D-, L-nomenclature is currently limited to three classes of optically active substances: carbohydrates, amino acids and hydroxy acids.
b) R -, S-nomenclature (nomenclature of Kahn, Ingold and Prelog)
To determine the R (right) or S (left) configuration of an optical isomer, it is necessary to arrange the substituents in the tetrahedron (stereochemical formula) around the asymmetric carbon atom in such a way that the youngest substituent (usually hydrogen) has the direction “away from the observer.” If the transition of the three remaining substituents from senior to middle and junior in seniority occurs clockwise, this is an R-isomer (the decrease in seniority coincides with the movement of the hand when writing the upper part of the letter R). If the transition occurs counterclockwise, it is S - isomer (declining precedence coincides with the movement of the hand when writing the top of the letter S).
To determine the R- or S-configuration of an optical isomer using the projection formula, it is necessary to arrange the substituents by an even number of permutations so that the youngest of them is at the bottom of the projection. The decrease in seniority of the remaining three substituents clockwise corresponds to the R-configuration, and counterclockwise to the S-configuration.
Optical isomers are obtained by the following methods:
a) isolation from natural materials containing optically active compounds, such as proteins and amino acids, carbohydrates, many hydroxy acids (tartaric, malic, almond), terpene hydrocarbons, terpene alcohols and ketones, steroids, alkaloids, etc.
b) splitting of racemates;
c) asymmetric synthesis;
d) biochemical production of optically active substances.
DO YOU KNOW THAT
The phenomenon of isomerism (from the Greek - isos - different and meros - share, part) opened in 1823. J. Liebig and F. Wöhler using the example of salts of two inorganic acids: cyanic H-O-C≡N and explosive H-O-N= C.
In 1830, J. Dumas extended the concept of isomerism to organic compounds.
In 1831 the term “isomer” for organic compounds was proposed by J. Berzelius.
Stereoisomers of natural compounds are characterized by different biological activities (amino acids, carbohydrates, alkaloids, hormones, pheromones, medicinal substances of natural origin, etc.).