2. Organic compound structure

Question 16. What is the definition of the term: organic compound structure?

Answer. Term organic compound structure is molecule structure complete description. It comprises two levels at molecule description: chemical structure and space (three-dimensional) structure. Chemical structure is order (succession) of bonds between atoms according to their valency. Chemical structure does not take into consideration space localization of atoms making a molecule. Space structure is studied at special chemistry branch which is called stereo-chemistry. Modern development level of science and technology allows stating space structure of very complex molecules. Space structure is connected to physical, chemical properties and biologic activity.

Question 17. What substances could be called isomers?

Answer. Isomers are compounds with the same qualitative and quantitative content but which differ from each other in chemical or space structure and which have different physical, chemical or biological properties.

Question 18. What substances are called structural isomers or isomers on structure?

Answer. Structural isomers or isomers on structure are compounds with the same qualitative and quantitative content but which differ from each other in chemical structure (succession between atom bonds).

Question 19. What substances are called stereo-isomers?

Answer. Stereo-isomers (three-dimensional isomers) are compounds in which molecules there is the same succession of chemical bonds between atoms but they have different atom localization in relation to each other in space.

Question 20. What are the types of structural isomerism?

Answer. There are three types of structural isomerism:

1 – Carbon chain isomerism (isomers on carbon chain): butane and 2-methyl-propane;

2– isomerism on position of functional group (or position of multiple bonds):

1-butanol and 2-butanol or 1-butene and 2-butene;

3– isomerism on functional group (interclass): di-ethyl ether and 1-butanol.

Question 21. What are the types of space isomerism (stereo-isomerism)?

Answer. Stereo-isomerism could be subdivided into conformation and configuration.

C onfiguration isomers include geometric (cis- and trans-) isomerism: cis-2-butene and trans-2-butene, and enantiomerism (optical and mirror reflection isomerism): D- and L-lactic acids:

Conformation isomers comprise isomers received at molecule rotation in relation to single bond. They are called eclipsed and staggered conformations:

Question 22. What is molecule configuration?

Answer. Configuration is atom localization succession at space without taking into consideration differences which appear as a result of rotation around single s-bonds. For example, 2-butene has two configuration isomers: cis-2-butene and trans-2-butene. Lactic acid could also exist in the form of two configuration isomers: D-lactic and L-lactic acid.

Question 23. What is molecule conformation?

Answer. Molecule conformations are its different geometric forms which appear as a result of its rotation around single s-bonds. Conformation isomers are stereo-isomers which are formed at molecule separate parts rotation around single bonds.

Question 24. What is homologues series?

Answer. Homologues series is rank of similar structured compounds which have similar chemical properties. In the series members differ from each other on one or more –СН₂– group.

At members of homologues series majority of reaction goes on in the similar way (exception is only the first members of the series). Thus, if you know chemical reactions for only one member of the homologues series you can state that similar transformations are typical of other members of homologues series.

A general formula could be written for any homologues series. It reflects relation between carbon atoms and hydrogen atoms at the members of the homologues series; this formula is called homologues series general formula. For example, С_пН_2п+2 is formula for alkanes, С_пН_2п+1ОН is formula for aliphatic mono-alcohols.

Question 25. Write down butanoic acid homologue, it should have two more carbon atoms.

Answer. First, let’s write butanoic acid formula. The root is “butane”, so in the molecule there are four carbon atoms.

Step one: write down main carbon chain consisting of four carbon atoms.

Step two: define what functional group should be in the compound. According to the ending “-oic acid” it is carboxylic group –COOH. Add hydrogen atoms to each carbon atom so that carbon atom valency is four. We receive the following formula:

Homologue should belong to the same class of organic compounds, i.e. it is aliphatic acid with one carboxylic group and un-branched hydrocarbon radical.

Step three: define what is the difference between initial acid and homologue. According to the task, homologue should have two more carbon atoms in the structure. It means that two –CH₂– groups should appear in the hydrocarbon radical, in other words homologue molecule should have not 4 but 6 carbon atoms. Let’s write main carbon chain comprising 6 carbon atoms, the first carbon atom belongs to carboxylic group –COOH. Let’s add hydrogen atoms to each carbon atom so that carbon valency is four.

Let’s give name to the compound. In the main (and the only) carbon chain there are 6 carbon atoms, then the root is “hexane”. There is one functional group in the compound, it is marked in the form of ending “-oic acid”. Carbon atom number is not marked here as it is clear that it could be only the first. Homologue name is hexanoic acid.

Question 26. Write down butanal homologue formula, it should have four hydrogen atoms less.

Answer. First of all let’s write butanal formula. The root is “butane”, it means that there are four carbon atoms in the molecule.

Step one: write main carbon chain comprising four carbon atoms.

Step two: define what functional group should be in the compound. Suffix “-al” shows that there is aldehyde group –COH. Now add hydrogen atoms to each carbon atom so that carbon valency is four. We receive the following formula:

Homologue should belong to the same class of organic compounds, i.e. it should be an aldehyde.

S tep three: define what is the difference between homologue and initial aldehyde. According to our task in its molecule there are four hydrogen atoms less. Four hydrogen atom decrease is the same as decrease in two –CH₂-groups, i.e. in homologue there are two carbon atoms less in the carbon chain. Thus, homologue molecule contains not 4 but 2 carbon atoms. Let’s write main carbon chain consisting of two carbon atoms, first carbon atom belongs to aldehyde group –CHO. Now we add hydrogen atoms to carbon atoms so that carbon atom valency is four.

Let’s give name to the compound. In the main (and the only) chain there are 2 carbon atoms, thus, the root is “ethane”. There is one functional group in the compound, we name it in the form of a suffix “-al”. Carbon atom number in the carboxylic group is not marked in this compound as it could be only the first. Homologue name is ethanol. This compound is also called acetic aldehyde or acetaldehyde.

Question 27. Write down pentanoic acid isomer on the carbon chain structure.

Answer. First of all write down pentanoic acid formula. We understand from the name that in the basis there is un-branched (direct) chain consisting of five carbon atoms. Initial acid formula is:

According to the term “isomer on the structure of the carbon chain”, pentanoic acid isomer chain should have different structure as compared to pentanoic acid carbon chain structure. In this situation it could be only branched chain. For five carbon atoms there are three variants of carbon chain structure:

Let’s give names to the compounds. In compounds I and II the longest carbon chain has four carbon atoms, thus, root is “butane”, as there is carboxylic group the ending is “-oic acid”. In both molecules there is one methyl radical which is attached to different carbon atoms. Radical name and number of its carbon atom is given as prefix. Carbon chain enumeration is started from functional group. Thus, compounds I and II are called 2-methyl-butanoic acid and 3-methyl-butanoic acid respectively. In compound III the longest carbon chain has three carbon atoms, so the root is “propane” and ending is “-oic acid”. This molecule has a more branched structure: there are two methyl radicals (it means that there is multiplying prefix “di-”), both radicals are attached to the second carbon atom (in the name this number is repeated twice). Name of the third isomer is 2,2-di-methyl-propanoic acid.

The content of all three isomers and initial compound is the same: С₅Н₁₀О₂, but chemical structure is different.

Question 28. Write down 1-pentene isomer on the position of the multiple bond.

Answer. Let’s start with writing 1-pentene formula. It is evident that in molecule basis there is un-branched (direct) chain consisting of five carbon atoms. Ending “-ene” shows that there is double bond. Number 1 shows double bond position: it is between the first and second carbon atoms. Initial alkene formula is:

A ccording to the term “isomer on the position of the multiple bond” in isomer formula multiple bond should be located between other different carbon atoms. Theoretically it could have the following form:

At first glance, we could conclude that these are different compounds. At giving names to compounds according to systematic nomenclature, it turns out that it is impossible to have the same chain enumeration as compared to initial compound, as chain enumeration is started from the end with double bond. Following this rule we get:

It is evident that formulas I and II correspond to one and the same compound: it is 2-pentene, and formula III corresponds to initial compound: it is 1-pentene.

Thus, there are only two isomers which follow the same general formula С₅Н₁₀ and differ on chemical structure, to be more particular they differ on the position of the double bond. These are 1-pentene and 2-pentene.

Question 29. Write down butanal isomer on functional group.

A nswer. Let’s start with butanal formula writing. It is clear from the name that in the basis there is un-branched (direct) chain with four carbon atoms. Suffix “-al” shows that one carbon atom belongs to aldehyde group:

A ccording to the term “isomer on functional group” at butanal isomer formula there is some other (not aldehyde) functional group. It could be, for example, keto-group:

Both isomers content is the same: С₄Н₈О, but chemical structure is different, in molecules there are different functional groups, in other words, these substances belong to different classes of organic compounds.

Let’s give name to new compound. The longest carbon chain has four carbon atoms so the root is “butane”, suffix is “-on” as there is keto-group. Name is 2-butanon. It is not necessary to mark number 2 as this is the only position of the functional group in the compound.

C arboxylic acid isomers on functional group are esters.

Alcohol isomers on functional group are ethers.

Amino acid isomers on functional group are esters and hydroxyl acid amides.

Question 30. Serine (2-amino-3-hydroxy-propanoic acid) in L-isomer form takes part at protein molecule formation. Write down structural formula of its isomer and enantiomer.

Answer. Let’s write down serine formula:

Second carbon atom is asymmetric (it is attached to four different substitutes: carboxylic group, amino group, hydrogen atom and third carbon atom in the carbon chain).This is first condition of optical activity. Second condition is asymmetry at molecule elements. Thus, compound is optically active and could exist in the form of two enantiomers – space isomers, which are each other mirror reflections.

T o show enantiomers relative configuration, carbon chain is drawn vertically, substitute of the highest priority is at the top; substitutes at asymmetric carbon atom are located to the right and to the left in relation to carbon chain:

СООН

L-serine

Н

Н₂N

СН₂ОН

If amino group is located to the left in relation to carbon chain, then isomer belongs to L-isomers. To write D-isomer, substitutes at asymmetric carbon atom are drawn in mirror reflection. In our situation they are at second carbon atom. Amino group is located on the right in relation to carbon chain. Asymmetric carbon atom is not marked by any chemical sign, it is supposed to be at line crossing which shows corresponding bonds.

Question 31. What types of isomers are oleic and elaidic acids.

Answer. We should write down formulas of these acids. Oleic and elaidic acids are un-saturated carboxylic acids with one double bond, they are cis- and trans-isomers.

Thus, these acids are space (geometric) isomers, as they differ in atom and group localization in space in relation to double bond.

Question 32. What is isomer type of butanoic acid and propanoic acid methyl ester?

A nswer. We should write formulas of these compounds. Root is “butane”, it means there is un-branched carbon chain containing four carbon atoms, one of which is in the structure of carboxylic group. In propanoic acid molecule there are three carbon atoms; as ester is methyl it is formed at interaction between propanoic acid and methyl alcohol. Compound formulas are:

General formula of both compounds is the same: С₄Н₈О₂, but chemical structure is different. These compounds belong to different classes as they have different functional groups. We can conclude that they are isomers on functional group.

Question 33. What is isomer type of 1-butanol and 2-butanol?

Answer. We should write these compound formulas. The root is “butane”, it means that in the molecule there is un-branched carbon chain having four carbon atoms. At both compounds there is suffix “-ol”, both of them belong to one class, it is alcohols. At alcohol molecule there is hydroxyl group (-OH). In the first compound this group is attached to first carbon atom, in the second compound it is attached to second carbon atom. Formulas of compounds are:

Both compounds general formula is the same: С₄Н₉ОН, but chemical structure is different. It means that they are structural isomers. They belong to the same class and differ from each other only by position of functional group. They are isomers on position of the functional group.

Question 34. Are leucine and iso-leucine isomers?

Answer. First of all, let’s write down these compounds formulas. These are amino acids in the structure of proteins. Their chemical names are 2-amino-4-methyl-pentanoic acid and 2-amino-3-methyl-pentanoic acid.

T he root at both substances is “pentane”, it means that there are five carbon atoms in the main carbon chain, one of those carbon atoms is in the structure of carboxylic group. In both molecules there are amino groups attached to second carbon atom (next to carboxylic group) and methyl radical (-CH₃) attached to fourth and third carbon atoms respectively. Compound formulas are:

General formula of both compounds is the same: С₆Н₁₃О₂N, but carbon chain structure is different. These compounds are isomers on carbon chain structure.