Introduction

Aim of bioorganic chemistry as a subject is system knowledge formation about interconnection between structure and chemical properties of biologically important classes of organic compounds, biopolymers and their structural components as basis to understand life processes on modern molecular level.

At bioorganic chemistry teaching much attention is paid to students’ independent work. This Higher school-book has detailed answers on students’ questions at bioorganic chemistry. Using this detailed information students study on their own to have general approach and logic which facilitates writing tests.

1. Biologically important organic compounds classification and nomenclature

Question 1. What is the basis to classify organic compounds?

Answer. At organic compounds classification basis there is carbon skeleton structure and functional groups.

According to carbon skeleton organic compounds are subdivided into hydrocarbons (in which structure there are only carbon and hydrogen atoms) and hetero-cyclic compounds (in molecule cyclic part in addition to carbon atoms there are some other atoms, for example, nitrogen, sulfur or oxygen). Hydrocarbons, in their turn, could be aliphatic (with open carbon chain) and cyclic (with closed carbon chain). Both hydrocarbon types and hetero-cyclic compounds could be saturated (in the molecule there are only single bonds) and unsaturated (in the molecule there is one or more double or triple bonds).

Aliphatic hydrocarbons:

saturated unsaturated

alkanes alkenes (double bond) alkynes (triple bond)

C yclic hydrocarbons:

saturated unsaturated heterocyclic compound

Classification according to functional groups:

– halogen-derivatives R–Hal: СН₃СН₂Cl (chloro-ethane);

– alcohols and phenols R–ОН: СН₃СН₂ОН (ethanol);

– thiols R–SH: СН₃СН₂SН (ethane-thiol);

– ethers R–О–R: СН₃–О–СН₃ (di-methyl ether),

– esters R–СОО–R: СН₃СН₂СООСН₂СН₃ (acetic acid ethyl ester; ethyl acetate);

– carbonyl compounds: aldehydes R–СНО: ketones R–СО–R:

(ethanal, aldehyde),СН₃СОСН₃ (propanon, ketone),

– carboxylic acids R - СООН: (acetic acid);

– sulpho-acids R–SО₃Н: СН₃SО₃Н (methane-sulpho acid);

– amines R–NH₂: СН₃СН₂NH₂ (ethyl-amine);

– nitrocompounds R–NO₂: СН₃СН₂NО₂ (nitro-ethane);

– metalorganic compounds: СН₃СН₂Nа (ethyl sodium).

Question 2. What organic compound nomenclature types are used at present?

Answer. There are three types of organic compounds nomenclature: trivial (historical), rational and systematic. Trivial nomenclature is the system of historical names for compounds. People who know Greek could easily understand that acetic acid means something sour, and glycerol means something sweet. These names do not give any information about structure of these compounds. Due to increased synthesis of organic compounds new nomenclatures were created, they reflected organic substance structure.

Rational nomenclature makes compound name from easier compound structure (first member of homologues series). СН₃ОН – carbinol, СН₃СН₂ОН – methyl-carbinol, СН₃СН(ОН)СН₃ – di-methyl-carbinol etc.

At present this nomenclature type is very rare.

According to international (systematic) nomenclature organic substance name is based on the name of corresponding hydrocarbon with prefixes and suffixes which belong to this compound class (homologues series).

Question 3. In what way is it proper to give a name to organic compound according to systematic nomenclature?

Answer. To give proper name to an organic compound according to systematic nomenclature it is necessary:

1) to choose the longest carbon chain as main carbon skeleton and give it a name taking into consideration saturation or unsaturation degree of the compound;

2) to find all functional groups;

3) to denote what group is the group of the highest priority (table 1), this group name is reflected in compound name as suffix and it is located at the end of the name; other groups are used as prefixes;

Table 1. Names of functional groups (groups are presented according to their priority degree from top to bottom according to priority decrease)

Compound class	Functional groups	Prefix	Suffix or ending
Acids	СООН	Carboxy-	-oic acid
Sulpho-acids	– SO3H	Sulpho-	Sulphonic acid
Nitryls	– C  N	Cyan-	-nitryl
Aldehydes	 СНО	Oxo-	-al
Ketones	– С = О	Oxo-	-on
Alcohols	 ОН	Hydroxy-	-ol
Thiols	– SH	Mercapto-	-thiol
Amines	NН2	Amino-	-amine
Alkenes	– С = С –		-ene
Alkynes	 С ≡ С 		-yne
Halogen-derivatives	Br, I, F, Cl	Bromo-, iodo-, fluro-, chloro-	-bromide, iodide, fluoride, chloride
Nitro compounds	NO2	Nitro-	

4) to give numbers to carbon atoms in the main chain, giving smaller number to a group of the highest priority (i.e. chain is enumerated from the end which functional group with the highest priority is closer to, in the case of hydrocarbons it is enumerated from the end to which multiple bond or branch is closer);

5) to enumerate prefixes in alphabetic order (multiplying prefixes as di-, tri-, tetra- etc. are not taken into consideration);

6) to make complete name of the compound.

You should remember:

– in alcohols, aldehydes, ketones, carboxylic acids, amides, nitryls, halogen-anhydrides names, suffix defining class follows unsaturation degree suffix, for example, 2-butenal;

– compounds containing other functional groups are called as hydrocarbon derivatives. These functional group names are placed as prefixes at the beginning of hydrocarbon name, for example, 1-chloro-propane.

Names of acid groups, for example, sulpho-acids or phosphine acid are placed after hydrocarbon skeleton name, for example, benzene-sulpho-acid.

Aldehyde and ketone derivatives are named according to the name of initial carbonyl compound.

Esters are named as initial acid derivatives. Ending -oic acid is changed into -oate, for example, methyl-propionate is propionic acid methyl ester.

To mark the fact that substitute is attached to nitrogen atom from primary structure capital N is used at the beginning of the substitute name, for example, N-methyl-alanine.

Thus, start with main structure, you should know the names of first 10 members of alkane homologues series by heart (methane, ethane, propane, butane, pentane, hexane, heptane, octane, nonane, decane). You should also know radicals formed from them, ending -ane is changed into -yl.

Question4. What principles are at the basis of organic substance classification?

Answer. At the basis of organic substance classification there are two principles: structure of the carbon chain and functional group origin.

Question 5. What class of organic compounds does hydrocarbon belong to and what is its proper name according to international (systematic) nomenclature?

Answer. 1) It is necessary to find what bond types are there in the compound: single bonds, double bonds or triple bonds. In our case there are only single bonds, it means that our compound belongs to alkanes (saturated) hydrocarbons.

2) We should find the longest carbon chain. In this compound there are two equally long chains containing 5 carbon atoms. They could be enumerated in the following way:

E numeration is started from the end to which radical (branch) is closer. That is why the following enumeration is a mistake:

In the longest chain there are 5 carbon atoms, thus, the root is “pentane”.

3) We find radicals attached to the main carbon chain. In the compound there are same two methyl radicals. In the name they are reflected by prefix “methyl” with multiplying prefix “di-”. Thus, the compound name is di-methyl-pentane.

4) It is necessary to mark numbers of carbon atoms to which hydrocarbon radicals are attached.

As a result we have 2,3-di-methyl-pentane.

Do not forget that between numeral and letter part of the name you should write a dash.

Question 6. What is the name of this hydrocarbon and what is its class of organic compounds?

Answer. 1) It is necessary to define bond types at the compound: single, double, triple. In the molecule there is a triple bond, it means that compound belongs to alkynes.

2) We should find the longest carbon chain. It contains 6 carbon atoms. Enumeration is started from the end with multiple bond. In this compound triple bond is located in the centre of the molecule. In this situation enumeration is started from the end to which hydrocarbon radical is closer.

Alkane name with 6 carbon atoms is “hexane”. In the molecule there is a triple bond, that is why, ending “-ane” is changed into ending “-yne”.

3) We define carbon atom number with the triple bond. We receive 3-hexyne.

4) We define radicals attached to main chain. In this compound there is one methyl radical. In the name it is used as prefix “methyl-” adding carbon atom number to which it is attached. As a result we receive: 2-methyl-3-hexyne.

Question 7. What is following compound class and name according to systematic nomenclature:

Answer. To define the class of organic compound we need to find functional groups. In this case there is hydroxyl group, thus, the class of compounds is alcohols. In the molecule there is only one such group then, it is mono-alcohol.

In the chain there are 5 carbon atoms, there are no multiple bonds, then the root is “pentane”. Functional group is only one and we name it as suffix “-ol” and note carbon atom number to which hydroxyl group is attached. Chain enumeration starts from the end with functional group.

We receive 2-pentanol. What is left is to add radical name –СН₃ (methyl) and carbon atom number to which it is attached. We receive 4-methyl-2-pentanol.

Let’s give a name to di-alcohol.

In the chain there are 4 carbon atoms, there are no multiple bonds, thus, the root is “butane”. Two the same functional groups are marked in the form of suffix “-ol” and add multiplying prefix “di-”, mark carbon atom numbers to which hydroxyl groups are attached. Thus, we receive 1,3-butane-diol.

Algorithm is the same for tri-alcohol but we should take into consideration the fact that multiplying prefix for three groups is “tri”, then we receive the following name: 1,2,3-butane-triol.

Question 8. What is the class of organic compounds and name according to systematic nomenclature for the following compound:

Answer. To define class of organic compounds we should find functional groups in the compound. In the compound there are two functional groups: carboxylic and amino group, then, the class of organic compounds is amino acids. Carboxylic group is the group of the highest priority, then, we start enumeration of the carbon chain from carbon atom attached to carboxylic group:

All in all there are 4 carbon atoms in the chain, then, the root is “butane”. Group of the highest priority is marked in the form of ending “-oic acid”. Second functional group is used as prefix “amino-” marking carbon atom number to which this group is attached. We receive the name 2-amino-butanoic acid. What is left, is to give the name to radical –СН₃ (methyl) attached to the third carbon atom. Radical name is used as prefix. Prefixes are enumerated in alphabetic order. The name is 2-amino-3-methyl-butanoic acid.

Question 9. What is the class and name according to systematic (international) and rational nomenclatures for the following compounds:

Answer. To define the class of organic compounds we should analyze the structure. Both substances have similar structure: two hydrocarbon radicals are attached to each other through oxygen atom. This structure is typical of ethers. According to systematic (international) nomenclature these compounds are alkoxy-alkanes. Root is defined by the longest alkyl group. According to this nomenclature rules the first compound is called metoxy-methane and the second is called metoxy-ethane.

According to rational nomenclature rules ethers are called on radicals attached to oxygen atom and the word “ether” is added. First compound is called di-methyl ether, the second compound is called methyl-ethyl ether. We could enumerate radicals according to their priority or alphabetically.

Question 10. What is the formula and name of secondary alcohol which contains 4 carbon atoms?

Answer. 1) It is necessary to write main carbon chain. In this situation it should comprise 4 carbon atoms.

2) We define what functional group should be at this compound. For alcohols it is hydroxyl group: –ОН.

3) We state what carbon atom is hydroxyl group attached to. According to the wording of the task our alcohol is secondary, thus, hydroxyl group is attached to secondary carbon atom (this is atom directly attached to two other carbon atoms).

1

2

3

4

С

С

С

С

ОН

In this situation this is second carbon atom.

4) We add hydrogen atoms to carbon atoms so that each carbon atom valency is four. As a result we receive the following formula:

We make the name of the compound. In the chain there are 4 carbon atoms, there are no multiple bonds, thus, the root is “butane”. There is one functional group, that is why, it is marked as suffix “ol” and mark carbon atom number with hydroxyl group. Name is 2-butanol.

If we were asked to write down primary alcohol structure, according to the same algorithm we receive 1-butanol:

Question 11. Write down the formula and name (according to international nomenclature) to thiol containing 3 carbon atoms and mark secondary carbon atom in the structure.

Answer. 1) First we write down main carbon chain. In this case it has 3 carbon atoms.

2) We define what functional group should be in the compound. Thiols and thio-alcohols are characterized by –SН group. According to the wording of the task it is clear what carbon atom should have thiol group. There are two formulas corresponding to the task:

First formula is primary thiol as thiol group is attached to primary carbon atom, in the second formula we have secondary thiol as thiol group is attached to secondary carbon atom (atom which is directly attached to two other carbon atoms). It means that secondary carbon atom here is carbon atom number 2.

Now we make compound name. In the chain there are 3 carbon atoms, there are no multiple bonds so the root is “propane” for both substances. There is only one functional group so we mark it in the form of suffix “-thiol” and mark carbon atom number to which it is attached. We receive 1-propane-thiol and 2-propane-thiol.

Question 12. What is the formula and acid amide name which has 4 carbon atoms?

Answer. First we should write down the formula of carboxylic acid containing 4 carbon atoms.

1) Write down main carbon chain. In this situation it should have 4 carbon atoms.

2 ) Define functional group in this compound. This is carboxylic group –СООН. We should add hydrogen atoms to carbon atoms so that each carbon atom valency is four. We receive the following formula:

Amide is carboxylic acid derivative in which –ОН group at carboxylic group is substituted by amino group. Let’s write down corresponding formula:

This compound is called butanoic acid amide (butane-amide). According to trivial nomenclature it is called butter acid (butyric) (as it is at butter in the form of ether with glycerol, in free form it is at rotten butter), correspondingly, amide could be called butyric acid amide (butyramide).

Amide bond is at biologically important polymers, for example, proteins, in which structure it is called peptide bond.

Question 13. Choose formula with mistake in it.

A nswer. It is necessary to analyze compound structure and define whether written atom valency corresponds to real atom valency. Let’s analyze two compounds:

You can see that in alkyne (a) three first carbon atoms valency is four (each of them forms four bonds with other carbon and hydrogen atoms) and fourth carbon atom has only three bonds (one is formed with the third carbon atom and two are formed with hydrogen atoms). It means that 1-butyne formula is not correct. In compound (b) all carbon atom valency is four but oxygen atom valency is three but not two. In this compound hydrogen attached to carbonyl carbon atom is extra and if it is omitted then we receive correct butanal formula. So correct formulas are:

Question 14. Choose name with a mistake:

а) 1-methyl-butane; b) 2-methyl-butane; c) 2,2-di-methyl-butane; d) 2,3-di-methyl-butane

Answer. You should write down formulas of the compounds according to their names and compare formula and suggested name.

All four names have the same root “butane”, it means that main chain consists of four carbon atoms. Let’s enumerate chain and add methyl radical to the first carbon atom in the first compound, methyl radical is added to the second carbon atom in the second compound. In two last compounds according to multiplying prefix “di-”, there are two methyl groups. In the third compound, two groups are attached to second carbon atom, in the fourth case they are attached to second and third carbon atoms. Now we should add hydrogen atoms according to carbon atom valency (4).

At formula and name analysis it is clear that there is a mistake in the first compound name as the longest (the only) chain has not four but five carbon atoms and proper name is “pentane”.

Question 15. Does the name 2-methyl-4-butanol correspond to international nomenclature?

Answer. To define whether the name is correct, let’s try to make formula of the compound. The root is “butane” and it means that there are 4 carbon atoms in the main chain. Let’s enumerate chain and add hydroxyl group (-OH) corresponding to suffix “-ol” and radical methyl (–СН₃) at corresponding carbon atoms. We receive the following compound:

A t analysis we could see a mistake: main carbon chain is not properly enumerated. Numeration should start with the end to which functional group is closer.

Correct name of this compound is 3-methyl-1-butanol.