3. General laws of organic compound reaction ability

Question 35. What type of reactions is more typical of compounds belonging to some class of organic compounds?

Answer. We are speaking not about all reactions but about more typical of some class.

Alkanes practically reject one type of chemical reactions. It is addition. Other reactions are quite possible for alkanes (taking into consideration reaction conditions and reagent type). Most widely-spread reactions for alkanes are substitution reactions. In alkane molecules there is equal electron density distribution. That is why, possibility of electrophilic particle or nucleophilic particle attack is very low. Most typical reactions for alkanes are radical substitution reactions.

Molecules containing p-bonds in the structure of double or triple bond there is increased electron density which is higher and lower in relation to molecule plane. It leads to the fact, that positively charged ions or polar molecules by their positive pole (electrophilic particles) are attracted to double bond. As a result of following interaction there is reagent addition on the place of multiple bond. Most typical reactions for alkenes and alkynes are electrophilic addition reactions.

In molecules containing functional groups there is electron density displacement from carbon atom to hetero-atom from functional group (for example, to halogen atom at halogen derivatives or to oxygen atom at alcohols, etc.). As a result there is partial positive charge on carbon atom. Thus, reagent attacking carbon atom, should have partial or complete negative charge (nucleophilic particle). As a result of interaction there is functional group substitution by nucleophilic reagent. Halogen derivatives and alcohols are characterized by nucleophilic substitution reactions.

Question 36. What is the type of chemical reaction according to its equation?

Answer. It is necessary to know differential signs of reactions belonging to different types.

1. According to reagent electron nature reactions are subdivided into nucleophilic, electrophilic, free radical reactions.

Nucleophilic reagents are mono- and multi-anions or molecules which have centres with increased electron density. They include such anions and molecules as HO^-, RO^-, Cl^-, Br^-, RCOO^-, CN^-, R^-, NH₃, C₂H₅OH, etc.

Electrophilic reagents are cations, simple or complex molecules which have increased affinity to electron pair or affinity to negatively charged molecule centres. They include cations H⁺, Cl⁺, ⁺NO₂, ⁺SO₃H, R⁺ and molecules with free orbitals AlCl₃, ZnCl₂, etc.

Free radicals are electro-neutral particles which have un-shared electron, for example, Cl^·, ^·NO₂.

2. According to particle number change during reaction there are substitution, addition, elimination, degradation and redox reactions.

In the case of substitution reaction in the molecule one atom (or atom group) is substituted by another atom (or atom group), new compounds are formed as a result. In this situation number of initial substances equals number of reaction products.

СН₃–СН₃ + С1₂ ® СН₃–СН₂С1 + НC1

In the case of addition reaction from two (or several) molecules, one new substance is formed:

CH₂ = CH₂ + HBr → CH₂Br–СH₃

As a result of elimination reaction new organic substance is formed with a multiple bond:

СН₃–СН₂С1 + NaOH_{(alcohol solution)} ® СН₂ = СН₂ + NaC1 + Н₂О

Degradation reaction leads to one substance formation from two or more substances with more simple structure:

НСООН → СО₂ + Н₂

If in the molecule of initial substance there are two or more carbon atoms, degradation reaction usually goes with carbon-carbon bond breaking.

For example, under enzymatic action malic acid could degrade and lactic acid is formed:

НООС­СН₂­СН(ОН)­СООН ® СН₃­СН(ОН)­СООН + СО₂

3. In organic chemistry there are a lot of reactions which are defined according to panial (additional) principles:

- hydration is water addition on the place of double bond carbon-carbon or carbon-hetero-atom:

СН₃–CH = CH₂ + H₂О ® СН₃–CH(ОН)–CH₃;

- dehydration is water elimination with unsaturated compound formation:

СН₃–CH(ОН)–CH₃ ® СН₃–CH = CH₂ + H₂О;

- hydrogenation is hydrogen addition on the place of the double bond carbon-carbon or carbon-hetero-atom:

СН₃–CH = CH₂ + H₂ ® СН₃–CH₂–CH₃,

СН₃–CО–CH₃ + H₂ ® СН₃–CH(ОН)–CH₃;

- dehydrogenation is hydrogen elimination with unsaturated compound formation:

СН₃–CH₂–CH₃ ® СН₃–CH = CH₂ + H₂

- hydrolysis is interaction between complex substrate with water with more simple compounds formation:

- nitration is interaction with nitric acid as a result hydrogen atom at substrate molecule is substituted by nitro-group:

СН₃–CH₃ + HNO₃ ® СН₃–CH₂NO₂ + H₂O;

- sulfonation is interaction with sulfuric acid or oleum as a result hydrogen atom at substrate molecule is substituted by sulfo-group:

СН₃–CH₃ + H₂SO₄ ® СН₃–CH₂SO₃H + H₂O;

- halogenation is interaction with halogen as a result hydrogen atom at substrate molecule is substituted by halogen atom or halogen atoms are added on the place of multiple bond:

СН₄ + С1₂ ® СН₃С1 + НC1

CH₂ = CH₂ + Br₂ ® CH₂Br – CH₂Br;

- dehalogenation is halogen molecule elimination with multiple bond or cyclic compound formation:

CH₂Br – CH₂Br ® CH₂ = CH₂ + Br₂;

- hydro-halogenation is interaction with halogen-hydrogen as a result hydrogen and halogen atoms are added on the place of the multiple bond:

CH₂ = CH₂ + HBr ® CH₂Br–СH₃;

- dehydro-halogenation is halogen-hydrogen molecule elimination with multiple bond formation:

CH₂Br–СH₃ ® CH₂ = CH₂ + HBr;

- carboxylation is reaction which result is carboxylic group formation. For example, if there is ATP then pyruvic acid is transformed into aceto-acetic acid in the organism:

СН₃–СО–СООН+СО₂+АТP ® НООС–СН₂–СО–СООН+АDP+ Н₃РО₄;

- decarboxylation is reaction which result is carboxylic group elimination from substrate, it is in the form of CO₂:

R–CH(NH₂)–COOH ® R–CH₂NH₂ + CO₂;

- deamination is amino group elimination from substrate molecule, as a result unsaturated, hydroxy- or oxo-compounds could be formed according to reaction conditions or enzymatic conditions. Thus the following reactions could be in vivo (in the organism):

- oxidative deamination, as a result amino group is substituted by oxo-group:

R–CH(NH₂)–COOH + [O] ® R–СО–CООН + NH₃;

- hydrolytic deamination, as a result amino group is substituted by hydroxyl group:

R–CH(NH₂)–COOH + H₂O ® R–СН(ОН)–CООН + NH₃;

- intra-molecular deamination, as a result unsaturated compound is formed:

R–CH₂–CH(NH₂)–COOH ® R–CH=CH–CООН + NH₃;

- reductive deamination, as a result amino group is substituted by hydrogen atom:

R–CH(NH₂)–COOH + [Н] ® R–CH₂–CООН + NH₃;

- deamination in vitro (in the test-tube) is under nitrous acid action with hydroxyl-compounds formation:

R–CH(NH₂)–COOH + НNО₂ ® R–СН(ОН)–CООН + N₂­ + H₂O;

- isomerisation is substrate transformation into its isomer which has the same quantitative or qualitative content but different structure:

In organic chemistry reduction reactions include hydrogenation reaction (hydrogen addition), and oxidation reaction includes dehydrogenation reaction (hydrogen elimination) and reactions which result is oxygen atom addition to substrate molecules.

Enzyme names, catalyzing reactions, are connected to reaction classification according to additional principles. Enzymes catalyzing redox reactions are called oxido-reductases; enzymes catalyzing hydrogen elimination (dehydrogenation) are called dehydrogenases; decarboxylation reactions are catalyzed by decarboxylases; deamination reactions are catalyzed by deaminases; isomerization reactions are catalyzed by isomerases; hydrolysis reactions are catalyzed by hydrolases.

Question 37. Write down equation of reaction properly and define its mechanism.

Answer. You should know organic substance classification and be able to write their formulas (questions 1, 9-11) and study characteristic principles of different reaction types (questions 35, 36).

For example you should write reaction between 2-butenoic acid and hydrogen. In the first turn we should write formulas of substrate and reagent: СН₃–СН=СН–СООН and Н₂

Let’s analyze substrate molecule structure. In molecule structure there is carboxylic group (this is acid) and hydrocarbon radical containing double bond (butEnoic but nor butAnoic acid). Unsaturated compounds are characterized by addition reaction on the place of the multiple bond. In this situation the most possible reaction is hydrogen addition on the place of the double bond. And p-bond is broken, hydrogen atoms add to the second and the third carbon atoms:

СН₃–СН=СН–СООН + Н₂ ® СН₃–СН₂–СН₂–СООН

Reaction product is butanoic acid.

In such compounds it is easier to attack electron density of the p-bond and attacking particle should have positive charge. Thus, this reaction mechanism is electrophilic addition. This reaction could belong to reduction reactions, according to panial principles it is hydrogenation reaction.

Let’s analyze one more example. Let’s try to write reaction between bromo-ethane and sodium hydroxide aqueous solution and define its mechanism.

Let’s write formulas of reacting substances: CH₃–CH₂Br and NaOH.

Let’s analyze bromo-ethane molecule structure. Bromine atom is more electro-negative than carbon atom. Carbon-bromine bond is polarized, carbon atom has partial positive charge. It means that attack should be done by negatively charged reagent. Such negatively charged particle is in the solution: it is hydroxyl-anion ОН^–, which is formed at sodium hydroxide dissociation in water. This anion attacks carbon atom with partial positive charge and forces out bromine anion from bromo-ethane molecule. Thus, there is nucleophilic substitution reaction with ethanol formation:

CH₃–CH₂Br + NaOH ® CH₃–CH₂OH + NaBr

Question 38. What product is predominantly formed at bromination reaction of 3-methyl-pentane?

Answer. All carbon atoms at 3-methyl-pentane are in the state of sр³-hybridisation. In the molecule there are non-polar C-C and low polar C-H s-bonds. It determines their participation at radical substitution reactions S_R: if there are primary, secondary and tertiary carbon atoms, bromination predominantly goes on tertiary carbon atom (the least solid C-H bond and the most solid intermediate bond – tertiary radical). It means that predominant reaction product is 3-bromo-3-methyl-pentane.

Question 39. Give name to reaction product at reaction between propene and 2-butenoic acid with bromo-hydrogen if there is hydrogen peroxide and there is no hydrogen peroxide in the mixture.

Answer. Halogen-hydrogen electrophilic addition goes according to Markovnikov rule: hydrogen atoms are added mainly to the most hydrogenated carbon atom. For example, at interaction between propene and bromo-hydrogen there is mainly 2-bromo-propane formed.

СН₃СН=СН₂ + НВr ® СН₃СНВr–СН₃

propene 2-bromo-propane

At 2-butenoic acid both carbon atoms with double bond have equal hydrogen atoms number. In this case Markovnikov rule is used in more general form: non-symmetric reagent addition to non-symmetric alkene according to ionic mechanism goes in the direction of more stable cation formation.

Electron density of the p-bond is attacked by positively charged particle Н⁺. As a result of hydrogen addition to the double bond in the acid СН₃СН=СНСООН two carbo-cations could be formed:

СН₃®⁺СН–СН₂®СООН (I) and СН₃СН₂®⁺СН®СООН (II)

In cation (I) near to positively charged carbon atom there is electron donor methyl group which has positive inductive effect which leads to positive charge delocalization and cation stability. In cation (II) methyl group is farther from positively charged carbon atom, its effect is weaker (inductive effect decreases going along the chain). Electron acceptor carboxylic group due to its negative inductive effect, destabilizes cation; but in cation (I) it is located farther and its destabilizing action is lower than in secondary cation (II). As a result charge distribution in cation (I) is more equal and this particle is more stable, reaction goes on along the direction of its formation. Addition would follow Markovnikov rule in its general form:

СН₃СН=СНСООН + НВr ® СН₃СНВrСН₂СООН

2-butenoic acid 3-bromo-butanoic acid

If there is peroxide, addition reaction has radical mechanism and does not follow Markovnikov rule (goes against Markovnikov rule):

СН₃СН=СН₂ + НВr(there is Н₂О₂) ® СН₂ВrСН₂СН₃

propene 1-bromo-propane

СН₃СН=СНСООН + НВr(there is Н₂О₂) ® СН₃СН₂СНВrСООН

2-butenoic acid 2-bromo-butanoic acid

Question 40. In the metabolic process in living organisms fumaric acid is transformed into malic acid. In what way is it possible to receive malic acid from fumaric acid in vitro?

Answer. Fumaric acid is unsaturated di-based acid (trans-butene-diovic acid). Malic acid is saturated di-based hydroxyl acid (hydroxyl-butane-diovic acid).

Fumaric acid transformation in to malic acid could be performed with the help of water addition on the place of the multiple bond, i.e. with the help of hydration reaction. Alkene hydration is performed at strong acid aqueous diluted solution, for example, it could be sulfuric acid. Acid is the source of electrophilic particle – proton Н⁺. Reaction mechanism is electrophilic addition А_Е.

Electron density of carbon-carbon p-bond at fumaric acid molecule is decreased due to two carboxylic groups electron acceptor action. That is why fumaric acid hydration is performed at comparatively severe conditions (heating with diluted aqueous solution at temperature 150-200 °С).

Fumaric acid hydration in vitro leads to racemic mixture formation (racemic mixture is mixture of equal amount of malic acid enantiomers). In organism this reaction is catalyzed by an enzyme which is called fumarase which has strict space specificity which leads to only one stereo-isomer formation, it is L-malic acid.

Question 41. R-2-bromo-butane could react to sodium hydroxide aqueous solution according to S_N1 and S_N2 mechanisms. Describe stereo-chemical reaction result in each case.

Answer. Secondary halogen derivative could have mono- or bi-molecule mechanism. Reaction product is 2-butanol, difference is only in the optical activity of products. If reaction follows mono-molecular mechanism (S_N1), intermediate product is secondary carbo-cation СН₃НС⁺СН₂СН₃. Nucleophilic reagent (hydroxyl group) attack on cation has equal direction possibility, as a result racemic mixture is formed which contains R- and S-2-butanols with equal amount in moles. Mixture has optically active components but generally racemic mixture is optically inactive.

If reaction follows bi-molecular mechanism (S_N2), intermediate product is complex containing substrate and reagent molecules and nucleophilic reagent attack is possible only at the side opposite to halogen atom. Result is rotation of the configuration, we get S-2-butanol which is optically active.

Question 42. At high doses of hydrasine or its derivatives action on organism there are nervous disorders. What is the chemical basis of hydrasine action if it is known that it reacts with co-enzyme pyridoxalphosphate?

Answer. Pyridoxalphosphate is hetero-cyclic compound which has aldehyde group in the cycle together with other substitutes. Hydrasine as nucleophilic reagent reacts with carbonyl carbon atom.

Final product at addition-elimination reaction is pyridoxalphosphate hydrason.

Hydrason formation leads to aldehyde group blocking at pyridoxalphosphate which breaks its interaction as co-enzyme with glutamic acid amino group. This reaction is one stage of glutamic acid transformation in the organism into g-amino-buteric acid. Co-enzyme blocking by hydrosine leads to g-amino-buteric acid deficiency which slows nervous impulse transmission.

Question 43. Methyl alcohol coming into organism leads to severe poisoning which is accompanied by eye-sight loss. Its reason is methanol oxidation product influence on eye cornea. Write down the scheme of methanol oxidation.

Answer. At methanol poisoning in the organism under alcohol-dehydrogenase action there is formaldehyde (methanol) and formic acid formation which are more toxic than methanol. This is an example of lethal synthesis (when less toxic substances due to metabolic process are transformed in to more toxic compounds).

Ethyl alcohol C₂H₅OH has better reaction with the enzyme which is called alcohol-dehydrogenase. It slows methanol transformation into methanal and formic acid. Methanol is discharged at initial form. That is why ethanol drinking immediately after methanol poisoning decreases the severity of poisoning.

Question 44. What are the signs of interaction between sodium oleate and bromine water?

Answer. Let’s analyze substrate structure in this reaction. Sodium oleate is oleic acid sodium salt. Oleic acid is higher unsaturated acid. Bromine water is bromine solution in water. Bromine is brown liquid, bromine water is coloured from brown to light-yellow according to concentration. Double bond in oleic acid radical participates in reaction with bromine:

СН₃(СН₂)₇СН=СН(СН₂)₇СООН+Br₂®СН₃(СН₂)₇СНBr–СНBr(СН₂)₇СООН

As a result of the reaction bromine concentration in water is decreased and bromine water turns lighter or there is complete discoloration. Bromine water discoloration is the sign of qualitative reaction on unsaturated compounds.