Topic 10: Organic Chemistry

 10.1 Introduction

10.1.1 Describe the features of a homologous series.

The compounds of a homologous series:

1.      Have the same general formula

e.g. alkanes – C_nH_2n+2, alkenes – C_nH_2n

2.      If neighbors, differ by one CH₂ group

e.g. methane (CH₄) and ethane (C₂H₆)

3.      Have similar chemical properties

4.      Gradually changing physical properties

e.g. the b.p. of alkanes increases with the number of carbons and with their length

10.1.2 Predict and explain the trends in boiling points of members of a homologous series.

The boiling point increases as the chain of a hydrocarbon gets longer. The larger the boiling point is, the smaller the effect of each added carbon on the boiling point.

10.1.3 Distinguish between empirical, molecular and structural formulas.

Structural formula is one, which shows unambiguously how the atoms are arranged together. A full structural formula shows every atom and bond:

Structural formula of ethane

Sometime, we can use condensed structural formula, which shows only atoms and omits bonds. In such formulas we can use:

1.      R as the symbol for an alkyl group;

2.      for representing the benzene ring.

Empirical formula shows the overall number of each atom in the compound.

 

C₂H₆

Molecular formula of ethane

10.1.4 Structural isomers

Structural isomers are compounds with the same molecular formula but different structural formula (different arrangement of atoms).

The isomers of butane

10.1.5 Deduce structural formulas for the isomers of the non-cyclic alkanes up to C₆.

10.1.6 Apply IUPAC rules for naming the isomers of the non-cyclic alkanes up to C_6.

Number of carbons	IUPAC Prefix
1	Meth
2	Eth
3	Prop
4	But
5	Pent
6	Hex

To name alkanes we use an IUPAC prefix and the suffix –ane.

Ex: Empirical formula – C₆H₁₄, hexane

10.1.7 Deduce structural formulas for the isomers of the straight-chain alkenes up to C₆.

Each carbon can have only four single bonds. If a carbon is double-bonded to another carbon, it can have only two more single bonds.

10.1.8 Apply IUPAC rules for naming the isomers of the straight-chain alkenes up to C₆.

To name alkenes we use an IUPAC prefix and the suffix –ene.

Ex: Empirical formula – C­₃H₆, this can be prop-1-ene or prop-2-ene, i.e. the double bond is the second or the first bond in the alkene. 

We always use the smallest numbers in IUPAC nomenclature. E.g. it is prop-1-ene and not prop-3-ene.

10.1.9 Deduce structural formulas for compounds containing up to six carbon atoms with one of the following functional groups: alcohol, aldehyde, ketone, carboxylic acid and halide.

Single hydrocarbon can contain more than one functional group. However, the functional groups are treated as different entities only if they are separated by a -CH₂- group!

The functional groups have certain formulas, which are unique for all of them. These formulas are used to represent the functional groups in the organic compounds. The formulas of the functional groups are:

1.      Alcohol – OH;

2.      Aldehyde – CHO;

3.      Ketone – CO;

4.      Carboxylic acid – COOH;

5.      Halide – F/Cl/Br/I.

10.1.10 Apply IUPAC rules for naming the compounds containing up to six carbon atoms with one of the following functional groups: alcohol, aldehyde, ketone, carboxylic acid and halide.

Single hydrocarbon can contain more than one functional group. However, the functional groups are treated as different entities only if they are separated by a -CH₂- group!

1.      Alcohol

We recognize primary, secondary and tertiary alcohols.

Primary alcohols – the OH functional group is attached to carbon with only one more carbon attached to it.

Secondary alcohols – the OH functional group is attached to carbon with two more carbon attached to it.

Tertiary alcohols – the OH functional group is attached to carbon with three more carbon attached to it.

Alcohols are shown in text by the suffix –ol.

2.      Aldehyde

Aldehydes have a hydrogen and alkyl group attached to a carbonyl function (RCHO). Aldehydes are shown in text by the suffix –al. They exist only on the ends of a compound.

3.      Ketone

Ketones have a pair of alkyl groups attached to a carbonyl function (RCOR). Ketones are shown in text by the suffix –one. They can never be on the ends of a compound, only on 2^nd and (n-1)^th carbon, where n is the number of carbons in the hydrocarbon. Thus, the smallest ketone is propanone.

4.      Carboxylic acid

Carboxylic acids have an alkyl group attached to a hydroxyl-carbonyl function (RCOOH). Carboxylic acids are shown in text by the suffix –anoic.

5.      Halide

Hydrogencarbons, which have one or more hydrogens replaced by a halogen (RX, where X=F, Cl. Br or I). Like with the alcohols halides can be primary, secondary or tertiary. For explanation refer to the point 1. 

10.1.11 Identify the following functional groups when present in structural formulas: amino (NH₂), benzene ring () and esters (RCOOR).

Benzene

With benzene resonance occurs.

Benzene permanently vacillates between two states, so that neither double bonds, nor single bonds occur.

It is important to remember that the bond in benzene is neither single nor double. Thus, we use the following symbol to show benzene:

Amines

R—NH₂

IUPAC nomenclature: Alkamine

Esters

  

IUPAC nomenclature: Alk₍₁₎yl Alk₍₂₎anoate

The alkyl (R') group is named first.

10.1.12 Identify primary, secondary and tertiary carbon atoms in alcohols and halogenoalkanes.

In primary alcohol or halogenalkane the functional group is bonded to a carbon, which has one carbon attached to it.

In secondary alcohol or halogenalkane the functional group is bonded to a carbon, which has two carbons attached to it.

In tertiary alcohol or halogenalkane the functional group is bonded to a carbon, which has three carbons attached to it.

10.1.13 Discuss the volatility and solubility in water of compounds containing the functional groups listed in 10.1.9.

Volatility and solubility of compounds depends on the intermolecular forces. We know three kinds of intermolecular forces, whose strength goes as following:

Van der Waals > dipole-dipole > hydrogen bonding

Volatility: If there are strong intermolecular forces (dipoles or hydrogen bonding), a compound is not very volatile. If there are only weak Van der Waal forces, the compound probably is volatile.

Solubility: Depends on the difference in electronegativity of two bonded atoms. If the difference is high, then the bonding is polar – if the difference is low, then it is non-polar. If we have a regular structure the polarity may cancel out. For instance, in tetrafluoromethane, the bond polarity cancels out, so the specie is non-polar.

 

The fundamental rule: likes dissolve likes

  

10.2 Alkanes

 

10.2.1 Explain the low reactivity of alkanes in terms of bond enthalpies and bond polarity.

Bond enthalpy: The bond enthalpy of C-C and H-H bond is very high – this means it takes a lot of energy to break the bonds. Hence, the C-C and H-H bonds are very strong. Alkanes are stable in air until sufficient activation energy is introduced.

Polarity: Alkanes have a very low polarity of the C-H bond (0.4). Thus, they are very stable and alkanes do not dissolve in polar solvents like water.

10.2.2 Describe, using equations, the complete and incomplete combustion of alkanes.

All alkanes may undergo combustion with oxygen to form carbon dioxide and water:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

If there is insufficient oxygen for complete combustion, alkanes will still burn, but undergo incomplete combustion (this is what occurs in internal combustion engines such as car engines).

Incomplete combustion may produce elemental carbon, carbon monoxide or both:

C₂H₄ + 2O₂ → 2CO + 2H₂O

C₂H₄ + O₂ → 2C +2H₂O

The combustion of unsaturated hydrocarbons is more likely to be incomplete. 

10.2.3 Describe, using equations, the reactions of methane and ethane with chlorine and bromine.

The significance of such reactions is that alkanes react readily only with halogens. The process is called free radical substitution.

Methane: CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g)

Ethane: CH₃CH₃(g) + Cl₂(g) → CH₃CH₂Cl(g) + HCl(g)

The process requires UV light, usually from the sunlight.

10.2.4 Explain the reactions of methane and ethane with chlorine and bromine in terms of a free-radical mechanism.

Alkanes react readily only with halogens. The process is called free radical substitution and has three steps:

1.      Initiation – UV light breaks the halogen gas to create a radical (homolitic fission)

Cl₂→2Cl∙

2.      Propagation – radicals used and re-created

Cl∙ + CH₃CH₃ → CH₃CH₂∙ + HCl

Cl₂ + CH₃CH₂∙ → CH₃CH₂Cl + Cl∙

3.      Termination – radicals removed and reaction proceeds to the end

Cl∙ + CH₃CH₂∙ → CH₃CH₂Cl

OR

Cl∙ + Cl∙ →Cl₂

OR

CH₃CH₂∙+ CH₃CH₂∙ → C₄H₁₀

The process requires UV light, usually from the sunlight.

Homolitic fission – a covalent bond is broken so that each atom retains one electron from the broken bond; these atoms are very reactive free radicals.

10.3 Alkenes

10.3.1 Describe, using equations, the reactions of alkenes with hydrogen and halogens.

Alkenes have a carbon-carbon double bond of which one bond, the π bond, can be easily broken. As this bond is broken, two new bonding positions within the compound are created. This enables alkanes to undergo addition reaction with:  

1.      Hydrogen - hydrogenation

·         Hydrogen gas reacts with alkenes to form alkanes

·         Ex: CH₃CHCH₂ + H₂ --> CH₃CH₂CH₃

2.      Halogen - halogenation

·         Halogens react with alkenes to produce dihalogeno compounds

·         Ex: CH₃CHCH₂ + Br₂ --> CH₃CHBrCH₂Br

10.3.2 Describe, using equations, the reactions of symmetrical alkenes with hydrogen halides and water.

Alkenes have a carbon-carbon double bond of which one bond, the π bond, can be easily broken. As this bond is broken, two new bonding positions within the compound are created. This enables alkanes to undergo addition reaction with:

1.      Hydrogen halides (Ex: HCl, HBr)

·         Hydrogen halides react with alkenes to produce halogenoalkanes

·         These reactions are very rapid

·         Reactivity: HI>HBr>HCl because the hydrogen-halide bond weakens down Group 7

2.      Water

·         The reaction between water and turns the alkene into alcohol

·         Addition reaction of water and halogen requires a catalyst to take place – such catalyst is concentrated sulfuric acid

10.3.3 Distinguish between alkanes and alkenes using bromine water.

Because halogens react with alkenes we can use bromine water to distinguish alkanes from alkenes. Bromine only reacts with alkenes and when it does, it loses its orange/yellow color.

This test is useful because the addition reaction with halogens proceeds very quickly at room temperature. Alkanes react with Br₂ only if UV light is present.

10.3.4 Outline the polymerization of alkenes.

Because the C=C bond in alkenes can be easily broken, alkenes like to undergo additional reactions. Two alkenes, whose double bond is broken, can react with each other in an addition polymerization reaction. 

The addition polymerization forms long hydrocarbons, usually containing 40,000 to 80,000 carbons, called polymers. The repeating building unit of a polymer is called monomer. 

The polymerization reactions, which we should know are:

1.      Additional polymerization of ethane to form poly(ethane).

2.      Additional polymerization of chloroethene to poly(chloroethane), aka, PVC.

3.      Additional polymerization of propene to poly(propene).

Propene                                    poly(propene)

Do not forget:

We identify the polymer as a repeating unit.

E.g. for polyethene -(-CH₂-CH₂-)_n-

10.3.5 Outline the economic importance of the reactions of alkenes.

• The hydrogenation of vegetable oils in the manufacture of margarine
• The process of hydrogenation is used in the margarine industry to convert oils containing many unsaturated hydrocarbon chains into more saturated compounds which have higher melting points
• This is done so that the margarine will be a solid at room temperature.
• Margarine

• The hydration of ethene in the manufacture of ethanol
• The hydration of ethene is of industrial significance because ethanol is a very important solvent and is therefore manufactured on a large scale.
• Solvent of substances intended for human contact or consumption, fuel, alcoholic beverages, antiseptic

• The polymerization in the manufacture of plastics
• Polyethene
• household containers, plastic bags, water tanks and piping
• Polypropene
• clothing and thermal wear for outdoor activities
• Polychloroethene
• construction materials, packaging, electrical cable sheeting
• it is one of the world's most important plastics

10.4 Alcohols

 

10.4.1 Describe, using equations, the complete combustion of alcohols.

All alcohols burn in presence of oxygen to make carbon dioxide and water.

Example: Combustion of ethanol

C₂H₅OH(l) + 3O₂ → 2CO₂(g) + 3H₂O(l)

10.4.2 Describe, using equations, the oxidation reactions of alcohols.

We distinguish partial oxidation and complete oxidation. Partial oxidation converts alcohol to aldehyde, while complete oxidation coverts alcohol to carboxylic acid.

Conditions for oxidation:

1.   Excess potassium dichromate catalyst;

2.   The catalyst must be acidified, for instance, by sulfuric acid (since it is inert);

3.   The mixture must be heated;

4.   Individual conditions:

§  Partial oxidation to aldehyde – distill as the reaction goes;

§  Complete oxidation to carboxylic acid – heat under reflux and then distill.

During the reaction the color of the solution changes from orange to dark green because orange chromium(VI) is reduced into green chromium(III).

10.4.3 Determine the products formed by the oxidation of primary and secondary alcohols.

Primary alcohols are oxidized to form aldehydes and carboxylic acids.

Secondary alcohols are oxidized to form ketones.

10.5 Halogenalkanes

 

10.5.1 Describe, using equations, the substitution reactions of halogenoalkanes with sodium hydroxide.

Halogenalkanes react with aqueous hydroxide ions from sodium hydroxide – the name of this process is nucleophilic substitution. It is called substitution because the hydroxide ion takes place of halogen in a halogenalkane. It is nucleophlic because the hydroxide ion is attracted to positive dipole of carbon atom.

The carbon atom has a positive dipole because it is bonded to a halogen, which has much higher electronegativity than carbon. The hydroxide ion is a nucleophile because it has a negative charge and thus it is attracted to positively charged particles.

Substitution does not occur with fluoroalkanes because the C-F bond is very strong for fluorine’s high electronegativity.

Example of nucleophilic substitution reaction:

Do not forget to use the curly arrow notation!

10.5.2 Explain the substitution reactions of halogenoalkanes with sodium hydroxide in terms of S_N1 and S_N2 mechanisms.

In nucleophilic substitution, a nucleophile gets attracted to the localized positive charged of a molecule, repulses halogen and takes its bonding position.

S_N1

The S_N1 mechanism is unimolecular nucleophilic substitution – this means that it starts with one molecule only. It occurs with tertiary halogenalkanes (tertiary – carbon to which halogen is attached has three more carbons attached to it). S_N1 occurs with tertiary halogenalkanes because alkyl groups prohibit nucleophiles from accessing the positive dipole (just imagine a tertiary halogen).

S_N1 occurs in two steps:

1.      Random collision causes heterolytic fission of the C-(halogen) bond. Halide ion and a carbocation are formed. This is the rate determining step because it is very slow.

2.      The carbocation reacts rapidly with the OH^- nucleophile.

S_N2

The S_N2 mechanism is bimolecular nucleophilic substitution. It occurs with primary halogenalkanes. It occurs in one step, but for a simplicity we can divide it in multiple phases (remember that these phases cannot be distinguished):

1.      Nucleophile becomes attracted to the localized positive charge created by uneven sharing of electrons between a carbon (α⁺) and halogen (α^-).

2.      Nucleophile comes to the localized positive charge and, at the same time, repulses the halogen. Because the halogen has higher electronegativity than the carbon, as it gets repulsed a heterolytic fission occurs – the halogen takes both electrons from the C-(halogen) bond.

3.      The halogen leaves the halogenalkane with both electrons from the C-(halogen) bond making a halide ion. The OH­^- nucleophile takes its bonding position.

The position, in which both nucleophile and halogen are equally strongly attracted to the carbon is called transition state.

Do not forger to use curly arrows to indicate the movement of electrons!

10.6.1 Deduce reaction pathways given the starting materials and the product.