# IB Chemistry/Organic Chemistry

Organic chemistry is the study of compounds which contain carbon. The name 'organic' refers to the historical link between these compounds and living organisms. 'Inorganic' compounds were those which could be derived from non-living sources. Confusingly, some inorganic compounds contain carbon.
This material in this chapter is restricted to that needed by SL Chemistry students
For further material see Organic Chemistry

# 10.1

10.1.1 Describe the features of a homologous series. This is how organic compounds are classified into 'families'. Successive members of a homologous series differ by a -CH$_2$. Molecular mass increases by a fixed amount as you go up the series. For example, alkanes. Methane (CH$_4$), ethane (C$_2$H$_6$, propane (C$_3$H$_8$), butane (C$_4$H$_{10}$). General formula: C$_n$H$_{2n+2}$ Other series have an additional 'functional group' which is also shown in the general formula. The functional group is usually a small group of atoms attached to a carbon atom in the molecule. For example, alcohols have the functional group -OH. General formula: C$_n$H$_{2n+1}$OH.
10.1.2 Predict and explain the trends in boiling points of members of a homologous series. Since successive members differ by -CH$_2$, their carbon chains will continue to increase in length. Boiling point increases with increasing carbon number. Note that if you know the chemical characteristics of a functional group, you can predict the chemical properties of the series.
10.1.3 Distinguish between empirical, molecular and structural formulas. The empirical formula is the lowest whole number ratio of the atoms. The molecular formula is the actual number of atoms. The structural formula is a representation of how the molecules are bonded together. Full structural shows all bonds, whereas condensed structural omits bonds where they can be assumed. Ethane... Empirical: CH$_3$, Molecular: C$_2$H$_6$, Condensed structural: CH$_3$CH$_3$ Full structural:
H — C — C — H
10.1.4 Describe structural isomers. Compounds with the same molecular formula but different arrangements of atoms.
10.1.5 Deduce structural formulas for the isomers of the non-cyclic alkanes up to C$_6$. C$_4$H$_{10}$: Butane and 2-methyl propane.
H — C — C — C — C — H H — C — C — C — H
C$_5$H$_{12}$: Pentane, 2-methylbutane, 2,2-dimethylpropane.
H — C — C — C — C — C — H H — C — C — C — C — H H — C — H
Naming Organic Compounds:
Organic chemistry is concerned with the compounds of carbon. Since there are more compounds of carbon known than all the other elements together, it is helpful to have a systemic way of naming them.
Identify longest carbon chain (-ane) 1 carbon = meth- 2 carbons = eth- 3 carbons = prop- 4 carbons = but- 5 carbons = pent- 6 carbons = hex- 7 carbons = hept- 8 carbons = oct-
Identify type of bonding in the chain All single bonds in the carbon chain = -an- One double bond in the carbon chain = -en- One triple bond in the carbon chain = -yn-
Identify functioning group
Formula Name Example R-H Alkane Methane R-OH Alcohol ethanol R-NH$_3$ Amine Ethylamine R-X Halogenoalkane Bromoethane (Where X = F, Cl, Br, I) R-C=O Aldehyde Ethanal
R-C-R Ketone Propanone
R-C-OH Carboxylix Acid Ethanoic Acid
R-C-OR Ester Ethyl Ethanoate
Numbers are used to give the positions of groups or bonds along the chain.
Homologous Series The alkanes form a series of compounds all with the same general formula e.g. Methane Ethane Propane
If one of the hydrogen atoms is removed what is left is known as an alkyl radical R- (e.g. methane) When other atoms or groups are attached to an alkyl radical they can form different series of compounds. These atoms or groups are known as functioning groups and the series formed are all homologous series. Homologous series have the same general formula with the neighbouring members of the series differing by by : for example the general formula of alcohols is OH. The chemical properties of the individual members of an homologous series are similar and they show a gradual change in physical properties.
Boiling Points As the carbon chain gets longer the mass of the molecules increases and the van der Waals' forces of attraction increase. A plot of boiling points against the number of carbon atoms show a sharp increase at first, as the percentage increase in mass is high, but as successive groups are added the rate of increase in boiling point decreases. When branching occurs the molecules become more spherical in shape, which reduces the contact surface area between them and lowers the boiling point. Other homologous series show similar trends but the actual temperatures at which the compounds boilwill depend on the type of attractive forces between the molecules. The volatility of the compounds also follows the same pattern. The lower members of the alkanes are all gases as the attractive forces are weak and the next few members are volatile liquids. Methanol, the first member of the alcohols is a liquid at room temperature, due to the presence of hydrogen bonding. Methanol is classed as volatile as its boiling point is 337.5K but when there are four or more carbon atoms in the chain the boiling points exceed 373K and the higher alcohols have low volatility.
Functional Group Strongest type of intermolecular attraction Alkane van der Waals' Alkene van der Waals' Alkyne van der Waals' Ester Dipole:Dipole Aldehyde Dipole:Dipole Ketone Dipole:Dipole Amine Hydrogen bonding Alcohol Hydrogen bonding Carboxylix Acid Hydrogen bonding
Low Reactivity of Alkanes Because of the strong C-H and C-C in alkanes they have low reactivity and only usually undergo combustion in the presence of Oxygen and substitution reactions with halogens in ultra-violet light.
MECHANISM OF SUBSTITUTION REACTION WITH HALOGENS Chemical bonds can break homolytically or heterolytically. Heterolytically is when both of the shared electrons go to one of the atoms resulting in a positive and one negative ion. When halogen are exposed to ultraviolet light the halogen molecule bond can break homololytically meaning each of the atoms retains one of the pair of previously shared electrons. The bond between two halogen atoms is weaker than C-H or C-C bond in alkanes and so is more likely to become a free radical due to homolytic fission of the covalent bond. When a Chlorine molecule is exposed to ultraviolet rays and becomes two free radical chlorine atoms it is called the intiation stage. Free radicals have an unpaired electron and are therefore very reactive.
Topic 20 is the additional HL material for Topic 11.
Determination of structure
This HL Sub-topic is required for Sub-topic 1 of SL Option A
Hydrocarbons
The material on benzene in this HL Sub-topic is required for Sub-topic 1 of SL Option A
Nucleophilic substitution reactions
This HL Sub-topic is Sub-topic 4 of SL Option
Alcohols

## Option G Syllabus

G.1.1: Describe and explain the electrophilic addition Mechanisms of the reactions of alkenes with halogens and hydrogen halides.
The following is an example of an electrophilic addition Mechanism describing the reaction of the alkene Ethene and the halogen Bromine, forming 1,2-dibromoethane and Hydrogen (not pictured)
The next mechanism describes the electrophilic addition reaction between Ethene and Hydrogen Bromide (a hydrogen halide).
G.1.2: Predict and explain the formation of the major products in terms of the relative stabilities of carbocations.
Markovnikov’s Rule(See Mechanism above as example):
In the addition of HX to an unsymmetrical alkene, the H attaches to the C with fewer alkyl substituents, producing the more stable carbocation (Tertiary carbocations are more stable than secondary which are more stable than primary). The X then attaches to the C with more alkyl groups.
G.2.1: Describe, using equations, the addition of hydrogen cyanide to aldehydes and ketones.
G.2.2: Describe and explain the mechanism for the addition of hydrogen cyanide to aldehydes and ketones.
G.2.3: Describe, using equations, the hydrolysis of cyanohydrins to form carboxylic acids.
G.3.1: Describe, using equations, the dehydration reactions of alcohols with phosphoric acid to form alkenes.
G.3.2: Describe and explain the mechanism for the elimination of water from alcohols.
G.4.1: Describe, using equations the reactions of 2,4-dinitrophenylhydrazine with aldehydes and ketones.
G.5.1: Describe and explain the structure of benzene using physical and chemical evidence.
Physical Evidence:
All bonds of benzene are of equal length—shorter than single bonds, but longer than double.
Chemical Evidence:
Benzene does not undergo addition reactions
Heat of hydrogenation of benzene is endothermic whereas the heat of hydrogenation of a triene would be extremely exothermic
G.5.2: Describe and explain the relative rates of hydrolysis of benzene compounds halogenated in the ring and in the side chain.
The rate of hydrolysis in the side chain is much faster than benzene compounds halogenated in the ring because benzene does not undergo addition reactions. Chlorobenzene is much slower than alkanes in hydrolysis and does not undergo nucleophilic substitution for two main reasons:
1. The nonbonding pair of election of the chlorine may react with the pi bond, and become delocalized, thus strengthening (and possibly depolarizing) the C-Cl bond.
2. The high electron density of the delocalized pi bond repels the nucleophile and blocks it from the slight positive carbon atom attached to the chlorine.
G.6.1: Outline the formation of Grignard Reagents.
G.6.2: Describe, using equations, the reactions of Grignard reagents with water, carbon dioxide, aldehydes, and ketones
G.7.1: Deduce the reaction pathways given the starting materials and the product.
G.8.1: Describe and explain the acidic properties of phenol and substituted phenols in terms of bonding.
Alcohols are not very acidic because they produce an anion which contains only one resonance structure, and localized electrons.
Phenol is acidic due to its electron-withdrawing groups. As an anion, it becomes more stable due to the delocalized electrons which produce various resonance structures.
2,4,6-trinitrophenol contains many electron-withdrawing groups which cause it to be very acidic due to further delocalization of the electrons.
G.8.2: Describe and explain the acidic properties of substituted carboxylic acid in terms of bonding.
The adjacent C=O bond weakens the normally strong O-H bond.
The ion formed by the removal of the H from a carboxylic acid is more stable because of the delocalized electrons forming several resonance structures.
G.8.3: Compare and explain the relative basicities of ammonia and amines.
Ammonia is a weak base
Amines are more basic than ammonia because the inductive effect of the alkyl group pushes the electrons towards the nitrogen atom, increasing the electron density of the nonbonding electron pairs of the nitrogen.
G.9.1: Describe, using equations, the reactions of acid anhydrides with nucleophiles to form carboxylic acids, esters, amides, and substituted amides.
G.9.2: Describe, using equations, the reactions of acyl chlorides with nucleophiles to form carboxylic acids, esters, amides, and substitutes amides.
G.9.3: Explain the reactions of acyl chlorides with nucleophiles in terms of an addition-elimination mechanism.
G.10.1: Describe, using equations, the nitration, chlorination, alkylation, and acylation of benzene.
G.10.2: Describe and explain the mechanism for the nitration, chlorination, alkylation, and acylation of benzene.
G.10.3: Describe, using equations, the nitration, chlorination, alkylation, and acylation of methylbenzene.
The methyl group is an electron supplier and an activator, so it is an ortho/para-director.
G.10.4: Describe and explain the directing effects and relative rates of reaction of different substituents on a benzene ring.
G.11.1: Deduce reaction pathways given the starting materials and the product.