Organic Chemistry Study Notes PDF

Preview

Summary

Study notes on organic chem...

Description

CHEMISTRY HSC MODULE 7 STUDY NOTES: ORGANIC CHEMISTRY

BY JENNIFER AZZI

INQUIRY QUESTION 1: HOW DO WE SYSTEMATICALLY NAME ORGANIC COMPOUNDS? 1. CARBON BONDING ● ● ● ●

It forms four bonds to achieve a stable electron configuration. C readily forms carbon-carbon bonds C- bonds can be single, double or triple covalent bonds Called ‘organic’ because they are derived from living things. However, chemists now synthesise these molecules in the laboratory.

2. HYDROCARBONS ● ●

●

●

Compounds containing only carbon and hydrogen atoms and are the building blocks of organic molecules. Homologous series ○ Series of compounds with the same functional group and similar chemical properties in which the members of the series differ by the number of repeating units they contain ■ Alkanes ■ Alkenes ■ Alkynes Saturated hydrocarbons ○ Molecules made entirely of single carbon-carbon bonds (C-C); they cannot incorporate additional atoms into their structure ○ Called alkanes ○ Are stable and not very reactive Unsaturated hydrocarbons ○ Alkenes (C=C) and alkynes (C≡C) are molecules that contain at least one double or triple carbon-carbon bond within their structure. ○ These molecules are highly reactive, and thus can incorporate other atoms into their structure.

3. NAMING & REPRESENTING ORGANIC MOLECULES Carbon Atoms 1 2 3 4 5 6 7 8 9 10 ● ● ●

Prefix MethEthPropButPentHexHept OctNonDec-

IUPAC name methane ethane propane butane pentane hexane heptane octane nonane decane

Formula CH4 C2H6 C3H8 C4H10 C5H12 C6H14 C7H16 C8H18 C9H20 C10H22

Alkyl group methylethylpropylbutylpentylhexylheptyloctylnonyldecyl-

Alkyl formula -CH3 -C2H5 -C3H7 -C4H9 -C5H9 -C6H13 -C7H15 -C8H17 -C9H19 -C10H21

A molecular formula such as C3H8 gives information about the number of atoms, but nothing about their arrangement. A structural formula shows the arrangement of atoms in space and all the bonds between the atoms. A condensed structural formula show the atoms that are connected to each carbon atom but do not show the bond.

●

The skeletal formula is more basic showing only a representation of the carbon chain arrangement, but not the atom detail.

Molecular formula:

Structural formula:

Condensed structural formula:

Ball and Stick:

Lewis Structure:

C3H8

Skeletal (line) formula:

4. ALKANES ● ● ● ● ● ●

Suffix: -ane All the bonds are single bonds Four bonds are tetrahedrally arranged Saturated (no more H can be added) Represented by the formula above (n can be 1, 2, 3 … etc.) Called straight-chain alkanes

CnH2n +2

● ● ●

Suffix: -ene Same prefixes used as for alkanes. The position of the double bond is indicated with the number of the C atom with the double bond. Numbered from the end giving the lowest number. A dash is used to separate numbers from words

NB: Methene does not exist. Isomers: have the same molecular formula but different structural formula. Have different physical properties. 1–butene and 2–butene are isomers of each other. There are no isomers of ethene and propene.

CnH2n

6. ALKYNES ● ● ● ● ●

Suffix: -yne Unsaturated hydrocarbons (ideal for synthesis reactions) At least one triple bond between C atoms Represented by formula above (n can be 2, 3, 4…) Isomers are possible

CnH2n-2 7. BRANCHED HYDROCARBONS

● ●

Suffix: -yl “Alkyl” groups

IUPAC Naming of Hydrocarbons 1. Identify the LONGEST carbon chain – be careful – it does not have to appear in a straight line 2. The long chain is assigned the usual alkane name 3. The branch is assigned its alkyl name depending on the number of carbons in the branch 4. The location of the branch off the main chain is assigned a number (the lowest possible) Rules ● Alphabetical order ● Prioritise the lowest sum 8. ISOMERS Isomers are compounds that have the same molecular formulas (same number and type of atoms) but differ in the way the atoms are arranged. Structural Isomers ● Isomers that differ in the way the atoms are bonded to each other. ● They may have similar but not identical properties. ● They are formed as a result of branching. Types of Structural Isomers Type

Definition

Chain Isomers

The arrangement of carbon atoms in the carbon chain differs

Example

The position of one or more functional groups differs Positional Isomers

Functional Group Isomers

The molecules have different functional groups and therefore belong to different homologous series

9. CARBOXYLIC ACIDS ● ● ●

Carbon chain which contains a -COOH group (single bonded alcohol and a double bonded oxygen). Suffix: -oic acid E.g. ethanoic acid, 1-butanoic acid

10. ESTERS ● ● ● ●

Alkyl alkanoates Formed from the reaction between an alkanol and an alkanoic acid, where a water molecule is lost (via a condensation reaction) with the use of a catalyst (usually conc. H2SO4). The water formed comes from the OH group of the alkanoic acid and the H from the OH group in the alkanol. All esters contain the functional group –COO–. Naming Esters ○ The alkanol forms the first part of the ester’s name with its ending changing from ‘anol’ to ‘yl’ ○ The alkanoic acid forms the second part of the ester’s name with its ending changing from ‘oic’ to ‘oate’.

11. ALDEHYDES & KETONES Aldehydes ● Contain the functional group =O (double bonded oxygen) at the end of the carbon chain (can’t be in the middle of the chain, only at the end) ● Suffix: -al

Ketones ● Contains the functional group =O located at any carbon other than the terminal carbons. ● Suffix: -one

12. AMINES ● ● ● ●

These compounds are ammonia with one or more of the hydrogen atoms replaced by R Contain the functional group NH2, NH3, NH Suffix: -amine Prefix: N-

Ammonia Primary, Secondary and Tertiary Amines TYPE OF AMINE PRIMARY

Amines

DESCRIPTION

Example

- have ONE C atom directly bonded to the N atom - Length of C continuous chain determines the end of the name. - Carbon bonded to the amino group assigned lowest locator number Group name is “amine”

Butan 2 - amine

SECONDARY

- have TWO C atoms directly bonded to the N atom - Amine named using the longest C chain - The other alkyl group is prefixed with N to show that it is part of the amine group N-methylpropan-1-amine

TERTIARY

- have THREE C atoms directly bonded to the N atom -As for secondary amines but N must be shown for both alkyl groups – listed alphabetically - N,N used when the 2 alkyl groups are attached to the nitrogen atom - If the alkyl groups are the same, “di” is also used.

N,N-ethyl methyl pentan-2amine

N-diethyl pentan-2-amine

13. AMIDES ● ● ● ● ● ● ●

Another ammonia derived compound The carbon bonded to the nitrogen is also bonded to an oxygen It is a carboxylic acid group (-COOH) with the NH2 group replacing the OH and so must be an end C on a carbon chain To name them – name the carboxylic acid and then change the end of the name from –oic acid to –amide If alkyl groups replace the hydrogens bonded to N – apply the same rule as for amines Suffix: -amide Prefix: N-

Types of Amides: ● PRIMARY - have ONE C atom directly bonded to the N atom ● SECONDARY - have TWO C atoms directly bonded to the N atom ● TERTIARY - have THREE C atoms directly bonded to the N atom

INQUIRY QUESTION 2: HOW CAN HYDROCARBONS BE CLASSIFIED BASED ON THEIR STRUCTURE AND REACTIVITY? 1. PROPERTIES OF HYDROCARBONS AND HALOALKANES

●

●

Chemical properties: similar chemical properties as compounds contain the same functional group ○ Covalent bonding is present within organic molecules ○ The same covalent bonds form between the carbons and the functional group. For the compound to react, energy is required to break the covalent bonds so that the compound can rearrange. As the functional group is the centre of reactivity of the compound, the energy required for molecules in homologous series to react will be similar ○ Therefore, the energy required for propanoic acid to react will be similar to the energy required for butanoic acid react. ○ E.g. reactivity, toxicity, oxidation and stability are all relatively constant across the homologous series Physical properties: different physical properties due to the difference in compound size and intermolecular forces ○ Melting point and boiling point increase as carbon chain length increases ○ The dispersion forces (Van der Waals forces) between the compounds increases as atomic mass increases – the greater the chain length, the greater the melting and boiling point

2. MOLECULAR SHAPE & HYBRIDISATION ● ● ●

Bonding between atoms requires the mixing of atomic orbitals to form molecular orbitals The mixing of orbitals is called hybridisation Hybridisation will determine the geometry around an atom

Single bonds: ● Form from the mixing of 1 s orbitals and 3 p orbital ● This forms a sp3 hybridised orbital ● Sp3 hybridised molecules have a tetragonal geometry ● Tetragonal: bond angles = 109.5o Double bonds: ● Form from the mixing of 1 s orbitals and 2 p orbital ● This forms a sp2 hybridised orbital ● Sp2 hybridised molecules have a trigonal geometry ● Trigonal: bond angles = 120o Triple bonds: ● Form from the mixing of 1 s orbital and 1 p orbital ● This forms a sp hybridised orbital ● Sp hybridised molecules have a planar geometry ● Planar: bond angles = 180o

3. SAFE HANDLING AND DISPOSAL Safe handling: ● Use the smallest amount of the chemical as possible ● Containers should be labelled properly ● Substances should be stored in appropriate containers ● Never return dispensed chemicals into the stock bottle to prevent contamination ● Must wear appropriate footwear (closed in shoes, no heels) ● Must wear appropriate eye protection (safety glasses) ● Must wear appropriate clothing that will cover skin (lab coat)

Disposal: ● Most organic compounds can be flushed down the sink with approximately 100x as much water as the volume of the substance ● Halogenated organic compounds need to be disposed into halogenated waste containers as mixing with other organic solvents can cause explosions ● Highly reactive organic compounds need to be disposed of separately and not in general waste containers ● Cyanide waste needs to be kept in a separate cyanide wastage container and kept alkaline 4. IMPLICATIONS OF OBTAINING AND USING HYDROCARBONS Hydrocarbons are mined to be used as fuels and the basis of polymers. For example, propane is used to fuel BBQs and heat houses and long chain alkanes can be converted to short chain alkanes and alkenes which can be made into plastics via polymerisation. Implications from obtaining and using hydrocarbons: - Environmental: o Clearing landscapes for efficient mining ▪ Removal of trees ● Increases the water table o Brings more salt to the surface o Water needs to be pumped out of the mine to prevent flooding ▪ Removal of animal habitats o Erosion: ▪ Aggravation of soil and exposure of rocks to the elements makes it easier for water to wash away dirt o Water pollution: ▪ Contamination of water by toxic compounds (sulfuric acid, mercury, arsenic, heavy metals) ▪ Contamination of water by run-off o Abandoned mines have changes in soil compositions: ▪ Decreases biodiversity ● Different plants grow back ▪ Infertile ground can have nutrients stripped o Air pollution: ▪ Particulates containing metals, sulfates, arsenic and suspended matter degrade air quality ▪ Can cause respiratory and skin disease in humans o Combustion for energy production: ▪ Produces: ● Complete combustion: water and carbon dioxide (CO2) ● Incomplete combustion: water, carbon dioxide (CO2), carbon (soot, C) and carbon monoxide (CO) ▪ CO2: ● Is a greenhouse gas – major contribution to global warming ▪ CO: ● Extremely poisonous ● Odourless and colourless such that it can build up in a room without detection and cause death ▪ C: ● Causes respiratory problems o Methane: ▪ Oxidises to CO2 o SO2 and NOX released from fuel containing contaminants due to combustion: ▪ SO2:

● irritates respiratory system causing breathing difficulties and triggering asthma and emphysema ● unpleasant odour ● acid rain formation ▪ NO2: ● irritates respiratory system causing breathing difficulties and extensive tissue damage at high concentrations ● creates photochemical smog which is a form of air pollution where sunlight reacts with nitrogen dioxide in the presence of hydrocarbons and oxygen to form ozone and other pollutants which can lead to poor visibility and harmful effects ● acid rain formation ▪ React with water vapour in the atmosphere to produce acid rain ● increasing acidity of lakes which has a detrimental effect on fish populations ● damage to pine forests ● erosion of the marble and limestone of building surfaces and decorations as they contain carbonates which readily react with acids ● severe damage to vegetation around mine and smelter sites, especially in town with high rainfall such as Queenstown in Tasmania -

-

Economic: o Australia is the greatest exporter of coal, iron ore, lead, diamonds, rutile, zinc and zirconium o Represents around 20% of the Australian stock exchange o Contributes around 6% of Australia’s gross domestic product o Workforce: ▪ Provides a small number of jobs – approximately 2% of the population o Highly subsidised by the government ($4 billion) o Most mining companies (83%) in Australia are owned by foreign companies o Businesses and power stations are powered by burning of fossil fuels o Businesses rely on transport powered by petrol Sociocultural: o Housing: ▪ High cost and shortage of housing in areas near mining site o Increase in “fly in and fly out” workforces o Unfairness in communities where there is a significant gap in the pay between mining staff and local industry workers o Mining communities are at risk of limited jobs and sustainability if mining businesses were to close down o People of Australia require fossil fuels for energy used in their everyday lives (e.g. car fuel, refrigerators, lights) o Other countries receive Australian fossil fuels and rely on it for energy production

INQUIRY QUESTION 3: WHAT ARE THE PRODUCTS OF REACTIONS OF HYDROCARBONS AND HOW DO THEY REACT? 1. ADDITION REACTIONS -

hydrogen (H2)

Hydrogenation reaction: - between an alkene and hydrogen gas - requires a non-soluble metal catalyst -> usually palladium as Pd-C, platinum as PtO 2 or nickel as Ra-Ni - the reaction is thermodynamically favourable, forming a lower energy product. Therefore, it is an exothermic reaction - Reaction: o In the presence of the catalyst, the H-H bond of the H2 is cleaved o The Hs bond to the catalyst surface forming a metal-hydrogen bond o The metal catalyst absorbs the alkene onto its surface, breaking the double bond o In series, the Hs are added to the same side of the alkene forming an alkene Metal catalyst

Alkene + H2 → Alkane Pd/C

E.g. C3H6 (aq) + H2 (g) → C3H8 (aq)

-

Halogens (X2)

Halogenation reaction: - Between an alkene and a halogen - Addition reaction: two Br, I or Cl are added to the alkene E.g. C4H8 (aq) + Br2 (g) → C4H8Br2 (aq) - hydrogen halides (HX) Reaction: - Between an alkene and a hydrogen halide - Addition reaction: first the hydrogen is added to the alkene and then the halogen is added

E.g. C2H4 (g) + HCl (aq) → C2H5Cl (aq) -

water (H2O)

Hydration reaction: - Between an alkene and water - Requires a dilute acid catalyst – usually, dilute H 2SO4 - Will form an alcohol Dilute H2SO4

Alkene + H2O → Alcohol Dilute H2SO4

E.g. C2H4 (aq) + H2O (l) → C2H5OH (aq) 2. SUBSTITUTION REACTIONS The reaction between an alkene and a halogen is an addition reaction, where the reactants are added together to form one product. The reaction between an alkane and a halogen is a substitution reaction, where part of one reactant is substituted for part of the second reactant to form multiple products. This is due to the alkane only having single bonds and causing it to be already saturated. Halogenation with an alkane: - The reaction is not as favourable and so requires UV to provide sufficient energy for the reaction to proceed UV light

E.g. C3H6 (aq) + Br2 (g) → C2H5Br (aq) + HBr (aq) PRACTICAL INVESTIGATION: IDENTIFYING UNSATURATED BONDS IN ALKENES USING BROMINE WATER (DEMONSTRATION) Aim: To determine a chemical test which will allow us to distinguish between saturated and unsaturated carbon compounds Materials ● small test tubes with rubber stopper to match ● test tube racks ● dropper bottles containing bromine water ● an example of a saturated carbon compound ● an example of an unsaturated carbon compound ● organic waste bottle ● Safety glasses ● Gloves ● Lab coat Method 1. 2. 3. 4. 5. 6.

Darken the room to eliminate the presence of U.V light. Place 5 mL of an unsaturated compound into a small test tube. Add several drops of bromine water to the test tube, and mix well. Place 5 mL of a saturated compound into a different test tube. Add 2 mL of bromine water to the test tube, and mix well. When you have finished the experiment, place the liquids in the waste bottle provided by your teacher.

7.

Make sure that your work area is left clean and any spills are cleaned up.

Results Testing Saturated Compound type

Initial Colour of bromine water Orange / brown

Saturated carbon compound (Hexane) Unsaturated carbon compound (Hexene)

Orange / brown

Final Colour of bromine water No colour change (will only react under the presence of UV light) Discoloured / clear

Results

NO REACTION (will only occur in presence of U.V light) Hexane layer (Colourless)

Bromine water layer (Brown)

Hexane/Cyclohexane (Saturated/single bond)

REACTION after shaking

Hexene layer (Colourless)

Bromine water layer decolourises by any alkene

Hexene/Cyclohexene (Unsaturated/double bond)

Discussion 1. Describe the results obtained and include relevant equations. An alkene (hexene) was distinguished from an alkane (hexane) by shaking with bromine water. Bromine water is

brown/yellow and it will lose its colour when it reacts with a double bond in an alkene. The alkane in bromine water will stay brown as there is no double bond. In an alkene, the Bromine water adds across the double bond. This is an example of an addition reaction.

Both test tubes contained two layers: a lower aqueous layer (bromine water) and an upper organic (alkene/alkane). In the test tube containing the hexane, the lower layer remained yellow/brown, whereas in the test tube containing the hexane, this discoloured. Hexene reacted with the bromine water as it is unsaturated. This is due to the very reactive double bond which “split open” leaving a single C-C bond and creating two new bond positions for the bromine atoms to attach. 2. Explain why alkanes and their corresponding alkene have similar physical properties but very different chemical properties. Include a relevant chemical equation. (4 marks) Alkanes and alkenes contain only C–H, single C–C and double C=C bonds. All of these bonds are non-polar and so the only intermolecular forces in these compounds are dispersion forces and their strengths will be very similar in corresponding alkanes and alkenes (because dispersion forces depend upon the number of electrons in the molecule). Physical properties such as melting and boiling points and solubility depend upon intermolecular forces and so corres...