4.2

Question 1

4.2.1 a) i) Explain, in terms of hydrogen bonding and polarity, the water solubility of alcohols compared with alkanes.

Answer 1

Alcohols are polar molecules, a result of the electronegative hydroxyl group which pulls the electron in the C-OH bond away from the carbon atom: C(δ+) - O(δ-) - H(δ+). The slightly positive hydrogen (a result of the electronegative oxygen drawing electron away from the hydrogen) can attract the lone pairs on a neighbouring oxygen, forming hydrogen bonds.
Small alcohols, such as methanol, ethanol and propanol, are completely soluble in water (forming hydrogen bonds between molecules). However, as alcohols increase in size, the solubility of the alcohol decreases: the non-polar alkyl chain grows longer, resulting in less hydrogen bonding between molecules.
In contrast, alkane molecules are non-polar (the electronegativity of hydrogen and carbon are very similar). This means that alkanes cannot form hydrogen bonds with water, and are therefore insoluble.

Question 2

4.2.1 a) i) Compare the volatility of alcohols to alkanes.

Answer 2

The boiling point of an alcohol tends be higher than the boiling point of its corresponding alkane. Alcohol molecules are polar, forming strong hydrogen bonds that require a large amount of energy to overcome. In contrast, alkanes are non-polar, forming weak London forces which require little energy to overcome.
The volatility of a substance is its tendency to evaporate at normal temperature - the higher the boiling point, the less volatile the substance (i.e. the less likely it is to evaporate). This means that alcohols have a lower volatility than corresponding alkanes - more energy is required to break the stronger intermolecular forces.

Question 3

4.2.1 a i) Explain the trend in alcohol boiling points.

Answer 3

The boiling point of alcohols increase as the molecules get longer - the mass, and therefore the chain length, increases, so the surface area contact between molecules increases. This results in stronger induced dipole-dipole interactions, which require more energy to break.

Question 4

4.2.1 a) ii) How are alcohols classified into primary, secondary and tertiary alcohols?

Answer 4

A primary alcohol has the functional group attached to a carbon with one other alkyl group (at the end of the chain).
A secondary alcohol has the functional group attached to a carbon with two other alkyl groups.
A tertiary alcohol has the functional group attached to a carbon with three other alkyl groups.

Question 5

4.2.1 b) Which process allows alcohols to be used as fuels?

Answer 5

Combustion is a rapid oxidation reaction that combines oxygen with another substance. Alcohols burn completely in a plentiful supply of oxygen, producing CO2 and H2O. This reaction allows them to be used as fuels: undergoing combustion (an exothermic process) enables them to transfer stored chemical energy into a more usable form (such as thermal energy). As the number of carbon atoms in the alcohol chain increases, the quantity of heat released per mole also increases.

Question 6

4.2.1 c) i) Outline the oxidation of primary alcohols, including how oxidation products are controlled with certain conditions.

Answer 6

When a primary alcohol is gently heated with acidified potassium dichromate (Kr2Cr2O7/H2SO4), an aldehyde (+H2O) is formed. The aldehyde is distilled out of the reaction mixture as it forms - this prevents any further reaction with the oxidising agent.
e.g. CH3CH2-OH + [O] → CH3CHO + H2O
When a primary alcohol is strongly heated under reflux, with an excess of acidified potassium dichromate (Kr2Cr2O7/H2SO4), a carboxylic acid (+H2O) is formed. Heating under reflux ensures that any aldehyde initially formed also undergoes oxidation to the carboxylic acid.
e.g. CH3CH2-OH + 2[O] → CH3COOH + H2O

[Note: the orange solution containing dichromate(VI) is reduced to a green solution containing chromium (III) ions: Cr2O2(2-) → Cr3+]

Question 7

4.2.1 c) ii) Explain how secondary alcohols are oxidised.

Answer 7

Secondary alcohols can be refluxed with acidified potassium dichromate (Kr2Cr2O7/H2SO4) to form ketones:
e.g. CH3CHOHCH3 + [O] → CH3COCH3 + H2O

[Note: the orange solution containing dichromate(VI) is reduced to a green solution containing chromium (III) ions: Cr2O2(2-) → Cr3+]

Question 8

4.2.1 c) iii) What happens when tertiary alcohols are refluxed with acidified potassium dichromate?

Answer 8

Tertiary alcohols resist being oxidised, and the acidified dichromate(VI) remains orange.

No change is observed when reacting ketones with Fehling’s solution, but aldehydes form a dark red precipitate. No change is observes when reacting ketones with Tollens’ reagent, but a silver mirror is produced with aldehydes.

Question 9

4.1.2 d) Explain how alkenes can be produced from alcohols.

Answer 9

Alkenes can be made from alcohols in an elimination reaction (in this case, water is eliminated, so it’s a dehydration reaction). This takes place in the presence an acid catalyst (e.g. concentrated H3PO4 or concentrated H2SO4) and heat.
The eliminated water molecule is made up from:
- the hydroxyl group
- and one hydrogen atom (from a carbon adjacent to the hydroxyl carbon).
Often there are two possible alkene products (as well as any E/Z isomers of these products).

Question 10

4.1.2 e) Explain how haloalkanes can be produced from alcohols.

Answer 10

Alcohols can be substituted with halide ions, in the presence of an acid (e.g. H2SO4), to form haloalkanes (i.e. the hydroxyl group is replaced by a halide) :
ROH + HX -> RX + H2O
[Note: for substitution with bromide, a salt such as sodium bromide (NaBr) is used. NaBr reacts with sulfuric acid to form HBr in situ: NaBr (s) + H2SO4 (aq) → NaHSO4 (aq) + HBr (aq). The HBr then reacts with the alcohol to form the haloalkane]

Question 11

4.2.2 b) What is a nucleophile?

Answer 11

A nucleophile is an electron pair donor.

Question 12

4. 2.2 a) i) Explain the fastest process that produces alcohols from haloalkanes.
   c) Outline the mechanism for this process.

Answer 12

Hydrolysis of a haloalkane by an aqueous alkali (where the OH- ion acts as a nucleophile), produces an alcohol and a halide ion. This reaction involves heating the haloalkane under reflux, but it is much faster than hydrolysis with water.
R-X + OH- → R-OH + X-
The Cδ+ atom in the polar C-X bond (a result of the halogen being more electronegative than carbon) attracts the lone pair of electrons on a hydroxide ion. This results in the :OH- nucleophile providing a pair of electrons to the electron-deficient Cδ+ atom - a new bond forms between the hydroxide ion and the carbon atom (represented in a curly arrow going from the :OH- ion to the Cδ+ atom).
The carbon-halogen bond, C-X, breaks by heterolytic fission (represented in a curly arrow going from the C-X bond to the Xδ-) which results in the products: a halide ion, X-, and an alcohol, R-OH.

Question 13

4.2.2 a) ii) How can the rate of hydrolysis of different carbon-halogen bonds be compared experimentally?

Answer 13

Hydrolysis of a haloalkane by water produces an alcohol, a halide ion and a H+ ion. The water molecule forms a weak nucleophile, and although the reaction is much slower than with aqueous alkali, nucleophilic substitution does occur:
R-X + H2O → R-OH + H+ + X-
The rate of reaction can be determined by adding aqueous silver nitrate, AgNO3. As the hydrolysis takes place, a halide ion (X-) is produced - this reacts with silver ions (Ag+) to form a silver halide precipitate.
Ag+(aq) + X-(aq) → AgX(s)
The time taken for the precipitate to appear can be recorded. An ethanol solvent is also added - this allows the water and haloalkane to mix, producing a single solution rather than two layers.

Compare the rate of hydrolysis of different haloalkanes:
1) Set up three test tubes containing 1 cm3 of ethanol. Add two drops of the following haloalkane to each test tube, placing them in a 60° water bath: 1-chlorobutane, 1-bromobutane, and 1-iodobutane.
2) Set up three test tubes containing 1 cm3 of 0.1 mol dm-3 aqueous silver nitrate, also placing them in a 60° water bath.
3) Once all of the test tubes have reached a constant temperature, add the silver nitrate to the haloalkanes. Immediately start a stop clock.
3) Record the time taken for each precipitate to form. 1-chlorobutane forms a white precipitate, 1-bromobutane forms a cream precipitate, and 1-iodobutane forms a yellow precipitate.
Iodoalkanes are the must reactive haloalkanes (forming the quickest precipitate and alcohol), followed by bromoalkanes, with chloroalkanes reacting the slowest.
The length of the alkyl chain, the amount of moles of haloalkanes used, the concentration and the volume of the reactants should all be kept constant.

Question 14

4.2.2 d) Explain the trend in the rate of hydrolysis of primary haloalkanes.

Answer 14

The rate of hydrolysis of primary haloalkanes is dependent on bond enthalpies. As you go down the group, the C-X bond enthalpies decrease. This is a result of:

1) Increasing atomic radius in the halogens - the successive addition of shells creates a larger distance between the bonding pair of electrons in the outer shell and the positive nuclei.
2) Increased electron shielding - as the number of inner shells increases, more electrons shield the bonding pair of electrons from the attraction of the positive nuclei.

So as you go down the group, the electrostatic force of attraction is weakened, and the covalent C-X bond is easier to break (less energy is required to overcome lower bond enthalpies). This is why the rate of hydrolysis increases as you go down group 7.

Question 15

4.2.2 e) i) How are halogen radicals produced from CFCs?

Answer 15

Chlorofluorocarbons (CFCs) are haloalkanes made solely of chlorine, fluorine and carbon. Halogen radicals are formed by the action of UV radiation on CFCs in the upper atmosphere, where the C-Cl bonds undergoes homolytic fission: C2F2Cl2 → C2F2Cl• + Cl• [Initiation]
[The reason the C-Cl bond breaks is because it has the lowest bond enthalpy]

Question 16

4.2.2 e) ii) How does the production of halogen radicals affect the ozone layer? Outline the reactions involved.

Answer 16

Ozone (O3) molecules form a protective layer in the atmosphere, preventing harmful UV radiation from reaching Earth. However chlorine radicals catalyse the breakdown of Earth's protective ozone layer (i.e. shrinking of the ozone layer) by forming diatomic oxygen molecules instead):
Cl• + O3 → O2 + ClO• [Propagation]
ClO• + O3 → Cl• + 2O2 [Propagation]
Overall: 2O3 → 3O2
OR
Cl• + O3 → O2 + ClO• [Propagation]
ClO• + O → Cl• + O2
Overall: O3 + O → 2O2
[Note: the second propagation step regenerates a chlorine radical - this chlorine radical will go on to catalyse the breakdown of thousands of ozone molecules (in a chain reaction)]

Question 17

4.2.2 e) ii) How is the breakdown of the ozone layer catalysed by other molecules, e.g. NO•?

Answer 17

Nitrogen oxide radicals also catalyse the breakdown of ozone:
NO• + O3 → NO2• + O2 [Propagation]
NO2• + O → NO• + O2 [Propagation]
Overall: O3 + O → 2O2

Question 18

4.2.3 a) i) How can a Quickfit apparatus be used to separate solutions?

Answer 18

Distillation: the separation of miscible liquids or solutions.
Chemical reactions may not go to completion or may produce by-products as well as the desired products -these solutions can be separated using distillation.
Distillation works by gently heating a mixture until it begins to boil. The different liquids in the mixture have different boiling points, and the substances evaporate out of the mixture in order of increasing boiling point (i.e. the liquid with the lowest boiling point is the most volatile and will boil first). This vapour will move out the flask (leaving behind the less volatile substances) and into the cold condenser, where it becomes a liquid and drips into the collecting flask.
[Note: the thermometer shows the boiling point of the substance that is evaporating - this allows you to know what substances are being evaporated and condensed at any given time]

Question 19

4.2.3 a) ii) How can a Quickfit apparatus ensure no volatile organic substances are lost during a reaction?

Answer 19

Reflux: a technique that prevents the reaction mixture from boiling away into the air.
Many organic reactions occur so slowly at room temperature that heat is required to overcome the activation energy and increase the rate of reaction. However, these substances also tend to be flammable and volatile (with low boiling points). If you heat them with a Bunsen burner, often they will evaporate or catch fire before they have time to react.
Refluxing a reaction enables it to be carried out at a fixed temperature. Heating the mixture in a flask using a vertical Liebig condenser (which continuously boils, evaporates and condenses the vapours) prevents the reactants from boiling away - this provides enough time for the mixture to react.
You can also avoid naked flames that ignite the reactants by using hot plates or water baths.

Question 20

4.2.3 a) iii) How can water be used to purify an organic liquid? How can an organic layer then be removed from an aqueous layer?

Answer 20

If you have an impure organic product that is insoluble in water, separation can be used to remove any impurities that do dissolve in water (such as salts or a water soluble organic layer). This is done by adding water to the organic liquid (so that any impurities will dissolve in aqueous solution) and then removing the water (containing the impurities) using a separating funnel.
This method is used because organic liquids and water are often immiscible (the organic layer, less dense than the aqueous layer, should float on top of the water). A separating funnel can be used to run off the aqueous layer (by opening a stopper) into a conical flask, leaving your organic product to then be collected in a different conical flask.

Question 21

4.2.3 a) iii) How can trace amounts of water be removed from an organic liquid?

Answer 21

If you use separation to purify a product, the organic layer will end up containing trace amounts of water. However, an anhydrous inorganic salt, such as magnesium sulfate (MgSO4) or calcium chloride (CaCl2), can be used as a drying agent: it binds to any water present, becoming hydrated. The solid drying agent can be filtered out.

Question 22

4.2.3 a) iii) How can redistillation purify a product?

Answer 22

Redistillation is the purification of a liquid by performing multiple distillations. If a product and its impurities have different boiling points, then redistillation can be used to separate them.

Question 23

4. 2.4 a) How does infrared radiation affect molecules?

c) How can an infrared spectrum of be used to identify an organic compound?

Answer 23

Atoms in molecules are joined by covalent bonds - these bonds possess energy and vibrate naturally. Infrared radiation causes covalent bonds to vibrate more and absorb energy.

Bonds between different atoms absorb different frequencies of IR radiation, which can be displayed as an infrared spectrum. An absorption peak can indicate which functional group a molecule displays, allowing you to identify organic compounds. For example, you can identify:
(i) an alcohol from an absorption peak of the
O–H bond
(ii) an aldehyde or ketone from an absorption
peak of the C=O bond
(iii) a carboxylic acid from an absorption peak of the C=O bond and a broad absorption peak
of the O–H bond

Question 24

4.2.4 b) How can infrared radiation impact energy usage?

Answer 24

When fossil fuels are burnt, ‘greenhouse’ gases are produced - these gases absorb IR radiation and re-emit it, raising the average temperature of the earth’s surface.
The absorption of infrared radiation by atmospheric gases containing C=O, O-H and C-H bonds (such as H2O (g), CO2 (g), and CH4 (g)) provided scientific evidence for the suspected link to global warming - a higher amount of infrared radiation is absorbed, suggesting an increased concentration of these gases in the atmosphere. This explained the rise in temperature and resulted in changes to energy usage (such as prompting government policies that increased renewable energy supplies).

Question 25

4.2.4 e) Give two uses of infrared spectroscopy.

Answer 25

It can be used to monitor gases causing air pollution (e.g. CO and NO from car emissions) and in modern breathalysers to measure ethanol in the breath.

Question 26

4.2.4 f) How can a mass spectrum be employed to determine the molecular mass of an organic compound? What is the M+1 peak?

Answer 26

When a compound is placed in a mass spectrometer, it is ionised (forming positive ions). The ions are detected on a mass spectrum as a mass-to-charge ratio (m/z). For an ion with a +1 charge, the m/z ratio is equivalent to the relative isotopic mass.
- When an organic compound is placed in the mass spectrometer, it loses an electron and forms a positive ion: this is the molecular ion.
e.g. CH3CH2CH2OH → CH3CH2CH2OH+ + e-
The m/z of a molecular ion is equivalent to the molecular mass of the organic compound. The mass spectrum of an organic compound can be used to identify the m/z of a molecular ion peak (M+ peak), and therefore the organic compound’s molecular mass (Mr). The M+ peak is often the one with the second highest mass/charge (m/z) ratio.
- The smaller peak after the M+ peak, with the highest mass/charge (m/z) ratio, is the M+1 peak - this arises from the small proportion of molecules with a carbon-13 isotope.

Question 27

4.2.4 g) How can certain parts of the structure of a molecule be identified using mass spectroscopy?

Answer 27

In the mass spectrometer, some molecular ions break down into smaller fragment ions. The other peaks in a mass spectrum represent these fragment ions, and the analysis of these fragmentation peaks allow certain parts of the structure of a molecule to be identified.
e.g. CH3CH2CH2OH+ → CH2OH+ + CH3CH2•
Any positive ions formed will be detected by the mass spectrometer (e.g. there will be a peak at m/z = 31 for CH2OH+), but the uncharged radicals are not detected.