NESA Chemistry Products of Reactions Involving Hydrocarbons
8 sample questions with marking guides and sample answers · Avg. score: 76.7%
Which of the following will decolourise a solution of bromine water?
1.0 mol L solution
1.0 mol L KCl solution
Reveal Answer
1.0 mol L solution
This is a solution of Iron(III) nitrate. Neither nor ions react with bromine water, so the solution would not be decolourised.
1.0 mol L KCl solution
Chloride ions () cannot reduce bromine () because chlorine is a stronger oxidising agent than bromine. Therefore, no reaction occurs to remove the bromine colour.
Pentanal (C₅H₁₀O) can cause a colour change with bromine because it is an aldehyde. Aldehydes are easily oxidised to carboxylic acids, and in this reaction, bromine (Br₂), which is orange-brown, is reduced to colourless bromide ions (Br⁻). As a result, the orange colour of bromine water disappears when it reacts with pentanal.
The condensed formula represents 2-pentene, which contains a carbon-carbon double bond (). Alkenes undergo a rapid addition reaction with bromine water, consuming the and turning the orange solution colourless.
One mole of methane, , reacts with one mole of halogen, . X can be fluorine, F, chlorine, Cl, or bromine, Br. The general equation for the reaction is given below.
Which one of the following statements is true?
The strength of the bonds from weakest to strongest is .
Since hydrogen has the smallest atomic radius, the bond is the weakest bond.
The bond is stronger than the bond because of the size of the bromine atom.
The , and bonds are equal in strength because Br, Cl and F are halogens.
Reveal Answer
The strength of the bonds from weakest to strongest is .
Bond strength generally increases as atomic radius decreases because the shared electrons are held closer to the nuclei. Since fluorine is the smallest and bromine is the largest of these halogens, the bond is the strongest and is the weakest.
Since hydrogen has the smallest atomic radius, the bond is the weakest bond.
A smaller atomic radius generally results in a shorter internuclear distance, which typically forms a stronger bond, not a weaker one.
The bond is stronger than the bond because of the size of the bromine atom.
The larger size of the bromine atom increases the bond length, which actually makes the bond weaker than the bond.
The , and bonds are equal in strength because Br, Cl and F are halogens.
Even though they belong to the same group (halogens), their differing atomic radii and electronegativities result in significantly different bond strengths.
Identify the major product when 2-methylbut-2-ene reacts with water under acidic conditions.
(CH)CHCOCH
(CH)C(OH)CHCH
(CH)CHCH(OH)CH
(CH)C(OH)CH(OH)CH
Reveal Answer
(CH)CHCOCH
This is a ketone (3-methyl-2-butanone). Acid-catalyzed hydration of alkenes produces alcohols, not carbonyl compounds; ketones are typically formed from the hydration of alkynes or oxidation of alcohols.
(CH)C(OH)CHCH
The reaction follows Markovnikov's rule, proceeding via the most stable tertiary carbocation intermediate. The hydroxyl group () attaches to the more substituted carbon, yielding the tertiary alcohol 2-methyl-2-butanol.
(CH)CHCH(OH)CH
This is the anti-Markovnikov product (3-methyl-2-butanol). The reaction does not favor the formation of the less stable secondary carbocation intermediate required to produce this secondary alcohol.
(CH)C(OH)CH(OH)CH
This is a vicinal diol. Acid-catalyzed hydration adds water () across the double bond to form a mono-alcohol, whereas diols are formed through oxidation reactions like dihydroxylation.
Consider the following organic reaction.
In the space provided, identify reaction condition X and name the organic product.
| Reaction condition X | IUPAC name of organic product |
|---|---|
Reveal Answer
| Reaction condition X | IUPAC name of organic product |
|---|---|
| UV light | 2-bromobutane |
| Descriptor | Marks |
|---|---|
Correctly identifies both the reaction condition AND product | 2 |
Correctly identifies either the reaction condition OR product | 1 |
None of the above | 0 |
Classify the type of reaction represented in the following equation:
addition
oxidation
combustion
condensation
Reveal Answer
addition
This is an addition reaction because two reactant molecules ( and ) combine to form a single product molecule () without losing any atoms.
oxidation
Oxidation typically involves the loss of electrons or the addition of oxygen, whereas this reaction simply adds across a carbon-carbon double bond.
combustion
Combustion requires oxygen () as a reactant and typically produces carbon dioxide and water, which are not present in this equation.
condensation
A condensation reaction joins two molecules together while eliminating a small molecule like water (), but here all reactant atoms are incorporated into the single product.
One mole of a triglyceride containing only one type of fatty acid reacts completely with 6 moles of iodine, .
Which one of the following could be the fatty acid?
linolenic acid
linoleic acid
oleic acid
stearic acid
Reveal Answer
linolenic acid
Linolenic acid contains 3 carbon-carbon double bonds per molecule. A triglyceride made of three linolenic acid chains would react with 9 moles of , not 6.
linoleic acid
A triglyceride contains 3 fatty acid chains. Since 1 mole of the triglyceride reacts with 6 moles of , each fatty acid chain must contain 2 double bonds (). Linoleic acid has exactly 2 double bonds.
oleic acid
Oleic acid contains only 1 carbon-carbon double bond per molecule. A triglyceride made of three oleic acid chains would only react with 3 moles of .
stearic acid
Stearic acid is a saturated fatty acid, meaning it has no carbon-carbon double bonds and would not react with at all.
Compound A has the molecular formula . The information below shows some chemical reactions beginning with this compound.
- Compound A reacts with to produce compounds B and C.
- Compound B does not react with .
- Compound C reacts with to produce compound D.
- Compound D does not react with .
Determine the structure of compound A. Justify your answer using all the data given.
Reveal Answer
Compounds B and C are alkanols.
Compound B cannot be oxidised – so B must be a tertiary alkanol, therefore there must be branching.
Compound C can be oxidised but not to a carboxylic acid (no reaction with carbonate) – so the oxidation product is a ketone, so compound C is a secondary alkanol.
To explain the alkanols produced above, the C=C in compound A is not in the terminal position.
Therefore compound A is 2-methylbut-2-ene (CH).
| Descriptor | Marks |
|---|---|
Correctly determines the structure of compound A AND Correctly justifies the structure using all the data provided | 4 |
Correctly determines the structure of compound A AND justifies the structure using some of the data provided OR Correctly interprets all of the data provided | 3 |
Correctly interprets some of the data provided | 2 |
Provides some relevant information | 1 |
None of the above | 0 |
Ethanol can be produced either from plant materials or from petrochemical sources.
When ethanol is produced from plant sources, the material is ground up. The starches and cellulose in the material are then converted into sugars. Yeast or zymase is mixed with the sugars at 25 to 37 °C and a pH of between 3 and 5 at atmospheric pressure. The products of the fermentation process are ethanol and carbon dioxide.
Ethanol can also be produced by the endothermic hydration of ethene. This is carried out at 250 to 300 °C and 6000 to 7000 kPa in the presence of an acid catalyst.
Justify the conditions used for fermentation.
Reveal Answer
Zymase are enzymes, which are only effective in a narrow pH band and temperature band.
| Descriptor | Marks |
|---|---|
Recognises that yeast or zymase are enzymes | 1 |
Recognises that enzymes are only effective in a narrow pH band and temperature band, or recognises that without the enzymes, the reaction either does not proceed or is too slow to be viable | 1 |
Write an equation for the fermentation process, using as the sugar. Use condensed structures in your equation.
Reveal Answer
| Descriptor | Marks |
|---|---|
Correct reactants and products | 1 |
Correct balancing | 1 |
Write an equation for the hydration of ethene. Use condensed structures in your equation.
Reveal Answer
| Descriptor | Marks |
|---|---|
Correct reactants | 1 |
Correct products | 1 |
Uses condensed structures in equation | 1 |
Justify the temperature and pressure used for the hydration of ethene.
Reveal Answer
high pressure increases rate of reaction as there are more particles per unit volume, therefore a greater frequency of collisions. It also increases yield due to it favouring the direction with the fewer number of gas particles which is the product side of the reaction. High temperature also means particles are moving more rapidly and collide more often, and more particles having sufficient energy for successful collisions, a greater proportion of collisions will be successful, increasing reaction rate.
Because hydration of ethene is endothermic, high temperature will favour the formation of the products.
High temperature favours both rate and yield, however, a moderate temperature will still produce an economical yield.
| Descriptor | Marks |
|---|---|
Recognition that high pressure increases rate of reaction as there are more particles per unit volume, therefore a greater frequency of collisions | 1 |
Recognition that high pressure also increases yield due to it favouring the direction with the fewer number of gas particles which is the product side of the reaction | 1 |
Recognition that high temperature will increase rate as the particles are moving more rapidly and collide more often, as well as more particles having sufficient energy for successful collision, so greater proportion of collisions will be successful | 1 |
Recognition that because hydration of ethene is endothermic, high temperature will favour the formation of the products | 1 |
Recognition that, although high temperature favours both rate and yield, a moderate temperature will produce economical/safe yield | 1 |
State three reasons why the fermentation process to produce ethanol is more common than the hydration of ethene.
Reveal Answer
Fermentation requires less energy input than hydration of ethene, fermentation costs less than hydration of ethene, and fermentation is a 'greener' process than hydration of ethene. Furthermore, fermentation uses a renewable feedstock while hydration of ethene does not.
| Descriptor | Marks |
|---|---|
1 mark for each correct point (any 3 of):
| 3 |