Catalytic cracking can be used to convert a long-chain hydrocarbon into smaller
molecules. For the hydrocarbon C16H34, one possible cracking reaction is given by the equation
C16H34 reacts to form C5H10 plus C4H8 plus C3H8 plus 2C2H4. Which product of this reaction is not an alkene? Tick one box. C5H10, C4H8, C3H8, C2H4.
The formula for the members of the homologous series alkenes, monoalkenes, is
C𝑛H2𝑛. Let’s test each answer against this general formula and see whether each one is an
alkene or not. For C5H10, the number of hydrogens, 10, is equal to twice the number of carbons,
five. Therefore it is an alkene. For C4H8, eight is equal to two times four. Therefore, C4H8 is also an alkene. For C3H8, eight is not equal to two times three. Therefore, this is not an alkene. This actually corresponds to the formula for an alkane, which is C𝑛H2𝑛+2. This is therefore the correct answer. Just to be safe, let’s check out C2H4. Four is equal to two times two. Therefore, C2H4 is another alkene.
The hydrocarbons C16H34 and C5H10 display very different physical properties. In which row of table one are the most flammable, most viscous, and most volatile of
these compounds correctly identified? Tick one box. A, B, C, or D.
To answer this question, you must recall the relationship between the length of a
hydrocarbon chain and its flammability, viscosity, and volatility. Let’s start with flammability. You should recall that methane is the smallest possible hydrocarbon, and it’s highly
flammable, whereas a long-chain alkane such as might be found in road tar is not
very flammable. Therefore, flammability decreases as chain length increases.
Now let’s look at viscosity. This is another word for how thick a liquid is. The longer the chain, the more likely these chains are to get tangled up. So the longer the chain, the more viscous the hydrocarbon.
Now let’s look at volatility. Volatility is a measure of how easy it is to vaporize something. Something that is highly volatile has a very low boiling point, like a gas, like
methane. As the hydrocarbon chain length increases, the volatility decreases.
Now let’s go back to the table. We have a short hydrocarbon, C5H10, and a long hydrocarbon, C16H34. It doesn’t matter that one is an alkene and one is an alkane. Because C16H34 is so much bigger than C5H10, the overall properties of hydrocarbons
apply. So we expect C5H10 to be more flammable than C16H34 because it’s shorter. Now let’s look at viscosity again. Because C16H34 is so much longer, we expect it to be much more viscous than
C5H10. Now let’s look at volatility. C5H10 is much smaller than C16H34. Therefore, we expect it to be much more volatile. The only row in the table with all correct answers is A. Therefore, the answer to this question is A.
The hydrocarbon C3H8 undergoes complete combustion. Which of the following is a balanced chemical equation for this reaction? Tick one box. 2C3H8 plus 7O2 react to form 6CO2 plus 8H2O, 2C3H8 plus 7O2 react to form 6CO plus
8H2O, C3H8 plus 5O2 react to form 3CO plus 4H2O, or C3H8 plus 5O2 react to form 3CO2
Now there are two conditions in this question that must apply to the answer. It must be a complete combustion. This means no carbon and no carbon monoxide, only carbon dioxide and water as the
products. And it must also be balanced. So let’s go through one by one.
Equation one is balanced in carbon, six on each side. It’s also balanced in hydrogen, 16 on each side. However, it is not balanced in oxygen. There are 14 on the left-hand side, seven times two, and 20 on the right-hand side —
that’s six times two plus eight. Equation one fails the balanced criterion, and therefore it is not the correct
Equation two has carbon monoxide as a product. It is therefore not a complete combustion. It is therefore not the correct answer. Now on to equation three. Equation three also is not a complete combustion reaction because it has carbon
monoxide as a product. It is therefore not the correct answer.
By a process of elimination, we have proven that the last answer must be correct, but
let’s check just to make sure. It is balanced in carbon, three on both sides. It is balanced in hydrogen, eight on both sides. And it is balanced in oxygen, 10 on both sides. It is also a complete combustion reaction because the only products are CO2 and
H2O. There is no carbon monoxide or carbon. Therefore, the balanced complete combustion reaction for the oxidation of C3H8 is
C3H8 plus 5O2 react to form 3CO2 + 4H2O.
Propyl ethanoate is an example of an ester. Which of the following is the displayed formula of propyl ethanoate? Tick one box. A, B, C, or D.
Let’s review how esters are named. The name of an ester is composed of two parts: the name of the alkyl chain connected
to the oxygen and the name for the remainder which is related to the chain
length. Recall that the name for an alkyl chain relates to the number of carbon atoms in that
chain: methyl, one; ethyl, two; propyl, three. So propyl ethanoate has a three-member carbon chain attached to the oxygen and a
carbon chain involving two carbon atoms at the other end. Let’s go through the answers one by one. A has a C2 alkyl chain and a C3 alkanoate segment. B is C2 and C2. C is C3 and C3. And D is C3 C2, propyl ethanoate. Therefore, the answer is D.
Propyl ethanoate can be formed by a condensation reaction involving which carboxylic
acid? Tick one box. Ethanol, ethanoic acid, propanoic acid, propanol.
The first thing we can do with this question is eliminate ethanol and propanol as
possible answers because they are not carboxylic acids. They are in fact alcohols, which you can tell by the fact they end in “ol.”
Now let’s look at how a condensation reaction produces an ester. When an alcohol and a carboxylic acid react, they produce an ester, with water as a
byproduct. The alcohol goes to form the alkyl chain of the alkyl alkanoate, and the carboxylic
acid goes to form the remainder. Here we have propyl ethanoate. Therefore, we must’ve produced it from ethanoic acid. Therefore, the correct answer is ethanoic acid. Propanoic acid would’ve made a propanoate, not an ethanoate.