Lesson video

Hello, I am Mrs. Adcock and welcome to today's lesson.

Today's lesson is on chemical tests involving alkanes, alkenes, and alcohols.

We are going to be looking at how we can distinguish between an alkane and an alkene and then looking at how we can use calorimetry and viscosity data to predict chain length of alkanes, alkenes, and alcohols.

Today's lesson outcome is, I can use data to identify the functional group and predict the chain length for an alkane, alkene, or alcohol.

Some of the keywords that we will be using in today's lesson include homologous series, functional group, calorimetry, and viscosity.

You can see those keywords used here in a sentence.

It would be a good idea to pause the video now and read through those sentences.

You might like to make some notes so that you can refer back to them later in the lesson if needed.

Today's lesson on chemical tests involving alkanes, alkenes and alcohols is split into three main parts.

First of all, we are going to be looking at distinguishing between alkanes and alkenes.

Then we're going to move on to look at calorimetry data and how we can use this to predict chain length.

And finally, we're going to look at how we can use viscosity data to predict the chain length of alkanes, alkenes and alcohols.

Let's get started on the first part of our lesson, distinguishing between alkanes and alkenes.

The alkanes are a homologous series of hydrocarbons.

Hydrocarbons are molecules that are made of hydrogen and carbon atoms only, and a homologous series, that's one of our key words from today's lesson, that is a group of organic compounds that have the same general formula, and they show similar chemical properties and physical property trends.

The general formula of alkanes is CNH2N plus two.

We will come back and have a look at that again in a moment.

The first four alkanes in the alkane homologous series are, methane, methane just contains one carbon atom.

The molecular formula is CH4, and you can see a molecule of methane there where each carbon atom forms four covalent bonds.

The next alkane in the homologous series is ethane and ethane contains two carbon atoms. The molecular formula, we can see there is C2H6.

We've got methane and ethane.

Next we have propane.

Propane is an alkane that contains three carbon atoms. It has the molecular formula, C3H8, and finally our fourth alkane in the homologous series is butane, and butane contains four carbon atoms. It has the molecular formula, C4H10.

If we go back and have a look at the general formula of alkanes, we can see that that's CNH2N plus two.

If we take butane as our example, N in this case equals four.

If we want to work out how many hydrogen atoms there are in a molecule of butane, we would do N times two.

So four times two, that's eight, and then we add two, so plus two and we get 10.

So the molecular formula for butane is C4H10.

What is the general formula of alkanes? Is it A, CNH2N, B, CNH2N plus one, or C, CNH2N plus two? The correct answer is C.

The general formula of alkanes is CNH2N plus two.

Well done if you got that one correct.

The alkenes are also a homologous series of compounds, and these are different to alkanes.

If we have a look here at the first four alkenes in the homologous series, we have, first of all, ethene and ethene contains two carbon atoms and it has the molecular formula, C2H4.

Then we have propene.

Propene has three carbon atoms, and that has the molecular formula, C3H6.

Next we have butene that has four carbon atoms and has the molecular formula, C4H8.

The fourth alkene in the homologous series is pentene, and pentene has five carbon atoms that has the molecular formula, C5H10.

Let's just recap the first four alkenes in the homologous series, it's ethene, propene, butene and pentene.

Alkenes differ to alkanes because they contain that carbon carbon double bond and that carbon carbon double bond is the functional group of alkenes.

A functional group is an atom or group of atoms that is responsible for the way a compound reacts.

Here are examples of alkenes showing that carbon to carbon double covalent bond, and that's their functional group.

We've got ethene and propene, and they both contain that carbon to carbon double bond.

That's the functional group of alkenes.

With alkanes, you may have noticed that we started with methane.

We don't have an alkene called methene because all alkenes need to have that carbon to carbon double bond.

So our simplest alkene is ethene.

The alkenes have a different general formula to alkanes.

The general formula for alkenes is CNH2N.

If we take ethene as an example, we can see that N is two, and then to work out the number of hydrogen atoms, we do two N, so we're going to times that two by two, and then we will get four.

So the molecular formula for ethene is C2H4.

Time for another question.

What is the name of this molecule? Is it A, propane, B, propene, C, pentane, or D, pentene? First of all, look closely to work out whether this is an alkane or an alkene.

Then you'll need to count the number of carbon atoms to work out the prefix for this molecule.

The correct answer is pentene.

Well done if you chose D, pentene.

It contains five carbon atoms and it's got a carbon carbon double bond, so therefore we know it's an alkene.

Let's have a go at another question.

What is the functional group present in alkenes? Is it A, a carbon to carbon double bond? Is it B an -OH group, or is it C, a -COO group? The correct answer is A.

Hopefully you identified that carbon carbon double bond as the functional group that's present in alkenes.

B, the -OH group, that is the functional group of alcohols, and C, the C-O-O group is the functional group that's present in a group of compounds called esters.

Alkenes are unsaturated, whereas alkanes are saturated molecules.

An unsaturated molecule is a molecule that contains one or more double carbon to carbon covalent bonds.

If a molecule contains that carbon to carbon double bond, then it's unsaturated, and if it only contains single bonds, then it will be a saturated molecule.

Here we can see an example.

We've got ethene.

It's an alkene.

It's got a carbon to carbon double bond, so it's an unsaturated molecule.

Here we have an example of an alkane.

It's ethane and it only contains single bonds.

Therefore, this is a saturated molecule.

Let's check that we've understood about saturated and unsaturated molecules.

You need to identify which of the following shows an unsaturated molecule.

The correct answer is B.

B shows an unsaturated molecule, because B is an alkene that contains a carbon to carbon double bond.

The molecules present in A and C do not contain a carbon to carbon double bond, and therefore they are saturated molecules.

The presence of that carbon to carbon double bond in our unsaturated alkenes can be detected using bromine water.

When we add bromine water to an alkene, the bromine reacts with the alkene.

We can see an example of that here.

We've got an alkene and we're going to react it with bromine.

There is bromine molecules present in bromine water and they react together.

You can see that those bromine atoms bond to the carbon atoms on either side of that carbon to carbon double bond, so the double bond opens up and those bromine atoms can attach.

We start with an alkene, we can react to that with bromine in bromine water, and then we will form a haloalkane.

In this case, our halogen is bromine, so we form a bromoalkane molecule.

There you can see those bromine atoms and how they've bonded to the carbon atoms on either side of that double bond as it's opened up.

Bromine water is an orange colour, and the solution turns colourless when it is added to an alkene as the dissolved bromine molecules react with the alkene.

Here we can see some bromine water and it turns from an orange colour to colourless when added to an alkene, and this is showing that a chemical reaction has happened.

We need to be really careful here that we don't use the word clear.

The solution goes from an orange colour to no colour, so it goes to colourless.

When we use the word clear, that means we can see through.

You will see that the bromine starts as a clear orange solution, and then it turns into a colourless solution.

We can use bromine water to detect for the presence of that carbon to carbon double bond.

When bromine water is added to an alkane, it does not react as alkanes do not contain a carbon to carbon double bond.

The orange colour of bromine water remains.

Here we have an alkane and we've added orange bromine water to that alkane, and the solution stays an orange colour.

Bromine water stays orange when added to an alkane as there is no chemical reaction.

You might notice in the beaker that there's something spinning round.

This is a magnetic stirrer that we've got at the bottom of the beaker, just mixing these two solutions together.

Just to recap, the bromine water turns from orange to colourless when added to an alkene, but it remains orange when added to an alkane.

So we can use bromine water to distinguish between an alkane and an alkene.

Time for a question.

What colour change occurs when an alkene is added to bromine water? Is it A, the bromine water turns from orange to clear? B, the bromine water turns from orange to colourless, or C, the bromine water remains orange? The correct answer is B.

The bromine water turns from orange to colourless.

It doesn't turn to clear.

The solution was already an orange clear solution, and it does not remain orange.

It would remain orange with an alkane, not with an alkene.

Time for our first practise task of today's lesson.

You need to first of all decide whether the following molecules are alkanes or alkenes.

Pause the video now, have a look at the formula you've been given from A through to H and decide whether they are alkanes or alkenes.

Let's see how you got on.

A is an alkane, that's methane.

B is an alkene.

We have a carbon to carbon double bond.

C is an alkene.

We could work that out using the general formula for alkenes, which is CNH2N.

So we've got double the number of hydrogen as we have carbon.

D is an alkane.

Again, we would need to use the general formula to work out that.

E is an alkene.

F is an alkane.

G, we can see has a carbon to carbon double bond.

That's an alkene, and H is a molecule of butane.

That's an alkane.

Well done if you were able to identify whether those molecules were alkanes or alkenes.

Time for us to move on to the second part of our task.

Question two, draw a molecule of ethene and ethane.

You need to draw the displayed formula showing the atoms that are present and the covalent bonds.

Question three, describe a chemical test which can be used to distinguish between ethene and ethane.

Question four, draw the product from reacting ethene with bromine water.

If you pause the video now, have a go answering these three questions.

Then when you come back, we'll go over the answers.

Let's see how you got on.

When drawing a molecule of ethene and ethane, you should have two carbon atoms in both of your molecules.

Ethene will contain a carbon to carbon double bond, whereas ethane will only contain single bonds.

Check that you've drawn the correct number of hydrogen atoms around your carbon atoms. Remember, each carbon atom will form four covalent bonds.

A chemical test, which can be used to distinguish between ethene and ethane.

That's adding bromine water, and you will add bromine water to the ethene and to the ethane.

The bromine water will turn from orange to colourless with ethene, whereas the bromine water will remain orange with the ethane.

Hopefully you got that one correct.

Question four was a difficult question, to draw the product when ethene reacts with bromine water.

You need to open up that carbon carbon double bond and add on bromine atoms, so your answer should look similar to that molecule there.

Time for us to move on to the second part of our lesson.

We're going to be looking at using calorimetry data to predict chain length.

Alcohols are a homologous series of compounds that contain the -OH functional group, and the first four alcohols in the alcohol homologous series are, methanol.

Methanol just contains one carbon atom, and you can see it's got that -OH functional group.

The structural formula for methanol is CH3OH.

Next we have ethanol.

Ethanol has the structural formula CH3CH2OH.

And remember, the structural formula shows how the atoms are bonded together in the molecule.

Next we have propanol.

Propanol is an alcohol that has three carbon atoms, and again, it's got that -OH functional group, and hopefully you can already work out what's coming next.

So we've had methanol, ethanol, propanol, and the next alcohol in the alcohol homologous series is butanol.

Butanol contains four carbon atoms. It has the structural formula there, which is CH3, attached to a CH2, attached to a CH2, and we've got another CH2, and then we've got our -OH functional group.

Let's see if we can remember what we've just learned.

The image shows a molecule of, is it A, methanol, B, ethanol, C, propanol, or D, butanol? The correct answer is D, butanol.

So well done if you got that one correct.

The molecule there contains four carbon atoms, so we know it starts with the prefix but and it's got the -OH functional group, so we know it's an alcohol.

So that's butanol.

Calorimetry is an experiment that can be used to measure the energy released from fuels.

Calorimetry can be used to indirectly measure the energy that's released from the combustion of different fuels, and they could be an alkane, an alkene or an alcohol.

Here we can see the apparatus used for calorimetry.

We have a thermometer and that will be used to measure the temperature change of the water.

There's the water held in our calorimeter.

These are quite often made of copper.

We've got our fuel burner there at the bottom, and we could add an alkane or an alkene or alcohol into that fuel burner.

That's on a heatproof mat.

Then we have a clamp and clamp stand that are holding the calorimeter in place above that fuel burner.

We can measure the mass of fuel that's burnt and also measure the temperature change of the water and use this to work out the amount of energy that is released when we burn different fuels.

Calorimetry data can be used to predict the chain length of alkanes, alkenes, and alcohols.

As that chain length increases, more energy is released during combustion.

For example, if we take the alkanes as an example, we can go from methane to ethane to propane to butane, and as that carbon chain length is increasing, then more energy is released during combustion.

So butane is going to release more energy than a shorter chained alkane such as methane.

When comparing calorimetry data, if a lower mass of an organic compound combusts to cause the same temperature rise on the same volume of water, then this fuel is releasing more energy.

We have a sample set of data there where we have used three different alcohols in our fuel burner.

They are methanol, ethanol, and propanol.

We would've used those to measure a set temperature rise on the same volume of water.

And we can see that we used 0.

37 grammes of methanol, 0.

28 grammes of ethanol, and 0.

25 grammes of propanol.

So less propanol was required to cause the same temperature rise on the same volume of water.

This data that propanol releases more energy during combustion than the shorter chained alcohols.

When using the same mass of alkene, which one will release the most energy during combustion? A, butene, B, ethene, C, propene, or D, pentene? Have a think about what we've just learned and consider whether a shorter or longer chained alkene will release the most energy during combustion.

The correct answer is pentene.

Pentene is the longest alkene out of those options.

Butene has four carbon atoms. Ethene has two carbon atoms. Propene has three carbon atoms. Pentene is our longest alkene out of those options with five carbon atoms, and the longer the chain length, the more energy that is released during combustion.

Well done if you chose D, pentene.

Time for our second practise task of today's lesson.

Question one.

Laura performed calorimetry experiments to measure the mass of alcohol that is burned to raise a 100 centimetres cubed of water by 20 degrees Celsius.

Use Laura's data to predict the mass of propanol required to raise 100 centimetres cubed of water by 20 degrees Celsius.

If you pause the video now, look carefully at the data you've been given in the table and answer that question.

The correct answer is 0.

51 grammes of propanol would need to be burned to raise 100 centimetres cubed water by 20 degrees Celsius.

You may not have chosen exactly 0.

51, but any answer between 0.

48 and 0.

54 would all be considered correct.

Hopefully, you worked out that as the chain length increases, then we need a lower mass of alcohol to be burned, and that is because the longer the chain length, the more energy is released during combustion.

Question two, Jacob collected the calorimetry data below from the combustion of different alcohols.

Write the alcohols in order of increasing chain length, starting with the shortest, and once you've done that, give a reason for your answer.

You can see in the table there you've got alcohols, A, B, C, and D.

You've been told the mass of alcohol that was burned in grammes.

Pause the video now, have a go at answering this question, then come back when you're ready to go over the answer.

Let's see how you got on.

In order of increasing chain length, the alcohols are A, D, C, and then B.

The longer the alcohol chain then, the lower the mass of alcohol combusted to raise the same volume of water by a set temperature.

We know the A must be our shortest alcohol because we needed the largest mass of alcohol to be burned, and B was our longest alcohol because we needed the lowest mass of alcohol to be burned.

This is because more energy is released during combustion of longer chained alcohols than shorter chained alcohols.

Well done if you were able to correctly order those alcohols and then give a correct reason as to why you'd put them into that order.

We've looked at a test to distinguish between alkanes and alkenes.

Then we've looked at how we can use calorimetry data to predict chain length.

Now we're going to move on to the final part of our lesson, which is using viscosity data to predict chain length.

Viscosity is how easily a liquid flows.

The higher the viscosity, the less easily a liquid flows.

We have water here as an example of a substance with a low viscosity, so water flows very easily.

It has a low viscosity.

Honey is an example of a substance with a high viscosity.

So remember, the less easily a liquid flows, the higher the viscosity.

As chain length increases the viscosity of alkanes, alkenes and alcohols increases.

For example, we've got the alkanes here and we've gone from methane to ethane to propane to butane to pentane.

So as we go down there, the chain length is increasing, and as the chain length increases, so does viscosity.

Is this statement true or false? As the chain length of alcohols increases, they become less viscous.

That means they have a lower viscosity.

That statement is false.

Can you justify your answer? Is it A, as the chain length of alcohols increases, they become more viscous? Or is it B, as the chain length of alcohols increases, the viscosity stays the same? The correct answer is A, hopefully you got that one correct.

So as chain length of alcohols increases, they become more viscous, the viscosity increases.

It is time for us to complete the final practise task of today's lesson.

First of all, you need to complete the table for the first four alkenes in the homologous series.

In the final column, you need to put a viscosity ranking with one being the substance that you think is the most viscous, and four being the substance, which is the least viscous.

So in that column, you will end up with the numbers one, two, three, and four each used once.

Pause the video now, have a go at completing this table.

Well done for having a go.

Let's see how you got on.

First of all, you should have ethene, and ethene has the molecular formula, C2H4.

Then we have propene.

Propene has the molecular formula, C3H6.

Next we have butene, and we've been given the molecular formula for butane.

And finally, we have pentene, and the molecular formula for pentene is C5H10.

Well done if you identified the first four alkenes in the homologous series and you worked out their molecular formula.

In terms of viscosity ranking, you should have four, three, two, and one.

Ethene is the shortest chained alkene, and therefore it's the least viscous.

Then the viscosity increases as the chain length increases.

Question two, Aisha had sealed tubes of equal length containing different alkanes.

She inverted the tubes and timed how long it took for the air bubble to reach the top.

And here are Aisha's results.

We've got alkane A, B, and C.

We can see the time that it took in seconds for the air bubble to reach the top of each of those tubes.

A, which alkane is the most viscous, and B, order the alkanes by chain length starting with the shortest, and give a reason for your answer.

Pause the video now, have a go answering this question, and then come back when you're ready to go over the answers.

Alkane B is the most viscous.

We know this because the air bubble took the longest time to travel through alkane B.

Ordering the alkanes by chain length, starting with the shortest.

The shortest is A, followed by C and then B, and the reason for this is that as the chain length increases, so does viscosity, and therefore the air bubble will take longer to travel through the more viscous, longer chained alkane.

Well done if you answered that question correctly.

We've reached the end of today's lesson on chemical tests involving alkanes, alkenes, and alcohols.

We're just going to summarise some of key points that we've covered in today's lesson.

A chemical test involving bromine water can identify the differences between alkanes and alkenes.

Data from calorimetry and viscosity experiments can be used to predict the chain length of alkanes, alkenes, and alcohols.

As the chain length increases, more energy is released during combustion, and as chain length increases the viscosity of alkanes, alkenes and alcohols increases.

Well done for all your hard work in today's lesson.

I hope that you've enjoyed the lesson, and I hope that you're able to join me for another lesson soon.