CL1810 Week 3

Editor's Notes

• #2: 30 Dec 2003 Last lecture we learnt about alkanes. Alkanes are compounds that made up of hydrogen and carbon atoms only. And they have only single bonds. In this lecture, we are going to learn 3 types of other organic compounds. They are cycloalkanes, alkenes and alkynes. Basically, you are going to learn how to identify these compounds, how to name them and some isomerism, in particular cis and trans isomers that you have heard about last semester in physical & inorganic chemistry.
• #3: What is cycloalkanes? Cycloalkanes are saturated hydrocarbon. Meaning that it is same as alkanes, where they are made up of hydrogen and carbon atoms and all the bonds are single bonds only. The difference between alkanes and cycloalkanes is that alkanes are open chain compounds whereas cycloalkanes are ring structured compounds. In this slide, cyclopropane which has 3 carbon atoms, cyclobutane has 4 carbon atoms and cyclopentane has 5 carbon atoms. Practice and see if you could draw all the simplify structure. Remember, all carbon atoms are represented as a dot and all hydrogen atoms are hidden. 30 Dec 2003
• #4: Now, we are to learn how to name cycloalkanes. To name a cycloalkane, just simply add a prefix “cyclo” in front of teh respective alkane’s name. In the first example, there are 3 carbon atoms , the respective name is propane. As the compound is a cycloalkane, add a prefix “cyclo” and the compound is called cyclopropane. In the second example, there are altogether 8 carbon atoms in the ring, therefore the name is cyclooctane. Next, how to name a cycloalkane that has only one substituent, the monosubstituted cycloalkane. We use the same naming concept that you have learnt in last lecture, which is identify the parent first follow by the substituent. In this example, the parent is the 5 carbon atoms ring which we called cyclopentane. The substituent is a 2 carbon atoms chain and we call it ethyl. The full name for this compound is ethyl cyclopropane. Note that, for cycloalkane that has only one substituent, you do not need to define the position of the substituent. Also, take note that, when writing the name, there should be a space between the substituent and the parent. 30 Dec 2003
• #5: Lastly, how are we to name a cycloalkane that has more than 1 substituent. Again, we use the same naming concept that we have learned in the last lecture. Which is, in this case, the position of the substituent must be stated, where the position must be the lowest possible number. We must use Greek prefix di, tri, tetra and so on if there are more than one identical substituent. If the substituent are different, we name them according to their alphabetical order. Note that, each substituent must have a number given to it to tell its position whether or not the subsituents are identical or different and whether or not the substituents are place on the same position. For example, if a cycloalkane has 3 substituents, your final answer for its name must have 3 numbers. Let’s look at the example given here. We first identify the parent. The parent is a 6 carbon atoms ring and it is called cyclohexane. Then we look at the substituents. There are 2 different substituents, one of them has 2 carbon atoms, it is called ethyl, the other one has only one carbon atom, it is called methyl. Now give the positions for both substituents. Both are placed on the same carbon atom of the ring, and the lowest possible number for this carbon atom is 1. Therefore, the position for ethyl is 1 and the position for methyl is also 1. After you have settle the name for the parent, the name and the positions for the subsituents, you can then join them together. Based on alphabetical order ethyl is before methyl, therefore the full name for this compound is 1-ethyl-1-methyl cyclohexane. In the second example, the parent is a 7 carbon atom ring, it is called cycloheptane. There are 2 identical substituents with one carbon atom each. They are called methyl. As there are 2 methyl groups, we add a Greek prefix “di” and it becomes dimethyl. The lowest possible number for the positions are 1 and 2 respectively, therefore the substituents are 1,2-dimethyl and the full name is 1,2-dimethyl cycloheptane. One last point on this slide is that note the hyphen and comma that I have highlighted in yellow. You use the hyphen or dash in between a number and an alphabet. You use a comma in between 2 numbers. Up till now, you can try answering Qn 1(a) and Qn 2(a) in your tutorial worksheet. 30 Dec 2003
• #7: 30 Dec 2003 Let’s look at the properties of cycloalkanes. In Lecture 1, atoms with single bonds are sp3 hybridised. In cycloalkanes, all the bonds are single bonds, so the carbon atoms are sp3 hybridised. Usually, sp3 hybridization will have a bond angle of 109 °. In cycloalkanes, there are exception. For example, propane with 3 sp3 hybridized carbon atoms have bond angle of 109 ° however, when these 3 sp3 hybrizied carbon atoms form a ring, the bond angle is much smaller than 109°.
• #8: 30 Dec 2003 Let’s look at some exaples. Use modeling to explain. Built from scratch . The Carbon-carbon-carbon bond in cyclopropane is 60 °, this makes the cyclopropane ring highly strain and the carbon-carbon bond are locked and cannot rotate freely without breaking the ring.
• #9: 30 Dec 2003 The same goes with cyclobutane, though the carbon atoms are sp3 hybridized the carbon-carbon-carbon bond angle is 90 ° which is a big difference from the standard 109°. So, cyclobutane is also highly strain and the carbon-carbon bond cannot rotate freely without breaking the ring.
• #10: In Cyclopentane, the condition is better, where the carbon-carbon-carbon bond angle is almost 109 ° and the ring has almost no strain however, the carbon-carbon bond still could not rotate freely without breaking the ring. 30 Dec 2003
• #11: 30 Dec 2003 What about cyclohexane. In cyclohexane, the sp3 carbon atom is exactly the same as the carbon atom of an open chain. The carbon-carbon-carbon bond angle is exactly 109 ° and there is no ring strain at all. Therefore, the carbon-carbon bond in the cyclohexane can somehow rotate and that causes ring flip. You will get more experience on this in your week 4 practical lesson
• #12: 30 Dec 2003 Identify the axial first then the equatorial. In this slide, it tells us that cyclohexane has a chair like conformation, that means the structure look like an arm chair. We have learned from last lecture that the staggered and eclipsed conrformers of alkanes are results of carbon-carbon bond rotation. The carbon-carbon bond in the cyclohexane can some how rotate, therefore it will have different conformation. The chair conformation of cyclohexane is the most stable conformation. In this conformation, note that there are 6 axial Hydrogen atoms. In this slides, the axial hydrogen atoms are labelled in red, note that they are pointing up and down from the ring in an alternate manner, therefore, there 3 hydrogen atoms pointing up and 3 hydrogen atoms pointing down. Then, there are the 6 equatorial hydrogen atoms, which are labelled as blue in this slides. They are pointing outward from the side of the ring.
• #13: As we mentioned earlier, due to the size of the cyclohexane ring, the carbon-carbon bond can some how rotate. Look at the slide, the rotation of the carbon-carbon bond can result in a ring-flip, for example, you perform a ring flip when you push carbon 4 up and simultaneously push carbon 1 down. This ring flip will convert the chair conformation on the left to the conformation on the right and vise versa. Both left and right structure are cyclohexane in different conformations, they are called conformers. You will perform a ring flip using a model in your week 4 practical lesson. 30 Dec 2003
• #14: One question you may want to ask, will there be any changes to the hydrogen atoms during ring flip from one chair conformation to the other? The answer is yes, there are changes to the hydrogen atoms. Look at the slide, the axial hydrogen atoms will be converted to the equatorial hydrogen atoms and vice versa. You will try this out in week 4 practical lesson. 30 Dec 2003
• #15: Let say, you substitute one of the hydrogen atom in the cyclohexane with subsituent X and note that subsituent X is always larger in size than hydrogen atom. Same as hydrogen atoms, the ring flip of the cyclohexane from one chair conformation to the other chair conformation will convert the subsituent X from equatorial to axial position and vice versa. Remember this point clearly, the chair conformation with the substituent X in its equatorial position is more stable. Therefore, in this slide, of these two chair conformations, the chair conformation on the left is more stable. 30 Dec 2003
• #16: This slide show you the example where one of the hydrogen atom has been substituted and the substituent is a methyl group, a group that made up of 1 carbon and 3 hydrogen atoms. From the previous slide you know that of the two chair conformations, the chair conformation on the left where the methyl group is in its equatorial position is more stable. 30 Dec 2003
• #17: 30 Dec 2003 Why is the chair conformation with the substituent at the equatorial position would always be the more stable conformation? The reason is because of 1,3-diaxial interaction. In the top figure of this slide, where the methyl group is in its axial position, the hydrogen atom of the methyl group is of very close proximity with the axial hydrogen atom from the ring. This close proximity made the environment very crowded and call this steric interaction 1,3-diaxial interaction. Why 1, 3 because the position of axial is always alternate. So, the first axial is at position 1 of the cyclohexane ring, the second axial will be at position 3 of the cyclohexane ring. So, with this 1,3-diaxial interaction, this conformation is not very stable as compared to the bottom drawing in this slide, where the substituent methyl group is at the equatorial position. In this position, the hydrogen atom of the methyl group are farther away from the rest of the hydrogen atoms, there no steric interaction, thus, it is more stable. So, remember this principle, the chair conformation with the substituent in its equatorial position is more stable because it has no 1,3-diaxial interaction.
• #18: 30 Dec 2003 Let’s look at the 2 nd example. trans -1- tert -butyl-3-methylcyclohexane. In this cyclohexane, 2 hydrogen atoms have been replaced by 2 substituent, 1 is a methyl group, the other one that looks like a tripod stand is called tert-butyl group. Tert-butyl group is much larger in size as compared to methyl and in turn they are larger than hydrogen atom. When the size of the substituent is larger, the steric interaction is greater. So, remember this principle, when the larger group is in the equatorial position, the conformation is more stable. Now, you can try answering Qn 3(b) in your tutorial worksheet. Ok, class, I do not expect you to reproduce these 2 drawing, so long as you know how to apply the prniciple.
• #20: Let’s continue with the 2 nd and 3 rd groups of compound in this lecture. Let’s look at alkene first. Alkene are compound that made up of hydrogen and carbon atoms, and they must have at least one carbon-carbon double bond. One example of alkene that occur naturally is beta carotene, which is an orange pigment and it is the precursor for making vitamin A. Can you count how many double bond are there in the structure of beta-carotene. There are altogether 11. 30 Dec 2003
• #21: What is alkyne. Alkynes are also compounds that made up of hydrogen and carbon atoms but instead of a double bond, it must at least has one carbon-carbon triple bond. For example, Norplant, the active ingredient in a registered contraceptive drug. Can you identify the carbon-carbon triple bond in the strcuture? 30 Dec 2003
• #22: To name the alkenes and alkynes, we are still using the same concept of prefix and parent that we used in naming the alkanes , just that in alkenes and alkynes, the position for the double bond and triple bond respectively must be specified. Also, the parent is the longest continuous carbon chain that must contains the multiple bonds. 30 Dec 2003
• #23: 30 Dec 2003 Use overhead projector for illustration. Use more examples Let’s name the structure given in this slide. Step 1: Identify the parent chain. The parent chain must be the longest carbon chain that contains the double bond. In the drawing on the left, the longest chain includes the double bond and it has 5 carbon atom. This way, the parent chain has been correctly identified. Whereas, in the drawing on the right, the longest chain which contains 6 carbon atom does not include the double bond. This is incorrect. Step 2: For alkene, change the “ane” to “ene”. Therefore the parent is called pentene.
• #24: 30 Dec 2003 Use more examples Step 3: In naming the parent, not only the number if carbon is indicate, you must also indicate the position of the carbon-carbon double. To do this, you must number the parent chain in a sequence that would give the smallest possible number to the double bond. In this slide, the numbering system on the left has the double bond on carbon 1 and carbon 2. This is correct as these are the smallest possible number. However, the numbering system on the right is not correct because the double bond falls on carbon 4 and 5 which are not the smallest possible number. Therefore, by following the numbering sequence on the left, of carbon 1 and carbon 2 where the double bond falls in between, we choose to give the smallest number for the position of the double bond. Thus, the name of the parent with the position of double bond is 1-pentene.
• #25: Step 4: After settling the parent, we continue to look at the substituent, in this example, there is one substituent and it is circled in blue. It has 2 carbon atoms and it’s called ethyl. Then, we have to indicate the position of the substituent. Please take note, in alkene and alkyne, for determining the position of the substituent, you must reuse the numbering system that you have used for determining the double bond position earlier. Therefore, the position for ethyl is at position 2 following the numbering system in the previous step. The IUPAC name is 2-ethyl-1-pentene. Note the use of hyphen here, the always put a hyphen between a number and an alphabet. 30 Dec 2003
• #26: 30 Dec 2003 Use more examples When an alkene has more than 1 double bond, a Greek prefix is used to tell the number of double bond. For example, when there is 2 double bond, use a prefix “di”, when there are 3, use a prefix “tri” and so on. In addtion, similar to naming the substituent, for every double bond you must give a number to tell its position. Let say, in the example here, the structure has 2 double bonds, so the name must have 2 numbers to indicate the position of each double bond. Our last concern is, how do you number the parent chain. The numbering should give both the double bond the smallest possible number. Let’s look at the example given here. The parent has 6 carbon therefore hexa. Because it has 2 double bonds, we call it diene. “Di” means 2 and “ene” means double bond. So, 6 carbon atom with 2 double bond is called hexadiene. Now we have to number the chain so that the both double bond will have the smallest possible number. In the example on the left, the numbering give positions 1 and 4. The total sum for 1+4 is 5. Whereas, the numbering on the right gives positions 2 and 5. The sum for 2+5 is 7. So, the correct name for this compound is 1,4-hexadiene. Again. Note that, a must be placed in between 2 numbers and a hyphen in between a number and an alphabet. Now, you can try Qn1(e) adn Qn2(d).
• #28: 30 Dec 2003 Use more examples. For naming clcloalkenes, means alkenes in a ring, we basically applies the same rules for naming alkenes. Just that when you do numbering of he ring, you must give number 1 and 2 to the 2 carbon atoms that have the double bonds and the substituents will receive the smallest possible number. In the first example, let’s find out the parent first. There are 6 carbon in a ring with 1 double bond, therefore, the parent name is 1-cyclohexene. There is one substituent with 1 carbon atom, it is called methyl. Number the ring such that the carbon with double bond must be 1 and 2. There are 2 ways to give the double carbon atoms number 1 and 2. the red way on the left and the blue way on the right. The red way will give the substituent position number 1 but the blue way will ended up giving a larger position to the substituent, therefore the red way is the correct way of numbering. So, the name for example one is 1-methyl-1-cyclohexane. Remember, always use a hyphen in between a number and an alphaber. Let’s look at Example 2. There are 5 carbon in a ring with 1 double bond, therefore, the parent name is 1-cyclopentene. There are two substituents with 1 carbon atom each, the substituent are called dimethyl. Now, number the ring such that the carbon with double bond must be 1 and 2. There are 2 ways to give the double carbon atoms number 1 and 2. the red way on the left and the blue way on the right. The red way will give the substituent positions number 2 and 3, the total sum of 2+3 is 5. The numbering by the blue way will give the substituents positions 1 and 5 and the sum for 1 and 5 is 6 which is larger than the red coloured numbering system. Therefore the red way is the correct way of numbering. So, the name for example two is 2,3-dimethyl-1-cyclopentene. Remember, always a comma is used in between two numbers and a hyphen in between a number and an alphabet. You can now answer Qn2(e)
• #30: We use the same method for naming alkane in naming alkyne, the hydrocarbon with at least one triplr bond. In this case, we change the “ane” to “yne”. In this example, the parent chain is the longest continuous chain that include the triple bond. It is a chain with 7 carbon atoms and one triple bond and it is called heptyne. Then, we have to determine the position of the triple bond. To do this, the parent chain need to be numbered in such a way that will give the smallest position for the triple bond and in this case it is position 2. So, the parent name is 2-heptyne. There is a substituent with 1 carbon atom and it is at position 4 according to the numbering system used earlier. Therefore, the prefix is 4-methyl. The full name for this alkyne is 4-methyl-2-heptyne. 30 Dec 2003
• #31: 30 Dec 2003 May show the advertisement that Jacque has. The car transformed into a robots. Probably not this leacture, should be the first lecture, well it depentds on timing. From this slides onwards, we come to the last part of this lecture, which is the cis-trans isomerism in cycloalkanes and alkenes. First of all, let’s compare the Carbon-carbon single bond and the carbon-carbon double bond. The carbon-carbon single bond can rotate freely . The 2 models on the top are inteconvertable and they are 2 models of the same molecule. However, unlike the carbon-carbon single bond, the carbon-carbon double bond cannot rotate freely. Therefore, the two models cannot interconvert and they represent two different molecules. The carbon-carbon double bond is said to be restricted.
• #32: 30 Dec 2003 Try with Vanaja or Thomas on single bond (one hand) holding each other and turn. And then try with both hands and turn. Alkenes you have 2 bonds 1. sigma 2. pi bonds Let’s try to understand why the carbon-carbon double bond is constrained. In lecture 1, you learn that a double bond consists of 1 sigma bond and 1 pi bond. To rotate the bond about the sigma bond, we have to break the pi bond which need a very high energy, and making the double bond cannot rotate freely. Therefore, the carbon-carbon souble bond is constrained. Besides alkene, cycloalkane is the other system that has constrained carbon carbon bond. The carbon-carbon bond cannot rotate without breaking the ring. So, remember this. There are 2 systems that have retricted carbon-carbon bond, which are alkene and cycloalkane.
• #33: Let us learn about the cis and trans isomerism. There are 2 requirements for cis-trans isomers to occur. A compound must fulfil these 2 requirements in order to have cis and trans isomers. The cis isomer and trans isomer of a compound are 2 different molecules. The first requirement is that the compound must have restricted carbon-carbon bond. From the previous slide, the systems that have restricted carbon carbon bond are the alkenes and the cycloalkanes. For example, 1,2-dichloroethene which is an alkene has a restricted carbon-carbon double bond. With the formula given, you can arrange the atoms in two different ways as shown in the two drawings on the left. In one of the drawing the two hydrogen atoms are on the same side. In the other drawing the 2 hydrogen atoms are on the opposite side. These structures cannot interconvert because the double bond is restricted. Therefore, they are 2 different molecules, one is the cis isomer the other is the trans isomer. For 1,2-dichloroyclopropane, if we draw the structure in 3 dimensions, we also can have 2 different arrangement of the atoms as shown in both the drawing on the right. In one of the drawing, the 2 hydrogen atoms are on the same side, in the other drawing the 2 hydrogen are on the opposite side. The structures cannot interconvert because the carbon-carbon bond in a ring is restricted. These 2 molecules ere 2 different molecules, one is cis the other one is trans. 30 Dec 2003
• #34: 30 Dec 2003 Examples for trans However, not all the alkenes and cycloalkanes will have cis and trans isomers. In order for an alkene or a cycloalkane to have cis and trans isomer they must fulfil the second requirements. For alkenes, the 2 nd requirement states that the 2 carbon atoms with the double bond must have different groups attach to each of them. So, remember this, when you want to decide a given alkene can or cannot have cis and trans isoemrs, you just need to focus on the 2 carbon atoms that have the double bond. Let’s look at all 3 examples given in this slides and decide which compound will have cis and trans isomers. In the first example, the 2 carbon atoms that you should be focusing on are the ones with the double bond which I have marked with stars. Let’s examine these carbon atoms one by one. The double bond carbon on the left are attached to different groups, one is the methyl group, the other one is a hydrogen atom. Then, we look at the carbon atom on the right, this carbon atom also attach to different groups, one is an ethyl group, the other one is a hydrogen atom. In conclusion, the two double bond carbon atoms in the first example have different groups attached to each of them. Therefore, this alkene can have cis and trans isomer. In the 2nd example, the 2 carbon atoms with the double bond have been marked with stars. Let’s examine these carbon atoms one by one. The double bond carbon on the left are attached to different groups, one is the methyl group, the other one is a hydrogen atom. Then, we look at the carbon atom on the right, this carbon atom are attached to two identical groups, both are methyl groups, so, this alkene does not fulfil the 2 nd requirement because not both the double bond carbon atoms are attached to different groups. In conclusion, the 2 nd alkene in this slide cannot have cis and trans isomers. In the last example, the 2 carbon atoms with the double bond are again marked with stars. Let’s examine these carbon atoms one by one. The carbon atom on the left are attached to identical group, which are the hydrogen atoms. The double bond carbon on the right also attach to identical group, which are the methyl groups. Therefore, both the double bond carbon in this example are attached to identical group and the 2 nd requirement is not fulfilled. In conclusion, this alkene cannot have cis and trans isomer.
• #35: 30 Dec 2003 We apply the same requirement when we want to decide whether a cycloalkane can or cannot have cis and trans isomers. Just that instead of looking at the double bond carbon in alkene, you just need to look out for any 2 carbon atoms in the ring that have different groups attach to them. Let’s look at the 3 examples given in this slides and decide which compound will have cis and trans isomers. In the first example, carbon on the left are attached to different groups, one is the methyl group, the other one is a hydrogen atom. There is another carbon atoms also attach to different group, one is the methyl group the other one is a hydrogen atom. Therefore, this cycloalkane will have cis and trans isomer, and in fact it is a trans isomer. In the 2nd example, one carbon atom is attach to identical group, which are the chloro group. Another carbon atom also attach to identical group which are bromo groups. In this cycloalkane, we cannot find 2 carbon atoms attach to different groups, so it cannot have cis and trans isomers. In the last example, there is only 1 carbon atom attach to different groups, which is a methyl group and a fluoro group. This does not fulfil the requirement also, because the requirement for cyclloalkane is to have 2 carbon atoms attach to different groups. In conclusion, this cycloalkane also cannot have cis and trans isomer.
• #36: In conclusion, for one compound to have cis and trans isomer. Firstly, it must be an alkene or a cycloalkane. Secondly, the alkenes or cycloalkanes must have 2 carbon atoms attach to different groups. Now, if a compound is examined to have cis and trans isomers, how do we determine which one is the cis isomer and which one is the trans isomers? To do this, you can draw a dotted line to join any of the identical groups from different carbon atoms. Let’s look at Example 1, both double bond carbon atoms have hydrogen atoms and methyl groups. We can draw a dotted line to join the identical groups together. Let’s say we choose the hydrogen atoms. In the structure on the left, the dotted line is crossing the double bond, identical groups are on the opposite side, it is a trans isomer. In the structure on the right, the dotted line that join the identical groups does not cross the double bond, and the identical groups are on the same side, it is the cis isomer. Let’s look at Example 2, the cycloalkane system. The identical groups are again the hydrogen atoms and the methyl groups. Let’s draw a dotted line to join any of the identical groups together. Let’s say we choose the methyl groups. In the structure on the left, the dotted line is crossing the ring, the identical groups are on the opposite side, it is a trans isomer. In the structure on the right, the dotted line that join the identical groups does not cross the ring, and the identical groups are on the same side, it is a cis isomer. 30 Dec 2003
• #37: Let’s try some exercises. The first one, draw the structure of trans-2-butene. I suggest that you try drawing now using paper and pen. Step 1, for butene, draw 4 carbon atoms in a zig-zag manner and join them using single bonds first. Step 2, for double bond at position 2, draw a double bond in between the carbon 2 and carbon 3. Step 3, circle the identical groups that are attached to both double carbon atoms, which is the second and third carbon. Step 4, for trans isomer, check that the identical groups are on the opposite side, or you can draw a dotted line to join the identical groups, and see that the dotted line are crossing the double bond. Now, try to draw cis-2-butene. Step 1, for butene, draw 4 carbon atoms in a zig-zag manner and join them using single bonds first. Step 2, for double bond at position 2, draw a double bond in between the carbon 2 and carbon 3. Step 3, circle the identical groups that are attached to both double carbon atoms, which is the second and third carbon. Step 4, for cis isomer, redraw one of the identical group such that the identical groups are on the same side, you can then draw a dotted line to join the identical groups and see that the dotted line does not cross the double bond. 30 Dec 2003
• #38: 30 Dec 2003 Of the 2 isomers that you have drawn for 2-butene, the trans isomer is more stable. This is because in the cis isomer, both methyl groups are closer to each other than in trans isomer, therefore, the cis isomer has steric interaction, making it less stable.
• #39: Now, draw the structure of cis-1,2-diethylcyclopentane. Again I suggest that you try drawing using paper and pen. Step 1, for cyclopentane, draw 5 carbon atoms in a pentagon ring. Step 2, for diethyl at position 1 and 2, draw a two ethyl substituents on 2 side by side carbon atoms. Step 3, for cis isomers, make sure that both the ethyl groups are on the same side Now, try to draw trans-1,3-dichlorocyclobutane Step 1, for cyclobutene, draw 4 carbon atoms in a square. Note that the 2 parallel lines are of different length. Step 2, for dichloro at positions 1 and 3 draw 2 chloro groups at position 1 and 3. Step 3, for trans isomer, ensure that the chloro groups are on the opposite side. Note, that it is alright, if you choose not to draw the hydrogen atoms. With this you can answer the rest of the Qns in your tutorial worksheet. 30 Dec 2003