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Hello guys, we're going to continue with a series of videos about how to name organic compounds

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following IUPERC rules. At the end of this video you should be able to identify a name correctly,

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alkyens and alkyens. So let's start. Alkyens are hydrocarbons with a double bone.

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To make the name of analkene, we need to use the root, which is related to the number of carbon atoms present in the molecule, and the ending "-ene".

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If we take the first one as an example, we can see C2H4. It has two carbon atoms, F, the ending "-ene", ethene.

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With three carbon atoms, it is CH2 double bond, CH, CH3. The root is propene, the ending in propene. However, when we reach to C4H8, there are two possible compounds, depending on where the double bond is placed.

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Then, we need to indicate the position of the double bond with the number, which is called locant.

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In the first case, the double bond is near the terminal carbon.

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So, put, because it has 4 carbon atoms, and 1 in.

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If we started numbering from the other end of the molecule, the double bone would be in 3. So, it is chosen the first option.

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In the second hydrocarbon, the double bond is placed in between. It doesn't matter from where we start numbering, because in both cases, the double bond would be in 2.

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Then, the name is root because it has four carbons, to, in. Remember to put hyphons between the root, the locant and the suffix.

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Let's do the next one, the following example. C5H10. It has five carbon atoms, so the root is pent.

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In the first molecule, the double bone is placed near the terminal carbon, so bent 1.

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In the second case, we need to start numbering by the end that assigns the lowest locant to the double bone.

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In this case, it's from left to right, then the name would be bent 2 in.

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When the compound has two double bones, it is a diene.

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They are named. Here we have two examples.

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In the first one, there are four carbon atoms, so the prefix is but.

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And then there are two double bones in one and three.

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The name is Buddha 1,3 Da-in.

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In the second case, there are two double bones, but we need to choose from which end to start numbering.

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If we go from left to right, the first double bone would be in 3.

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Whereas, if we start from right to left, the first double bond is found in 1.

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Then, let's take the second option.

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Hexa-1,3-diene.

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All kinds of hydrocarbons with a triple bond.

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They are named exactly in the same way as alkenes, but using the suffix "-ine".

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For example, the easiest is CH, triple bond, CH, F because it has two carbon atoms and the ending I, in time.

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It has a special name, acetylene.

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Let's do a couple of examples.

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In the first example, there are five carbon atoms, so the root is pent.

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The triple bone is placed in the first, in position 1, so pent, 1, iron.

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In the second case, the 6 carbon atoms, so hex.

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The double bone, if we start from numbering, from left to right, is in 2, hex, 2, iron.

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When there are two or three triple bonds in the molecule, the prefixes di-, tri- or even

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tetra- will be used.

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In the first example, there are five carbon atoms, so the root is penta and then if we

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start numbering from left to right. The triple bones would be in one and four.

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Penta, one, four. Di, iron. In the second case there are seven carbon atoms so

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hepta. It will start from left to right. The triple bones will be in one and

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in five however it will start numbering from right to left the first triple bone is in two

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then using the lowest possible lock hint rule that we will start from left to right the name is penta

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1,5-di-ion. In the last example, there are nine carbon atoms, so the prefix is Nona.

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Starting from left to right, the triple bonds are found in 1, 4 and 7. The name is Nona 1,4,5-tri-ion.

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Alkenines are acyclic hydrocarbons which contain both double and triple bonds.

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In this case the rules are a bit more complicated, firstly we need to identify the main chain

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which is the longest chain that contains all unsaturated bonds.

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Now the question arises from which end we need to start numbering.

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It's always going to start numbering by the end that assigns the lowest lock-end to the unsaturated bones, regardless whether they are double or triple bone.

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So, let's try to solve the first example.

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There are 7 carbon atoms.

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It will start numbering from left to right.

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The double bone is in 2 and the triple in 4.

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It will start numbering from right to left.

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The double bone is in five and the triple is in three. We're gonna choose from left to right because in this way the lowest locants are assigned. Helped because it has seven carbon atoms. Two in, five iron.

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Let's try to solve the next example. Here there are 6 carbon atoms. If we start numbering

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from left to right, the triple bone is in 1 and the double bone is in 4. However, if

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we start the other way round, the double bone is in 2 and the triple in 5. Then, from left

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to right is preferred because lowest locants are assigned to the unsaturated bones .

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In the following example, if we start numbering from left to right, the triple is in 1 and

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the double in 5. However, from right to left, the double is in 1 and the triple in 5. In

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case it is possible to choose and according to the rule we need to select the numbering scheme

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which assigns the lowest lock end to the double bone then from right to left hex one in five on

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finally in the last example the molecule has seven

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The double bone is in 1 and the triple bone is in 5.

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So hepta 1,5 diene 6 iron.

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Thanks a lot for watching this video, see you next time, bye.