Chapter 13.16 - Polymerisation of Alkynes

In the previous section, we saw the preparation of alkynes. We also saw some chemical properties of alkynes. In this section, we will see two more chemical properties.

Addition of water

Let us see the reaction between an alkyne and water. It can be written in 5 steps:
1. Alkynes react with water. Mercuric sulphate and concentrated sulphuric acid is also required for the reaction. The reaction mixture should be warmed to a temperature of 333 K. Carbonyl compounds will be obtained as products. An example is shown in fig.13.92 below:

 

Fig.13.92

 

• First the water molecule splits to give two parts:
    ♦ H⁺ ion, which is the +ve part.
    ♦ OH^- ion, which is the -ve part.
2. The H⁺ ion thus produced, will get attached to one of the two C atoms of the ethyne molecule. This is shown in fig.a.
• The H⁺ ion does not have any electrons. Both electrons required for the bond, are obtained by breaking the triple bond in the alkyne. That is why both electrons in the bond are shown in red color.
3. When the H⁺ ion leaves the water molecule, the remaining portion will have an extra electron. So it will have a -ve charge.
• This remaining portion is the -ve part of the addendum. This -ve part (OH^-) gets attached to the other C atom.
• Both electrons required for the bond, are supplied by this -ve part. They are the green and yellow electrons.
4. The alcohol thus formed will undergo isomerisation. As a result we get ethanal. This is shown in fig.b.
• Isomerisation is a reaction in which a molecule gets transformed into it's isomer.
• We saw a type of isomerisation in an earlier section [see fig.13.38 in section 13.6].
5. In this example, we see that, the two species H⁺ and OH^- get added to the original alkyne molecule.
• So we can write:
The reaction between alkynes and water is an addition reaction.

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Let us see another example. It is shown in fig.13.93 below. It can be written in 6 steps.

 

Fig.13.93

1. First the water molecule splits to give two parts:
    ♦ H⁺ ion, which is the +ve part.
    ♦ OH^- ion, which is the -ve part.
2. The H+' ion thus produced, will get attached to the first C atom of the propyne molecule. This is shown in fig.a.
• The H⁺ ion does not have any electrons. Both electrons required for the bond, are obtained by breaking the triple bond in the alkyne. That is why both electrons in the bond are shown in red color.
3. When the H⁺ ion leaves the water molecule, the remaining portion will have an extra electron. So it will have a -ve charge.
• This remaining portion is the -ve part of the addendum. This -ve part (OH^-) gets attached to the second C atom.
• Both electrons required for the bond, are supplied by this -ve part. They are the green and yellow electrons.
4. Here we see that, the -ve part gets attached to the C atom with the least number of H atoms.
• So we can write:
The reaction between alkenes and water obeys Markovnikov’s rule.
5. The alcohol thus formed will undergo isomerisation. As a result we get propanone. This is shown in fig.b.
6. In this example, we see that, the two species H⁺ and OH^- get added to the original alkyne molecule.
• So we can write:
The reaction between alkynes and water is an addition reaction.

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Polymerization of alkynes

• First we will see linear polymerization. It can be written in 6 steps:
1. Consider the molecule of ethyne ($\rm{CH ☰ CH}$).
• We know that in ethyne, the triple bond is necessary to satisfy the valencies of C and H atoms.
2. If we break the triple bond, the structure would look like this: $\rm{-CH = CH -}$
• The ‘${}-{}$’ on the sides indicate that, the structure is looking for electrons.
3. Consider the structure shown in the above step (2).
• If there are a large number of such structures, they can join together to satisfy the valencies. The joined structure would look like this:
$\rm{-CH = CH - CH = CH - CH = CH - CH = CH -}$
• The process of making the joined structure is known as polymerization. We saw this in the case of alkenes also.
4. In our present polymerization, we subject ethyne molecules to high pressure and high temperature. Presence of a suitable catalyst is also necessary.
• Due to the high pressure and temperature, ethyne molecules will change into $\rm{-CH = CH -}$
• These structures will join together to form a large molecule.
   ♦ The large molecule is called a polymer.
   ♦ The simple molecule from which polymer is obtained is called a monomer.
• In our present case,
   ♦ Ethyne is the monomer.
   ♦ The polymer obtained has a common name: polyacetylene.
        ✰ It's IUPAC name is polyethyne.
5. Under special conditions, polyethyne conducts electricity. Thin films of this polymer can be used as electrodes in batteries. These films are good conductors. They are lighter and cheaper than metal conductors. Some images can be seen here.
6. The joined structure in step (3) can be written in short form as: $\rm{-(CH = CH - CH = CH )_n -}$
• So the process of this polymerization to make polyethyne can be written as:
$\rm{n(CH ☰ CH)~ \color {green}{\xrightarrow[{\text{Catalyst}}]{{\text{High temp./pressure}}}} ~ -(CH = CH - CH = CH )_n -}$

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Now we will see cyclic polymerization. It can be written in 4 steps:

1. We have already seen the details about the structure of the benzene ring [see fig.12.53 of section12.8].
• We see that, there are six CH groups in benzene. Can three ethyne molecules join together to form a benzene ring?
2. Fig.13.94(a) below shows three ethyne molecules aligned together in favorable positions.

Fig.13.94

• If two electrons in the triple bonds can shift, new single bonds will be formed between the three ethyne molecules. The shifting of electrons are indicated by the three curved red arrows in fig.b.
3. This is a cyclic polymerization. For this polymerization to take place, we allow the ethyne gas to pass through red hot iron tube kept at 873 K.
4. So we have an easy method to prepare benzene. Once we obtain benzene, we can make a variety of useful products like benzene derivatives, dyes, drugs etc.,
• This method helps us to enter the world of aromatic compounds from the world of aliphatic compounds.
• We know that:
    ♦ Aliphatic compounds have an open chain structure.
    ♦ Aromatic compounds have a closed chain structure.
• So we enter the world of closed chain structures from the world of open chain structures.

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Let us see a solved example.

Solved example 13.14
How will you convert ethanoic acid into benzene ?
Solution:
1. We can start from benzene and think in a reverse direction.
• To obtain benzene, we must have ethyne (CH☰CH). Because, ethyne when passed through a red hot iron tube will give benzene. See fig.13.94 above.
• This process is marked as the last step (X) in fig.13.95 below.
• So our aim must be to convert the given ethanoic acid to ethyne.

Fig.13.95

2. Ethyne can be obtained from alkenyl halides. We saw this in the preparation of alkynes. See the topic at the beginning of the previous section.
• In our present case, the alkenyl halide CH₂=CHBr can be used to obtain ethyne.
• This process is marked as (IX) in fig.13.95 above.
• So our aim must be to convert the given ethanoic acid to the alkenyl halide CH₂=CHBr.
3. Alkenyl halides can be obtained from vicinal dihalides. We saw this in the preparation of alkynes. See the topic at the beginning of the previous section.
• In our present case, CH₂Br-CH₂Br can be used to obtain the alkenyl halide CH₂=CHBr.
• This process is marked as (VIII) in fig.13.95 above.
• So our aim must be to convert the given ethanoic acid to the vicinal dihalide CH₂Br-CH₂Br.
4. The above vicinal dihalide can be prepared by the addition reaction between ethene and Br₂. We saw this in the chemical properties of alkenes. See fig.13.68 of section 13.11.
• This process is marked as (VII) in fig.13.95 above.
• So our aim must be to convert the given ethanoic acid to ethene.
5. Ethene can be obtained from the alkyl halide CH₃-CH₂Cl. We saw this in the preparation of alkenes from alkyl halides. See fig.13.65 of section 13.10.
• This process is marked as (VI) in fig.13.95 above.
• So our aim must be to convert the given ethanoic acid to the alkyl halide CH₃-CH₂Cl.
6. The above alkyl halide can be prepared from the alkane CH₃-CH₃. We saw this in the addition reaction of alkanes. See substitution reactions in section 13.4.
• This process is marked as (V) in fig.13.95 above.
• So our aim must be to convert the given ethanoic acid to the alkane CH₃-CH₃.
7. The above alkane can be obtained using CH₃Cl by Wurtz reaction, We saw this in the preparation of alkanes from alkyl halides. See section 13.3.
• This process is marked as (IV) in fig.13.95 above.
• So our aim must be to convert the given ethanoic acid to the alkyl halide CH₃Cl.
8. The above alkyl halide can be obtained from CH₄. We saw this in the chemical properties of alkanes. See substitution reactions in section 13.4.
• This process is marked as (III) in fig.13.95 above.
• So our aim must be to convert the given ethanoic acid to the alkane CH₄.
9. CH₄ can be obtained from the sodium salt of ethanoic acid. We saw this in the preparation of alkanes. See preparation of alkanes from carboxylic acids in section 13.3.
• This process is marked as (II) in fig.13.95 above.
• So our aim must be to convert the given ethanoic acid to the sodium salt of the ethanoic acid.
10. The preparation of the above sodium salt is a simple process.
• Ethanoic acid is an acid. It will react with the base NaOH to give the sodium salt.
• This process is marked as (I) in fig.13.95 above.
• So we have worked in the reverse order from benzene to ethanoic acid.

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In the next section we will see aromatic hydrocarbons.

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Chapter 13.15 - Preparation and Properties of Alkynes

In the previous section, we saw nomenclature and isomerism in alkynes. We also saw the structure of the triple bond. In this section, we will see preparation of alkynes.

• We will see two methods for preparing ethyne.
   ♦ From calcium carbide
   ♦ From vicinal dihalides

From calcium carbide

This can be written in 3 steps:
1. First, lime stone is heated to obtain quick lime (CaO).
• The equation is:
$\rm{CaCO_3~ \color {green}{\xrightarrow[{}]{Δ}} ~ CaO~+~CO_2}$
2. The quick lime is heated with coke to obtain calcium carbide (CaC₂)
• The equation is:
$\rm{CaO~+~3C~ \color {green}{\xrightarrow[{}]{}} ~ CaC_2~+~CO}$
3. The calcium carbide is treated with water to obtain ethyne.
• The equation is:
$\rm{CaC_2~+~2H_2O~ \color {green}{\xrightarrow[{}]{}} ~ Ca(OH)_2~+~C_2H_2}$

From vicinal dihalides

This can be written in 2 steps:
1. The vicinal dihalide is first treated with alcoholic potassium hydroxide.
• One atom of hydrogen and one atom of halogen will be removed from the vicinal dihalide. This removal is known as dehydrohalogenation.
• The product will contain a double bond between the two carbon atoms.
• So the product is not an alkyl halide, but an alkenyl halide.
• The equation is:
$\rm{CH_2 Br - CH_2 Br~+~KOH~ \color {green}{\xrightarrow[{-KBr~and~-H_2O}]{alcohol}} ~ CH_2 = CHBr}$
2. The alkenyl halide is treated with sodamide to obtain the alkyne.
• The equation is:
$\rm{CH_2 = CHBr~ \color {green}{\xrightarrow[{-NaBr~and~-NH_3}]{Na^{+} {NH_2}^{-}}} ~ CH ≡ CH}$

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Physical properties of alkynes.

This can be written in 5 steps:
1. We have already seen the physical properties of alkanes [see section 13.4] and those of alkenes [see section 13.11].
• The physical properties of alkynes follow a similar trend as alkanes and alkenes.
2. The first three members of the alkyne series are gases.
• The next eight members are liquids.
• The members coming after that are solids.
3. All alkynes are colorless.
• All alkynes except ethyne are odour less. Ethyne has a characteristic odour.
4. Alkynes are insoluble in water. But they are soluble in non-polar solvents like carbon tetrachloride, benzene and petroleum ether.
(Petroleum ether is obtained from petroleum. It is used as a laboratory solvent)
5. The members of the alkyne series show a regular increase in melting point, boiling point and density.

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Chemical properties of alkynes

• We have to learn about three chemical properties of alkynes. They are:

A. Acidic character

B. Addition reaction

C. Polymerization

A. Acidic character

• Before discussing about the acidic character of alkynes, we must consider the electronegativity of C atom.
• For that, we make a statement related to the electronegativity of C. The statement can be written in 3 steps:
(i) Consider a molecule containing C atom.
(ii) Suppose that, the C atom is sp hybridized.
Then that C atom will be highly electronegative.
(iii) If that C atom is sp² hybridized or sp³ hybridized, then it will not be so much electronegative.

 

• Now we will see the proof for the statement. It can be written in 5 steps:
1. Consider the sp³ hybridized orbitals. Each of those orbitals will be having 25% s-characteristics.
2. Consider the sp² hybridized orbitals. Each of those orbitals will be having 33% s-characteristics.
3. Consider the sp hybridized orbitals. Each of those orbitals will be having 50% s-characteristics.
4. We have already seen the above details in an earlier section [see fig.4.135 in section 4.24].
• So it is clear that, sp hybridized orbitals have greater s-characteristics.
5. s-orbitals are closer to the nucleus. So the electrons in the s-orbitals will be attracted more towards the nucleus of the atom.
• So in our present case, if the C atom is sp hybridized, then the shared electrons around that C atom will be attracted more towards the nucleus of the C atom.
• Consequently, such a C atom will be more electronegative.

Now we can discuss about the acidic character of alkynes. It can be written in 4 steps:

1. Consider a C☰H bond in ethyne. The C atom here is sp hybridized. So it will pull the electrons in the bonds. The H will become +ve charged.

2. The H can be released as H⁺. That means, ethyne can act as a proton donor. We know that acids are proton donors. So now we understand why ethyne is acidic.

3. In alkanes, the C atoms are sp³ hybridized. Those C atoms do not have high electronegativity. So alkanes are not acidic.

4. In alkenes, the C atoms are sp² hybridized. Those C atoms do not have high electronegativity. So alkenes are not acidic.

Note:
In an alkyne, there may be more than one triple bonds. There may be other double and single bonds also. Only those H atoms in the triple bonds are available for release as protons. We must not expect the other H atoms to contribute to the acidic character of that alkyne.

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Let us see two reactions where alkynes show their acidic character.

Reaction 1:

This can be written in three steps

1. We know that, acids react with sodium (Na) to release hydrogen gas.

• An example is:
$\rm{HCl~+~Na~ \color {green}{\xrightarrow[{}]{}} ~ Na^{+} Cl^{-}~+~\frac{1}{2}H_2}$

2. In a similar way, ethyne reacts with Na to release hydrogen gas. The equation is:

$\rm{HC☰CH~+~Na~ \color {green}{\xrightarrow[{}]{}} ~ HC☰C^{-}Na^{+}~+~\frac{1}{2}H_2}$

• HC☰C^-Na⁺ obtained above is monosodium ethynide.

3. This monosodium ethynide reacts with another atom of Na to give disodium ethynide. The equation is:

$\rm{HC☰C^{-}Na^{+}~+~Na~ \color {green}{\xrightarrow[{}]{}} ~ Na^{+}C^{-} ☰ C^{-}Na^{+}~+~\frac{1}{2}H_2}$

 

Reaction 2:

• Here we consider the reaction between propyne and sodamide. The equation is:

$\rm{CH_3 - C☰CH~+~Na^{+} {NH_2}^{-}~ \color {green}{\xrightarrow[{}]{}} ~ CH_3 - C☰C^{-} Na^{+}~+~NH_3}$

• $\rm{CH_3 - C☰C^{-} Na^{+}}$ is sodium propynide 

• The above reactions are not shown by alkanes and alkenes. So we can test a sample of unknown hydrocarbons by adding Na or sodamide. If a reaction takes place, we can confirm that, the sample contains alkynes.

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Let us compare the acidic characters of but-1-yne and but-2-yne. The comparison can be written in 3 steps:
1. First we write the structures:
   ♦ The structure of but-1-yne is: HC☰C-CH₂-CH₃
   ♦ The structure of but-2-yne is: CH₃-C☰C-CH₃
2. Consider the structure of but-1-yne.
• There are two C atoms on either sides of the triple bond.
   ♦ Both of them will be sp hybridized.
• One of those C atoms have a H atom.
   ♦ This particular C atom can pull the electrons away from the H atom.
   ♦ As a result, there is one H atom available to be donated as a proton.
• Thus but-1-yne gets it’s acidic character.
3. Consider the structure of but-2-yne.
• There are two C atoms on either sides of the triple bond.
   ♦ Both of them will be sp hybridized.
• None of those C atoms have any H atoms.
   ♦ So the C atoms cannot pull electrons.
   ♦ As a consequence, there are no H atoms available to be donated as protons.
• Thus but-2-yne gets no acidic character.

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B. Addition reaction

Some basics can be written in 2 steps:
1. We know that, each triple bond in alkynes consist of two π-bonds. The electrons in the π-bonds are loosely held. So those electrons are easily available for ‘electron seeking species’ (electrophiles).
2. Due to this availability of electrons, the electrophiles will get attached to the alkynes, resulting in new compounds. We call such reactions as addition reactions. We saw this situation in the case of alkenes also. Now we will see different types of addition reactions in alkynes.

I. Addition of dihydrogen
This can be written in 6 steps:
1. Each triple bond in an alkyne can add two molecules of dihydrogen.
2. The dihydrogen molecule first splits into two H atoms. This is indicated by the blue dashed curves in fig.13.88(a) below:

Fig.13.88

• Each newly formed H atom will have only one electron (yellow dot). So each H atom will be looking for one more electron to complete octet.
3. The loosely held electrons at the π-bonds will supply the required electrons for the H atoms.
• That means, two of the six red dots in the triple bond, will leave the triple bond. Those two red dots will help the new H atoms to form single bonds with C atoms.
• Thus we get two new C-H bonds.
4. So we now know how the H atoms add up to the alkyne. The alkyne then will no longer require the triple bond. It will be converted to an alkene.
5. The newly formed alkene has a π-bond. So two more atoms of H can be added. The result will be an alkane. This is shown in fig.b above. We saw this process in the case of alkenes [see fig.13.67 of section 13.11].
6. In an earlier section, we saw this process as a method for preparing alkanes [see section 13.3].
• For this process, finely divided nickel, palladium or platinum is required as catalyst.

II. Addition of halogens
This can be written in 7 steps:
1. Each triple bond in an alkyne can add two molecules of halogen.
2. First the alkyne gets converted into an alkene. Two individual halogen atoms (F, Cl, Br or I) are required for this process. This is indicated by the two Br atoms in fig.13.89(a) below:

Fig.13.89

• Each individual X atom will have only seven electrons in the outer most shell.
• This is indicated by the seven dots around the Br atoms.
   ♦ Seven grey dots for the first Br atom
   ♦ Seven yellow dots for the second Br atom.
• So each X atom will be looking for one more electron to complete octet.
3. The loosely held electrons at the π-bond will supply the required electrons for the X atoms.
• That means, two of the six red dots in the triple bond, will leave the triple bond. Those two red dots will help the two new X atoms to form single bonds with C atoms.
• Thus we get two new C-Br bonds to form 1,2-Dibromopropene.
4. So we now know how the X atoms add up to the alkyne. The alkyne then will no longer require the triple bond. It will be converted to an alkene.
5. The newly formed alkene has a π-bond. So two more atoms of Br can be added.
• The result will be $\rm{CH_3 - CBr_2=CHBr_2}$. This is shown in fig.b above. We saw this process in the case of alkenes [see fig.13.69 of section 13.11].
6. The equations of the reactions in fig.13.89 above can be written as:
• $\rm{CH_3 - C☰CH~+~Br-Br~ \color {green}{\xrightarrow[{}]{}} ~ CH_3 - CBr=CHBr}$
• $\rm{CH_3 - CBr=CHBr~+~Br-Br~ \color {green}{\xrightarrow[{}]{}} ~ CH_3 - CBr_2 - CHBr_2}$
7. Let us see how this reaction can be used as a test for the presence of double or triple bonds (test for unsaturation). It can be written in 4 steps:
(i) Bromine solution has a reddish orange color. This color is due to the presence of Br^- ions.
(ii) We add this bromine solution to a solution which is to be tested. If the solution contains any triple bonds, those triple bonds will break. Each triple bond will take up four Br atoms.
(iii) Thus all the Br^- ions will be used up. The reddish orange color will disappear.
(iv) So, if the reddish orange color disappear, we will get an indication that, triple bonds are present.
• We saw these details in the case of alkenes also.

III. Addition of hydrogen halides
We will consider the reaction between but-2-yne and HBr. It can be written in 8 steps:
1. First the HBr molecule splits into two parts. We get H⁺ and Br^-. It is shown in fig.13.90 (a) below:

Fig.13.90

2. The H⁺ attacks the alkyne molecule. This is shown in fig.13.90 (b) above.
• At the product side, we see that, the H⁺ is attached to the third C atom.
• Note that, the H⁺ did not bring any electrons. Both electrons required for the bond is made available from the triple bond. That is why we see two red dots in the new C-H bond.
3. But now, the triple bond is broken (the π-electrons in the triple bond were utilized for the new C-H bond).
• When the triple bond is broken in this way, the second C atom looses an electron.
• This C atom now has sextet only. Thus a vinylic cation is formed.
• A vinylic cation is a carbocation in which the +ve charge is possessed by a C atom in a double bond. 
4. The Br^- now attacks the newly formed vinylic cation. This is shown in fig.13.90 (c) above.
• The Br^- can donate two electrons to form a bond. So it attaches to the second C atom. Thus we get a molecule of 2-Bromobut-2-ene.
5. Now we have a molecule with a double bond. This double bond has to be converted into a single bond.
• For that, a new H⁺ attacks the 2-Bromo-but-2-ene. This is shown in fig.d above.
• At the product side, we see that, the H⁺ is attached to the third C atom.
• Note that, the H⁺ did not bring any electrons. Both electrons required for the bond is made available from the double bond. That is why we see two red dots in the new C-H bond.
6. But now, the double bond is broken (the π-electrons in the double bond were utilized for the new C-H bond).
• When the double bond is broken in this way, the second C atom looses an electron.
• This C atom now has sextet only. Thus a carbocation is formed.
7. The Br^- now attacks the newly formed carbocation. This is shown in fig.13.90 (e) above.
• The Br^- can donate two electrons to form a bond. So it attaches to the second C atom. Thus we get a molecule of 2,2-Dibromobutane.
8. We note an interesting point here. It can be written in 5 steps:
(i) In this reaction, there are two stages.
• In the first stage, the triple bond is converted to a double bond.
• In the second stage. the double bond is converted to a single bond.
(ii) A molecule of HBr is added in each stage.
• In the first stage, a molecule of HBr is added to the triple bond. 
• In the second stage, another molecule of HBr is added to the double bond.
(iii) H atoms are being added to the same C atom
• In the first stage, the H atom is added to the third C atom. This is shown in fig.13.90(b)  
• In the second stage, the H atom is added to the same third C atom. This is shown in fig.13.90(d)
(iv) Br atoms are being added to the same C atom
• In the first stage, the Br atom is added to the second C atom. This is shown in fig.13.90(c)  
• In the second stage, the Br atom is added to the same second C atom. This is shown in fig.13.90(e)
(v) So we get an end product in which two halogen atoms are attached to the same C atom. A dihalide in which two halogen atoms are attached to the same C atom is called gem dihalide.

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• In the above example, the alkyne was symmetric. Now we will consider an unsymmetrical alkyne. For that, we will see the reaction between propyne and HBr. It can be written in steps:
1. First the HBr molecule splits into two parts. We get H⁺ and Br^-. It is shown in fig.13.91 (a) below:

Fig.13.91

2. The H⁺ attacks the alkyne molecule. This is shown in fig.13.91 (b) above.
• At the product side, we see that, the H⁺ is attached to the first C atom.
• Note that, the H⁺ did not bring any electrons. Both electrons required for the bond is made available from the triple bond. That is why we see two red dots in the new C-H bond.
3. But now, the triple bond is broken (the π-electrons in the triple bond were utilized for the new C-H bond).
• When the triple bond is broken in this way, the second C atom looses an electron.
• This C atom now has sextet only. Thus a vinylic cation is formed.
4. The Br^- now attacks the newly formed vinylic cation. This is shown in fig.13.91 (c) above.
• The Br^- can donate two electrons to form a bond. So it attaches to the second C atom. Thus we get a molecule of 2-Bromopropene.
5. Now we have a molecule with a double bond. This double bond has to be converted into a single bond.
• For that, a new H⁺ attacks the 2-Bromopropene. This is shown in fig.d above.
• At the product side, we see that, the H⁺ is attached to the first C atom.
• Note that, the H⁺ did not bring any electrons. Both electrons required for the bond is made available from the double bond. That is why we see two red dots in the new C-H bond.
6. But now, the double bond is broken (the π-electrons in the double bond were utilized for the new C-H bond).
• When the double bond is broken in this way, the second C atom looses an electron.
• This C atom now has sextet only. Thus a carbocation is formed.
7. The Br^- now attacks the newly formed carbocation. This is shown in fig.13.90 (e) above.
• The Br^- can donate two electrons to form a bond. So it attaches to the second C atom. Thus we get a molecule of 2,2-Dibromopropane.
8. We note an interesting point here. It can be written in 5 steps:
(i) In this reaction, there are two stages.
• In the first stage, the triple bond is converted to a double bond.
• In the second stage. the double bond is converted to a single bond.
(ii) A molecule of HBr is added in each stage.
• In the first stage, a molecule of HBr is added to the triple bond. 
• In the second stage, another molecule of HBr is added to the double bond.
(iii) H atoms are being added to the same C atom
• In the first stage, the H atom is added to the first C atom. This is shown in fig.13.91(b)  
• In the second stage, the H atom is added to the same first C atom. This is shown in fig.13.91(d)
(iv) Br atoms are being added to the same C atom
• In the first stage, the Br atom is added to the second C atom. This is shown in fig.13.91(c)  
• In the second stage, the Br atom is added to the same second C atom. This is shown in fig.13.91(e)
(v) So we get a gem dihalide.

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• Based on the two examples, we can write:
An alkyne may be symmetric or unsymmetric. When the addition of hydrogen halides occur, we always get a gem dihalide.

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We have seen that Markovnikov's rule is applicable when addition of hydrogen halides to unsymmetrical alkenes occur. Let us see whether the rule is applicable to unsymmetrical alkynes. It can be written in steps:
1. In example 2, HBr is being added to CH₃ - C☰CH
2. The product is: CH₃ - CBr₂ - CH₃
3. It is clear that, the -ve part (Br^-) is being added to the C atom with the least number of H atoms.
4. So Markovnikov's rule is applicable here.

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In the next section we will see addition of water and polymerisation.

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Chapter 13.14 - Nomenclature and Isomerism in Alkynes

In the previous section, we completed a discussion on ozonolysis and polymerisation of alkenes. In this section, we will see alkynes.

Let us recall some properties of alkynes that we have seen in earlier chapters. They can be written in 3 steps:
1. Alkynes are unsaturated hydrocarbons. They contain at least one triple bond.
2. If there is one triple bond in an alkyne,
   ♦ it will contain four H atoms less than the corresponding alkane.
   ♦ it will contain two H atoms less than the corresponding alkene.
• For example:
    ♦ Molecular formula of butane is C₄H₁₀
    ♦ Molecular formula of butene is C₄H₈
    ♦ Molecular formula of butyne is C₄H₆
• We know that, the general formula for alkenes is C_nH_2n
• We also know that, the corresponding alkyne has two H atoms less. So the general formula for alkenes is C_nH_2n-2
3. Ethyne is the IUPAC name of the first member. It’s common name is acetylene.
• Acetylene is used for arc welding. In this process, a mixture of acetylene and oxygen is subjected to combustion. As a result, a flame with high temperature and heat is produced. This flame can be used for fusing metal parts together.

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Structure of triple bond

Alkynes have at least one triple bond. We have seen the details about that triple bond in the previous chapters. [see fig.4.156 of section 4.27 ] Let us recall those details. They can be written in 6 steps:
1. The triple bond consists of:
    ♦ One sigma (σ) bond
    ♦ Two pi (𝜋) bonds.
2. The sigma bond is formed by the head-on overlapping of the sp hybridized orbitals.
    ♦ The pi bonds are formed by the sideways overlapping of the 2p orbitals.
3. The sigma bond is a strong bond.
    ♦ Bond enthalpy of a sigma bond is 397 kJ mol^-1
• Pi bond is a weak bond.
    ♦ Bond enthalpy of a pi bond is 284 kJ mol^-1
4. The triple bond is shorter in bond length.
    ♦ The C-C single bond in alkanes has a bond length of 154 pm.
    ♦ The C=C double bond in alkenes has a bond length of 134 pm.
    ♦ The C☰C triple bond in alkynes has a bond length of 120 pm.

5. The triple bond has greater strength.
    ♦ The C-C single bond has a bond enthalpy of 348 kJ mol^-1.
    ♦ The C=C double bond has a bond enthalpy of 681 kJ mol^-1.
    ♦ The C☰C triple bond has a bond enthalpy of 823 kJ mol^-1.
6. We have seen that, ethyne has a linear structure. The electron clouds between the two C atoms are cylindrically symmetrical around the internuclear axis.

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Nomenclature of Alkynes

• We have seen the rules for writing the IUPAC names in an earlier chapter [see section 12.3]
• Let us recall some basic rules which are applicable to alkynes. It can be written in 3 steps:
1. The longest chain should be selected in such a way that, it contains the triple bonds.
2. Numbering must be done in such a way that, the triple bonds get the lowest possible numbers.
3. The suffix ‘yne’ is used instead of ‘ane’ of alkanes.
• Let us see some examples:

Example 1:
Write the IUPAC name of the structure: CH₃-C☰CH
Solution:
1. Applying rule 1, we see that, there are three C atoms in the main chain.
2. Applying rule 2, we see that, ‘prop’ must be used.
3. Applying rule 3, we get propane.
• But since it is an alkyne, we must remove ‘ane’ and replace it with ‘yne’.
• We get: propane – ane + yne = propyne
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that the correct way of numbering is from right to left.
    ♦ Numbering from right to left will give number ‘1’ to the triple bond.
    ♦ Numbering from left to right will give number ‘2’ to the triple bond.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
Prop-1-yne.
7. Here we note an interesting point:
• In propyne, the triple bond has only two possible positions. Both those positions will give the same name: Prop-1-yne. So we can write the name simply as: Propyne.

Example 2:
Write the IUPAC name of the structure: CH₃-CH₂-C☰CH
Solution:
1. Applying rule 1, we see that, there are four C atoms in the main chain.
2. Applying rule 2, we see that, ‘but’ must be used.
3. Applying rule 3, we get butane.
• But since it is an alkyne, we must remove ‘ane’ and replace it with ‘yne’.
• We get: butane – ane + yne = butyne
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that the correct way of numbering is from right to left.
    ♦ Numbering from right to left will give number ‘1’ to the triple bond.
    ♦ Numbering from left to right will give number ‘3’ to the triple bond.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
But-1-yne.

Example 3:
Write the IUPAC name of the structure: CH₃-CH☰CH-CH₃
Solution:
1. Applying rule 1, we see that, there are four C atoms in the main chain.
2. Applying rule 2, we see that, ‘but’ must be used.
3. Applying rule 3, we get butane.
• But since it is an alkyne, we must remove ‘ane’ and replace it with ‘yne’.
• We get: butane – ane + yne = butyne
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that numbering can be done either from left to right or from right to left.
    ♦ Numbering from left to right will give number ‘2’ to the triple bond.
    ♦ Numbering from right to left will give number ‘2’ to the triple bond.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
But-2-yne.

Example 4:
Write the IUPAC name of the structure: CH☰C-C☰CH
Solution:
1. Applying rule 1, we see that, there are four C atoms in the main chain.
2. Applying rule 2, we see that, ‘but’ must be used.
3. Applying rule 3, we get butane.
• But since it is an alkyne, we must remove ‘ane’ and replace it with ‘yne’.
• We get: butane – ane + yne = butyne
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that numbering can be done either from left to right or from right to left.
    ♦ Numbering from left to right will give numbers '1,3' to the triple bonds.
    ♦ Numbering from right to left will give number ‘1,3’ to the triple bonds.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
Buta-1,3-diyne.
• Note that, since more than one triple bond is present, we write 'buta' instead of 'but'
• Similarly, 'penta' is written instead of 'pent', 'hexa' is written instead of hex', so on . . .

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Isomerism in Alkynes

Some basics can be written in 6 steps:
1. Consider the three structures in fig.13.86 below:

Fig.13.86

   ♦ Structure (a) is Pent-1-yne.
   ♦ Structure (b) is Pent-2-yne.
   ♦ Structure (c) is 3-Methylbut-1-yne.
2. Let us compare the number of atoms:
   ♦ In (a), there are five C atoms and eight H atoms.  
   ♦ In (b) also, there are five C atoms and eight H atoms.
   ♦ In (c) also, there are five C atoms and eight H atoms.
• So all the three structures have the same molecular formula C₅H₈
• It is clear that pent-1-yne, pent-2-yne and 3-Methylbut-1-yne are isomers.
3. We have discussed about isomers in a previous section [see section 12.9]. Also we have discussed about the various isomers among alkanes and alkenes.
• Based on those discussions, we can write:
   ♦ Structures (a) and (c) in fig.13.86, are chain isomers.     
   ♦ Structures (b) and (c) in fig.13.86, are chain isomers.
   ♦ Structures (a) and (b) in fig.13.86, are position isomers.     
(Recall that position isomers have the same carbon chain. But the positions of the functional groups will be different. In our present case, the functional group is the alkyne group)
4. In the case of alkanes, we have seen that, only those which have more than three C atoms will exhibit isomerism.
• In the same way, the first two alkynes (ethyne and propyne) do not exhibit isomerism.
• This is because, a minimum of four C atoms are required to obtain different structural arrangements.
5. We have seen that the first two members (ethyne and propyne) of the alkyne series do not exhibit isomerism.
• In examples 2 and 3 above, we have seen the isomers of the third member butyne.
• In fig.13.86 above, we have seen the isomers of the fourth member pentyne. Next let us see the isomers of the fifth member.
• It is explained as as solved example below:

Solved example 13.13
Write all the structures and IUPAC names of the structural isomers of alkynes corresponding to C₆H₁₀.
Solution:
Seven isomers are shown in fig.13.87 below:

Fig.13.87

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In the next section we will see preparation of alkynes.

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Chapter 13.8 - Nomenclature of Alkenes

In the previous section, we completed a discussion on conformations of ethane. In this section, we will start a discussion on alkenes.

Let us recall some properties of alkenes that we have seen in earlier chapters. They can be written in 4 steps:
1. Alkenes are unsaturated hydrocarbons. They contain at least one double bond.
2. If there is one double bond in an alkene, it will contain two H atoms less than the corresponding alkane.
• For example:
    ♦ Molecular formula of butane is C₄H₁₀
    ♦ Molecular formula of butene is C₄H₈
• We know that, the general formula for alkanes is C_nH_2n+2
• We also know that, the corresponding alkene has two H atoms less. So the general formula for alkenes is C_nH_2n
3. Chemists in the 19^th century noticed that, the first member of the alkene series (ethene) formed a oily liquid on reaction with chlorine. So the alkenes were generally known as olefins. The word olefin means ‘oil forming’.
4. Ethene is the IUPAC name of the first member. It’s common name is ethylene.

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Structure of double bond

Alkenes have at least one double bond. We have seen the details about that double bond in the previous chapters. [see fig.4.147 of section 4.26 ] Let us recall those details. They can be written in 8 steps:
1. The double bond consists of:
    ♦ One sigma (σ) bond
    ♦ One pi (𝜋) bond.
2. The sigma bond is formed by the head-on overlapping of the sp² hybridized orbitals.
    ♦ The pi bond is formed by the sideways overlapping of the two 2p orbitals.
3. The sigma bond is a strong bond.
    ♦ Bond enthalpy of a sigma bond is 397 kJ mol^-1
• Pi bond is a weak bond.
    ♦ Bond enthalpy of a pi bond is 284 kJ mol^-1
4. The double bond is shorter in bond length.
    ♦ The C-C single bond in alkanes has a bond length of 154 pm.
    ♦ The C=C double bond in alkenes has a bond length of 134 pm.
    ♦ The C-H single bond has a bond length of 110 pm.
• The bond angle between C-H single bond and C=C double bond is 121.7^o
• The bond angle between the two C-H single bonds in a CH₂ group is 116.6^o
• These details are shown in fig.13.46 below:

Fig.13.46

5. In a pi bond, there is sideways overlapping of the p-orbitals. The two p-orbitals combine together to form a 𝜋-cloud. We saw this in figs.4.148 and 4.149 of section 4.26.
6. The electrons in the 𝜋-cloud are loosely held.
• Those loosely held electrons are easily available for reagents which are in search for electrons (reagents in search for electrons are known as electrophilic reagents).
• So electrophilic reagents attack alkenes easily.
7. We can say that:
• Alkenes have lesser stability when compared to alkanes.
• Alkenes can be easily converted into alkanes by combining with electrophilic reagents.
8. In step (3), we wrote that:
    ♦ Bond enthalpy of the sigma bond in a double bond is 397 kJ mol^-1
    ♦ Bond enthalpy of the pi bond in a double bond is 284 kJ mol^-1
• So the bond enthalpy of a C=C double bond as a whole will be (397+284) = 681 kJ mol^-1
• The bond enthalpy of a C-C single bond is 348 kJ mol^-1

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Nomenclature of Alkenes

• We have seen the rules for writing the IUPAC names in an earlier chapter [see section 12.3]
• Let us recall some basic rules which are applicable to alkenes. It can be written in 3 steps:
1. The longest chain should be selected in such a way that, it contains the double bonds.
2. Numbering must be done in such a way that, the double bonds get the lowest possible numbers.
3. The suffix ‘ene’ is used instead of ‘ane’ of alkanes.
• Let us see some examples:

Example 1:
Write the IUPAC name of the structure: CH₃-CH=CH₂
Solution:
1. Applying rule 1, we see that, there are three C atoms in the main chain.
2. Applying rule 2, we see that, ‘prop’ must be used.
3. Applying rule 3, we get propane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: propane – ane + ene = propene
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that the correct way of numbering is from right to left.
    ♦ Numbering from right to left will give number ‘1’ to the double bond.
    ♦ Numbering from left to right will give number ‘2’ to the double bond.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
Prop-1-ene.
7. Here we note an interesting point:
• In propene, the double bond has only two possible positions. Both those positions will give the same name: Prop-1-ene. So we can write the name simply as: Propene.

Example 2:
Write the IUPAC name of the structure: CH₃-CH₂-CH=CH₂
Solution:
1. Applying rule 1, we see that, there are four C atoms in the main chain.
2. Applying rule 2, we see that, ‘but’ must be used.
3. Applying rule 3, we get butane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: butane – ane + ene = butene
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that the correct way of numbering is from right to left.
    ♦ Numbering from right to left will give number ‘1’ to the double bond.
    ♦ Numbering from left to right will give number ‘3’ to the double bond.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
But-1-ene.

Example 3:
Write the IUPAC name of the structure: CH₃-CH=CH-CH₃
Solution:
1. Applying rule 1, we see that, there are four C atoms in the main chain.
2. Applying rule 2, we see that, ‘but’ must be used.
3. Applying rule 3, we get butane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: butane – ane + ene = butene
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that numbering can be done either from left to right or from right to left.
    ♦ Numbering from left to right will give number ‘2’ to the double bond.
    ♦ Numbering from right to left will give number ‘2’ to the double bond.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
But-2-ene.

Example 4:
Write the IUPAC name of the structure: CH₂=CH-CH=CH₂
Solution:
1. Applying rule 1, we see that, there are four C atoms in the main chain.
2. Applying rule 2, we see that, ‘but’ must be used.
3. Applying rule 3, we get butane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: butane – ane + ene = butene
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that numbering can be done either from left to right or from right to left.
    ♦ Numbering from left to right will give numbers '1,3' to the double bonds.
    ♦ Numbering from right to left will give number ‘1,3’ to the double bonds.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
Buta-1,3-diene.
• Note that, since more than one double bond is present, we write 'buta' instead of 'but'
• Similarly, 'penta' is written instead of 'pent', 'hexa' is written instead of hex', so on . . .

Example 5:
Write the IUPAC name of the structure shown in fig.13.47(a) below:

Fig.13.47

Solution:
1. Applying rule 1, we see that, there are three C atoms in the main chain containing double bond.
2. Applying rule 2, we see that, ‘prop’ must be used.
3. Applying rule 3, we get propane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: propane – ane + ene = propene
4. Applying rule 4, we see that, there is one branch: methyl.
5. Applying rule 5, we see that numbering should be done from left to right. This is shown in fig.b.
    ♦ Numbering from left to right will give number '1' to the double bond.
    ♦ Numbering from right to left will give number ‘2’ to the double bond.
6. Applying rule 6, we get: 2-methyl.
7. Rule 7 is not applicable here because, there is only one branch.
8. Applying rule 8, we get: 2-Methylprop-1-ene.

Example 6:
Write the IUPAC name of the structure shown in fig.13.48(a) below:

Fig.13.48

Solution:
1. Applying rule 1, we see that, there are four C atoms in the main chain containing the double bond.
2. Applying rule 2, we see that, ‘but’ must be used.
3. Applying rule 3, we get butane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: butane – ane + ene = butene
4. Applying rule 4, we see that, there is one branch: methyl.
5. Applying rule 5, we see that numbering should be done from left to right. This is shown in fig.b.
    ♦ Numbering from left to right will give number '1' to the double bond.
    ♦ Numbering from right to left will give number ‘3’ to the double bond.
6. Applying rule 6, we get: 3-methyl.
7. Rule 7 is not applicable here because, there is only one branch.
8. Applying rule 8, we get: 3-Methylbut-1-ene.

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Now we will see some solved examples:

Solved example 13.7
Write the IUPAC names of the three compounds shown in fig.13.49 below:

Fig.13.49

Solution:
Part (i):
We are given the condensed formula. It is shown in fig.13.50(a) below:

Fig.13.50

• The expanded form is shown in fig.b. Based on the expanded form, we can write 8 steps: 
1. Applying rule 1, we see that, there are ten C atoms in the main chain containing both the double bonds.
2. Applying rule 2, we see that, ‘dec’ must be used.
3. Applying rule 3, we get decane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: decane – ane + ene = decene
4. Applying rule 4, we see that, there are two branches: two methyl branches.
5. Applying rule 5, we see that numbering should be done from left to right. This is shown in fig.b.
    ♦ Numbering from left to right will give numbers '3,6' to the double bonds.
    ♦ Numbering from right to left will give numbers ‘4,7’ to the double bond.
6. Applying rule 6, we get: 2-methyl and 8-methyl.
7. Applying rule 7, we get: 2,8-dimethyl.
8. Applying rule 8, we get: 2,8-Dimethyldeca-3,6-diene.

Part (ii):
We are given the bond line formula. It is shown in fig.13.51(a) below:

Fig.13.51

• The expanded form is shown in fig.b. Based on the expanded form, we can write 6 steps: 
1. Applying rule 1, we see that, there are eight C atoms in the main chain.
2. Applying rule 2, we see that, ‘oct’ must be used.
3. Applying rule 3, we get octane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: octane – ane + ene = octene
4. Applying rule 4, we see that, there are no branches.
5. Applying rule 5, we see that numbering can be done either from left to right or from right to left. This is shown in fig.b.
    ♦ Numbering from left to right will give numbers '1,3,5,7' to the double bonds.
    ♦ Numbering from right to left will give the same numbers ‘1,3,5,7’ to the double bonds.
6. Rules 6 and 7 are not applicable here because, there are no branches.
• So we can apply rule 8 to assemble the IUPAC name. We get:
Octa-1,3,5,7-tetraene.

Part (iii):
We are given the condensed formula. It is shown in fig.13.52(a) below:

Fig.13.52

• The expanded form is shown in fig.b. Based on the expanded form, we can write 8 steps: 
1. Applying rule 1, we see that, there are five C atoms in the main chain containing the double bond.
2. Applying rule 2, we see that, ‘pent’ must be used.
3. Applying rule 3, we get pentane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: pentane – ane + ene = pentene
4. Applying rule 4, we see that, there is one branch: propyl.
5. Applying rule 5, we see that numbering should be done from left to right. This is shown in fig.b.
    ♦ Numbering from left to right will give number '1' to the double bond.
    ♦ Numbering from right to left will give number ‘4’ to the double bond.
6. Applying rule 6, we get: 2-propyl.
7. Rule 7 is not applicable because, there is only one branch.
8. Applying rule 8, we get: 2-Propylpent-1-ene.

Part (iv):
We are given the expanded form. It is shown in fig.13.53(a) below:

Fig.13.53

• Based on the expanded form, we can write 8 steps: 
1. Applying rule 1, we see that, there are ten C atoms in the main chain containing the double bond.
2. Applying rule 2, we see that, ‘dec’ must be used.
3. Applying rule 3, we get decane.
• But since it is an alkene, we must remove ‘ane’ and replace it with ‘ene’.
• We get: decane – ane + ene = decene
4. Applying rule 4, we see that, there are three branches: two methyl branches and one ethyl branch.
5. Applying rule 5, we see that numbering should be done from right to left. This is shown in fig.b.
    ♦ Numbering from left to right will give number '6' to the double bond.
    ♦ Numbering from right to left will give number ‘4’ to the double bond.
6. Applying rule 6, we get: 2-methyl, 4-ethyl and 6-methyl.
7. Applying rule 7, we get: 4-ethyl-2,6-dimethyl.
(Note that, ethyl gets preference over methyl when we consider the alphabetical order)
8. Applying rule 8, we get: 4-Ethyl-2,6-dimethyldec-4-ene.

Solved example 13.8
Calculate the number of sigma (𝜎) and pi (π) bonds in the four structures in the above solved example 13.7
Solution:
Part (a):
1. From the expanded form in fig.13.50(b), we can count the number of bonds:
   ♦ Counting the number of single bonds, we get: 31 Nos.
        ✰ Number of C-H single bonds = 22 Nos. 
        ✰ Number of C-C single bonds = 9 Nos. 
   ♦ Counting the number of double bonds, we get: 2 Nos. 
        ✰ Number of C=C double bonds = 2 Nos.
2. Number of σ bonds can be calculated in 3 steps:
(i) We know that, all single bonds will be σ bonds.
• So we can write:
    ♦ Number of $\sigma_{C-H}$ = 22
    ♦ Number of $\sigma_{C-C}$ = 9
(ii) We know that, all double bonds will be have one σ bond each.
• The double bond is between two C atoms. In our present case, there are two double bonds.
• So we can write:
    ♦ Number of $\sigma_{C=C}$ = 2
(iii) Now we can write the total numbers:
• Total number of σ bonds:
    ♦ $\sigma_{C-H}$ =22
    ♦ $\sigma_{C-C}$ = 9
    ♦ $\sigma_{C=C}$ = 2
3. Number of π bonds can be calculated in 2 steps:
(i) We know that, all double bonds will be have one π bond each.
• We see that the two double bonds in this structure are between C atoms.
    ♦ So the number of $\pi_{C=C}$ from the two double bonds = 2
(ii) Now we can write the total numbers:
• Total number of π bonds:
    ♦ $\pi_{C=C}$ = 2

Part (b):
1. From the expanded form in fig.13.51(b), we can count the number of bonds:
   ♦ Counting the number of single bonds, we get: 13 Nos.
        ✰ Number of C-H single bonds = 10 Nos. 
        ✰ Number of C-C single bonds = 3 Nos. 
   ♦ Counting the number of double bonds, we get: 4 Nos. 
        ✰ Number of C=C double bonds = 4 Nos.
2. Number of σ bonds can be calculated in 3 steps:
(i) We know that, all single bonds will be σ bonds.
• So we can write:
    ♦ Number of $\sigma_{C-H}$ = 10
    ♦ Number of $\sigma_{C-C}$ = 3
(ii) We know that, all double bonds will be have one σ bond each.
• The double bond is between two C atoms. In our present case, there are four double bonds.
• So we can write:
    ♦ Number of $\sigma_{C=C}$ = 4
(iii) Now we can write the total numbers:
• Total number of σ bonds:
    ♦ $\sigma_{C-H}$ =10
    ♦ $\sigma_{C-C}$ = 3
    ♦ $\sigma_{C=C}$ = 4
3. Number of π bonds can be calculated in 3 steps:
(i) We know that, all double bonds will be have one π bond each.
• We see that the four double bonds in this structure are between C atoms.
    ♦ So the number of $\pi_{C=C}$ from the four double bonds = 4
(ii) Now we can write the total numbers:
• Total number of π bonds:
    ♦ $\pi_{C=C}$ = 4

Part (c):
1. From the expanded form in fig.13.52(b), we can count the number of bonds:
   ♦ Counting the number of single bonds, we get: 22 Nos.
        ✰ Number of C-H single bonds = 16 Nos. 
        ✰ Number of C-C single bonds = 6 Nos. 
   ♦ Counting the number of double bonds, we get: 1 Nos. 
        ✰ Number of C=C double bonds = 1 No.
2. Number of σ bonds can be calculated in 3 steps:
(i) We know that, all single bonds will be σ bonds.
• So we can write:
    ♦ Number of $\sigma_{C-H}$ = 16
    ♦ Number of $\sigma_{C-C}$ = 6
(ii) We know that, all double bonds will be have one σ bond each.
• The double bond is between two C atoms. In our present case, there is one double bond.
• So we can write:
    ♦ Number of $\sigma_{C=C}$ = 1
(iii) Now we can write the total numbers:
• Total number of σ bonds:
    ♦ $\sigma_{C-H}$ =16
    ♦ $\sigma_{C-C}$ = 6
    ♦ $\sigma_{C=C}$ = 1
3. Number of π bonds can be calculated in 3 steps:
(i) We know that, all double bonds will be have one π bond each.
• We see that the one double bond in this structure is between C atoms.
    ♦ So the number of $\pi_{C=C}$ from the one double bonds = 1
(ii) Now we can write the total numbers:
• Total number of π bonds:
    ♦ $\pi_{C=C}$ = 1

Part (d):
1. From the expanded form in fig.13.53, we can count the number of bonds:
   ♦ Counting the number of single bonds, we get: 13 Nos.
        ✰ Number of C-H single bonds = 28 Nos. 
        ✰ Number of C-C single bonds = 12 Nos. 
   ♦ Counting the number of double bonds, we get: 1 No. 
        ✰ Number of C=C double bonds = 1 No.
2. Number of σ bonds can be calculated in 3 steps:
(i) We know that, all single bonds will be σ bonds.
• So we can write:
    ♦ Number of $\sigma_{C-H}$ = 28
    ♦ Number of $\sigma_{C-C}$ = 12
(ii) We know that, all double bonds will be have one σ bond each.
• The double bond is between two C atoms. In our present case, there is one double bond.
• So we can write:
    ♦ Number of $\sigma_{C=C}$ = 1
(iii) Now we can write the total numbers:
• Total number of σ bonds:
    ♦ $\sigma_{C-H}$ =28
    ♦ $\sigma_{C-C}$ = 12
    ♦ $\sigma_{C=C}$ = 1
3. Number of π bonds can be calculated in 3 steps:
(i) We know that, all double bonds will be have one π bond each.
• We see that the one double bond in this structure is between C atoms.
    ♦ So the number of $\pi_{C=C}$ from the one double bonds = 1
(ii) Now we can write the total numbers:
• Total number of π bonds:
    ♦ $\pi_{C=C}$ = 1

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In the next section we will see isomerism in alkenes.

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Chapter 13.2 - Problems in Nomenclature of Alkanes

In the previous section, we saw the nomenclature and isomerism in alkanes. In this section, we will see some advanced examples.

Example 1:
Write the IUPAC name of the structure shown in fig.13.13 below:

Fig.13.13

Solution:
• We have seen the 8 rules in section 12.3
1. Applying rule 1, we see that, there are six C atoms in the main chain.
2. Applying rule 2, we see that, 'hex' must be used.
3. Applying rule 3, we get hexane.
4. Applying rule 4, we see that, there are two branches: methyl and ethyl
5. Applying rule 5, we see that, the correct way of numbering is indeed from left to right as shown in fig.13.13(a).
• If we number the C atoms from left to right as in fig.b, the branches will get the numbers 2,4. The sum is (2+4) = 6.
• If we number the C atoms from right to left, the branches will get the numbers 3,5. The sum is (3+5) = 8, which is larger and hence not acceptable.
6. Applying rule 6, we get: 2-methyl and 4-ethyl
7. Applying rule 7, we get: 4-ethyl-2-methyl
• This is because, 'e' comes before 'm' in the alphabetical listing.
8. Applying rule 8, we get: 4-Ethyl-2-methylhexane.

Example 2:
Write the IUPAC name of the structure shown in fig.13.14 below:

Fig.13.14

Solution:
In this problem, there are branches within a branch. So we must be ready to apply the rules that we saw in section 12.4 also. Fortunately, there is a common name available for that branch. So we can first name the main carbon chain.
1. Applying rule 1, we see that, there are eight C atoms in the main chain.
2. Applying rule 2, we see that, 'oct' must be used.
3. Applying rule 3, we get octane.
4. Applying rule 4, we see that, there are four branches: two ethyl branches, one methyl branch and one isopropyl branch.
5. Applying rule 5, we see that, the correct way of numbering is from right to left as shown in fig.13.14(b).
• If we number the C atoms from right to left, the branches will get the numbers 3,3,4,5. The sum is (3+3+4+5) = 15.
• If we number the C atoms from left to right (fig.a), the branches will get the numbers 4,5,6,6. The sum is (4+5+6+6) = 21, which is larger and hence not acceptable.
6. Applying rule 6, we get: 3-ethyl, 3-ethyl, 4-methyl and 5-isopropyl
7. Applying rule 7, 7A and 7B, we get: 3,3-diethyl-5-isopropyl-4-methyl
• This can be explained in 2 steps:
(i) 'e' comes before 'i' and 'm' in the alphabetical listing.
'di' is not considered as part of the name. So 'd' should not be considered for alphabetical listing.
(ii) 'i' comes before 'm' in the alphabetical listing.
'iso' is considered as part of the name.
8. Applying rule 8, we get: 3,3-diethyl-5-isopropyl-4-methyloctane.

Example 3:
Write the IUPAC name of the structure shown in fig.13.15 below:

Fig.13.15

Solution:
In this problem, there are branches within a branches. So we must be ready to apply the rules that we saw in section 12.4 also. Fortunately, there are common names available for those branches. So we can first name the main carbon chain.
1. Applying rule 1, we see that, there are ten C atoms in the main chain.
2. Applying rule 2, we see that, 'dec' must be used.
3. Applying rule 3, we get decane.
4. Applying rule 4, we see that, there are two branches: one isopropyl branch and and one sec-butyl branch.
5. Applying rule 5, we see that, the correct way of numbering is from left to right as shown in fig.13.15(a).
• If we number the C atoms from left to right, the branches will get the numbers 4,5. The sum is (4+5) = 9.
• If we number the C atoms from right to left (fig.b), the branches will get the numbers 6,7. The sum is (6+7) = 13, which is larger and hence not acceptable.
6. Applying rule 6, we get: 4-isopropyl and 5-sec-butyl
7. Applying rule 7, 7A and 7B, we get: 5-sec-butyl-4-isopropyl
• This can be explained in 3 steps:
(i) 'b' comes before 'i' in the alphabetical listing.
(ii) 'sec' is not considered as part of the name. So 's' should not be considered for alphabetical listing.
(iii) 'iso' is considered as part of the name.
8. Applying rule 8, we get: 5-sec-butyl-4-isopropyl decane.

Example 4:
Write the IUPAC name of the structure shown in fig.13.16 below:

Fig.13.16

Solution:
In this problem, there are branches within a branch. So we must be ready to apply the rules that we saw in section 12.4 also. The branch do not have any common name. So we first name the branch.
1. Applying rule 1, we see that, there are three C atoms.
2. Applying rule 2, we see that, 'prop' must be used.
3. Applying rule 3, we get propane.
4. Applying rule 4, we see that, there are two branches: two methyl groups.
5. Applying rule 5, we see that, the correct way of numbering is from the main branch as shown in fig.13.16(a).
6. Applying rule 6, we get: 2-methyl and 2-methyl
7. Applying rule 7, we get: 2,2-dimethyl
8. Applying rule 8, we get: 2,2-dimethylpropyl.
• This name must be written within parenthesis.

Now we can name the main chain.
1. Applying rule 1, we see that, there are nine C atoms.
2. Applying rule 2, we see that, 'non' must be used.
3. Applying rule 3, we get nonane.
4. Applying rule 4, we see that, there is only one branch.
5. Applying rule 5, we see that, numbering can be done in both ways. Both will give the number 5.
6. Applying rule 6 we get: 5-(2,2-dimethylpropyl)
7. Applying rule 7 is not required because there is only one branch.
8. Applying rule 8, we get: 5-(2,2-dimethylpropyl)nonane.

Example 5:
Write the IUPAC name of the structure shown in fig.13.17 below:

Fig.13.17

Solution:
• We have seen the 8 rules in section 12.3
1. Applying rule 1, we see that, there are seven C atoms in the main chain.
2. Applying rule 2, we see that, 'hept' must be used.
3. Applying rule 3, we get heptane.
4. Applying rule 4, we see that, there are two branches: methyl and ethyl
5. Applying rule 5:
   ♦ We see that, the numbering can be done in both ways.
   ♦ In both fig. a and b, the numbers are 3 and 5.
• So we apply rule 5A:
If two branches are present in equivalent positions, then the lower number should be given to that branch which comes first in the alphabetical order.
   ♦ Here, ethyl comes first in the alphabetical order.
   ♦ Numbering in fig.a is correct.
6. Applying rule 6, we get: 3-ethyl and 5-methyl
7. Applying rule 7, we get: 3-ethyl-5-methyl
• This is because, 'e' comes before 'm' in the alphabetical listing.
8. Applying rule 8, we get: 3-Ethyl-5-methylheptane.

Solved example 13.3
Write the IUPAC names of the following compounds:
(i) (CH₃)₃CCH₂C(CH₃)₃
(ii) (CH₃)₂C(C₂H₅)₂
(iii) tetra-tert-butylmethane
Solution:
Part (i):
• We are given the condensed formula. Based on the condensed formula, we can draw the structural formula. It is shown in fig.13.18 below:

Fig.13.18

• We have seen the 8 rules in section 12.3
1. Applying rule 1, we see that, there are five C atoms in the main chain.
2. Applying rule 2, we see that, 'pent' must be used.
3. Applying rule 3, we get pentane.
4. Applying rule 4, we see that, there are four branches: four methyl branches.
5. Applying rule 5:
   ♦ We see that, the numbering can be done in both ways.
   ♦ In both fig. a and b, the numbers are 2 and 4.
6. Applying rule 6, we get: 2-methy, 2-methyl, 4-methy, 4-methyl
7. Applying rule 7, we get: 2,2,4,4-tetramethyl
8. Applying rule 8, we get: 2,2,4,4-Tetramethylpentane.

Part (ii):
• We are given the condensed formula. Based on the condensed formula, we can draw the structural formula. It is shown in fig.13.19 below:

Fig.13.19

• We have seen the 8 rules in section 12.3
1. Applying rule 1, we see that, there are five C atoms in the main chain.
2. Applying rule 2, we see that, 'pent' must be used.
3. Applying rule 3, we get pentane.
4. Applying rule 4, we see that, there are two branches: two methyl branches.
5. Applying rule 5:
   ♦ We see that, the numbering can be done in both ways.
   ♦ In both fig. a and b, the number is 3.
6. Applying rule 6, we get: 3-methy, 3-methyl
7. Applying rule 7, we get: 2,2-dimethyl
8. Applying rule 8, we get: 2,2-Dimethylpentane.

Part (iii):
• We have seen the 8 rules in section 12.3
• In this problem, there are branches within a branch. So we must be ready to apply the rules that we saw in section 12.4 also.
• We are given the common name tetra-tert-butylmethane.
   ♦ This is similar to the common name: tetra-chloromethane.
• In tetrachloromethane, the four H atoms of methane are replaced by four Cl atoms.
• In the same way, in tetra-tert-butylmethane, the four H atoms of methane are replaced by four tert-butyl groups.
• We saw the structure of tert-butyl in an earlier section [see fig.12.33 in section section 12.4]. It is shown again in fig.13.20(a) below:

Fig.13.20

• Based on the structure of tert-butyl, the structure of tetra-tert-butylmethane will be as shown in fig.b. Now we can write the IUPAC name.
1. Applying rule 1, we see that, there are five C atoms in the main chain.
2. Applying rule 2, we see that, 'pent' must be used.
3. Applying rule 3, we get pentane.
4. Applying rule 4, we see that, there are six branches: four methyl branches and two tert-butyl branches.
5. Applying rule 5:
   ♦ We see that, the numbering can be done in two ways.
   ♦ In both fig. b and c, the numbers are 2, 3 and 4.
6. Applying rule 6, we get: 2-methy, 2-methyl, 4-methy, 4-methyl, 3-tert-butyl, 3-tert-butyl
7. Applying rule 7, we get: 3,3-di-tert-butyl-2,2,4,4-tetramethyl
• Remember that, 'tetra' is not considered as part of name. So 't' cannot be considered in the alphabetical listing.
8. Applying rule 8, we get: 3,3-Di-tert-butyl-2,2,4,4-tetramethylpentane.

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• In the above discussion, we were given the structures of hydrocarbons. We wrote the corresponding IUPAC names.
• We must be able to do the reverse also. That is., we will be given the IUPAC names. We must draw the corresponding structures.
• Let us see an example:
Draw the structure of 3-Ethyl-2,2-dimethylpentane
Solution:
1. In the given name, we have ‘pent’ as the root. Then there will be five C atoms in the longest chain. So we first draw a chain of five C atoms. This is shown in fig.13.21(a) below:

Fig.13.21

2. Next we number the C atoms from 1 to 5. This is shown in fig.b
3. Attaching the branches:
‘3-Ethyl’ indicates that, there is an ethyl group at the C atom number 3
So we draw an ethyl group at that C atom.
‘2,2-dimethyl’ indicates that, there are two methyl groups at the C atom number 2
So we draw two methyl groups at that C atom
This is shown in fig.c
4. Now we put the required number of H atoms to satisfy the valencies of the C atoms.
Thus we get the final structure shown in fig.d

Solved example 13.4
Write the structural formulas for the following compounds:
(i) 3,4,4,5-Tetramethylheptane
(ii) 2,5-dimethylhexane
Solution:
Part (i):
1. In the given name, we have ‘hept’ as the root. Then there will be seven C atoms in the longest chain. So we first draw a chain of seven C atoms. This is shown in fig.13.22(a) below:

Fig.13.22

2. Next we number the C atoms from 1 to 7. This is shown in fig.b
3. Attaching the branches:
‘3,4,4,5-tetramethyl’ indicates that, there are four methyl groups at the three C atoms as written below:
   ♦ One methyl group at C atom number 3
   ♦ Two methyl groups at C atom number 4
   ♦ One methyl group at C atom number 5
• This is shown in fig.c
4. Now we put the required number of H atoms to satisfy the valencies of the C atoms.
• Thus we get the final structure shown in fig.d

Part (ii):
1. In the given name, we have ‘hex’ as the root. Then there will be six C atoms in the longest chain. So we first draw a chain of six C atoms. This is shown in fig.13.23(a) below.

Fig.13.23

2. Next we number the C atoms from 1 to 6. This is shown in fig.b
3. Attaching the branches:
‘2,5-dimethyl’ indicates that, there are two methyl groups at the two C atoms as written below:
   ♦ One methyl group at C atom number 2
   ♦ One methyl group at C atom number 5
• This is shown in fig.c
4. Now we put the required number of H atoms to satisfy the valencies of the C atoms.
• Thus we get the final structure shown in fig.d

Solved example 13.5
Write structures for each of the following compounds. Why are the given names incorrect. Write the correct IUPAC names.
(i) 2-Ethylpentane
(ii) 5-Ethyl-3-methylheptane
Solution:
Part (i):
1. In the given name, we have ‘pent’ as the root. Then there will be five C atoms in the longest chain. So we first draw a chain of five C atoms. This is shown in fig.13.24(a) below:

Fig.13.24

2. Next we number the C atoms from 1 to 5. This is shown in fig.b
3. Attaching the branches:
‘2-ethyl indicates that, there is an ethyl group at the C atom number 2.
• This is shown in fig.c
4. Now we put the required number of H atoms to satisfy the valencies of the C atoms.
• Thus we get the final structure shown in fig.d
5. For this structure, the name 3-Ethylpentane is wrong. The correct name can be obtained in 4 steps:
(i) For this structure, the C atoms should be numbered as shown in fig.e
(ii) Now we see that, there are six C atoms in the longest chain.
(iii) We also see that, the branch is methyl, not ethyl. The branch is at the third C atom.
(iv) So the correct IUPAC name is: 3-Methylhexane.

Part (ii):
1. In the given name, we have ‘hept’ as the root. Then there will be seven C atoms in the longest chain. So we first draw a chain of seven C atoms. This is shown in fig.13.25(a) below:

Fig.13.25

2. Next we number the C atoms from 1 to 7. This is shown in fig.b
3. Attaching the branches:
   ♦ ‘5-ethyl' indicates that, there is an ethyl group at the C atom number 5.
   ♦ ‘3-methyl' indicates that, there is a methyl group at the C atom number 3.
• This is shown in fig.c
4. Now we put the required number of H atoms to satisfy the valencies of the C atoms.
• Thus we get the final structure shown in fig.d
5. For this structure, the name 5-Ethyl-3-methylheptane is wrong. The correct name can be obtained in 4 steps:
(i) For this structure, the C atoms can be numbered as shown in fig.e also
(ii) Both methods of numbering will give the same numbers 3 and 5.
• That means, the branches are at equivalent positions.
(iii) So we must apply rule 5A. [see section 12.3]
Then 'ethyl' gets the lower number because, it comes first in the alphabetical order.
(iv) So the numbering in fig.e is the correct method.
• Based on this numbering, the correct IUPAC name is: 3-Ethyl-5-methylheptane.

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In the next section we will see preparation of alkanes.

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