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.
♦ 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
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:
$\rm{CH_2 = CHBr~ \color {green}{\xrightarrow[{-NaBr~and~-NH_3}]{Na^{+} {NH_2}^{-}}} ~ CH ≡ CH}$
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.
Chemical properties of alkynes
A. Acidic character
• 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.
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:
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.
$\rm{HCl~+~Na~ \color {green}{\xrightarrow[{}]{}} ~ Na^{+} Cl^{-}~+~\frac{1}{2}H_2}$
• 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.
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.
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.
•
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.
•
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.
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.
In the next section we will see addition of water and polymerisation.
Copyright©2022 Higher secondary chemistry.blogspot.com
No comments:
Post a Comment