In the previous section, we completed a discussion on the preparation of alkenes. In this section, we will see properties of alkenes.
First we will see the physical properties. This can be written in 6 steps:
1. We have already seen the physical properties of alkanes [see section 13.4].
• The physical properties of alkenes and alkanes are similar, except in isomerism and polar nature. This can be explained in 2 steps:
(i) Polar nature:
• Earlier in section 13.4, we saw that alkanes are almost non-polar.
• In section 13.9, we saw that cis-isomers of alkenes show polar nature [see fig.13.58 of section 13.9].
(ii) Isomerism:
• Earlier in section 13.7, we saw that, alkanes can have different conformations.
• But since rotation is restricted in alkenes, they cannot have different conformations. Instead, they show cis/trans isomerism.
2. The first three members of the alkene series are gases.
• The next fourteen members are liquids.
• The members coming after that, are solids.
3. The first member ethene is colorless. But it has a faint sweet smell.
• All other members are colorless and odour less.
4. Alkenes are insoluble in water. But they are soluble in non-polar solvents like benzene and petroleum ether.
(Petroleum ether is obtained from petroleum. It is used as a laboratory solvent)
5. The members of the alkene series show a regular increase in boiling point.
• We know that, for each successive member, the size increases by one CH2 unit.
• For each successive member, the boiling point increases by 20-30 K
6. Difference in boiling points of isomers:
• Consider two isomers of an alkene. Let one of them be of straight chain type and the other branched chain type.
• The straight chain type will have a higher boiling point than the branched chain type.
• We saw the reason in the case of alkanes. The same reason is applicable here also.
Now we will see the chemical properties. Some basics can be written in 2 steps:
1. We know that, each double bond in alkenes consist of a 𝞹-bond. 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 alkenes, resulting in new compounds. We call such reactions as addition reactions. In this section, we will see different types of addition reactions.
A. Addition of dihydrogen
This can be written in 6 steps:
1. Each double bond in an alkene can add one molecule of dihydrogen.
2. The dihydrogen molecule first splits into two H atoms. This is indicated by the blue dashed curves in fig.13.67 below:
 |
| Fig.13.67 |
• 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-bond will supply the required electrons for the H atoms.
• That means, two of the four red dots in the double bond, will leave the double 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 on the sides of the alkane.
4. So we now know how the H atoms add up to the alkene. The alkene then will no longer require the double bond. It will be converted to an alkane.
5. In an earlier section, we saw this process as a method for preparing alkanes [see section 13.3].
6. For this process, finely divided nickel, palladium or platinum is required as catalyst.
B. Addition of halogens
This can be written in 6 steps:
1. Each double bond in an alkene can add one molecule of halogen.
2.
Two individual X atoms (F, Cl, Br or I) must be available for the reaction to take place. This is indicated by two Br atoms in fig.13.68 below:
 |
| Fig.13.68 |
• 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 four red dots in the double bond, will leave
the double bond. Those two red dots will help the new X atoms to form
single bonds with C atoms.
• Thus we get two new C-Br bonds on the sides to form 1,2-Dibromoethane.
4. So we now know how the X atoms add up to the alkene. The alkene then will no longer require the
double bond. It will be converted to a dihalide.
5. Note that, two X atoms will be required to convert a double bond to two single bonds. Obviously, the two X atoms will be attached to two adjacent C atoms. Thus the resulting dihalide will be a vicinal dihalide.
6. The equation of the reaction in fig.13.68 above can be written as:
$\rm{CH_2 = CH_2~+~Br-Br~ \color {green}{\xrightarrow[{}]{CCl_4}} ~ CH_2 Br- CH_2 Br}$
7. Another example is shown in fig.13.69 below:
 |
| Fig.13.69 |
• The equation is:
$\rm{CH_3 - CH=CH_2~+~Cl-Cl~ \color {green}{\xrightarrow[{}]{}} ~ CH_3 - CHCl - CH_2 Cl}$
4. Under normal conditions, iodine does not take part in this type of reactions.
5. 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 double bonds, those double bonds will break. Each double bond will take up two 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, double bonds or triple bonds are present.
6. This type of reaction involves the formation of cyclic halonium ions. We will see those details in higher classes.
C. Addition of hydrogen halides
• In this reaction, the hydrogen halide first splits into two ions.
♦ For example, HBr splits into H+ and Br-
• An intermediate product is formed due to the action of the H+ ion.
• So we will see the reaction mechanism in detail. It can be written in 4 steps:
1. First the HBr molecule splits into two parts.
• Since Br is more electronegative, it will retain both electrons in the bond.
• So it is a heterolytic cleavage. It is shown in fig.13.70 (a) below:
 |
| Fig.13.70 |
• Note that, H has lost both the electrons and thus became H+ ion.
• The Br has acquired an extra electron. It is indicated by the gray dot. It has become Br- ion.
2. The H+ attacks the ethene molecule. This is shown in fig.13.70 (b) above.
• At the product side, we see that, the H+ is attached to the left side 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.
3. 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, the right side C atom looses an electron.
• This C atom now has sextet only. Thus a carbocation is formed.
• In our present case, it is the ethyl carbocation. We have seen the details about carbocation in a previous section [see fig.12.69 in section 12.10].
4. The Br- now attacks the newly formed carbocation. This is shown in fig.13.70 (c) above.
• The Br- can donate two electrons to form a bond. So it attaches to the right side C atom. Thus we get a molecule of bromoethane.
Let us see another example. It can be written in 4 steps:
1. First the HBr molecule splits into two parts. We get H+ and Br-. It is shown in fig.13.71 (a) below:
 |
| Fig.13.71 |
2. The H+ attacks the alkene molecule. This is shown in fig.13.71 (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 double bond. That is why we see two red dots in the new C-H bond.
3. 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, the second C atom looses an electron.
• This C atom now has sextet only. Thus a carbocation is formed.
4. The Br- now attacks the newly formed carbocation. This is shown in fig.13.71 (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-Bromobutane.
• Based on the above two examples, we can write:
♦ The H atom will attach to one of the two C atoms of the double bond.
♦ The X atom will attach to the other C atom of the double bond.
• In both of the two examples that we saw, the original alkene was symmetric. That is., Same groups are attached to either sides of the double bond. So the incoming H atom can attach to any one of the two C atoms of the double bond. Similarly, the incoming X atom can attach to any one of the two C atoms of the double bond. The resulting product will be the same.
• Now let us consider an unsymmetrical alkene. We will take CH3-CH=CH2 (prop-1-ene). We want to know the result when prop-1-ene reacts with HBr
• Let us assume that,
♦ the incoming H atom attaches to the second C atom.
♦ the incoming Br atom attaches to the first C atom.
✰ Then the product will be: CH3-CH2-CH2Br (1-Bromopropane)
• Next, let us assume that,
♦ the incoming H atom attaches to the first C atom.
♦ the incoming Br atom attaches to the second C atom.
✰ Then the product will be: CH3-CHBr-CH3 (2-Bromopropane)
◼ We want to know this:
In an actual reaction, what will be the product? 1-Bromopropane or 2-Bromopropane?
• To find the answer, we must analyze the mechanism of this reaction.
◼ First we will analyze the mechanism of that reaction which produces 1-Bromopropane. It can be written in 4 steps:
1. First the HBr molecule splits into two parts. We get H+ and Br-. It is shown in fig.13.72 (a) below:
 |
| Fig.13.72 |
2. The H+ attacks the propene molecule. This is shown in fig.13.72 (b) above.
• At the product side, we see that, the H+ is attached to the second 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.
3. 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, the first C atom looses an electron.
• This C atom now has sextet only. Thus a carbocation is formed.
4. The Br- now attacks the newly formed carbocation. This is shown in fig.13.72 (c) above.
• The Br- can donate two electrons to form a bond. So it attaches to the first C atom. Thus we get a molecule of 1-Bromopropane.
◼ Next we will analyze the mechanism of that reaction which produces 2-Bromopropane. It can be written in 4 steps:
1. First the HBr molecule splits into two parts. We get H+ and Br-. It is shown in fig.13.73 (a) below:
 |
| Fig.13.73 |
2. The H+ attacks the propene molecule. This is shown in fig.13.73 (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 double bond. That is why we see
two red dots in the new C-H bond.
3. 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, the second C atom looses an electron.
• This C atom now has sextet only. Thus a carbocation is formed.
4. The Br- now attacks the newly formed carbocation. This is shown in fig.13.73 (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-Bromopropane.
◼ We want to know this:
In an actual reaction, what will be the product? 1-Bromopropane or 2-Bromopropane?
• We can now write the answer. It can be written in 3 steps:
1. We have seen the two possibilities. Let us compare them.
• In fig.13.72, we see that, the carbocation formed, is a primary carbocation. (C+ is attached to only one other C atom)
• In fig.13.73, we see that, the carbocation formed, is a secondary carbocation. (C+ is attached to two other C atoms)
• We saw the details about primary, secondary, tertiary carbocations in an earlier section [see fig.12.69 in section 12.10].
2. We have learnt that, secondary carbocations are more stable than primary carbocations.
• So in the reaction mixture, there will be a greater quantity of secondary carbocations.
• Consequently, the Br- will be reacting more with secondary carbocations. This will result in a greater quantity of 2-bromopropane.
3. So the answer is:
In an actual reaction, the principal product will be 2-Bromopropane.
Let us now try to formulate a general rule to find the product. It can be written in 6 steps:
1. We have seen that, regardless of whether it is a primary carbocation or a secondary carbocation, the Br- will be always attaches to the C+
2. Let us compare the number of H atoms held by the C+:
♦ The C+ in a primary carbocation will hold two H atoms.
♦ The C+ in a secondary carbocation will hold one H atom.
♦ The C+ in a tertiary carbocation will hold zero H atoms.
(The above number of H atoms can be easily obtained by drawing Lewis structures)
3. We know the order of stability of carbocations. The order is:
Tertiary > Secondary > Primary.
• So in a reaction mixture,
♦ Quantity of tertiary carbocations will be high.
♦ Quantity of primary carbocations will be low.
4. The Br- will be looking for C+ portion.
• If tertiary carbocations are available, those carbocations will be present in greater quantity. The Br- will then attach to those tertiary carbocations.
• Based on step (2), the observer will get the impression that, Br- selects that C atom with the least number of H atoms.
5. Similarly we can write:
The Br- will be looking for C+ portion.
• If only secondary carbocations and primary carbocations are available, then secondary carbocations will be present in greater quantity. The Br- will then attach to those secondary carbocations.
• Based on step (2), the observer will get the impression that, Br- selects that C atom with the least number of H atoms.
6. The Russian scientist Markovnikov made the above findings while studying such reactions in detail. He framed a rule called Markovnikov’s rule. The rule states that:
The negative part of the addendum gets attached to that carbon atom which possesses lesser number of hydrogen atoms.
(In our present example, the addendum is HBr. Negative part of the addendum is Br-)
In the next section we will see Anti Markovnikov addition.
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