In the previous section, we completed a discussion on the mechanism of halogenation. We also saw the basics of complete and incomplete combustion. In this section, we will see controlled oxidation. Later in this section we will also see isomerisation, aromatization and reaction with steam.
III. Controlled oxidation
This can be written in 4 steps:
1. We have seen two types of reaction of alkanes with oxygen:
♦ Reaction when dioxygen supply is sufficient.
♦ Reaction when dioxygen supply is insufficient.
2. Next we will see the reaction when supply of dioxygen is regulated.
• That means, we supply measured quantities of oxygen so that the desired products are obtained.
3. In addition to regulating the quantity of oxygen, we need to take care of some other factors also:
♦ Adjust the temperature to the suitable value.
♦ Adjust the pressure to the suitable value.
♦ Provide suitable catalysts.
4. Let us see some examples of such reactions:
Example 1:
This can be written in 4 steps:
(i) In the previous section, we have seen complete combustion of methane:
$\rm{CH_4 (g)~+~2O_2 (g)~ \color {green}{\xrightarrow[{}]{combustion}} ~
CO_2 (g)~+~2H_2 O (g)}$
• So for complete combustion, if we take two moles of methane, there must be four moles of dioxygen.
(ii) We also saw incomplete combustion of methane:
$\rm{CH_4 (g)~+~O_2 (g)~ \color {green}{\xrightarrow[{combustion}]{Incomplete}} ~
C (s)~+~2H_2 O (g)}$
• So for incomplete combustion, if we take two moles of methane, there must be only two moles of dioxygen.
(iii) Now consider the following reaction:
$\rm{2 CH_4 ~+~O_2 ~ \color {green}{\xrightarrow[{}]{Cu/523 K/100 atm}} ~
2CH_3 OH}$
• CH3OH is methanol
• Here, we are taking two moles of methane but only one mole of dioxygen.
• So we can write:
If we regulate the supply of dioxygen in such a way that, the ratio of methane to dioxygen is 2:1, the oxidation of methane will give methanol.
(iv) Also note that,
♦ Temperature needs to be kept at 523 K .
♦ Pressure needs to be kept at 100 atm.
♦ Cu must be used as catalyst.
Example 2:
This can be written in 2 steps:
(i) In the previous section, we saw incomplete combustion of methane:
$\rm{CH_4 (g)~+~O_2 (g)~ \color {green}{\xrightarrow[{combustion}]{Incomplete}} ~
C (s)~+~2H_2 O (g)}$
(ii) Now consider the following reaction:
$\rm{CH_4 ~+~O_2 ~ \color {green}{\xrightarrow[{Δ}]{Mo_2 O_3}} ~ HCHO}$
• HCHO is methanal
• Here, we are taking the same quantities as in incomplete combustion but Mo2O3 is used as catalyst.
• So we can write:
If
we regulate the supply of dioxygen in such a way that, the ratio of
methane to dioxygen is 1:1, and use Mo2O3 as catalyst, the oxidation of methane will give methanal.
Example 3
Controlled oxidation of ethane in the presence of (CH3COO)2Mn as catalyst, will give ethanoic acid. The equation is:
$\rm{2CH_3 CH_3 ~+~3 O_2 ~ \color {green}{\xrightarrow[{Δ}]{(CH_3 COO)_2 Mn}} ~ 2CH_3 COOH~+~H_2 O}$
Example 4
This can be written in 2 steps:
1. Normally, alkanes resist oxidation. But if there is a tertiary H atom in the alkane, that H atom can be oxidized to the corresponding alcohol. We must use potassium permanganate also in the reaction. The equation is:
$\rm{(CH_3)_3 CH ~+~3 O_2 ~ \color {green}{\xrightarrow[{oxidation}]{KMnO_4}} ~ (CH_3)_3 COH}$
2. Using Lewis structures, the reaction can be explained as shown in fig.13.37 below:
 |
| Fig.13.37 |
IV. Isomerisation
This can be written in 3 steps:
1. n-Alkanes can be heated in the presence of anhydrous aluminium chloride and hydrogen chloride gas.
• [The letter “n” in n-Alkanes stands for “normal”. They are straight chain alkanes. They do not have any branches]
• [The word “anhydrous” indicates that, no water molecules are present. If water molecules are present, they will interfere with the reaction and we will not get the desired products]
2. When heated in this way, the n-alkanes isomerise to branched chain alkanes.
• There may be more than one products.
♦ But each product will be an isomer of the n-alkane taken initially.
♦ Each product will contain branches.
3. Let us see an example:
 |
| Fig.13.38 |
• In fig.13.38(a) above, the n-alkane taken initially is n-Hexane.
• The two products are 2-Methylpentane and 3-Methylpentane.
♦ They are both isomers of n-Hexane .
♦ These are the major products.
• Some minor products may also form during the reaction. But minor products are usually not reported in organic reactions. In our present case, a possible minor product is shown in fig.b. It is also an isomer of n-Hexane.
V. Aromatization
This can be written in 6 steps:
1. Take a n-alkane and subject it to 773 K temperature and 10-20 atm pressure.
• Also use any one of the three catalysts given below:
♦ Oxide of vanadium
♦ Oxide of molybdenum
♦ Oxide of chromium supported over alumina.
• The n-alkane taken, should have at least six C atoms.
2. During this process, the n-alkane will get dehydrogenated. That means, the n-alkane will lose some H atoms.
• When H atoms are lost, some of the single bonds will be converted to double bonds. This is to satisfy the valencies of C atoms.
3. Also during this process, the straight chain of the n-alkane will get cyclised. That means, the straight chain will change into a ring.
4. If the n-alkane taken initially has six C atoms, then the result will be benzene. This is shown in fig.13.39 below:
 |
| Fig.13.39 |
• Though the n-Hexane has a straight chain structure, in the fig above, it is shown in a bent form. While in this bent form, if the two C atoms at the ends of the chain can join together, we will get the cyclic structure benzene.
5. If the n-alkane taken initially has more than six C atoms, then the extra C atoms will form methyl groups around the benzene ring.
• Toleune has one methyl group around a benzene ring. So to obtain toleune, we must subject n-heptane to aromatization.
6. Consider the bent form of n-hexane (C6H14) in fig.13.39 above. While in this bent form, if only two H atoms from the ends are lost, we will be getting cyclohexane (C6H12)
• But since the catalysts shown in fig.13.39 are used and also since the specified temperature and pressure are applied, a total of eight H atoms will be lost. Thus we will be getting benzene (C6H6) instead of cyclohexane.
VI. Reaction with steam
This can be written in 3 steps:
1. Methane will react with steam if a temperature of 1273 K is available.
• Nickel should be also present as a catalyst.
2. The products are carbon monoxide and dihydrogen.
3. This method is used for the industrial preparation of dihydrogen gas.
$\rm{CH_4 ~+~H_2 O ~ \color {green}{\xrightarrow[{Δ}]{Ni}} ~ CO~+~3H_2}$
VII. Pyrolysis
This can be written in 5 steps:
1. When higher alkanes are subjected to high temperature, they decompose into lower alkanes, alkenes etc.,
2. Such a decomposition reaction into smaller fragments by the application of heat is called pyrolysis. Another name for this process is cracking.
• An example is shown below:
 |
| Fig.13.40 |
♦ In the first case, hexane undergoes pyrolysis to give hexene and dihydrogen.
♦ In the second case, hexane undergoes pyrolysis to give butene and ethane.
♦ In the third case, hexane undergoes pyrolysis to give propene, ethene and methane.
3. Pyrolysis of alkanes is believed to be a free radical reaction.
(A free radical reaction is a reaction which involves free radicals. For example, in the previous section, we saw that halogenation involves chlorine free radicals and methyl free radicals)
4. Pyrolysis of kerosene will give oil gas.
• In this process, dodecane, a constituent of kerosene, decomposes into heptane and pentene.
• A temperature of 973 K is required for this process.
• Also nickel or palladium must be present as a catalyst.
$\rm{C_{12} H_{26} ~ \color {green}{\xrightarrow[{973 K}]{Pt/Pd/Ni}} ~ C_7 H_{16}~+~C_5 H_{10}~+~\text{other products}}$
♦ C12H26 is Dodecane.
♦ C7H16 is Heptane.
♦ C5H10 is Pentene.
5. Pyrolysis of petrol gives petrol gas.
We have completed a discussion on the chemical properties of alkanes. In the next section we will see conformations.
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