In the previous section, we completed a discussion on the preparation of alkanes. In this section, we will see the properties of alkanes.
Physical properties of Alkanes
The first physical property that we will discuss, is about the non-polar nature of alkanes. It can be written in 3 steps:
1. We have seen that some molecules are polar in nature. We know the reason for such polarity [see fig.4.223 in section 4.40]
2. In the case of alkanes, there are only two types of bonds: C-C bonds and C-H bonds.
• The difference in electronegativity between C and H atoms is very small. So neither C nor H can pull the electron clouds. Thus both C-C bonds and C-H bonds are non-polar.
3. So we can write:
Alkanes are almost non-polar molecules.
The second physical property that we will discuss, is about the state of various alkanes. It can be written in 7 steps:
1. Alkanes have weak van der Waal’s forces. We have seen van der Waals’s forces in an earlier section. It is the collection of four forces: (i) London force (ii) Dipole-dipole force (iii) Dipole-induced dipole force (iv) Hydrogen bond [see section 5]
2. When the number of C atoms in the chain is small, there will be a more uniform distribution of charge.
3. If the charge distribution is uniform, there will not be +ve and -ve regions.
• Consequently, there will not be much attraction between the molecules.
4. So, if the number of C atoms is small, the alkanes molecules will be in the gaseous state at room temperature (298 K).
• The first four members (C1 to C4) of the alkane series are gases.
5. When the number of C atoms in the chain is larger, there will be a non- uniform distribution of charge.
6. If the charge distribution is non-uniform, there will be +ve and -ve regions.
• Consequently, there will be attraction between the molecules.
7. So, if the number of C atoms is large, the alkane molecules will be in the liquid or solid states.
• C5 to C17 are liquids at room temperature (298 K).
• C18 and higher are solids at room temperature (298 K).
The third physical property that we will discuss, is about the solubility of alkanes. It can be written in 7 steps:
1. Consider a solution in which the solvent is polar and solute is non-polar.
• Let us name the polar solvent as A and non-polar solute as B
2. Since A is polar, the molecules of A will be attracting each other.
• Consider any two molecules of A, which are close together. They will be acting like a chain.
• B being non-polar, cannot form an electrostatic attraction with the two A molecules.
• So the molecule B will not be able to break the chain and occupy the space between the two A molecules.
• Even if we forcibly put the B between the two A molecules, the B will be expelled.
• So it is clear that, non-polar substances will not dissolve in polar solvents.
3. On the other hand, if B is a polar substance just like A, then B can occupy space in between the two A molecules to form a new chain.
• So if B is polar, it will dissolve in A.
• We can write: Polar substances will dissolve in polar solvents.
4. Consider the situation when both A and B are non-polar.
• In such a situation, neither A or B has permanent polarity. But London forces is applicable to both of them.
• Due to London forces, both A and B have equal chances of forming electrostatic attraction with each other. So B will dissolve in A
5. Thus we see that:
♦ Non-polar substances do not dissolve in polar solvents.
♦ Polar substances dissolve in polar solvents.
♦ Non-polar substances dissolve in non-polar solvents.
• Based on this, we can write:
Like dissolves like.
6. We have seen that alkanes are non-polar. We already know that water is polar.
So it is clear that, alkanes are insoluble in water.
7. Grease is composed of alkanes. Petrol is also composed of alkanes. Both being non-polar, grease will dissolve in petrol. That is why petrol is used in dry cleaning, to remove grease stains from clothes.
The fourth physical property that we will discuss, is about the boiling points of alkanes. It can be written in 3 steps:
1. The boiling point (b.p) of different alkanes can be obtained from the table in the data book. The table is also shown in the text book.
2. We see that, higher alkanes have higher boiling points.
This can be explained in two steps:
(i) As the number of C atoms increase, the size of the molecules increase. As a result, the magnitude of the inter molecular van der Waals forces will be higher.
(ii) When the magnitude of the forces increase, it will be difficult to separate the molecules from each other.
(iii) Consequently, the b.p increases for higher alkanes.
3. From the table we see another interesting point:
Pentane, 2-Methylbutane and 2,2-Dimethylpropane have the same molecular mass but different boiling points.
• This can be explained in 4 steps:
(i) Pentane, 2-Methylbutane and 2,2-Dimethylpropane are isomers. So they have the same molecular mass.
♦ Pentane has a straight chain structure.
♦ 2-Methylbutane has one branch.
♦ 2,2-Dimethylpropane has two branches.
(ii) When the number of branches increase, the shape of the molecule becomes more and more spherical.
(iii) Spherical shape has lower surface area.
• Due to the lower surface area, the area of contact between the spheres will be small. Consequently, the inter molecular forces will be small.
(iv) If the inter molecular forces are low, the molecules can be easily separated from each other.
• So we can write:
When the shape becomes more and more spherical, the b.p becomes lower and lower.
The fifth and final physical property needs a mention only:
Alkanes are colourless and odourless
Chemical properties of Alkanes
• Alkanes are generally inert towards acids, bases, oxidising agents and reducing agents. The reason can be written in 3 steps:
(i) In alkanes, all four valencies of C atoms are satisfied. The single valency of H atoms are also satisfied. So an alkane molecule as a whole, is stable.
(ii) All bonds in alkanes are sigma bonds. We know that, sigma bonds involve linear overlap of orbitals. This creates bonds which are very strong. So it is difficult to break the bonds in alkanes.
(iii) Alkanes are non-polar. So they do not have any additional +ve or -ve charges. This makes it difficult to attack alkane molecules.
• However, alkanes can undergo reactions like substitution, combustion, controlled oxidation etc., We will now see those reactions in detail
I. Substitution reactions
This can be written in 9 steps:
1. In this reaction, one H atom is first removed from the alkane.
• Then a halogen atom takes the place of that H atom.
• Instead of halogen atom, nitro group or sulphonic acid group can also take the place of the H atom.
♦ If it is a halogen, the reaction is called halogenation.
♦ If it is a nitro group, the reaction is called nitration.
♦ If it is a sulphonic acid group, the reaction is called sulphonation.
2. Some times more than one H atoms can be substituted in this way.
3. Halogenation requires higher temperatures (573-773 K)
• However, instead of higher temperatures, we can use diffused sunlight or ultraviolet light also.
• Diffused sunlight is obtained when direct sunlight is scattered by clouds or some artificial means.
• Ultraviolet light can be produced by passing electricity through a suitable gas like mercury vapour.
4. Lower alkanes do not undergo nitration and sulphonation. They can undergo halogenation only.
5. Let us see some examples of halogenation:
(i) $\rm{CH_4~+~Cl_2~ \color {green}{\xrightarrow[{}]{h \nu}} ~ CH_3 Cl~+~HCl}$
• CH3Cl is Chloromethane.
[$h \nu$ represent the energy provided by radiation (in our present case, radiation is sunlight or ultraviolet light). Here h is the plank's constant and 𝜈 is the frequency. We saw those details in chapter 2]
(ii) $\rm{CH_3 Cl~+~Cl_2~ \color {green}{\xrightarrow[{}]{h \nu}} ~ CH_2 Cl_2~+~HCl}$
• CH2Cl2 is Dichloromethane.
(iii) $\rm{CH_2 Cl_2~+~Cl_2~ \color {green}{\xrightarrow[{}]{h \nu}} ~ CH Cl_3~+~HCl}$
• CHCl3 is Trichloromethane.
(iv) $\rm{CH Cl_3~+~Cl_2~ \color {green}{\xrightarrow[{}]{h \nu}} ~ C Cl_4~+~HCl}$
• CCl4 is Tetrachloromethane.
(v) $\rm{CH_3 - CH_3~+~Cl_2~ \color {green}{\xrightarrow[{}]{h \nu}} ~ CH_3 - CH_2 Cl~+~HCl}$
• CH3-CH2Cl is Chloroethane.
6. The rate of reaction of alkanes with the various halogens varies.
♦ Rate of reaction with F is the highest.
♦ Rate of reaction with I is the lowest.
• The order is: F2 > Cl2 > Br2 > I2
(Recall that, rate of reaction is the quantity of products formed in unit time)
7. Reaction with F is violent. We will need special equipment to control the reaction.
8. Reaction with I is a very slow reaction. It is a reversible reaction. The equation is:
CH4 + I2 ⇌ CH3I + HI
• We have discussed about reversible reactions in an earlier chapter [see fig.7.3 in section 7.3]
• We can convert more quantities of CH4 into CH3I if we remove the HI.
• For that, we use oxidising agents like HIO3 or HNO3. The HI will react with the oxidising agent and form I2. The equation is:
HIO3 + 5HI ⟶ 3I2 + 2H2O
9. We have seen 1o carbon atom, 2o carbon atom etc., in a previous section of this chapter [see step 11 below fig.13.8 in section 13.1]
• Based on that, we can now write an important information. It can be written in 2 steps:
(i) During halogenation of alkanes, the H atoms attached to the 3o carbon atoms are removed more readily than those attached to the 2o carbon atoms.
(ii) Similarly, the H atoms attached to the 2o carbon atoms are removed more readily than those attached to the 1o carbon atoms.
In the next section we will see the mechanism of halogenation.
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