Priyam Study Centre Chemistry

The molecules are composed of partially charged nuclei and negatively charged electrons distributed in space. The structural arrangement of these particles is different in different molecules.

When the center of gravity of the positive charge due to coincides with the center of gravity of the negative charge due to electrons, the molecules become non-polar. Examples are,

H₂, CO₂, BCl₃, CCl₄, PCl₅, SF₆, C₆H₆, etc.

When the center of gravity of the positive charge does not coincide with the center of gravity of the negative charge, polarity arises in the molecules and the molecules are called polar.

HCl, H₂O, NH₃, CH₃Cl, C₆H₅Cl, etc.

The polar character of the molecules has quantified a term, called dipole moment (µ). The molecule is neutral and hence if (+ q) amount of charge separates at the positive charge center, (- q) will be accumulated at the negative charge center of the molecule. If l is the distance between two centers of the polar molecule, then the dipole moment,

For the non-polar molecules, l=0 and hence µ=0. Higher the value of µ of a molecule, higher will be its polarity.

Let us take an example of HCl. Due to the greater electronegativity of Cl-atom, the bonding electron pair is shifted towards Cl-atom and it acquires small negative charge (- q) and hydrogen atom acquires small positive charge (+ q). If l is the distance of the charge separation usually taken in bond length, then,

µ = q × l

The dipole moment is a vector quantity and it has both, magnitude and direction. The direction is represented by an arrow pointing towards the negative end. The length of the arrow is directly proportional to the magnitude of µ.

The dipole moment of Water

Unit of the dipole moment

In the C.G.S. system, the charge is expressed as esu and the length in cm.

Thus the unit of µ is esu cm

The charge is the order of 10⁻¹⁰ esu and the distance of separation of charge is in the order of 10⁻⁸ cm.

Hence the unit of µ is coulomb × meter (c х m).

Dimension of the dipole moment

Clausius mossotti equation

When a non-polar substance is placed between two parallel plates and an electric field is applied, the field tends to attract the negatively charged electrons towards the positive plate and positive charge towards a negative plate.

Under this condition, there will be the electrical distortion of the molecule and an electric dipole is created. Such distortion in a molecule is called the electric polarization.

Polarization, however, disappears as soon as the field is withdrawn and the molecule comes back to its original state.

It is thus the induced polarization (Pi) and the electric dipole is created in the molecule due to the presence of the electric field is called induced dipole moment (μ_i).

The induced dipole moment or simply the induced moment is directly proportional to the strength of the electric field applied(F).

That is, μ_i ∝ F When F is low, otherwise, hyperpolarization may occur.

α_i measures the case with which the electronic configuration of the molecule can be distorted by an applied electric field. It may also be defined as the amount of induced moment in the molecule when the unit field strength is applied. The polarizability has the dimension of the volume (L³).

It can also be shown that, αi = r³ where r = radius of the molecule assuming it to have a spherical shape.

For atoms also, distortion occurs when it is placed between the two charged plates. The polarizability of the atom increases with the atomic size (r), atomic number (Z) and the case of excitation (I.P.). Thus atom behaves like a dipole and this dipole moment is induced by the applied electric field.

Clausius Mossotti derived from electromagnetic theory, a relation between the polarizability (αi) and the dielectric constant (D) of the non-polar medium between two plates as,

Clausius mossotti equation in dipole moment

It gives the distortion produced in the 1 mole of the substance by a unit electric field. αi is constant for the molecule and independent of temperature. Hence Pi is also constant for the molecule and independent of temperature.

D = dielectric constant of the medium = C/C₀ where C = capacitance of the condenser containing the substance and C₀ is the vacuum.

D is the dimensionless quantity and it is unity for vacuum. For other substances, it is greater than unity.

Pi of the substance can be calculated by measuring dielectric constant (D), density (ρ) and knowing the molar mass (M) of the substance.

Debye modification of Clausius Mossotti equation

For polar molecules like CH₃Cl, H₂O, HF, etc, molar polarization is not constant but decreases with temperature.

Thus, Clausius Mossotti Equation fails very badly for polar molecules. The reason for the failure was put forward by P Debye. According to him, when an electric field is applied between two parallel plates containing polar molecules (gaseous state), two effects will occur.

• Induced polarization 
• Orientation polarization 

Induced polarization

The field will tend to induced polarization in the molecules and the induced polarization in the molecules,

Pi= (4/3) π N₀ αi

Orientation polarization

The dipolar molecules will be oriented in the field producing a net dipole moment in the direction of the field.

Considering the two tendencies that polar molecules tend to orient in the direction of the applied field (applied) and thermal orientation tends to destroy the alignment of the molecules.

Polarization of molecules

Debye equation

However the polar molecules in the substance are fixed and unable to orient in the fixed direction, the orientation polarization is taken zero.

This is true for the condensed system where strong intermolecular forces prevent the free rotation of the molecules. 

This is due to the fact that at high temperatures, the polar molecules rotate at such high speed that orientation polarization vanishes.

Problems solutions

How can convert 1 Debye to the coulomb meter?

(i)P-F, (ii)S-F, (iii)Cl-F, (iv)F-F which of the following compound has the lowest dipole moment?

The electronegativity difference between two F atoms is zero. Thus the dipole moment of F-F compound is zero.

The heat capacity of a substance is defined as the amount of heat required to rise of the temperature by one degree. Heat capacity per gram of substance is called specific heat and per mole called molar heat capacity.

Where Cp and Cv are the molar heat capacities at constant pressure and constant volume respectively. c_p and c_v are their specific heats.

The specific heat at constant pressure and constant volumes are 0.125 and 0.075 Cal gm⁻¹K⁻¹ respectively. Calculate the molecular weight and the atomicity of the gas. Name the gas if possible.

Solution

M = 40 and ⋎ = 1.66(mono-atomic), Ar(Argon). For gases, there are two heat capacities at constant volume and constant pressure. These are Represented as,

Cv = (dq/dT)v = (dU/dT)v

Cp = (dq/dT)p = (dU/dT)p + P(dV/dT)p

Heat capacities of an Ideal Gas

If the gas is assumed to be ideal, then,
PV = nRT and (dU/dT)v = (dU/dT)p
Again, P(dV/dT)P = nR

Thus for an ideal gas,
PV = nRT
or, P(dV/dT)p = nR

Again, Cp = (dU/dT)P + P(dV/dT)P
or, Cp = Cv + {P×(nR/P)}
or, Cp = Cv +nR

For 1 mole ideal gas
Cp = Cv +R

Molar heat capacity

From the above two descriptions, it is clear that Cp and Cv. Since for Cp, some mechanical work is required as additional energy to absorbed for definite piston from volume V₁ to V₂.

Cp - Cv = Mechanical work

or,PdV = P(V₂-V₁) = PV₂-PV₁
= R(T+1) - RT
= R

Now let us find the expression of Cv from the point of view of Kinetic theory.

Cv = Energy required to increase the translational kinetic energy of 1-mole gas for a rise of 1° temperature + energy required to increase the intermolecular energy of 1-mole gas for the rising temperature of 1°.

Increase of transnational K.E. = (3/2)R(T+1) - (3/2)R
= (3/2) R
For 1-mole gas for 1°C rise in temperature.

Mono-atomic gases

Cv = (dU/dT)v = (3/2)

Cp = (5/2)R.

Thus γ = Cp/Cv = 5/3 ≃ 1.667

Poly-atomic gases

For linear Cv = (dU/dT)v = (3/2)R + R +(3N - 5)R

For non-linear Cv = (dU/dT)v = (3/2)R + (3/2)R +(3N - 6)R
where N is the number of particles.

Molar heat capacities for diatomic molecules

For diatomic molecule N = 2.
Thus Cv = (3/2)R + R + R = (7/2)R

Cp = (9/2)R.

∴ ⋎ = Cp/Cv = 9/7 ≃ 1.286

Molar heat capacities for tri-atomic molecules

For triatomic molecule N = 3

Physical properties of alkenes

Numbers containing two to four carbon atoms are gases; five to seventeen is liquid; eighteen on words solid at room temperature and they burn in air with a luminous flame.

In general, the physical properties of alkenes are similar to those of alkenes, since the alkenes are only weak van der walls attractive forces.

Stabilities of alkenes

One way of measuring the stability of an alkene is the determination of its heat of hydrogenation.

CH₂=CH₂ (ΔH = -137 kJ mol⁻¹), MeCH=CH₂ (ΔH = -125.9 kJ mol⁻¹), MeCH₂CH=CH₂(ΔH = -126 kJ mol⁻¹), MeCH₂CH=CH₂(ΔH = -126.8 kJ mol⁻¹), cis- MeCH=CHMe(-119.7 kJ mol⁻¹), trans- MeCH=CHMe(-119.7 kJ mol⁻¹), Me₂C=CH₂(ΔH = -188.8 kJ mol⁻¹).

Since the reaction is exothermic, the smaller ∆H is (numerically), the more stable is the alkene relative to its parent alkane.

Thus, it is only possible to compare the stabilities of different alkenes which produce the same alkane on hydrogenation.

This arises from the fact that the enthalpy of formation of alkenes is not purely additive properties; it also depends on steric effects and these tend to vary from molecule to molecule. Since three n- butenes all give the n-butane on reduction.

This order may be explained in terms of steric effect and hyperconjugation.

In but-1-ene, steric repulsion is virtually absent.

In but-2-enes, the two methyl groups in cis isomer being closer together than in the trans isomer, experience greater steric repulsion and consequently the cis form is under greater strain than the trans. Thus steric repulsion destabilizes a molecule.

On the other hand, hyperconjugation stabilizes a molecule, and is small in but-1- ene but much larger in but-2-enes.

Since trans-but-2-ene is the most stable isomer, it follows that hyperconjugation has a greater stabilizing effect then steric repulsion a destabilizing effect.

Problem

Arrange the following Alkenes in order of increasing stability and give your reasons.(i) Me₂C=CH₂, (ii) cis-MeCH=CHMe, (iii) trans-MeCH=CHMe (iv) MeCH₂CH=CH₂

Solution

MeCH₂CH=CH₂(but-1-ene)ㄑcis-MeCH=CHMe(cis but-2-ene)ㄑtrans-MeCH=CHMe(trans but-2-ene) ㄑMe₂C=CH₂(isobutene)

In general, the order of stability of alkenes are R₂C=CR₂ 〉R₂C=CHR 〉R₂C=CH₂ ~ RCH=CHR 〉RCH=CH₂ 〉CH₂=CH₂

Chemical properties of alkenes

Owing to the presence of a double bond, the alkenes undergo a large number of addition reactions, but under special conditions, they also undergo substitution reactions.

The high reactivity of the olefinic bond is due to the presence of two π- electrons, and when addition reaction occurs, the trigonal arrangement in the alkene changed to the tetrahedral arrangement in the saturated compound produced.

Reactions of alkenes

Combustion reactions of alkenes

Alkenes are flammable substances they burn in air with a luminous smoky flame to produce carbon dioxide and water.

Addition reactions of alkenes

An addition reaction, organic chemistry, is in its simplest terms an organic reaction where two or more molecules combine to form the larger one and the product is called additive compound.

Catalytic hydrogenation of alkenes

Alkenes are readily hydrogenated under pressure in the presence of a catalyst.

Finely divided platinum and palladium are effective at room temperature; nickel on support requires a temperature between 200⁰C and 300⁰C; Raney nickel is effective at room temperature and atmospheric pressure.

Addition of halogens to alkenes

Alkenes from addition compounds with chlorine or bromine.

Halogen addition can take place either by a heterolytic (polar) or a free-radical mechanism.

Halogen addition radially occurs in solution, in the absence of light or peroxides and is catalyzed by inorganic halides as for example aluminum chloride or by polar surfaces. These facts lead to the conclusion that reaction occurs by a polar mechanism.

The free radical mechanism has generally accepted that the addition of halogen to alkenes in the absence of light is polar. Stewart showed that the addition of chlorine to ethylene is accelerated by light and this suggested the free radical mechanism.

Reaction of alkenes

Addition of halogen acids

Ethylene adds hydrogen bromide to form ethyl bromide.

The conditions for the addition are similar to those for halogens, only the addition of hydrogen fluoride occurs under pressure.

In the case of unsymmetrical alkenes, it is possible for the addition of the halogen acid to take place in two different ways, 

For example, propane might add on hydrogen iodide to form propyl iodide or isopropyl iodide.

Markownikoff studied many reactions of this kind, and as a result of his work, formulated the following rule.

Markownikoff rule for alkenes

The negative part of the addendum acids on to the carbon atom that is joined to the less number of hydrogen atoms.

In the case of the halogen acids, the halogen atom is the negative part. So according to Markownikoff rule, isopropyl halide is obtained.

Markownikoff’s rule is empirical but may be explained theoretically on the basis that the addition occurs by a polar mechanism.

The addition of halogen acid is an electrophilic reaction, the proton adding fast, followed by halide ion. Also, the addition is predominantly trans, and this may explain in terms of the formation of a bridge carbonium ion.

Markownikoff rule

Since the methyl group has a +I effect, the π electrons are displaced towards the terminal carbon atom which, in consequence, acquires a negative charge. Thus, the proton added on to the carbon fastest from the methyl group, and the halide ion then adds to the carbonium ion.

An alternative explanation for Markownikoff's rule is in terms of the stabilities of carbonium ions. Represent as primary, secondary and tertiary carbonium ion.

Which of the following carbocation is more stable? (i) CH₃CH₂⁺ (ii) CH₃(CH₃)CH⁺ (iii) CH₃(CH₃)C⁺CH₃.

The number of hydrogen atoms available for hyperconjugation is 3 for (i), 6 for (ii) and 9 for (ii). Consequently, (iii) would be expected to be the most stable.