November 2018

Dipole moment of benzene and its derivatives

Zero dipole moment of the benzene suggested that it has regular hexagonal planer structure and it confirms Kekule’s form.
Applications of Dipole Moment for Benzene Derivatives
Regular Planer Hexagonal Structure of
Benzene having Centre of Symmetry
 
However, if hydrogen atom is substituted by another atom or group, it acquires polar character.
Examples of such derivative of benzene are C₆H₅Cl, C₆H₅NO₂, C₆H₅OH, etc. 
When di-substituted benzene are considered it can be shown that o-isomer will have the highest value of dipole moment then other two isomers. p-derivative has the lowest value while m-derivative has the value between the two. 
The values can be calculated using the vector addition principle 
μ = m12 + m22 + 2 m1m2 cosθ 
provided the group must also lie on the same plane of the benzene ring.

Di-substituted derivative of benzene:

Applications of Dipole Moment for Benzene Derivatives
Di-substituted derivative of benzene
(a) For o-Isomer:
μ² = m12 + m22 + 2 m1m2 cosθ
Here, m1 = m2 = m and θ = 60°
Thus, μ² = 2 m²(1 + cos60°) 
= 2 m² (1 + 1/2) 
∴ μ= √3 m
(b) For m-Isomer:
μ² = m12 + m22 + 2 m1m2 cosθ
Here, m1 = m₂ = m and θ = 120°
Thus, μ² = 2 m²(1 + cos120°) 
= 2 m² (1 - 1/2)
∴ μ = m
For p-Isomer:
μ² = m12 + m22 + 2 m1m₂ cosθ
Here, m1 = m2= m and θ = 180°
Thus, μ² = 2 m²(1 + cos180°) 
= 2 m² (1 - 1)
∴ μ = 0
Thus μ of o-isomer〉
μ of m-isomer〉μ of p-isomer
(this has been confirmed for C₆H₄Cl₂ and C₆H₄(NO₂)₂.
This determination of dipole moment helps to determine the orientation of the groups in the benzene ring.
  • Problem 1:
p - dinitro benzene has μ = 0 but p - dihydroxy benzene μ ≠ 0. Explain.
  • Answer:
Through the p-dinitro benzene has μ = 0. p-dihydroxy benzene has dipole moment(μ ≠ 0). This is due to the fact that the two substituted hydroxy group are not on the same plane of the benzene ring but are inclined to the ring. 

(v) Cis and Trans forms of Geometrical Isomerism:

Presence of double bond in C - C link restricts the free rotation and geometrical isomerism develops.
Applications of Dipole Moment for Cis and Trans Isomer
Dipole Moment of Cis and Trans Isomers
The Measurement of dipole moment helps to distinguish the two forms of geometrical isomers. 

(vi) Change of Dipole Moment with Temperature:

For molecule CH₂Cl - CH₂Cl, several confirmations are also changed and so dipole moment (μ) of the molecule is also changed.
  • Problem 2:
The dipole moment of o-xylene = 0.693 D. Finds the dipole moment of toluene.
  • Answer:
For o - xylene the θ = 60° and μ = √3 m. Thus, m = 0.693/√3 = 0.4 D
Again θ = 120° for toluene and dipole moment(μ) of toluene = m.
Thus the dipole moment of toluene is 0.4 D.  

Evaluation and interpretation of the dipole moment of covalent molecules provide an important tool in the attack of molecular structure. It helps to determine the size and shape of the molecules, spatial arrangements of bonds, bonds partial ionic character, residue charge on the atoms of the molecules, etc.

Dipole Moment

    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,
µ = q × l
    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.
Dipole Moments and its calculation
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 order of μ is, 10⁻¹⁰ × 10⁻⁸ = 10⁻¹⁸ esu cm.
This magnitude is called one Debye.
That is, 1 Debye = 10⁻¹⁸ esu cm.

For example, µ of HCl is 1.03 D means, µ of HCl is 1.03 х 10⁻¹⁸ esu cm.
In the SI system, the charge is expressed in coulomb(c) and length is a meter(m).
    Hence the unit of µ is coulomb × meter (c х m).

Dimension of the dipole moment

Unit of µ = Unit of Charge × Unit of Length.
Thus, the unit of µ = esu × cm in the CGS system.

But from the Coulomb’s Low, F = q₁q₂/Dr²
Thus, (esu)² = dyne × cm² = gm cm sec⁻² × cm²
Hence, esu = gm1/2 sec ⁻¹ cm3/2
Hence the dimension of µ = M1/2 T-1 L5/2

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 (μi).
    The induced dipole moment or simply the induced moment is directly proportional to the strength of the electric field applied(F).
    That is, μi ∝ F When F is low, otherwise, hyperpolarization may occur.
μi = αi F
where αi is proportionality constant called induced polarizability of the molecule.
    αi 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).
αi = μi/F
= (esu × cm)/(esu × cm⁻²)
= cm³
    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,
Dipole moment and its application
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.
The orientation polarization,
P₀ = (4/3) π N₀ α₀

Where α₀ = orientation polarisability.
Debye calculated the value of α₀ = μ²/3KT
    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.
Debye modification of Clausius Mossotti equation
Polarization of molecules
Therefore, for the polar molecules, the total polarization,
Pt = Pi + P₀
And the total polarization for polar molecules
Pt = (D-1)/(D+2) M/ ρ
Debye modification of Clausius Mossotti equation
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. 
Again if 1/T =0 that is temperature is very high tending to infinity.
Thus 1/T = 0 and Pt=0
∴ Pt=Pi
    This is due to the fact that at high temperatures, the polar molecules rotate at such high speed that orientation polarization vanishes.

Problems solutions

Problem
    How can convert 1 Debye to the coulomb meter?
Answer
The dipole moment in CGS system is µ = 4.8 × 10⁻¹⁰ × 10⁻⁸ esu cm
= 4.8 D
In the SI system, 1.6 ×10⁻¹⁹ × 10⁻¹⁰ coulombs × meter
= 1.6 × 10⁻³⁰ C × m
Thus, 4.8 Debye = 1.6 ×10⁻³⁰ C × m

or, 1 Debye = (1.6 × 10⁻³⁰)/4.8 C × m
or, 1 Debye = 3.336 × 10⁻³⁰ C × m

Problem
    (i)P-F, (ii)S-F, (iii)Cl-F, (iv)F-F which of the following compound has the lowest dipole moment?
Answer
    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.
Thus, Molar heat capacity = Molar mass × Specific heat
Cp = M × cp, Cv = M × cv
    Where Cp and Cv are the molar heat capacities at constant pressure and constant volume respectively. cp and cv are their specific heats.
Problem
    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 capacities at constant volume and pressure
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.
Thus the order of stabilities
trans but-2-ene 〉 cis but-2-ene 〉 but-1-ene
    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.
2CnH₂n + 3H₂O → 2nCO₂ + 2nH₂O, ΔH = -ve

CH₂=CH₂ + 3O₂ → 2CO₂ + 2H₂O, ΔH = -ve

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.
CH₃CH = CH₂(propene) → CH₂CH₃ - CH₃(propane)
    Addition of halogens to alkenes
    Alkenes from addition compounds with chlorine or bromine.
CH₂ = CH₂(ethylene) + Br₂ → BrCH₂ - CH₂Br(ethylene dibromide)
    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.

Properties of Alkenes
Reaction of alkenes
    Addition of halogen acids
    Ethylene adds hydrogen bromide to form ethyl bromide.
CH₂=CH₂ + HBr → CH₃⎯CH₂Br

The order of reactivity of the halogen acids is,
hydrogen iodide 〉hydrogen bromide 〉hydrogen chloride 〉hydrogen fluoride
(this is also the order of acid strength).
    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.
CH₃ - CH = CH₂ + HI → CH₃ - CH(H) - CH₂(I)

or it might be,
CH₃ - CH = CH₂ + HI → CH₃ - CH(I) - CH₂(H)
    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.
Chemical properties of alkenes
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.
Problem
    Which of the following carbocation is more stable? (i) CH₃CH₂⁺ (ii) CH₃(CH₃)CH⁺ (iii) CH₃(CH₃)C⁺CH₃.
Solution
    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.
Thus the stability order is
CH₃(CH₃)C⁺CH₃ 〉CH₃(CH₃)CH⁺  〉CH₃CH₂⁺

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