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Application of Dipole Moment for Molecules

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:

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.

Bond polarity and dipole moments

Inorganic chemistry , Polar bond

Study polarity of the bond and dipole moments

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, bond polarity, residue charge on the atoms of the molecules.

The molecules are composed of partially charged nuclei and negatively charged electrons distributed in space. The structural arrangement of these bonds 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.
Hydrogen, carbon dioxide, boron trichloride, phosphorus pentachloride, benzine are the example of non-polar molecules.

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.
Hydrogen chloride, water, ammonia, methyl chloride, chlorobenzene are examples of polar molecules.

What is the dipole moment of a molecule?

The polarity of a bond in a molecule has quantified a term, called dipole moment (µ).

If +q amount of charge separates at the positive charge center, -q will be accumulated at the negative charge center of the molecule and l is the distance between two centers of the molecule.

∴ Dipole moment, µ = q × l
Non-polar molecules, l=0 and hence µ=0.

Higher the value of µ of a molecule, higher will be its polarity.

The polarity of bond in hydrochloric acid

In hydrochloric acid, due to the greater electronegativity of a chlorine atom, the bonding electron pair shifted towards chlorine atom. Thus chlorine atom acquires a small negative charge and hydrogen atom acquires a small positive charge. l is the distance of the charge separation usually taken in bond length.

∴ 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.

Polarity of bond

Problem
P - F, S - F, Cl - F, and F - F which of the following compound has the lowest dipole moment?

Answer
The electronegativity difference between the two F atoms is zero. Thus the dipole moment of the fluorine molecule is zero.

CGS and SI unit of the dipole moment

In the CGS system, the charge expressed as esu and the length in cm. Thus the unit of dipole moment is esu cm.
The charge in order of 10⁻¹⁰ esu and distance of separation of charge in order of 10⁻⁸ cm.

∴ Order of dipole moment = 10⁻¹⁰ × 10⁻⁸
= 10⁻¹⁸ esu cm.
This magnitude is known as Debye or simply D.

∴ 1 Debye = 10⁻¹⁸ esu cm.

The dipole moment of hydrochloric acid = 1.03 D = 1.03 × 10⁻¹⁸ esu cm.

In the SI system, the charge is expressed in coulomb and length is a meter.

∴ SI unit of dipole moment = coulomb × meter (c × m).

Problem
How can convert 1 Debye to the coulomb meter?

Solution

Dipole moment = q × l.

∴ The dipole moment of a molecule in the CGS unit
= 4.8 × 10⁻¹⁰ × 10⁻⁸ esu cm
= 4.8 D.

The dipole moment of a molecule in the SI unit
1.6 × 10⁻¹⁹ × 10⁻¹⁰ coulombs × meter
= 1.6 × 10⁻³⁰ C × m.

∴ 4.8 Debye = 1.6 ×10⁻³⁰ coulomb meter
or, 1 Debye = (1.6 × 10⁻³⁰)/4.8 coulomb meter
= 3.336 × 10⁻³⁰ coulomb meter.

The dimension of the dipole moment

Unit of dipole moment = unit of charge × unit of length.
∴ CGS unit of µ = esu × cm.

Coulomb’s Law, F = q₁q₂/Dr²
∴ (esu)² = dyne × cm² = gm cm sec⁻² × cm²
or, esu = gm^¹/₂ sec⁻¹ cm^³/₂

∴ Dimension of µ = M^¹/₂ T^-1 L^⁵/₂

What is the polarization?

When a non-polar substance placed between two parallel plates and an electric field 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 called the electric polarization.

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

Induced polarization of a molecule

Induced polarization (Pi) and the electric dipole created in the molecule due to the presence of the electric field 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).

∴ μi ∝ F
When F is low, otherwise, hyperpolarization may occur.
μi = αi F
where αi is proportionality constant called induced polarizability of the molecule.

Induced dipole moment 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.

αi = μi/F
∴ Dimension of αi = (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, atomic number and the case of ionization energy. Thus atom behaves like a dipole and this dipole moment is induced by the applied electric field.

Clausius Mossotti equation from electromagnetic theory

Clausius Mossotti derived from electromagnetic theory, a relation between the polarizability and the dielectric constant of the non-polar medium between two plates.

Clausius Mossotti equation

Study online for schools and college-level

It gives the distortion produced in the 1 mole of the substance by a unit electric field. Induced polarizability is constant for the molecule and independent of temperature. Hence induced dipole moment 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.

The induced dipole moment of the substance can be calculated by measuring dielectric constant, density (ρ) and knowing the molar mass (M) of the substance.

Polar bonding molecules - Debye modification

Polar bonding molecules like methyl chloride, water, hydrogen fluoride, etc, molar polarization not constant but decreases with temperature.

Clausius Mossotti equation fails very badly for polar bonding 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 (gas molecules), two effects will occur.

Induced polarization of molecules

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 of molecules

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

∴ 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.

Polarization of bond

Total polarization of bond

For the polar bonding molecules, the total polarization,

Pt = Pi + P₀

Debye equation

Temperature dependence on polarizability

A polar bond is fixed and unable to orient in the fixed direction, the orientation polarization has taken zero. This is true for the condensed system where strong intermolecular forces prevent the free rotation of the molecules.

When the temperature is very high tending to infinity.

1/T = 0 and Pt = 0
∴ Pt = Pi

At high temperatures, the polar molecules rotate at such high speed that orientation polarization vanishes.

Molar heat capacity of gases

Physical chemistry , Properties of gases

Heat capacity definition chemistry

The heat capacity of a substance is defined as the amount of heat required to raise the temperature by one degree.

Usually, the rise in temperature measured in centigrade and heat capacity expressed in calories. Heat capacity depends upon the amount of material in the system.

Heat capacity per gram of substance is called specific heat and per mole called molar heat capacity or simply heat capacity of the substances.

Heat capacity at constant volume and constant pressure

The magnitude of heat capacity depends on the pressure and volume, especially in the cases of properties of gases. The temperature of a gm -mole of gas raised by one degree at constant volume. Again in another operation same raise of temperature allowing the volume to very.

The quantity of heat in the two operations would be different. Hence, in mentioning the heat capacity of gas, the condition has to specify.

Heat capacity at constant volume denoted by Cv.
Heat capacity at constant pressure denoted by Cp.

Molar heat capacity definition

The amount of heat required to raise the temperature of one gram of a substance by 1°K called specific heat.
The heat input required to rise by 1°K the temperature of one mole of the substances is called its molar heat capacity or simply heat capacity. The heat capacity denoted by C.

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⁻¹ K⁻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).

How is specific heat capacity measured?

Let dq heat required to increase the temperature dT for one mole of the substances and its heat capacity = C.

∴ C = dq/dT

dq is a path function and values of dq depend on the actual process followed. We can place certain restrictions to obtain precise values of heat capacity. The usual restrictions are at constant pressure and at constant volume.

Cv = [dq/dT]v and Cp = [dq/dT]p

Definition of internal energy and enthalpy,
dU = dqv and dH = dqp.

∴ Cv = [dq/dT]v = [dU/dT]v
Cp = [dq/dT]p = [dH/dT]p

Mechanical work in chemistry

A gas can be heated at constant pressure and constant volume in cylinder fitted with a piston. The gas expands against the piston gives the mechanical work.

In order to attain 1° rise in temperature, the heat supplied should be sufficient to provide the kinetic energy to the molecules and also able to do extra mechanical work.

Heat capacity from mechanical work

Cp is some mechanical work is required as additional energy to absorb for lifting piston from volume V₁ to V₂.

Cp - Cv = Mechanical work = PdV
= P(V₂-V₁) = PV₂-PV₁.

If we consider the gas to be an ideal gas
PV = RT.

Cp - Cv = R(T+1) - RT
∴ Cp - Cv = R = 2 calories.

Determination of Cv from kinetic energy

Consider the monoatomic gases like argon, helium. If such gases heated at constant volume, the heat supplied will be entirely utilized for increasing the kinetic energy of the translation of the molecules.

Since the monoatomic gas molecules can not any absorption in vibrational or rotational motion.

Volume remaining constant, no energy can be used to do any mechanical work of expansion. Kinetic energy for one-mole gas at the temperature T denoted by E.

E = (3/2)PV = (3/2)RT

Increase of kinetic energy for 1° rise in temperature for monoatomic gas helium, argon

Δ E = (3/2) [(T + 1) - T]R = (3/2) × R
= 3 calories.

The heat supplied at a constant volume equal to the rise in kinetic energy per unit degree rise in temperature.

∴ Cv = ΔE = 3 calories.

For a mole of monoatomic gas the ratio of two heat capacities universally expressed by the symbol ɣ.

ɣ = Cp/Cv = (Cv + R)/Cv
= (3 + 2)/3
= 1.66.

The total translational kinetic energy of gas molecules

For polyatomic molecules, the haet supplied used up not only in increasing kinetic energy but also increasing vibrational or rotational energies.

Let x calories of heat used for increasing vibrational or rotational purposes.

∴ Cv = ΔE + x = (3/2)R + x.

Cp = ΔE + mechanical work + x
= (3/2)R + R + x
= (5/2)R + x.

Heat capacity difference remains the same, two calories, as monoatomic gas, but heat capacity ratio

ɣ = {(3/2)R + x}/{(5/2)R + x}
= (5 + x)/(3 + x).

Heat capacity of gases from energy

Let molar energy of a gas E.
For monoatomic gas molecules

E = (3/2)RT
∴ Cv = [dE/dT]v = (3/2)R

For polyatomic linear gas molecules

E = (3/2)R + R + (3N - 5)R.
∴ Cv = [dE/dT]v = (3/2)R + R + (3N - 5)R

For polyatomic non-linear gas molecules

E = (3/2)R + R + (3N - 5)R.
∴ Cv = [dE/dT]v = (3/2)R + (3/2)R + (3N - 6)R

Organic properties of alkenes or olefins

Hydrocarbon , Organic chemistry

Physical properties of alkenes or olefins

The name olefin coming from the ethylene was called oil-forming gas or olefiant gas. Ethylene forms an oily liquid when treated with chlorine or bromine. The name olefine had granted wide usage and it was decided to compromise and call the series of olefins.

carbon atoms

solid

properties of alkenes

Van der Waals

Hydrogenation of alkenes

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

ΔH = -137 kJ mol⁻¹

ΔH = -125.9 kJ mol⁻¹

ΔH = -126 kJ mol⁻¹

ΔH = -126.8 kJ mol⁻¹

-119.7 kJ mol⁻¹

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.

Enthalpy of formation of alkenes

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)

R₂C=CR₂ 〉R₂C=CHR 〉R₂C=CH₂ ~ RCH=CHR 〉RCH=CH₂ 〉CH₂=CH₂

Organic synthesis 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.

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

organic chemistry

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 require 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.

Organic properties and synthesis of alkenes

Organic synthesis 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,

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.

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Markownikoff rule for alkenes

hydrogen

In the case of the halogen acids, the halogen atom is the negative part. 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 reaction of alkenes and Markownikoff rule

The chemical reaction of alkenes

electrons

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⁺

Organic methods of preparation of alkenes

Hydrocarbon , Organic chemistry

Study of alcohol dehydration reaction

The study of dehydration of alcohol in organic chemistry is important for school and college courses for examination. Primary alcohols when heated with concentrated sulfuric acid at 170° to 180°C to produce alkenes.

C₂H₅OH + H⁺
↓
C₂H₅OH₂⁺
↓
H₂O + C₂H₅⁺
↓
CH₂=CH₂ + H⁺

Secondary and tertiary-alcohol are best carried out by using dilute sulfuric acid. Tertiary alcohol can polymerize under the influence of concentrated sulfuric acid.

Acid-catalyzed dehydration of primary alcohol gives 1-alkenes but secondary and tertiary alcohol gives a mixture of alkenes due to rearrangement of carbocation form by an alcohol dehydration reaction.

CH₃–CH₂–CH(CH₃)–OH
↓
CH₃–CH₂–CH=CH₂ + CH₃–CH=CH–CH₃

Rearrangement reaction avoided by dehydration of alcohol over alumina in pyridine.

Alcohol dehydration reaction

Saytzeff's Rule for the alcohol dehydration reaction

Rearrangement often occurs with acid-catalyzed dehydration. All the three types of alcohol may behave in this way via a carbonium ion that may undergo methyl or hydride ion 1,2 shift. The major product is in accordance with Saytzeff's rule.
This rule may be stated in two ways

The predominant product is the most substituted alkene that is the one carrying the largest number of alkyl substituents.
Hydrogen is eliminated preferentially from the carbon atom joined to the least number of hydrogen atoms.

Cyclic elimination reaction of alkenes

Most eliminations occur by polar mechanism, whereas cyclic eliminations are uni-molecular non-polar reactions which take place in one step. Most occur when the compound is subjected to pyrolysis, and proceed via a cyclic transition state.

This mechanism is supported by the fact that these chemical reactions show a negative entropy of activation.

Pyrolysis of ester reaction, the ester usually acetate.

R₂CHCH₂OCOMe ⟶ RCH＝CH₂ + HOCOMe.

Ethanolic potassium hydroxide on alkyl halides

Propene can be prepared from propyl bromide by the action of ethanolic potassium hydroxide.

CH₃CH₂CH₂Br + KOH → CH₃CH＝CH₂.

Dehalogenation of 1,1 - di-halogen derivatives of alkanes by means of zinc dust and methanol produces alkenes.

CH₃CH₂CHBr₂ → CH₃CH＝CH₂.

Zinc dust and methanol also dehalogenation of 1,2- di halogen derivative of alkanes. Propene from propene dibromide.

CH₃CHBrCH₂Br+Zn → CH₃CH＝CH₂

Heating of quaternary ammonium hydroxide

(C₂H₅)₄N⁺ OH⁻ → CH₃CH＝CH₂

Boord has prepared alkenes by conversion of an aldehyde into its chloro - ether. This chloro - ether reacts with bromine followed by a Grignard reagent and finally treating the product with zinc and n-butanol.

This method is very useful for preparing alkenes of definite structure, and an interesting point about it is the replacement of the ∝-chlorine atom by bromine when the ɑ-chloro ether undergoes bromination in the β-position.

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Triphenylphosphonium ylide

The witting reagent, alkylidene-triphenyl phosphorane prepared by treating triarylphosphine usually the latter with an alkyl halide in ether solution. The resulting phosphonium salt is treated with a strong base such as C₆H₅Li, BuLi, NaNH₂, NaH, C₂H₅ONa.

Ph₃P + CH₃Br
↓
Ph₃PCH₃⁺Br⁻ + PhLi
↓
Ph₃P=CH₂

Wittig reaction affords an important and useful method for the synthesis of alkenes by the treatment of aldehyde or ketones with triphenylphosphonium ylide or phosphorane.

Ph₃P = CH₂ + Ph₂C=O
↓
Ph₂C = CH₂ + Ph₃P = O

Triphenylphosphonium ylide

2018

Application of Dipole Moment for Molecules

Dipole moment of benzene and its derivatives

Di-substituted derivative of benzene:

(v) Cis and Trans forms of Geometrical Isomerism:

(vi) Change of Dipole Moment with Temperature:

Bond polarity and dipole moments

Study polarity of the bond and dipole moments

What is the dipole moment of a molecule?

The polarity of bond in hydrochloric acid

CGS and SI unit of the dipole moment

The dimension of the dipole moment

What is the polarization?

Induced polarization of a molecule

Clausius Mossotti equation from electromagnetic theory

Polar bonding molecules - Debye modification

Induced polarization of molecules

Orientation polarization of molecules

Total polarization of bond

Temperature dependence on polarizability

Molar heat capacity of gases

Heat capacity definition chemistry

Heat capacity at constant volume and constant pressure

Molar heat capacity definition

How is specific heat capacity measured?

Mechanical work in chemistry

Determination of Cv from kinetic energy

The total translational kinetic energy of gas molecules

Heat capacity of gases from energy

Organic properties of alkenes or olefins

Physical properties of alkenes or olefins

Hydrogenation of alkenes

Enthalpy of formation of alkenes

Organic synthesis of alkenes

Combustion reactions of alkenes

Addition reactions of alkenes

Catalytic hydrogenation of alkenes

Addition of halogens to alkenes

Addition of halogen acids

Markownikoff rule for alkenes

Organic methods of preparation of alkenes

Study of alcohol dehydration reaction

Saytzeff's Rule for the alcohol dehydration reaction

Cyclic elimination reaction of alkenes

Ethanolic potassium hydroxide on alkyl halides

Heating of quaternary ammonium hydroxide

Triphenylphosphonium ylide

Chemistry 1

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