1.1.chemical thermodynamics ii -...
TRANSCRIPT
1.1.Chemical Thermodynamics II:
1.1.1. Free energy functions: Helmholtz free energy, Gibbs free energy, Variation of Gibbs free
energy with pressure and temperature, Gibbs-Helmholtz equation
1.1.2. Thermodynamics of open system: Partial molal properties, chemical potential and its
variation with pressure and temperature, Gibbs Duhem equation.
1.1.3. Concept of fugacity and activity
1.1.4. Chemical equilibrium and equilibrium constant: Equilibrium constant, KP and KC and
their inter relation, van’t Hoff reaction isotherm, van’t Hoff reaction isochore.
1.1. Free Energy and Equilibrium II
According to the first law of thermodynamics, when heat is supplied to the system, some part of
it is converting into work and remaining part is rejected or used to increase internal energy of the
system. How much quantity of heat is used to increase the internal energy or rejected to
surrounding is determined from the entropy change. But how much quantity of heat is used to
performed work done is determined by mew thermodynamic state function is called as free
energy. The heat energy is supplied in the form of internal energy or enthalpy.
[Heat energy (E or H)] = (Heat available to perform work done) +
(Heat unavailable to perform work done)
If heat energy is supplied in the form of internal energy, the part of internal energy available to
perform work is called Helmholtz free energy. It is denoted by the latter A. The internal energy
unavailable to perform the work done is TS, S-is entropy at temperature T.
Heat energy (E)] = (E available to perform work done) + (E unavailable to perform work done)
E = A + TS
A = E - TS …………………………………… 1
If heat energy is supplied in the form of enthalpy, the part of enthalpy available to perform work
is called Gibbs free energy. It is denoted by the latter G or F. The enthalpy unavailable to
perform the work done is TS, S-is entropy at temperature T.
[Heat energy (H)] = (H available to perform work done) + (H unavailable to perform work done)
H = G + TS
G = H - TS …………………………………… 2
1.1.1. Helmholtz free energy:
Helmholtz free energy of any system A is defined as-
A = E - TS
Where, E and S are internal energy and entropy of the system respectively. A, E, S and T are
state functions. The change in A during any process is given as-
ΔA = A2 - A1
= (E2 – T2S2) – (E1 – T1S1)
= (E2 - E1) – (T2S2 – T1S1)
= ΔE – ΔTS …………………………………….. 3
Equation (3) is the general definition of ΔA. For isothermal process T2 = T1, then equation (3)
becomes-
ΔA = ΔE – TΔS …………………………………… 4
Equation (4) explain physical interpretation (significance) of ΔA, under isothermal conditions –
TΔS = qr (according to the definition of entropy).
ΔA = ΔE – qr ………………………………….. 5
According to the first law of thermodynamics for reversible and isothermal process-
qr = ΔE + Wm or Wm = qr - ΔE ………………………………. a
From equation (5) and (a)-
ΔA = -Wm …………………………………………. 6
The equation (6) indicates that, at constant temperature the maximum work done by the system is
due to expense of (decrease in) Helmholtz free energy of the system. Therefore, A is called as
Work function of a system.
Variation of A with V and T:
According to the definition of A [i.e. from equation (1)]-
A = E - TS
Differentiate it completely as- dA = dE - TdS - SdT ……………… 7
According to the definition of entropy- dqr = TdS
Therefore equation (7) becomes- dA = dE - dqr - SdT …………. 8
According to the first law of thermodynamics- dqr = dE + PdV (for reversible process)
Or -PdV = dE - dqr
Therefore equation (8) becomes - dA = -PdV - SdT …………………… 9
Equation (9) indicates that A is most conveniently expressed in terms of a T and V as the
independent variables, we have -
A = f (T,V)
Differentiate it completely -
dA =
dT +
dV ......................... 10
Comparing equation (9) and (10) we have -
= -P ……………………….. 11
= -S ……………………….. 12
An alternative equation for the variation of A with T can be obtained as follows - Differentiate
with respect to T at constant V gives -
=
= –
=
=
……………………………………… 13
Equation (9) shows that the dependence of A for a pure substance on both T and V, these can be
shown by (12) and (13) for T and (11) for V.
Gibb’s free Energy:
If heat energy is supplied in the form of enthalpy, the part of enthalpy available to perform work
is called Gibbs free energy. It is denoted by the latter G or F. The enthalpy unavailable to
perform the work done is TS, S-is entropy at temperature T.
[Heat energy (H)] = (H available to perform work done) + (H unavailable to perform work done)
H = G + TS
G = H - TS …………………………………………… 1
Where, H and S are enthalpy and entropy of the system respectively. G, H, S and T are state
functions. The change in G during any process is given as-
ΔG = G2 - G1
= (H2 – T2S2) – (H1 – T1S1)
= (H2 - H1) – (T2S2 – T1S1)
= ΔH – ΔTS ………………………………….. 2
Equation (2) is the general definition of ΔG. For isothermal process T2 = T1, then equation (2)
becomes -
ΔG = ΔH – TΔS ………………………………….. 3
Relationship between G and A:
Consider equation (3) is- ΔG = ΔH – TΔS
But H is also state function- ΔH = ΔE + PΔV (at constant P)
Therefore, ΔG = ΔE + PΔV – TΔS
= (ΔE – TΔS) + PΔV (But, ΔA = ΔE – TΔS)
ΔG = ΔA + PΔV ……………………….. 4
In general- G = A + PV ……………………….. 5
The equation (4) and (5) shows the relationship between G and A.
Physical significance of G:
At constant temperature, TΔS = qr and at constant pressure, ΔH = ΔE + PΔV. Put these values
into equation (3) -
ΔG = (ΔE + PΔV) – qr
= - (qr - ΔE - PΔV)
But according to the first law of thermodynamics,
qr = ΔE + Wm or Wm = qr - ΔE
Therefore, ΔG = - (Wm - PΔV) ………………………….. 6a
Or -ΔG = (Wm - PΔV) …………………………… 6b
Equation (6b) shows that ΔG is the available free energy for the work done other than P-V type
at constant T and P. the work is done due to expense of Gibb’s free energy.
Variation of G with P and T:
According to the definition of G [i.e. from equation (1)] -
G = H - TS
Differentiate it completely as - dG = dH - TdS - SdT …………………. 7
According to the definition of entropy and first law of thermodynamics -
dqr = TdS = dH + PdV (for reversible process) …………….. 8
According to the definition of enthalpy - H = E + PV
Differentiate it completely - dH = dE + PdV + VdP ……………………. 9
Therefore equation (8) and (9) dH = VdP + SdT ……………………. 10
From equation (7) and (10) - dG = VdP + SdT -TdS -SdT
dG = VdP - SdT ………………………………………… 11
Equation (11) indicates that G is function of a T and P as the independent variables, we have -
G = f (T,P)
Differentiate it completely -
dG =
dT +
dP ………………… 12
Comparing equation (11) and (12) we have-
= V
= - S ……………………………………….. 13
An alternative equation for the variation of G with T can be obtained as follows- Differentiate
G/T with respect to T at constant P gives-
=
= –
=
=
……………………………………… 14
Equation (11) shows that the dependence of G for a pure substance on both T and P, these can be
shown by (14) and (13) for T.
ΔG for reaction:
For any chemical reaction such as -
ΔG for such reaction is sum of the G’s for the products minus a similar sum of for the reactants.
aA + bB + ............ cC + dD + ..................
ΔG = ΣGp – ΣGR ..................................................... 1
ΔG = [Hp – HR] – T[Sp - SR]
ΔG = ΔH - TΔS at constant T ………………….. 2
ΔH and ΔS are change in enthalpy and entropy of the reaction respectively at constant
temperature T and pressure P.
Also ΔG is function of T and P as -
d(ΔG) =
dT +
dP ……….. 3
These derivatives are also obtained by differentiate equation (2) with respect to T at constant P
and with respect to P at constant T.
=
- T
– ΔS
But
= Cp therefore
= ΔCp and
= ΔCp -
– ΔS
= – ΔS ……………………………… 4
Also,
= Vp – VR = ΔV ………………………………… 5
Where, ΔV is change in molar volume during the reaction. From equation (3), (4) and (5), we
have-
d(ΔG) = ΔV dP – ΔS dT ……………………….. 6
From equation (4) and (2) -
ΔG = ΔH +
…………………… 7
The equation (7) is called as Gibb’s-Helmholtz equation where (δΔG/δT)p is called as
temperature coefficient of a reaction.
Significance of Gibb’s-Helmholtz equation
Gibb’s-Helmholtz equation is - ΔG = ΔH +
………… 7
Consider the reaction: Zn + CuSO4 (sol) → ZnSO4 (sol) + Cu.
When this reaction is carried out in open beaker by adding zinc to a solution of CuSO4, heat of
reaction is equal to ΔH. If same reaction is carried out reversibly in a cell by applying the
external voltage which is infinitely smaller than that of cell emf. The work is done denoted by
the later ΔG instead if heat evolved. The difference between these two i.e. (ΔG - ΔH) is equal to
a either TΔS or T[δ(ΔG)/δT]p.
From equation (2) and (7) -
- TΔS =
……………….. 8
But by the definition of the entropy - TΔS = qr.
Therefore, -qr =
. ………………………… 9
And equation (7) becomes - ΔG = ΔH - qr
Or qr = ΔH – ΔG ………………………… 10
The equation (8) represents the heat interchange between the system and its surroundings when
process is conducted isothermally and reversibly.
Case - I: when ΔH > ΔG, qr is positive; heat is absorbed from the surrounding.
Case - II: when ΔH < ΔG, qr is negative; heat is evolved to the surrounding.
Case - III: when ΔH = ΔG, heat is neither absorbed nor evolved with the surrounding.
1.1.2. Thermodynamics of open system
We need a set of concepts that enable us to apply thermodynamics to mixtures of variable
composition. For a more general description of the thermodynamics of mixtures we have to
introduce other ‘partial’ properties, each one being the contribution that a particular component
makes to the mixture.
The equilibrium thermodynamic state of a simple one-component open system can be specified
by T, P, and n, the amount of the single component. This gives the differential relation for a
general extensive quantity, X, in a one-component system:
In a one-component system the molar quantity Xm is given by –
The molar quantity Xm is an intensive quantity. Because an intensive quantity cannot depend on
an extensive quantity, Xm depends only on T and P. Therefore –
In a one-component system any partial molar quantity is equal to the corresponding molar
quantity. The most important examples are –
(for one component system)
For ideal gas system,
, therefore above equation become –
Where, is the molar Gibbs energy in the standard state. It is equal to μ◦, the chemical
potential in the standard state. The standard state for the Gibbs energy of an ideal gas is the ideal
gas at pressure (exactly 1 bar).
Also, we derived the equation,
, therefore, equation for Sm is –
where
Also the partial molar enthalpy of a one-component ideal gas is obtained from –
=
=
=
The molar enthalpy of an ideal gas does not depend on pressure. The equation for molar volume
is -
(for one component system)
The equation for molar Helmholtz energy can be derived from the equation – A = G – PV,
=
–
According to Dalton’s law of partial pressures, each gas in a mixture of ideal gases behaves as
though it were alone in the container. The equation for applies to any substance in an ideal gas
mixture:
Where, is the chemical potential of substance i in the standard state at pressure and is its
partial pressure. All of the other equations for one-component ideal gases apply as well.
Partial Molal Quantities: A partial molar property is the contribution (per mole) that a
substance makes to an overall property of a mixture. Dependence of one variable with respect to
particular variable by keeping other variables constants is called as partial molal quantities.
Consider a solution composed of ‘j’ constituents and n1, n2, n3, …….,nj be the number of moles
of various constituents present. Let X is any extensive property of the system is said to be
function of P and T and the number of moles of the various constituents present as -
X = f (P, T, n1, n2, n3, …….,nj ) …………………… 1
Partial differentiation of equation (1) gives dX as -
dX = (dX∕dT)P,ni dT + (dX∕dP)T,ni dP + (dX∕dn1)P,T,ni dn1 + (dX∕dn2)P,T,ni dn2
+…….. + (dX∕dnj)P,T,ni dnj …………………………… 2
Where, ni represents in each instance all the n’s except the one with respect to which X is
differentiated then,
(dX∕dni)P,T,nj is called as partial molal quantity and is denoted by drawing bar over the symbol of
that property.
Therefore, X1 = (dX∕dn1)P,T,nj , X2 = (dX∕dn2)P,T,nj ,…….., Xj = (dX∕dnj)P,T,ni ….. 3
Put eq. (3) in eq. (2),
dX = (dX∕dT)P,ni dT + (dX∕dP)T,ni dP + X1 dn1 + X2 dn2+…….. + Xj dnj ……… 4
At constant temperature and pressure, eq. (4) becomes-
(dX)T,P = X1 dn1 + X2 dn2+…….. + Xj dnj ……………………….. 5
On integration equation (5) become- (by Euler’s theorem for homogeneous function)-
X = X1 n1 + X2 n2+…….. + Xj nj ………………………………… 6
Equation (6) shows that any extensive property of a solution at constant T and P, is expressed as
a sum of X x n for individual components of solution. In each product, the ‘n’ represents a
capacity factor and X represents a partial molal quantity represent an intensive factor i.e. partial
molal quantities are intensive properties of a solution.
Differentiating equation (6) completely-
(dX)T,P = X1 dn1 + X2 dn2+…….. + Xj dnj + n1 dX1 + n2 dX2+…….. + nj dXj
(dX)T,P = (dX)T,P + n1 dX1 + n2 dX2+…….. + nj dXj
Or n1 dX1 + n2 dX2+…….. + nj dXj = 0 ……………………… 7
For binary solution, j = 1 and 2.
n1 dX1 + n2 dX2 = 0
or dX1 = -n2/n1 dX2 …………………………………… 8
The equation (7) and (8) is called as Gibb’s-Duhem equation. This equation shows that a partial
molal quantity depends on each others. These are determined by variety of methods both
graphical and analytical.
When X = G, n1 dG1 + n2 dG2 = 0
Partial molal volume: Imagine a huge volume of pure water. When a further 1 mol H2O is
added, the volume increases by 18 cm3. However, when we add 1 mol H2O to a huge volume of
pure ethanol, the volume increases by only 14 cm3. The quantity 18 cm
3 mol
−1 is the volume
occupied per mole of water molecules in pure water; 14 cm3 mol
−1 is the volume occupied per
mole of water molecules in virtually pure ethanol. In other words, the partial molar volume of
water in pure water is 18 cm3 mol
−1 and the partial molar volume of water in pure ethanol is 14
cm3 mol
−1. In the latter case there is so much ethanol present that each H2O molecule is
surrounded by ethanol molecules and the packing of the molecules results in the water molecules
occupying only 14 cm3.
Partial molal volume of the ith component is denoted by later Vi and defined as-
Vi = (dV∕dni)P,T,nj
The subscript nj means that the number of moles of each component except ith
component is
constant. “Partial molal volume is the change in volume (V) when one mole of component is
added to large quantity of system at constant temperature and pressure”.
For binary solution,
V = f (P, T, n1, n2)
Partial differentiation of V gives dV as-
dV = (dV∕dT)P,n1,n2 dT + (dV∕dP)T, n1,n2 dP +( dV∕dn1)P,T,ni dn1 +( dV∕dn2)P,T,ni dn2
At constant temperature and pressure,
(dV)T,P = V1 dn1 + V2 dn2
On integrating, (V)T,P = V1 dn1 + V2 dn2
Example: What is the total volume of a mixture of 50.0 g of ethanol and 50.0 g of water at
25°C?
Solution: Calculate the amounts by using - nJ = mJ /MJ. The molar masses of CH3CH2OH and
H2O are 46.07 g mol−1
and 18.02 g mol−1
, respectively. Therefore the amounts present in the
mixture are -
= 1.09 mol
= 2.77 mol
For a total of (1.09 + 2.77) 3.86 mol. Hence xethanol = 0.282 and xwater = 0.718. The partial molar
volumes of the two substances in a mixture of this composition are 56 cm3 mol
−1 (density of
ethanol is 0.789 g/lit) and 18 cm3 mol
−1, respectively, so from the total volume of the mixture is -
V = (1.09 mol) × (56 cm3 mol
−1) + (2.77 mol) × (18 cm
3 mol
−1)
= 1.09 × 56 + 2.77 × 18 cm3 = 110 cm
3
Partial molal free energy:
The partial molar Gibbs energy has exactly the same significance as the partial molar volume.
For instance, ethanol has a particular partial molar Gibbs energy when it is pure (and every
molecule is surrounded by other ethanol molecules), and it has a different partial molar Gibbs
energy when it is in an aqueous solution of a certain composition (because then each ethanol
molecule is surrounded by a mixture of ethanol and water molecules). The partial molar Gibbs
energy is so important in chemistry that it is given a special name and symbol. From now on, we
shall call it the chemical potential and denote it μ (mu).
The name ‘chemical potential’ is very appropriate, for it will become clear that μi is a measure
of the ability of ith
component to bring about physical and chemical change. A substance with a
high chemical potential has a high ability, in a sense we shall explore, to drive a reaction or some
other physical process forward.
Partial molal free energy of the ith
component is denoted by later Gi (called chemical potential,
denoted by later μi) and defined as-
Gi = μi = (dG∕dni)P,T,nj
The subscript nj means that the number of moles of each component except ith
component is
constant. “Partial molal free energy or chemical potential is the change in free energy when one
mole of component is added to large quantity of system at constant temperature and pressure”.
Suppose a substance J occurs in different phases in different regions of a system. For instance,
we might have a liquid mixture of ethanol and water and a mixture of their vapours. Let the
substance J have chemical potential μj(l) in the liquid mixture and μj(g) in the vapour. We could
imagine an infinitesimal amount, dμj, of J migrating from the liquid to the vapour. As a result,
the Gibbs energy of the liquid phase falls by μj(l)dμj and that of the vapour rises by μj(g)dμj. The
net change in Gibbs energy is -
dG = μj(g)dμj − μj(l)dμj. = { μj(g) − μj(l)} dμj.
There is no tendency for this migration (and the reverse process, migration from the vapour to
the liquid) to occur if dG = 0. The argument applies to each component of the system. Therefore,
for a substance to be at equilibrium throughout the system, its chemical potential must be the
same everywhere.
For binary solution,
G = f (P, T, n1, n2)
Partial differentiation of G gives dV as-
dG = (dG∕dT)P,n1,n2 dT + (dG∕dP)T, n1,n2 dP +(dG∕dn1)P,T,ni dn1 +(dG∕dn2)P,T,ni dn2
At constant temperature and pressure,
(dμi)T,P = (dG)T,P = G1 dn1 + G2 dn2
On integrating, (μi)T,P = (G)T,P = G1 dn1 + G2 dn2
It is extensive property, is independent on the size of system. This equation is called as Gibb’s-
Duhem equation.
Variation of chemical potential with temperature and pressure (Chemical potential for
gaseous):
The general expression for the differential change of the Gibbs energy, G, is given by:
dG = - S dT + V dP ………………………………… 1
Where, S is the entropy, V the volume, P is the pressure, and T is the absolute temperature. From
equation (1) we have a relation between the change of G with respect to a change of pressure at
constant temperature, i.e.,
………………………………………….. 2
If we rewrite Equation (2) as a differential relation, we can calculate the Gibbs free energy as a
function of pressure, namely -
………………………. 3
Where, the temperature is held constant.
Since PV = nRT fo r an ideal gas, with R the gas constant and n the number of moles, we find
=
= nRT ln
…….. 4
In equation (4) an arbitrary pressure reference, Pref , corresponds to the pressure where Gibbs free
energy is equal to G0. Thus, the Gibbs free energy of an ideal gas at constant T is given by:
……………………. 5
In practice the values of Pref defined a set of conditions called standard state. Usually one
atmosphere or one bar is chosen for all gases; in the present discussion we will leave Pref as a
value to be defined. In this case the molar Gibbs free energy or chemical potential, μ, is given by:
…………………………… 6
According to equation (6), when P is equal to the reference pressure, Pref, the molar Gibbs free
energy is equal to the standard state molar Gibbs free energy . If one considers interactions
between particles, i.e., non-ideal gases, one has to replace the argument of the natural logarithm
by the fugacity, f. Thus we have -
…………………………… 7
where,
f(P) = P exp
……………………. 8
where,
=
…………………………………… 9
Equation (8) is the chemical potential for an arbitrary standard state defined by the reference
pressure, Pref. When Pref = 1, then equation (6) become –
…………………………… 10
The chemical potential becomes negatively infinite as the pressure tends to zero. As the pressure
is increased from zero, the chemical potential rises to its standard value at 1 bar (because ln 1 =
0), and then increases slowly (logarithmically, as ln p) as the pressure is increased further. This
conclusion is consistent with the interpretation of the chemical potential as an indication of the
potential of a substance to be active chemically: the higher the partial pressure, the more active
chemically the species. In this instance the chemical potential represents the tendency of the
substance to react when it is in its standard state (the significance of the term μ) plus an
additional tendency that reflects whether it is at a different pressure. The molar Gibbs energy of a
pure substance is the same in all the phases at equilibrium. A system is at equilibrium when the
chemical potential of each substance has the same value in every phase in which it occurs.
The total Gibbs energy of the two substances A and B mixture by using an expression –
G = nAGA + nBGB
Or G = nAμA + nBμB
At constant temperature and pressure, when an amount nA of A and nB of B of two gases
mixture –
ΔG = nRT {xA ln xA + xB ln xB}
(
Where n = nA + nB, The crucial feature is that because xA and xB are both less than 1, the two
logarithms are negative (ln x < 0 if x < 1), so ΔG < 0 at all compositions. Therefore, perfect
gases mix spontaneously in all proportions.
Chemical potential for solvent:
In the case of solutions, where the solvent is at equilibrium with its vapor, we know that -
μsolvent = μvapour. ………………………………………….. 11
Since equation (11) applies to the chemical potential of an ideal vapor (from equation 6), we
have –
μsolvent = μvapour μ
…………… 12
For a pure solvent, P is the equal to the pure solvent vapor pressure, P*. Thus the pure liquid
chemical potential is given by:
solvent μ
………………… 13
Solving this equation for μ and substituting in equation (12) we get –
μsolvent
………………… 14
In the case of an ideal solution, Raoult's law is satisfied. Therefore, the following relation
between the vapor pressure and composition is observed:
= …………………………… 15
Where, X is the mole fraction. Therefore for ideal solutions we have notice that now the
reference point is the pure solvent.
μsolvent ………………… 16
For a non-ideal solution, the following relation is defined:
……………… 17
Where, is defined as the activity and as the activity coefficient. In this case the
solvent chemical potential of a non-ideal solution is given by:
μsolvent ………………… 18
Equation (18) is the most general expression for the solvent's chemical potential where the non-
ideality is taken in consideration in the activity coefficient. Also notice that for pure solvent the
activity and the activity coefficient are equal to unity.
1.1.3. Concept of activity:
For an ideal gas, either pure or in a mixture, equation for chemical potential is -
……………………. 1
For a component of an ideal solution or for the solvent in a dilute solution, we have –
=
2
Where, is the chemical potential of substance i in the appropriate standard state and ai, the
activity of substance i.
The activity is -
(pressure) = (mole fraction) =
(molality) =
(concentration)
The activity is a dimensionless quantity that is equal to unity if the substance is in its standard
state. Each measure of the activity corresponds to a different standard state. For example, the
standard state for a component in an ideal solution or a solvent in an arbitrary solution is the pure
substance. The standard state for a solute that obeys Henry’s law is the hypothetical pure
substance that has a vapor pressure equal to ki, the Henry’s law constant. The standard state of a
dilute solute in the molality description is a hypothetical solute with a molality of 1 mol kg−1
such that Henry’s law is obeyed at this composition. The standard state of a dilute solute in the
concentration description is a hypothetical solute with a molar concentration of 1 mol m−3
or 1
mol L−1
such that Henry’s law is obeyed at this composition.
The standard state of a pure solid or liquid is the pure substance at pressure . If the substance
is at a pressure P not equal to ,
= ……….. 3
Where we assume that the solid or liquid has a nearly constant volume. By comparison with
equation (2),
……………………. 4
The activity of a pure liquid or solid at pressure P is given by –
(For pure solid or liquid) …………… 5
The exponent in equation (5) is generally quite small unless P differs greatly from . For
ordinary pressures we will assume that the activity of a pure solid or liquid is equal to unity.
The activity of a non-ideal gas to be the ratio of the fugacity to :
(for non-ideal gas) …………………… 6
We define the activity coefficient γi of a non-ideal gas as –
……………………… 7
So that the activity coefficient of a gas equals the ratio of the fugacity to the pressure:
=
……………………… 8
The activity coefficient of an ideal gas equals unity. The chemical potential of a non-ideal gas
can be written –
=
=
…. 9
The activity coefficient of a gas is also known as the fugacity coefficient and is sometimes
denoted by φi instead of by γi. If the value of the activity coefficient of a real gas is greater than
unity, the gas has a greater activity and a greater chemical potential than if it were ideal at the
same temperature and pressure. If the value of the activity coefficient is less than unity, the gas
has a lower activity and a lower chemical potential than if it were ideal.
Concept of fugacity:
For an ideal gas, we have –
………………….. 1
Where, is the value of the chemical potential for a particular pressure . When a gas
requires corrections for non-ideality we write a new equation in the same form as equation (1),
replacing the pressure by the fugacity, f, which has the dimensions of pressure. The fugacity
plays the same role in determining the molar Gibbs energy of a real gas as does the pressure in
determining the molar Gibbs energy of an ideal gas. The quantity (T) is the molar Gibbs
energy of the gas in its standard state. The standard state of a real gas is defined to be the
corresponding ideal gas at pressure .
For a real gas, one defines the chemical potential as –
…………………… 2
The formula in following equation is equivalent to integrating from the standard - state pressure
down to zero pressure with the ideal gas, and then integrating back up to pressure P with the
real gas. This procedure gives the difference between the real gas at pressure P and the ideal gas
at pressure . Now, the difference between the real and ideal chemical potential can be written
as –
…………………. 3
Where, we have used the compressibility, Z(P) =
. From equation (3), we get following
equation for -
……………………….. 4
The real gas and the corresponding ideal gas become identical in the limit of zero pressure. Since
real gases at low pressures behave ideally, equation (3) reduces to –
………………………. 5
Therefore we pick an and such that –
………………………………….. 6
The equation (6) indicates that –
……………………………….. 7
But on the other hand, the value of the chemical potential is independent of the reference
pressure i.e.,
…….. 8
Where Pref is the reference pressure that defines the standard state. Hence, we find the following
relation between reference chemical potential.
…………………………. 9
–
………………………. 10
But,
……………………… 11
Where the fugacity is defined by –
……………………… 12
The final expression for the chemical potential for a real gas is given by:
…………………….. 13
Finally, we have to choose . From equation (12), we notice that –
………………… 14
Consequently we pick = 0.
1.1.4. Chemical equilibrium:
Many chemical reactions do not go to completion but instead attain a state of chemical
equilibrium. Chemical equilibria are important in numerous biological and environmental
processes. For example, equilibria involving O2 molecules and the protein hemoglobin play a
crucial role in the transport and delivery of O2 from our lungs to our muscles. Similar equilibria
involving CO molecules and hemoglobin account for the toxicity of CO. The another example is
- When a liquid evaporates in a closed container, molecules with relatively higher kinetic energy
escape the liquid surface into the vapour phase and number of liquid molecules from the vapour
phase strike the liquid surface and are retained in the liquid phase. It gives rise to a constant
vapour pressure because of an equilibrium in which the number of molecules leaving the liquid
equals the number returning to liquid from the vapour.
Chemical equilibrium is state in which the rates of the forward and reverse reactions are equal
and the concentrations of the reactants and products remain constant. Equilibrium is a dynamic
process in which the conversions of reactants to products and products to reactants are still going
on, although there is no net change in the number of reactant and product molecules. The mixture
of reactants and products in the equilibrium state is called an equilibrium mixture.
Based on the extent to which the reactions proceed to reach the state of chemical equilibrium,
these may be classified in three groups.
(i) The reactions that proceed nearly to completion and only negligible concentrations of the
reactants are left. In some cases, it may not be even possible to detect these experimentally.
(ii) The reactions in which only small amounts of products are formed and most of the reactants
remain unchanged at equilibrium stage.
(iii) The reactions in which the concentrations of the reactants and products are comparable,
when the system is in equilibrium.
For a better comprehension, let us consider a general case of a reversible reaction,
A + B C + D
With passage of time, there is accumulation of the
products C and D and depletion of the reactants A and B
(as shown in following fig.). This leads to a decrease in
the rate of forward reaction and an increase in the rate of
the reverse reaction. Eventually, the two reactions occur
at the same rate and the system reaches a state of
equilibrium.
Similarly, the reaction can reach the state of equilibrium even if we start with only C and D; that
is, no A and B being present initially, as the equilibrium can be reached from either direction.
There are number of important questions about the composition of equilibrium mixtures: What is
the relationship between the concentrations of reactants and products in an equilibrium mixture?
How can we determine equilibrium concentrations from initial concentrations? What factors can
be exploited to alter the composition of an equilibrium mixture? The last question in particular is
important when choosing conditions for synthesis of industrial chemicals such as H2, NH3, CaO
etc. To answer these questions, let us consider a general reversible reaction:
A + B C + D
where A and B are the reactants, C and D are the products in the balanced chemical equation. On
the basis of experimental studies of many reversible reactions, the Norwegian chemists Cato
Maximillian Guldberg and Peter Waage proposed in 1864 that the concentrations in an
equilibrium mixture are related by the following equilibrium equation,
KC x C
C x C=
A B
LM
C
a b
lm
C C C CA B LM, , , are the molar concentrations of A, B, C andD respectively.
where Kc is the equilibrium constant and the expression on the right side is called the equilibrium
constant expression. The equilibrium equation is also known as the law of mass action because
in the early days of chemistry, concentration was called “active mass”.
Equilibrium constant : The law of mass action gives relationship between concentration of
reactant and rate of reaction at any temperature T. “It is also stated as- the rate of a chemical
reaction at any instant is proportional to the molar concentration of the reacting substance at that
instant.” Consider the simplest reaction –
A → product.
If CA is the concentration of the reactant at time t, then-
Rate =
α
An equation which shows that how the rate of chemical reaction is related to the concentration is
called as rate law or rate equation. “The proportionality constant in the above rate equations is
called rate constant or velocity constant or specific rate constant.”
From above equation
α CA
= k CA
If CA = 1, then rate =
= k i.e. Rate constant of reaction is the rate of reaction when all the
reactants are at unit concentration or activity. “The chemical reactions shows (exhibiting) the
tendency to reverse themselves are called as reversible or opposing reaction.”
In reversible or opposing reactions, the products formed are also react and gives back the
reactants. Initially rate of product formation (forward reaction) is very large and is goes on
decreasing with time whereas initially rate of opposing reaction or backward reaction (reactant
formation) is zero and is increasing with time. After some time t, a stage is reached when two
rates are equals. It is called as equilibrium stage. It is dynamic in nature i.e. all the species are
reacting at the rate at which they are being formed. At equilibrium stage, overall rate of reaction
is zero i.e. = 0.
Consider following reversible reaction-
k
k
f
b
A B
Where, kf and kb are the rate constants for the forward and backward reactions respectively.
Depending on the physical state of the reactants and products, two physical constants for
reversible reaction is defined as KP and KC. At gaseous state, the equilibrium constant KP is in
terms of partial pressures of reactants and products while in other state, it is denoted by KC which
is in terms of partial concentrations.
Consider general reaction-
aA + bB lL + mMA & B are reactants (behaves ideally)L & M are products (behaves ideally)
a & b and c & d are number of moles of reactants and products respectively.
When concentrations of both reactants and products are expressed in molar terms, then KC is-
Where, are the molar concentrations of A, B, C and D respectively.
Characteristics of Kc:
1. Expression for equilibrium constant is applicable only when concentrations of the reactants
and products have attained constant value at equilibrium state.
2. The value of equilibrium constant is independent of initial concentrations of the reactants
and products.
3. Equilibrium constant is temperature dependent having one unique value for a particular
reaction represented by a balanced equation at a given temperature.
4. The equilibrium constant for the reverse reaction is equal to the inverse of the equilibrium
constant for the forward reaction.
5. The equilibrium constant K for a reaction is related to the equilibrium constant of the
corresponding reaction, whose equation is obtained by multiplying or dividing the equation
for the original reaction by a small integer.
Magnitude of Kc:
i. If the Kc value is large (Kc >> 1), the equilibrium lies to the right and the reaction mixture
contains mostly products.
ii. If the Kc value is small (Kc <<1), the equilibrium lies to the left and the reaction mixture
contains mostly reactants.
iii. If the Kc value is close to 1 (0.10 < Kc < 10), the mixture contains appreciable amounts of
both reactants and products.
Because gas pressures are easily measured, equilibrium equations for gas-phase reactions are
often written using partial pressures rather than molar concentrations. For the gaseous state, the
partial pressure is also measure the activity of the component, KP is-
Where, are the molar concentrations of A, B, C and D respectively.
Relationship between KP and KC :
Consider general reaction -
aA + bB lL + mMA & B are reactants (behaves ideally)L & M are products (behaves ideally)
a & b and c & d are number of moles of reactants and products respectively.
Therefore,
KP x P
P x P=
A B
LM
p
a b
lm
P P P PA B LM, , , are the partial molar pressure of A, B, C andD respectively.
For n mole of ideal gas- pV = nRT
p =
=
But n/V = C, therefore, p = C R T.
Substituting this value of p in above KP equation, then-
Kx
x=
A B
LM
p
a b
lm
(C RT)
(C RT) (C RT)
(C RT)
K =p
C x C
C x C
A B
LM
a
lm(RT)
(l+m)-(a-b)
But (l + m)-(a + b) = ∆n, therefore,
K =p K C (RT)n
This is the relationship between KP and KC. When ∆n = 0, then KP = KC.
The Reaction isotherm (Van’t Hoff’s Isotherm):
A quantitative relation of the free energy change during the chemical reaction has been
developed which is known as Reaction isotherm Van’t Hoff’s isotherm.
Van’t Hoff’s isotherm gives the net work done that can be obtained from gaseous reactants at
constant temperature, when both reactants and products are at suitable orbitary pressure.
Consider a general reaction-
aA + bB + ............ cC + dD + ..................
Where a, b, c and d are number of moles of reactants A and B and products C and D
respectively. If aA, aB, aC and aD are activities of reactants A, B and products C, D respectively.
The free energies of each of these substances per mole at T is-
GA = + RT ………………. 1
GB = + RT ………………. 2
GC = + RT ………………. 3
GD = + RT ………………. 4
Where, ,
, and
are the free energies at unit activity of the A, B, C and D respectively
or are the standard free energies of A, B, C and D respectively.
Therefore the free energy change of the chemical reaction is difference between sum of free
energies of product and sum of free energies of reactants.
ΔG = ΣGp – ΣGR
= [c GC + d GD + …….] – [a GA +b GB + ……..]
= [c + RT
+ d + RT
+ …….] –
[a + RT
+ b + RT
+ ……..]
= [(c + d
+ ……) – (a + b
+……)] +
RT[( +
+ …….) – ( + RT
+ ……..)]
ΔG = ΔGo + RT ln
………… 5
Where ΔGo is the free energy change of the reaction in standard condition or state i.e. free energy
change involved when the starting materials and the product formed having unit activity at
constant T.
The equation (5) is called as the Van’t Hoff’s isotherm or reaction isotherm. When activities of
all reactants and products are equal to one them ΔG = ΔGo. For any reaction, ΔG
o is constant at
any given temperature and is completely independent on the pressure. The variation of ΔGo with
temperature is explain by using following equation-
ΔGo = ΔH
o - TΔS
o
At equilibrium step of the reaction- ΔG = 0
0 = ΔGo + RT ln
At equilibrium, according to the law of mass action, Ka =
Therefore, 0 = ΔGo + RT ln Ka
Or ΔGo = - RT ln Ka ………….. 6
Ka = e-ΔGo/RT
……………… .. 7
Form equation (5) and (6) -
ΔG = RT ln
- RT ln Ka
= RT ln Qa - RT ln Ka where Qa =
= RT ln (
) ………………… 8
Where Qa is called reaction quotient, is the resulting value when we substitute reactant and
product concentrations into the equilibrium expression.
1. If Q > K, the reaction will go to the left. The ratio of products over reactants is too large & the
reaction will move toward equilibrium by forming more reactants.
2. If Q < K, the reaction will go to the right. The ratio of products over reactants is too small &
the reaction will move toward equilibrium by forming more products.
3. If Q = K, the reaction mixture is already at equilibrium, so no shift occurs.
For gases activities are proportional to the partial pressures of the components of reaction
mixture hence the activities of the equation (5) can be replaced by the partial pressures,
therefore-
ΔG = ΔGo + RT ln
……………. 9
= RT ln Qp - RT ln Kp ………………… 10
Where Kp is the equilibrium constant and Qp =
At equilibrium state, ΔG = ΔGo and Qp = 1.
ΔGo = - RT ln Kp = -W or Kp = e
-ΔGo/RT ……………….. 11
i.e. net work done of the reaction is equal to the decrease in free of the system and it is calculated
by using expression- RT ln Kp or 2.303RT ln Kp.
It will be observed that, ΔG is positive when Kp is less than unity. If ΔG is negative when Kp is
greater than unity and is zero when Kp =0.
Van’t Hoff Isochore:
The Van’t Hoff isochore is obtained by comparing the Van’t Hoff isotherm with Gibbs-
Helmeholtz equation.
The Gibbs Helmeholtz equation is ΔG = ΔH + T
Or - ΔH = T
- ΔG
Dividing both side by T2 gives:-
=
The right hand side of the above equation is obtained by differentiating ΔG/T w.r.t. T at constant
P-
=
=
……………… 12
According to Van’t Hoff isotherm- ΔG = - RT ln Kp ……………….. 13
Dividing both side by T and differentiating it w.r.t. T at constant P-
= R
………. 13a
Comparing equation (13) and (12), we have-
= R
Or
= d (ln Kp) ………………….. 14
The above equation is known as Van’t Hoff isochore. For applying isochore to any chemical
reaction, it is essential to integrate it. If ΔH remains same over the range of temperature, we have
on integration-
ln Kp =
=
+ constant of integration.
Appling the limits T1 and T2 at the equilibrium constants Kp1 and Kp2 respectively, we have-
lnKp2 - lnKp1 =
ln
=
ln
=
…………. 15
By using this equation, equilibrium constant at any temperature or heat of reaction is calculated.
In reversible reaction, the sign of ΔGo indicates whether the forward or reverse reaction is
spontaneous as- ΔGo = - 2.303RT log Kp
i) If ΔGo is negative; log K must be positive & K is greater than one and reaction proceeds
spontaneously in forward direction.
ii) If ΔGo is positive; log K must be negative & K is less than one and reaction proceeds
spontaneously in reverse direction.
iii) If ΔGo is zero; log K must be zero & K is equal to one and reaction is in equilibrium state.
Van’t-Hoff equation in terms of Kc:
We known that equilibrium constant in terms of partial pressure Kp and in terms of concentration
Kc are related to each other by the equation as-
Kp = Kc (RT)Δn
Taking logarithm of both side-
ln Kp = ln Kc + Δn ln (RT)
Differentiate it with respect to temperature, we get-
=
+
…………… 16
From equation (16) and (14)-
=
+
=
-
………………….. 17
=
-
………………….. 18
=
But (ΔH – ΔnRT) = ΔE,
=
………………. 19
Where, ΔE is the heat of reaction at constant volume.
1.1.Mechanism of organic reactions
1.1.1. Thermodynamic and Kinetic control reactions. Concept with mechanism of the
following reactions: Addition of HX to butadiene; Sulfonation of naphthalene.
Nucleophilicity/Electrophilicity vs Basicity/acidity.
1.1.2. Mechanism of elimination reaction with stereochemistry: E1 and E2 reactions,
Regioselectivity (Saytzeff and Hofmann rule).
1.1.3. Mechanism of reactions of carbonyl compounds with nucleophile: Formation of
acetal/ketal from aldehyde and ketone; reaction of aldehyde and ketone with primary and
secondary amines; Acyl nucleophilic substitution (tetrahydral mechanism) – Acid
catalyzed esterification of carboxylic acid, base catalyzed hydrolysis of ester.
1.1.4. Mechanism of rearrangement with examples and stereochemistry: Migration to electron
deficient carbon – Pinacol, Benzilic acid; Migration to electron deficient nitrogen –
Beckmann, Hofmann.
1.1.5. Mechanism of the following reactions with synthetic applications: Claisen condensation,
Michael addition.
Mechanism of organic reactions
Mechanism with stereochemistry of reactions
Mechanism of chemical reaction is the step by step theoretical representation of chemical
reaction i.e. theoretical actual series of discrete steps involve during the transformation of
reactants to products.
During the organic reactions, breaking of co-valent bonds/shifting of atoms, etc takes place. The
breaking of atoms takes place either homolytically or heterolyticaly forming free radicals or ions
respectively. Depending on nature of reaction or intermediate formed there are various types of
reactions:
1. Addition reaction; 2. Elimination reaction; 3. Substitution reaction; 4. Rearrangement; 5.
Pericyclic reaction; 6. Photochemical and free radical reaction; etc
Elimination reactions
The reaction in which the site of unsaturation goes on increasing are called as elimination
reactions.
The reaction in which two atoms or group of atoms (of same or adjacent or any two carbon atom)
are removed from the molecule result the formation of double bond or ring. During the
elimination reaction, in most of cases, one eliminating atom is hydrogen.
According to the position of eliminating atom, the elimination reaction has been calssify as 1,1-
or α-elimination, 1,2- or β-elimination, 1,3- or γ-elimination and so on. The 1,2- or β-elimination
is most common elimination.
Mechanism of β-elimination
Types of β-elimination:
There are two types of β-elimination:
One occurs in solution phase (forming carbonium or like intermediate i.e. atom leaves with
bonding electrons) and another occurs in gas phase (forming carbainon or like intermediate i.e.
atom leaves without bonding electrons; in most of cases hydrogen).
There are two mechanisms for β-elimination as E2 elimination and E1 elimination mechanism.
E2-elimination mechanism:
It is single step second order elimination reaction in which proton is removed by the base from
the β-carbon atom with simultaneous loss of leaving atom or group. This mechanism is proceeds
through transition state not through any ionic intermediate.
ClH
ClCl
Cl
ClCl
2
CH3
BrCH
3BrH
CH3
Br+1,1-elimination
:
1,2-elimination+
1,5-elimination
X
HH
X- +
In gas phase In solution phase
CarbanionCarbonium
CH3
H
H
H Br
H
CH3
H
H
H Br
H CH3
H H
H
OH2
e.g. Dehydrobromination of 1-bromopropane
HO HOd
d
Transition state
+ Br +slow fast
The first step of reaction is slow while second step is fast. Therefore rate of reaction is depends
on both concentration of base and the substrate.
It is stereospecific mechanism because this reaction proceeds with greater selectivity when both
the eliminating groups are anti to each others.
The stereochemistry of E2 mechanism can be understand by considering overlap of p- orbitals for
the formation of π bond. The p- orbitals of adjacent carbon should be parallel to each other so
that they are undergoes maximum overlapping. To get this orientation, the eliminating group
should be in one plane and anti- periplanar.
The anti- periplanar conformation is preferred for the elimination as- The electron pair retains on
carbon while that is loses along with leaving group are kept far away from each others.
The anti- periplanar conformation is the more stable staggered conformation whereas syn-
periplanar conformation is the less stable eclipsed conformation. Hence elimination
preferentially takes place from the more stable anti- periplanar conformation.
E1 Mechanism:
It is two step first order mechanism of elimination reaction. The first step (ionization step) is
slow forming carbonium ion while second step of reaction is fast in which base abstract the
proton forming elimination product.
H
X
H
X
anti-peri planar syn-peri planar(trans coplanar) (cis coplanar)
Staggered conformation eclipsed conformation
CH3
H
CH3
Br H
H CH3
H
CH3
H
H
CH3
CH3
H
H
OH2
CH3
H
CH3
H
H
e.g. Dehydrobromination of 1-bromopropaneHO
slow fast++ +
Step-I Step II
Carbonium ion
The first step is slow, therefore rate of reaction is only depends on concentration of substrate but
not on base concentration. During this mechanism forming intermediate planar carbonium ion.
As substrate changes from primary to secondary to tertiary, the reactivity of substrate via E1
mechanism goes on increasing. It does not depends on nature and concentration of base.
Factor affecting E1 and E2 mechanism:
Nature of substituent: It is very important factor decides whether reaction proceeds either E1 or
E2 mechanism. As substitution on the carbon of C-X increases, rate of elimination via E1
mechanism increases i.e. primary alkyl halides follow E2 mechanism while tertiary alkyl halide
follow E1 mechanism. In case of secondary alkyl halide, both the mechanism are possible
depending on the stability of TS or carbonium ion.
In case of E2 mechanism, base abstract proton in slow step; as strength and or concentration of
base increases, rate of reaction via E2 mechanism increases. But in case of E1 mechanism, base
abstract the proton from carbonium ion in fast step; the strength and or concentration of base
does not much more affect the rate of reaction via E1 mechanism.
E2 mechanism occurs via transition state (charge is dispersed); less polar solvent is favors it. E1
mechanism occurs via carbonium ion which is stabilized by solvation effect of polar solvent;
polar solvent is favors it.
Regioselectivity of elimination reactions:
The chemical reaction in which the starting materials reacts to give mixture of products in which
one is major product is called as stereoselective or regioselective reaction.
During the elimination reaction, there are more than one β-hydrogen atoms, so give the mixture
of products in which more substituted alkene is major product.
The Saytzeff rule:
Russian chemist Zaitsev in 1875 formulate this rule. Zaitsev’s name is transliterated as Saytzeff.
CH3
CH3
Br
H
CH3
CH3 CH
3
CH3
CH2
CH3+ +
NaOH
2-bromobutane trans-2-butene cis-2-butene 1-butene
“During the dehydrohalogenation reaction of alkyl halide, the hydrogen atom of the more
substituted carbon is removed preferentially i.e. hydrogen of the β-carbon carries less number of
hydrogen atom i.e. after elimination reaction forming more substituted alkene as major product.”
e.g. Dehydrobromination of 2-bromobutane forming 81% 2-butene and 19% of 1-butene.
2-Chloro-2-methylbutane forming a mixture of 2-methyl-2-butene and 1-chloro-2-methyl butene.
In the E1 mechanism, elimination of proton takes place after the elimination of leaving group.
Hence, elimination of it (leaving group) does not decides the position of double bond. Under
such conditions, the orientation and stability of carbonium ion decides the stability of alkene.
More substituted alkene is formed as major product.
Ingold (1941) explain the Saytzeff rule by considering example: Dehydrobromination of 2-
bromobutane via E2 mechanism forming two products as-
The sigma electrons of C-H bonds of –CH3 groups at β’ and γ position in the transition state (I)
can shows hyperconjugation with partially formed C=C bond between α,β-carbon atoms.
In transition state II, only the β-CH2 group shows hyperconjugation with the partially formed
C=C bond between α,β’-carbon atoms. Therefore T.S. I is more stable than T.S. II. Hence 2-
butene is obtained as major product.
Hofmann rule:
“The quaternary ammonium or sulphonium bases undergoes decomposition on heating, the
hydrogen atom attached to less substituted β-carbon atom(attached to more hydrogen atoms) is
prefencially removed forming less substituted alkene as major product i.e. less substituted alkene
is obtained as major product.”
CH3
Br
CH3 CH
3
CH3
CH2
CH3+
Strng acidic cond.
2-brombutane E/Z 2-alkene 1-alkene
CH3
Br
CH3
CH3
CH3
CH3
CH3
CH2
CH3
CH3
+Strng acidic cond.
2-bromo-2-methybutane E/Z 2-alkene 1-alkene
CH3
H
CH3
H
Br
HOH
CH3
H
Br
H
H
H
OHd-
d-
d-
d-
T.S.I T.S. II
e.g. 1. Ethyldimethylpropylammonium hydroxide on heating gives 98% ethene and 2% propene.
2. Ethyl-1,1-dimethylethyldimethylammonium hydroxide on heating gives 99% ethene and 1%
2-methylpropene.
Any factor which cans facilate the removal of proton can also affect the rate of reaction. When
there are more than one β protons, one can removed easily decides the structure of alkene (major
product).
Explanation of Hofmann rule:
The positive charged nitrogen atom of R3N- group shows strong +I effect on adjacent carbon and
therefore hydrogen atoms. The hydrogen atom attached to carbon having more positive charge
density is more acidic in nature and removed easily. As the number alkyl or other substituent on
the carbon which stabilizes the positive charge increases, acidity of proton decreases. Therefore
the hydrogen atom attached to less substituted carbon atom is more acidic and eliminate easily
forming less substituted alkene as major product.
The bond between carbon and nitrogen is difficult to break so that hydrogen can be departs first
and forming carbanion (carbon carries negative charge). As the substitution on the carbanion
goes on increases, decrease its stability i.e. primary > secondary > tertiary carbanion i.e.
hydrogen atom of less substituted β-carbon is departs easily. The carbanion acts as internal
nucleophile and departs R3N group forming alkene.
Mechanism of elimination reactions with stereochemistry:
CH3
N
CH3
CH3
CH3
CH2
CH2 CH
3
HO ++ -
98% 2%
CH3
N
CH3
CH3
CH3
CH3
CH3
CH2
CH2 CH
2
CH3
CH3
HO ++ -
99%1%
CH3
N
CH3
CH3
CH3
CH3
CH3
HO+ -
less acidic in nature due to +I effect of other two methyl groups.
more acidic in nature and departs easily.CH
3N
CH3
CH3
CH3
CH3
CH3
CH2
CH2
CH2
CH3
CH3
HO CH3
N
CH3
CH2
CH3
H2C
CH3
++ -
99% 1%
+
-
-
more stable
stabletwo methyl group decreases the acidity ofmethyl group.
During the chemical reaction, site of unsaturation goes on increasing by the removal of atoms or
groups of atoms or both is called elimination i.e. number of double or triple bonds goes on
increasing.
Br
CH3
CH3 CH3
CH2
CH3
CH3
+ OH2HO-
Br-+ +
The elimination reactions are further classified according to position of eliminating groups as
1,1- or α-elimination, 1,2- or β-elimination, 1,3-elimination, 1,4-elimation, 1,5-eliminations, etc.
If both atoms or groups are eliminated from same carbon atom and forming carbine is called as
1,1-elimination or α-elimination. The 1,2- or β-elimination is most common and gives alkene.
The 1,3-, 1,4-, 1,5-, etc are forming rings rather than double bond.
O
CH3
Cl OBase
According to the mechanism of 1,2- or β-elimination, it can be classified as E2, E1, E1CB, Ei, etc
elimination.
Br
CH3
CH3H
CH2
CH3
CH3
OH2
HO-
Br-+ +
Rate = k [t-BuBr] [HO-]
The hydroxide is behaving as base because it is attacking the hydrogen atom, instead of the
carbon atom. The hydrogen atom is not acidic, but proton removal can occur because bromide is
a good leaving group. As hydroxide ion attacks to hydrogen, the bromide is force to leave, taking
with it the negative charge (along with bonding electrons). It is the elimination reaction proceeds
through transition state. The formation of transition state is slow step of reaction. Therefore rate
of the reaction is depends on the concentration of substrate (tert-butyl bromide) and base
ClH
ClCl
Cl
ClCl
2
CH3
BrCH
3BrH
CH3
Br+1,1-elimination
:
1,2-elimination+
1,5-elimination
(hydroxide ion, base). This means that rate of the reaction affects both concentrations. Hence it is
bimolecular reaction.
Br
CH3
CH3H
CH2
CH3
CH3
OH2HO
-Br
-+ +HO
C Br
CH3
CH2 CH3
H
Transition state
slow fast
E1 elimination mechanism is the unimolecular 1,2- or β-elimination reaction. The rate of the
reaction is only depends on concentration of substrate but not on the concentration of base. It is
proceeds in two steps, first step is slow (rate determining step) forming carbonium ion
intermediate while second step is fast proceeds with elimination of proton (abstract by base).
e.g. Dehydration of tert-butanol to 2-methylpropene in presence of sulfuric acid.
H2SO4 H+
HSO4
-
+ Rate = k [t-BuOH]
OHCH3
CH3CH3
O+H2CH3
CH3CH3
CH3
CH3CH3 + Step - I
Generation of carbonium ion
2-methylpropan-2-ol
H+
Slow
Carbonium ion(Planar)
HSO4
-
CH3
CH3H
+ Fast
CH3
CH2CH3
2-methylprop-1-ene
Step - II
Deprotonation
The alcoholic –OH group is not good leaving group, therefore first convert it into good leaving
group by the protonation (proton comes from acid) i.e. water and then eliminated in the form of
water forming planar carbonium ion. It is slow step therefore rate of the reaction is depends only
on the concentration of tert-butanol but not on the acid which is used as catalyst or solvent. The
second step of reaction is fast proceeds with elimination of proton from the β-position in the
presence of HSO4-.
The HSO4- is not involved in the rate determining step – HSO4
- the is not at all basic and only
behaves as a base i.e. it removes the proton, because it is more even feeble as a nucleophile.
H+
OH
CH3
CH3
H
N
CH3
CH3
H
N
O+
H H
CH3
CH3
H
N+
CH3
CH3
N-H+-OH
2
E1 and E2 mechanism:
There are number of factors that affect whether elimination proceeds either E1 or E2 mechanism.
The E2 elimination mechanism is affected by the concentration of base so at high base
concentration favored E2 mechanism. The rate of E1 elimination mechanism is not even affected
by what base is present so E1 is just likely with weak base than the strong base while E2
mechanism goes faster with strong base than weak ones.
Br
CH3
CH3HCH2
CH3
CH3
OH2HO- Br
-+ +-H
2O
With less hindered alkyl halides, hydroxide would not be a good choice as a base for elimination
because it is rather small and still very good for SN2 substitution (even with tertiary alkyl halide,
substitution is major at low concentration of HO-). The tert-butoxide is very good alternative for
elimination as it is both bulky and a strong base (pKaH = 18).
Br
Br
Br2
Bromination
Br
Br
H
Br
H
E2
elimination
E2
elimination
H
HBr
BrH
Br
Among the most commonly used bases for converting alkyl halides to alkenes are DBU and
DBN. There are amidines – delocalization of lone pair of one nitrogen on to the other and the
resulting substitution of the protonated amidinium ion. Because of bulky nature / fused rings,
they pick off the easy and reach protons rather than attacking the carbon atoms in substitution
reactions.
NN
NN
DBN
DBU1,5-diazabicyclo[4.3.0]nonene-5 1,8-diazabicyclo
[5.4.0]undecene-7
NN
acid
N+
N
HNN
+
H
NN
CH3
CH3
CH2
I
OO
DBN / 80 °C
E2
CH3
CH3
CH2
I
OO
H CH3
CH3
CH2
CH2
OO
91%
E1 elimination can occur only with substrates that can ionize to give relatively stable carbonium
ion as tertiary, allylic or benzylic alkyl halides. Secondary alkyl halides may eliminates by E1 or
E2, while primary alkyl halides only ever eliminate by E2 mechanism because primary carbonium
is too unstable. The above substrates may under goes E2 elimination under appropriate
conditions. The alkyl halide does not contain any hydrogen on adjacent carbon atoms to the
leaving group are not undergoes elimination, but undergoes substitution reaction.
Substrate undergoes E1 elimination Substrate may
under goes E1
Never
undergoes E1
No elimination
reaction
30-alkyl halide, allylic halide,
benzylic halide, α-heteroalkylhalide,
etc
Sec.-alkyl
halide
Primary
alkylhalide
Methyl halide, benzyl
halide, R3C-CH2X, etc
May undergoes E2 elimination
The polar solvents also favor E1 reaction mechanism because of stabilization of intermediate
carbonium by salvation such as water, alcohol, etc.
OH O OH
Ph
Ph
H3PO
4/H
2O
165 °C/E1
PhMgBr H2SO
4
H2O/E
1
The role of the leaving group is very important in elimination reactions. In both eliminations,
leaving group departs in slow step, therefore any good leaving group will lead to a fast
elimination. The type of elimination is depends on nature of substrate and base used. This can be
explain with help of following example –
CH3
CH3
CH3
NMe 2
CH3
CH3
CH3
NMe 3I
H
CH3
CH3
CH3
CH2
MeI KOH
E2
Mechanism
MeICH3
NMe 2
CH3
CH3
CH3
NMe 3I
CH3CH3
HOH
CH3
CH2
CH3
Mechanism
E1
In first example, stabilized carbonium ion is not formed, but a strong base is used, allowing E2
mechanism. In the second example, a stabilized tertiary carbonium ion would be formed, but no
strong base is present so the mechanism must be E1.
Regioselectivity and Stereochemistry of elimination reactions:
During elimination, for the some reactions only-one product is possible, but for others, there may
be a choice of two or more alkene products that differ either in the location or stereochemistry of
the double bond. One product formed is major. There are number of factors that control the
stereochemistry (geometry) and regiochemistry (that is where the double bond is) of the alkene.
CH3 OH CH3CH2
+
Regioisomers
(Stable) (Less stable)
acid
acid
Ph
OH CH3Ph
CH3
Ph CH3
+ Stereo-isomers
(E, stable) (Z, less stable)
For the steric reason, E-alkenes (and transition states leading to E-alkene) are usually lower in
energy than Z-alkenes (and transition states leading to them) because of the substituents can get
further apart from one another. The geometry of the product is determined at the moment that the
proton is lost from the intermediate carbonium ion. The new pi(π)-bond can only form if the
vacant p-orbital of the carbonium and the breaking of C-H bond are aligned parallel. This can be
explain with the help of following example. There are two possible conformations of the
carbonium ion with parallel orientation, but one is more stable than other because of less steric
hindrance. The same is true for transition states on the route of the alkenes – the one leading to
the E-alkene is lower energy and therefore E-alkene is formed major product. It is example of
stereoselective elimination.
PhCH3
OH H
H H
acid
PhCH3
O+
H
H H
H
H
PhCH3
H
H H
-H2O
+Ph
CH3
-HX
Configuration of the product depends onconfiguration about this bond
Configuration of the carbonium with C-H
and vacant P-orbital aligned parallel
Ph
CH3
H
H
H+
Ph
CH3
H
H
H+
Ph
CH3
H
H
X
PhCH CH3
H
H Ph
CH3
H
H
XPh
CH
CH3
H
H
Higher energy intermediate, steric hindrance
Higher energy transition state
Z-alkene formed slowly
Lower energy intermediate
Carbonium ion
Lower energy transition
state
Stable E-alkene formedfaster
During E1 elimination, more than one regio-isomeric alkenes are formed, but more substituted
alkene is formed as major product.
CH3
CH3
H CH3
OH CH3
CH3 CH3
CH3
CH3
H CH2
+HBr/H
2O
(major) (minor)
The reason why more substituted alkene is more stable, is same as that of stability of more
substituted carbonium ion. The carbonium ion is stabilized when its empty p-orbital can be
interacting with the filled orbitals of parallel C-H and C-C bonds. The same is true for π-system
of the double bond – it is stabilized when the empty π* anti-bonding orbital interact with the
filled orbitals of parallel C-H and C-C bonds. The more C-C and C-H bonds, there are more
stable the alkene.
H
H H
H
H CH3
HCH3
H
H CH3
H
HH
H
H
π *HH
H
HH
H H
H
HH
H
H
H
H
C - H bond
C - C bond
This does not explain why more substituted alkene is formed as major product. It can be explain
from the following transition states. Both alkenes are formed from same carbonium ion, but
depend on which proton is lost. The removal of proton proceeds through transition state, and the
energy of the transition state decides the yield of alkene. The more substituted alkene proceeds
through lower energy transition state.
CH3
CH3
H
CH3
OH
CH3
CH3
H
CH3
O+
H
H
acid
CH3
CH3
CH3
-H2O
+
CH3CH3
H H
+
CH3
C
CH3
H H
OH2
+ CH3
C
CH3
H CH2 H
OH2
+
CH3
CH3 CH3
CH2
CH3
CH3
Carbonium ion intermediate
Product formed dependswhich proton is lost
Trisubstituted alkene, forms fasterMonosubstituted alkene, forms more slowly
Partial C=C bond hasthree substituentsmore stabilized
Low energy T.S.
Partial C=C bond has only one
substituent, less stabilized
High energy T.S.
In an E2-elimination, the new π-bond is formed by the overlap of the C-H σ-bond with the C-X
σ*-antibonding orbital. The two orbitals have to lie in the same plane for maximum overlap, and
now there are two conformation that are allow such overlapping, -H and –X are syn-periplanar
and –H and –X are antiperiplanar. The anti-periplanar conformation is more stable because it is
staggered [the syn-periplanar conformation is eclipsed] and C-H and C-X bonds are truly
parallel.
H X
H
X
X
H X
H Syn - periplanar (eclipsed) Anti - periplanar (staggered)
C-H & C-X are co-planar
C-H & C-X are fully parallel
Therefore, E2-elimination takes place from the anti-periplanar conformation. In anti-periplanar
conformation, the attacking electron rich base avoids repulsion with electron rich leaving group,
which is present in syn-periplanar conformation during elimination. E.g. 2-Bromobutane has two
conformation with –H and –Br are anti-periplanar, but one that is less hindered leads to more
stable product i.e. E-alkene.
CH3
CH3
Br
CH3
CH3
H
CH3
H
CH3
CH3
CH2
H+ +NaOEt
Major Minor19%
81%
C - H and C - Br bonds must be anti-periplanar to each others
CH3
CH3
Br
H
H
H
CH3
H3C
Br
H
H
H CH3
H
CH3
CH3
CH3
H
The stereochemistry of the product is depends on the proton present anti-planar to the leaving
group when the reaction proceeds.
In the following example, there is only one proton that can take part in the elimination. In such
cases, there is now choice of anti-periplanar transition state. It gives product either E or Z isomer
which is depends on which diastereoisomer is used as starting material by E2-elimination.
e.g. Base induced dehydrohalogenation of 1-bromo-1,2-diphenyl propane.
The 1-bromo-1,2-diphenyl propane has two chiral centers, therefore it can be exist in four
different diastereomeric forms – two enantiomers of erythro form and two are in threo form.
Ph
Ph
H CH3
H Br
Ph
Ph
CH3 H
H Br
Ph
Ph
CH3 H
Br H
Ph
Ph
H CH3
Br H
(I) (II) (III) (IV)
I & II, III & IV are enantiomers
I & II are erythro isomers
III & IV are threo isomers
(A)(B)
The 1-bromo-1,2-diphenyl propane undergoes dehydrohalogenation in the presence of base
forming 1,2-diphenylpropene which is also exist in two geometrical isomeric forms E and Z via
E2 mechanism of elimination reaction. The product formed was depends on the configuration of
starting halide.
PhPh
Br
CH3
* *
BaseThreo -
BaseErythro-
PhPh
CH3
PhCH3
Ph
E - isomer Z - isomer
When the diastereoisomer A (threo-) is drawn with the proton and bromine are anti-periplanar, as
required and in the plane of the page, the two phenyl groups have to lie as one is front and one
behind the plane of the paper. As the hydroxide attacks the C-H bond and with elimination of
bromide ion, this rearrangement is preserved and the two phenyl groups end up trans (E-alkene is
formed).
Ph
Ph
Ph
CH3 H
Br H
PhPh
CH3
H H
Br
Ph
CH3
Br
H
H
CH3
H
Ph
Ph
NaOH
[E - alkene]A
(Threo -)[H and Br are anti-periplanar]
The diastereo-isomer B (erythro-) forms Z-alkene, the two phenyl rings are now on the same side
of the H-C-C-Br plane in the reactive anti-periplanar conformation.
Ph
CH3 H
Br H
Ph CH3
Ph
H H
Br
Ph
Ph
H
Ph
CH3
H
Br
Ph
H
Ph
CH3
NaOH
[Z - alkene](erythro-)
[H and Br are anti-periplanar]B
Ph
H
Ph
H3C H
Br
Gauche interaction between Ph-groups, slow reaction
The diastereomers A and B gives different alkene geometry, with different rates. The
diastereomer A undergoes elimination 10 times faster than diastereomer B because, although this
anti-periplanar conformation is the only reactive one, it is not necessary that it is more stable.
The Newman projection for the reaction of B shows clearly that, the two phenyl groups have to
lie synclinal (gauche) to one another; the steric interaction between these bulky groups will
decrease its population and small portion of the molecules adopt this conformation for
elimination, slowing the process.
In such elimination reactions, the stereochemistry of the product is determined from the
stereochemistry of the starting material are called stereospecific reactions.
“The reaction gives mixture of products in which one is major are called as stereoselective
reactions while the reaction in which each diastereomer gives different product is called
stereospecific reaction.”
Consider the following reactions in which leaving group changes and the reaction conditions are
very different but the overall process is elimination of H-X to produce one of two alkenes with
different mechanisms.
CH3
OH
CH2
CH3
CH3
Cl
H3PO
4
120 °C
KOCEt3
E1 - Mechanism
E2 - Mechanism
The first example is acid catalyzed elimination of water molecule from a tertiary alcohol
produces trisubstituted alkene by E1 mechanism. The second is elimination of HCl from
corresponding tertiary alkyl chloride in presence of hindered alcoxide base gives less stable
disubstituted alkene by E2 mechanism. Why does E2 mechanism gives less substituted alkene
product? In this case, there is no problem getting C-H bond anti-periplanar to the leaving group;
in the conformation with the chlorine axial, there are two equivalent ring hydrogens available for
elimination and removal of either of these would lead to the trisubstituted alkene. Additionally
any of the three equivalent methyl hydrogens are in the position to undergoes E2 elimination to
form the trisubstituted alkene, whether the Cl is axial or equivalent and these are removed by
hindered base such as alcoxide.
CH3
ClCH3
Cl
HHH
Cl
HHH
H
H
H
RO-
RO-
O-R
CH2CH3
X
Two ring hydrogens are
anti -periplanar to ClSteric repulsion
Methyl hydrogens areanti -periplanar to Cl
The base attacks the methyl hydrogen atom because they are less hindered, attached to primary
carbon and well away from other axial hydrogens. The E2 eliminations with hindered bases
typically give less substituted double bond, because the faster E2 reaction involved deprotonation
at the least substituted site. The hydrogen atoms attached to less substituted carbon atom are also
more acidic.
CH3CH3
CH3
BrCH3
CH3
CH3
CH3
CH2
CH3
+
69%28%
31%
73%
Base
NaOEtt-BuOK
Kinetic control and thermodynamic control reactions:
The product that forms faster is called as kinetic product and the product formed that is more
stable during chemical reaction is called as thermodynamic product. The conditions that gives
rise to kinetic product are called as kinetic control while the conditions that gives rise to the
thermodynamic product are called thermodynamic control.
Typically the kinetic control involves at lower temperature and shorter reaction time which
ensures that only the fastest reaction has the chance to occur. Typically thermodynamic control
involves at higher temperature and longer reaction time to ensure that even the slower reactions
have a chance to occur and all the material is converted to the most stable compound.
Thermodynamic product has lower energy than the kinetic product also the thermodynamic
product forms via high energy transition state than kinetic product, the intermediate forms for
kinetic product is more stable than the intermediate of thermodynamic product. If temperature of
the reaction increases, and is enough for the kinetic product to get back to the starting material,
and is sufficient for the formation of thermodynamic control product. This energy is not
sufficient to convert thermodynamic product into starting material i.e. kinetic control product is
formed reversibly while thermodynamic control product is formed irreversibly.
e.g. Direct addition (1,2-addition) to the double bond of conjugated system is favored at low
temperature, but conjugate addition (1,4-addition) is favored at high temperature.
..
..
..
..
..
..
.. ..
..
..
Energy
Extent of reaction
Reactant
Transition state
Transition state
Kinetic product
Thermodynamic product
Intermediate
Intermediate
Eact
Eact
Heat of reaction
Heat of reaction
Addition of HX to butadiene:
At low temperature (-800C), during the addition of HBr to 1,3-butadiene, the major product
formed is 3-bromo-1-butene (by 1,2-addition) but at higher temperature (at 400C), the more
stable 1-bromo-2-butene is formed as major product. The 3-bromo-1-butene is kinetic control
product whereas 1-bromo-2-butene is thermodynamic control product.
CH2
CH2 CH3 Br CH2
Br
H+CH3 Br CH2
Br
H+
80%80%
20%20%
(1,4-addition product)(1,2-addition product)
-80 °C 40 °C
Mechanism:
CH2
CH2 CH3
CH2
+
CH2 H+
buta-1,3-diene
H+
Secondary carbonium ion
StablePrimary carbonium ion
Less Stable----Allylic carbonium ions ----
Addition of H+ to double bond
+
CH3
CH2
+
CH2 H
Primary carbonium ionSecondary carbonium ion
+ CH2. H+Stabilisation of carbonium ion
CH3
CH2
+
CH2 HCH3 BrCH2
Br
H
(1,4-addition product)(1,2-addition product)
Primary carbonium ion
Stable
Secondary carbonium ionLess Stable
+
Br-
Br-
1,3-Butadiene accept proton from HBr and forming carbonium ion, primary carbonium ion and
allylic carbonium ion (secondary carbonium ion). The allylic carbonium ion is more stable i.e.
stabilized by resonance effect of double bond. There are two carbons C2 and C4 are able to react
with bromonium ion (Br-) forming 1,2- addition and 1,4-addition product.
1. The 1,2-addition product is formed faster than 1,4-addition product because 1,2-addition
product is formed from secondary allylic carbonium ion which is more stable than primary
allylic carbonium ion generated for the formation of 1,4-addition product.
2. The 1,4-addition product formed is more stable (disubstituted alkene) product than the 1,2-
addition product (monosubstituted product).
CH2
CH2CH3 BrCH2
Br
H +
(1,4-addition product)(1,2-addition product)
CH2. H+
(I) (II)
(III)(IV)
Energy profile diagram:
Energy
Reaction co-ordinates
I
II
IIIIV
Stability of alkene:
The carbocation is stabilized when its empty p-orbital can be interact with the filled orbitals of
parallel C-H and C-C bonds. The same is true for the π-system of double bond, it is stabilized
when empty π* anti-bonding orbital can interact with the filled orbitals of parallel C-H and C-C
bonds. The number of C-C and C-H bonds increases, the stability of the double bond (alkene)
goes on increasing.
H
H CH3
HH
H H
H
H CH3
HCH3
H
H H
H H
H
H
H
H
H H
H
HH
H
HH H
No. C-H bonds parallelwith pi-antibonding
Increasing substitution allow s more C-H and C-C sigma bonding
orbitals to antibonding pi-orbital.
Sulfonation of naphthalene:
Sulfonation of naphthalene with concentrated sulfuric acid at 800C gives α or 1-naphthalene
sulfonic acid as major product while at 1600C gives β or 2-naphthalene sulfonic acid as major
product. The 1-naphthalene sulfonic acid is kinetic control product while 2-naphthalene sulfonic
acid is thermodynamic control product.
H
H
SO 3HH
SO3H
conc. H2SO4
80 °C
Kinetic control
160 °C
Thermodynamic control
naphthalene-1-sulfonic acid
naphthalene-2-sulfonic acidnaphthalene
(Less stable)
(Stable)
160 °C
Mechanism:
The 1-naphthalene sulfonic acid is formed more rapidly because α-position is more reactive
towards electrophilic substitution than β-position, it is explain with the help of following
mechanism.
H
H
SO3HH
SO3H
conc. H2SO4
80 °C
Kinetic control
160 °C
Thermodynamic control
naphthalene-1-sulfonic acidnaphthalene-2-sulfonic acid
naphthalene(Less stable) (Stable)
160 °C
+
-H+
SO3H
conc. H2SO4
+
SO3HH
+
-H+
SO3HH
1. The attack of electrophile (SO3) takesplace at α-position forming intermediate which is
stabilized by two resonance or canonical structures while intermediate β-substitution is
stabilized by one resonance or canonical structure without disturbing aromatic sextet
(character) of second benzene ring. The α-substitution takes place at 800C, as temperature
of the reaction increased to 1600C, equilibrium is reached and more stable 2-naphthalene
sulfonic acid obtained as major product.
2. 2-Naphthalene sulfonic acid is more stable than 1-naphthalene sulfonic acid is due to
steric interaction between very bulky –SO3H group and C-H atom present on the 8-
position in 1-naphthalene sulfonic acid whereas in 2-naphthalene sulfonic acid, such
interactions are absent.
naphthalene-1-sulfonic acidnaphthalene-2-sulfonic acid
SO3H
HH SO3HH
(No steric interaction) (Steric interaction)
8 1
Steric interaction
Nucleophilicity and electrophilicity:
The species having unshaired pair of electrons or excess non-bonding electron density which can
be easily donate to electron deficient species are called nucleophiles. They are either ionic
species or neutral molecules. e.g. HO-, HS
-, RO
-, NC
-, RCOO
-, H2O, NH3, RNH2, etc.
All nucleophiles are also called as Lewis bases and their relative strength can be explain by its
basicity. Basicity explain the extent to which the base accept proton.
The species having empty orbital or low electron density (required electron to complete its octet)
which can be easily Accept the electrons from electron rich species are called electrophiles. They
are either ionic species or neutral molecules. e.g. H+, NO2
+, SO3, BF3, SeO2, Fe
3+, etc.
All electrophiles are also called as Lewis acids and their relative strength can be explain by its
acidity. Acidity explain the extent to which the acid accept electron or donated the proton.
Distinguish between basicity and nucleophilicity:
Basicity Nucleophilicity
It is the extent to which base accept the proton It is the rate at which nucleophile forming
bond with substrate
It has absolute value in terms of Kb and pKb. It is relative term (ratio of K of nucleophile to
K of solvent)
It is thermodynamic term It is kinetic term
It affect the position of acid-base equilibrium It affect the rate of reaction
Nucleophile:
Most of the organic reactions are polar in nature i.e. electrons are flow from one molecule to
another molecule as the reaction proceeds. The electron donor reactant is called as nucleophile
(nucleus loving) while the reactant accept the electrons (electron acceptor, electron loving) is
called electrophile. The nucleophiles likes nuclei because they are positive charged and the
electrophile likes electrons because they are negatively charged. E.g.
P+Cl4 + HO
-Cl4P - OH POCl3
(Electrophile) (Nucleophile)
-HCl
In some reactions, electrons are transferred in the form of lone pair (completely filled orbital) to
an empty orbital. E.g. Reaction between amine (nucleophile) and BF3 (electrophile). The amine
is the nucleophile because of the lone pair of electrons on nitrogen of amine was shared with
empty p-orbital of boron of BF3 (electrophile).
F
B
F F
N
R
R
R B-
N+
R
R
R
F
F
F
+
Empty p-orbital
Completely filled orbital
The nucleophiles are of negatively charged atoms/molecules, having unshared pair of electrons
or having pi-electrons (e.g. alkene, aromatic rings, etc). The most common nucleophiles having
lone pair of electrons are usually containing heteroatom such as O, N, S, or P. e.g. H2O, NR3,
PR3, R2S, etc. The anions are also usually contains heteroatom bearing negative charge such as
O, N, S, or halogen; HO-, CH3S
-, Br
-, H2N
-, etc. There are few examples of carbon nucleophiles
with lone pairs of electrons. E.g. NC-. Though linear cyanide has a lone pair on nitrogen and one
on carbon, the nucleophilic atom is usually anionic carbon rather than neutral nitrogen atom, as
the sp-orbital on the carbon has higher energy than that on the more electronegative nitrogen. In
neutral molecules, the pi-bond act as the nucleophilic portion of the molecule. When the
molecule does not have high energy non-bonding electrons for donation, the next even lower
energy filled orbital (pi-electrons) are donated rather than lowest energy sigma-electrons. E.g.
Alkene are acts as weak nucleophiles. It is possible for the sigma-electrons acts as nucleophiles.
E.g. BH4- has the nucleophilic B-H bond and can donate those electrons into the antibonding
empty π* orbital of the carbonyl compound and breaking that bond and eventually giving
alcohol.
In most of the organic reactions, the orbitals of the nucleophile and that of electrophile (in
pericyclic reactions) are directional and so the molecular orbitals of the reacting molecules exert
important control. For the formation of new bond, the orbitals of the two species must be
correctly aligned in space.
Electrophiles are positively charged or neutral molecules with an empty atomic orbitals or a low
energy antibonding orbital. The simplest example of electrophile is proton (H+). The neutral
species such as BF3 or AlCl3 are the examples of electrophiles which has empty orbitals that are
usually metallic p-orbitals.
Cl
Al
Cl Cl
O
H
H
Al-
O+
R
R
Cl
Cl
Cl
+
Empty p-orbital
Completely filled orbital
Al2O3
The nucleophiles containing small electronegative atoms such as O or Cl tends to react under
predominantly electrostatic control, while nucleophile containing larger atoms (including the
sulfur of thiols, P, I and Se) are predominantly subject to control by orbital overlap. The terms
hard and soft are used to describe these two types of reagents. Hard nucleophiles are typically
from the early rows of the periodic table and having high charge density, while soft nucleophiles
are from the later rows of the periodic table, they are either uncharged or have higher atom with
higher energy, more diffuse orbital.
Hard Nucleophiles : F-, HO
-, RO
-, SO4
2-, Cl
-, H2O, ROH, ROR’, RCOR’, NH3, RMgX, RLi, etc
Borderline :N3-, CN
-, RNH2, RR’NH, Br
-, etc.
Soft Nucleophile :I-, RS
-, RSe
-, S
2-, RSH, RSR’, R3P, Alkene, aromatic rings, etc.
Nucleophilicity:
The reactivity or strength of nucleophile is called nucleophilicity. The negative charged
nucleophiles like HO-, RO
-, H2N
-, etc has more nucleophilicity than neutral nucleophiles such as
H2O, ROH, H3N. The RO- is stronger nucleophile than HO
- because of electron donating effect
of R group,
Electrophilicity:
The ractivity or capacity of an electrophile to accept an electron pair from a standard donor is
called electrophilicity strength of electrophile). The strength of electrophiles are depends on the
electrophilic character (nature of requirement of electrons) of the accepting atom.
Mechanism and reaction of carbonyl compound with nucleophile:
The carbonyl group is highly polar in which carbon having partial positive charge and oxygen
having partial negative charge. Therefore any nucleophile has been added to the carbon atom
forming alkoxy-anion which further undergoes protonation to get addition product.
During this addition reaction, hybridisation of carbonyl carbon changes from sp2 (trigonal
planar) to sp3 (tetrahedral). This mechanism has been change with change of the type of
nucleophile.
Formation of acetals/ketals from aldehyde and ketones:
O
R1
R2OR1
R2
z
OR1R2
z
z
OHR1R2
z
+ H+
T.S.nucleophile
aldehyde/ketone
Alkoxideintermediate
tetrahydral product
When aldehydes or ketones are reflux with an alcohol in presence dry HCl, alcohol shows
addition reaction to carbonyl group of aldehyde or ketone forming intermediate called hemicaetal
or hemiketal which further undergoes substitution of second alcohol molecule forming
acetal/ketal.
The aldehydes/ketones group can be protected by acetal/ketal formation during the treatment of
alkali, oxidation and other side reactions. These acetals/ketals are again converting to aldehyde
or ketone by the treatment of dilute acid. This is reversible reaction, so water can be removed by
continuous distillation.
Mechanism and stereochemistry of SN1 reaction:
Consider the reaction: Ph-CH2-Cl + Ph-S- → Ph-CH2-S-Ph + Cl
-
During this reaction, the benzyl group remains same but chlorine atom is replaced by Ph-S-
group; it is substitution reaction. During reaction, chlorine atom is replaced by Ph-S- group,
therefore the Cl- is called as leaving group and Ph-S- is called nucleophile. Therefore overall
reaction is nucleophilic substitution reaction. There two different mechanisms possible for such
reactions; the leaving group depart first and then nucleophile attack [SN1] and both departure of
leaving group and attack of nucleophile takes place simultaneously [SN2].
+
Ph
Cl
H
Ph H Ph
H H
SPh Ph
H H
PhS+
Leaving group depart first
-Cl, slow step
Nucleophile attack
PhS-/fast
Two step
Mechanism
The SN1 mechanism
Ph
Cl
Ph
H H
PhS
One step
Mechanism
The SN2 mechanism
Nucleophile at the same time as the leaving group depart
PhS- / slow, -Cl-PhS
-
O
R1
H
R'OHOHR1
H
OR'
OR'R1H
OR'R'OH
+
aldehydeAlcohol
Hemiacetal Acetal
anhyd.HCl
O
CH3
H
EtOHOHCH
3H
OEt
CH3
H
OEt
OEt
EtOH+
acetaldehydeAlcohol
Hemiacetal Acetal
anhyd.HCl
There are two possible mechanisms for nucleophilic substitution at saturated carbon; but how
these mechanisms are different to each other can be explain with help of following examples.
If we carried out nucleophilic substitution, the replacement of OH group by Br of following
substrates, two quite different reaction conditions are required. Tertiary alcohol reacts rapidly
with HBr to give tertiary alkyl bromides while primary alkyl halide reacts very slowly with HBr
but it reacts rapidly with PBr3 and convert into alkyl bromide. If we carried out same reaction of
allyl alcohol and benzyl alcohol with HBr, they reacts rapidly and convert into corresponding
bromide.
OH
CH3
CH3
CH3 CH2
OH
OH
Br
CH3
CH3
CH3 CH2
Br
Br
HBr
fast
HBr
fast
HBr
fast
CH3 OH CH3 Br
PBr3/fast
(HBr/slow)
Tertiary alcoholAllylic alcohol
Benzyl alcohol
Primary alcohol
Those alcohols (tertiary, allylic, benzyl) reacts rapidly with HBr to give good yield of
corresponding bromide, we find one common thing; they can all forms stable carbonium ions.
+OH
CH3
CH3
CH3 CH3
CH3
CH3
CH2
OHCH2
CH2
OHCH2 +
+
[Stabilized by +I effect][Stabilized by +R effect C=C bond]
[Stabilized by +R effect Ph ring]
But in case of primary alcohols, such stabilized carbonium ion intermediate is not formed.
Therefore, the substrate which are able to form stable carbonium ion shows SN1 mechanism and
those are not (forming transition state) are follow the SN2 mechanism.
The SN1 reaction proceeds through carbonium ion intermediate. It is planar, with empty p-
orbital having six electrons in its valance shell and have electrophilic nature.
+Empty p - orbital
Planar, sp2 hydrid carbon