Effects Strengths 2u4n4t

ORGANIC LECTURE NOTES GOC-II

PPPP-P Electronic effects (I, R, m, Hyperconjucation), Physical properties Acidic strength & basic strength

GOC-2 (LECTURE NOTES) ELECTRONIC EFFECTS (I & R EFFECTS)

ELECTRONIC EFFECT : The effect which appears due to electronic distribution is called electronic effect.

Types of electronic effect (1) Inductive Effect (2) Resonance (Mesomeric effect) (3) Hyper conjugation (4) Electromeric effect

(A)

Inductive Effect Inductive Effect When a sigma covalent bond is formed between two atoms of different electronegativity atom then the sigma () bond pair electron are shited towards more electronegative atom due to this shifiting of e– a dipole is created between the two atoms. Due to this dipole the  bonds e– in the chain are also shifted and the chain becomes polarised this effect of polorisation in the chain due to a dipole is called inductive effect. Diagram showing I effect

Features of Inductive effect (i) Inductive effect is a permanent effect. (ii) The effect is distance dependent and this effect (end up) negligible after three carbon atom in the change (iii) Inductive effect is operative only through  bond it does not affect or participate the  bond electron (iv) The C–H bond is the reference of inductive effect i.e. the electron movement or polarity in the C–H bond is considered to be neglible and the I effect of Hydrogen is taken to be zero.

2.

Types of inductive effect :

(a)

– I Effect : The group which withdraws electron cloud is known as – I group and its effect is called – I effect. Various groups are listed in their decreasing – I strength as follows. > – > – > – NO2 > –SO3H > –CN > – CHO > – COOH > – F > – Cl > – Br > – I > – OR > – OH > – NH2 > – C  CH > – C6H5 > – CH = CH2 > – H.

(b)

+ I Effect : The group which repel (or release electron) cloud is known as + I group & this effect is + I effect. >

> – C(CH3)3 > – CH (CH3)2 > – CH2 – CH3 > – CH3 > – D > – H

Example-1 Since – NO2 is – I group it pulls or withdraws e¯ from cyclohexane ring making it e¯ deficient

ExampleExampleDue to e¯ donating nature of carbon chain has become partially negative but – COOH is – I group therefore carbon chain has become partially positive. Example Page # 2

(a) CH3

CH2

(c)

CH2

(b)

CN

CH2

(d)

(e)

(f)

(g)

(h)

(i)

(j)

CH 3

+

CH

+

CH

–

NO2

O -

(k)

- -

3.

- -

Applications of Inductive effect : (i) Acidity (ii) Basicity (iii) Dipole moment (iv) Stability of intermediate (v) Reactivity of compound in chemical reaction

(a )

In deciding acidic strength of (i) Carboxylic acid – I effect  Acidic strength (presence of – I groups increases acidic character) + I effect  Basic strength (presence of + I groups increases basic character) Example : (I)

H3 C  H 2C  H2 C  CH  COOH | Cl

(II)

CH 3  CH 2  CH  CH 2  COOH | Cl

(III)

CH 3  CH  CH2  CH 2  COOH | Cl

(IV)

CH3 – CH2 – CH2 – CH2 – COOH

Acid strength order I  II  III  IV

Explanation : We know that – I effect increases acidic strength. Now morever it is distance based effect so where the – I group is nearest to – COOH, It enters strong effect and makes acid strong. Example : (I) O2N – CH2 – COOH (III) H3CO – CH2 – COOH

(II) (IV)

F – CH2 – COOH CH3 – CH2 – COOH

Acid strength order I  II  III  IV

Since NO2 has strong – I effect its influence will make corresponding acid strongest (– I effect  acid Page # 3

character). Example : Arrange for acidic strength order CH3COOH, ClCH2COOH, Cl2CHCOOH, Cl3CCOOH Ans. CH3COOH < ClCH2COOH < Cl2CHCOOH < Cl3CCOOH (b)

Stability of alkyl carbocation : Carbocations are electron deficient species and they are stabilised by + effect and destablised by –  effect. Because +  effect tends to decrease the positive charge and –  effect tends to increases the positive charge on Example :

(c)

Stability : CH3



(carbocation)

< CH3 CH2





 < (CH3 )2 C H < (CH3 )3 C

Stability of carbon free radical : Carbon free radical are stabilised by + I effect. Example :

 Stability : CH3 < CH3

CH2



< CH3



< CH3



 CH3

LECTURE # 2 (B)

Resonance Effect The delocalisation of adjacent (parallel) p-orbital electron is called resonanc. Through resonance the e– density in the molecule becomes uniform, due to uniformility of e– density the stability of molecule increases. Hence the phenomena of a resonance is a stabilising effect. In resonance only parallel p-orbital e– are involved. To explain the property of actual molecule different lewis structure are drawn with the movement of p-orbital electrons. This lewis structure are called resonating structure or canonical structure.

 Resonance hybrid

Examples : (I)



(II)





(III)



(IV) (V) (VI)

CH2 = CH –



– CH = CH2



Resonating structure are only hypothetical but they all contribute to a real structure which is called resonance hybride. The resonance hybride is more stable than any resonating structure. The P.E. difference between the most stable resonating structure and resonance hybride is called resonace energy. The stability of molecule is directly proportional to resonance energy.

Page # 4

The most stable resonating structure contribute maximum to the resonance hybride and less stable resonating structure contribute minimum to resonance hybride.

Rule for writing resonating structure : (i) In resonating structure only p-orbitals e– are shifted,  bond e– are not involved in resonance, therefore the  bond skeleton well remain same in two resonating structures. (ii) The atoms are not moved (iii) The no. of paired p e– are same and pair of unpaired p e– are same in two resonating structure. (iv) The octel rule is not violated (for second period element) (v) High energy structure are rejected as personating structure or their contribution to the resonance hybride is least. e.g. Opposite change on adjacent atoms and similar charge on adjacent atoms are cases of high energy. They are not accepted as resonating structures. e.g. 1. CH2 = CH – CH = CH2 2.

CH2 = CH – CH = CH2

3.

CH2 = CH

4.

CH2 = CH – CH = CH2

5.

CH2 = CH – CH = CH2

CH3 – CH = C = CH2

Condition for resonance : 1. All atoms participating in resonance must be sp or sp2 hybridised. 2. The parallel p-orbitals overlap to each other. 3. Molecule should be conjugated system (parallel p-orbitals systems are called conjugate system) e.g. 1. CH2 = CH – C  CH (Show resonance) 2. CH2 = CH – CH2 – CH = CH2 (Not show resonance) 3.

(Show resonance)

4.

(Not show resonance)

5.

CH2 = C = CH2

(Not show resonance)

Types of Conjugation:A given atom or group is said to be in conjugation with an unsaturated system if:(i) It is directly linked to one of the atoms of the multiple bond through a single bond. (ii) It has π bond, positive charge, negative charge, odd electron or lone pair electron. 1. 2.

CH2 = CH – CH = CH2 + CH 2 = CH – CH 2

3.

.. CH 2 = CH – NH 2

(Conjugation between C = C and C = C) (Conjugation between +ve charge and C = C) (Conjugation between lone pair and C = C) Page # 5

Q.



4.

CH2 = CH – CH2

(Conjugation between odd electron and C = C)

5.

– CH2 = CH – CH = CH – CH2

(Conjugation between negative charge and C = C)

Write resonating structure for each of the following molecules :

O || (a) CH3  CH  CH  C  CH3

(a)

.. (b) H 2 N CH  CH  C  N

(b)

(c)

(c)

(d) CH3 – O – CH = CH –

(d)

(e)

(e)



..

(f) CH3  C  O :

(f)

(g)

(g)

(h)

(h)

(i)

(i)

(j)

(j)

Q.

Draw resonating structure for following molecules ,

CH2 = C = O

,

+

+

Ans. + 

CH3 – C  O

..

..

..

..

CH 3– C = O

CH 2 = C = O





C H2 – C  O N = N = N

–2



N– N  N Page # 6

LECTURE # 3 Mesomeric Effect : When a group releases or withdraw p orbital electron in any conjugated system then it is called M effect group and effect is known as mesomeric effect. +M group : (electron releasing group) : A group which first atom bears -ve charge or lone pair always shows +M effect. Relative order of +M group : –

>–

>

> – NR2 > –OH > –OR > – NHCOR > – OCOR > – Ph > – x > –NO > –NC

e.g.

1.

2.

3.

4. – M group : (-electron withdrawing) Relative strength of –M group : – NO2 , –CN , –SO3H , –CHO, –CH = NH,

(G = H, R, OH, OR, NH2, NHR, NR2, X)

e.g. 1.

– + O – N = .O: .

– + . .– O – N – .O: .

– + . .– O – N – .O: .

+

2.

– + .–. O – N – .O: . +

– + .. O – N = O:

+

3. Q

H 2C = CH – C N:

+ – H 2C – CH = C = N: ..

By drawing resonating structures of following molecules, Judge whether the group attatched to ring exerts + m or – m effect.

,

Page # 7

Ans. NH–C–CH3

NH–C–CH3

O

O

NH–C–CH3

NH–C–CH3

NH–C–CH3

O

O

O

(a)

O C–CH3

(b)

–C–CH3 (–m effect group) O O C–CH3

Note: 1.

O

C

CH3

O

C

O

CH3

O

C

CH3

C

CH3

When a +m group and –m group are at meta-positions with respect to each other then they are not in conjugation with each other, but conjugation with benzene ring

etc. 2. Q.

+M group increases electron density in benzene ring while –M group decreases electron density in the benzene ring. Write electron density order in the following compound.

(a) Ans. III > I > II > IV

(b) Ans.

I > II > III . IV

Ans.

IV > I > III > II

(c)

Page # 8

Rules for stability of resonating structure : (1)

The resonating structure without any charge separation is more stable. e.g. 1.

CH2 = CH – CH = O I Stability order I > II

II

2.

Stability order I > II (2)

The resonating structure with more no. of  bonds is more stable.

(3)

Negative charge on more electronegative atom and positive charge on less electronegative atom are more stable. e.g.

1. Stability order II > I.

2.

(4)

Stability order II > I. Structure with complete octet of each atom is more stable. e.g.

Stability order II > I.

Note : The rule of electronegativity and rule of octet is are contradictory to each other then priority is given to the octet rule.

Stability order II > I.

Stability order III > II > I . (5)

Two different compounds in which one compound has more conjugation is more stability (provided nature of bonding is same). CH2 = CH – CH = CH2 CH2 = CH – CH = CH – CH = CH2 I II Stability order II > I.

Page # 9

(6)

In two compounds if one is aromatic and another is non aromatic and conjugation is equal in both the compound the aromatic compound is more stable (nature of bonding is same) (cyclic and conjugate) Stability order

(7)

II

.

Structure with linear conjugation is more stable than cross conjugation (nature of bonding is same). Cross conjugation : If two groups are in conjugation with a particular group but not conjugated with each other then the system is called cross conjugation.

CH2= CH–CH = CH – CH = CH2 linear conjugation

> LECTURE # 4

(C) Hyper conjugation : When a sigma C–H bond of sp3 hybridised carbon is in conjugation with  bond at p-orbital, half filled p-orbital or vacant p-orbital, then the bond pair e– of sigma C–H bond overlep with adjacent p-orbital. This phenomena is called hyperconjugation. Like resonance hyperconjugation is also a stabilising effect but the effect of resonance is more dominating than hyperconjugation, since in resonance only p-orbital e– overlape while in hyperconjugation  molecule orbitals overlape with - molecule orbital. * Hyperconjugation is also called no bond resonance. Condition : There must be at least one hydrogen at sp3 hybridised -carbon and the hydrogen for hyperconjugation are called hyperconjugation hydrogen atom.

2.

Hyperconjugation in alkene

3.

Hyperconjugation in carbocation

4.

Hyperconjugation in radical H • CH 2 – C H2

5.

• H CH2 = CH2

Hyperconjugation in toluene

Page # 10

(i) Stability of alkenes no. of hyperconjugative structures 

1 HHydrogenat ion

6.

Applications of hyperconjugation

(a)

Stability of Alkenes:- Hyperconjugation explains the stability of certain alkenes over other alkenes. Stability of alkenes  Number of alpha hydrogens and Number of resonating structures CH3

H3C

H3C

CH3

H3C Example : (b)

C = CH2

H3C

H3C

Stability in decreasing order

Heat of hydrogenation : Greater the number of  hydrogen results greater stability of alkene. Thus greater extent of hyperconjugation results lower value of heat of hydrogenation Example :

(c)

H3C C = CH – CH3

C=C

CH2 = CH2 < CH3 – CH = CH2 < CH3 – CH = CH – CH3

(Heat of Hydrogenation)

Bond Length : Bond length is also affected by hyperconjugation



Example :

(i) Bond length of C(II) – C(III) bond is less than expected (ii) Bond length of C(II) – C(I) bond is more than expected (iii) C – H bond is longer than expected (d)

Stability of carbocation : Greater number of ‘’ hydrogen results greater stability of carbocations. Example :

(a)

< CH3

< CH3

CH3 < (CH3)3 CH3

(b) CH3 –

> CH3 – CH2 – CH2 –

>

–

> CH3

C –

CH3

(due to resultant of inductive effect and hyperconjugation) (e)

Stability of free radical : Greater the number of -hydrogen results greater stability of carbon free radical Example : 7

(a)

< CH3 –

(b) CH3 –

< CH3 –

> CH3 – CH2 –

– CH3 < CH3 –

>

>

(due to resultant of inductive effect and hyperconjugation)

(D)

Electromeric effects (Deals this topic in very short) (i) Electromeric effect is a temporary effect in which a shared pair of electrons (-electron pair) is completely transfered from a double or a triple bond to one of the atoms ed by the bond at the requirement of attacking species. – CH2 – E (ii) If the attacking reagent or species is removed, charge disappears and substrate attains its original form. Thus this effect is reversible and temporary. (iii) The electromeric effect may be ed or opposed by other permanent effects in various conditions as follows. Case I : Electromeric effect may be ed or opposed by I effect as : (A) CH3

 CH3 –

–

Page # 11

(B) CH3 –

 CH3 –

–

In the condition of (A) E effect is ed by + I effect similarly in the case of (B) E effect is opposed by + I effect. That is why (A) is easily possible in comparison to (B). Case II : In the case of vinyl bromide : 



E 

(A)

– CH =

(B) CH2

E – Br 

–

– Br

(A) is the condition of E effect ed by +m effect and opposed by – I effect, (B) is the condition of E effect opposed by + m effect and ed by - I effect. Since + m > – I effect, so the (A) is valid case. Case III : If the multiple bond is present between two different atoms, the electromeric shift will take place in the direction of the more electronegative atom. (a)

(b)

+E and – E effect : The electromeric effect is represented by symbol E, and is said to be + E when the displacement of  electrons is away from the atom or group, and – E when displacement of  electrons is towards the atom or group. + E effect – E effect

+

NC

+

LECTURE # 5

Physical properties (A)

Hydrogen Bonding

1.

Definition : The hydrogen bond is an electrostatic attractive force between covalently bonded hydrogen atom of one molecule and an electronegative atom (such as F, O, N) of another molecule. eg:- Consider the hydrogen fluorine bond in hydrogen fluoride, HF. This bond is a polar covalent bond in which hydrogen is attached to a strongly electronegative element. δ δ– H F The positive charge on hydrogen will be attracted electrostatically by the negative charge on F atom by another molecule of HF. H – F ----- H – F ------ H – F Covalent hydrogen bond bond Hydrogen bond is a very weak bond (strength about 2 – 10 kcal/mol)

2.

Conditions for hydrogen bonding:(a) The molecule must contain a highly electronegative atom linked to hydrogen atom. (b) The size of electronegative atom should be small. F, O and N atoms only form effective hydrogen bonding. Example : 1 Greater the electronegativity and smaller the size of the atom (F, O, N), the stronger is the hydrogen bond. H – F ----- H > H – O ----- H > H – N ----- H 10 Kcal/mole 7 Kcal/mole 2 Kcal/mole

Page # 12

3.

Types of Hydrogen Bonding:-

(A)

Intermolecular hydrogen bonding:- In such type of linkages the two or more than two molecules of the same compound combine together to give a polymeric aggregate. This phenomena is also known as association. Example : 2 (I) Hydrogen bonding in formic acid (Dimerisation)

O – H ----- O C–H

H–C O ----- H – O (II) In m - Chlorophenol

O – H ------ Cl

--Cl

(III) In water H

H

OH ------

H

- - O – H ------- O – H ----- O – H O

+ N

– O --------- H – O

(IV) In p-Nitrophenol ----- H – O

O

N +

– O -------

(V) In p-Nitrophenol & water O

+ N

– O --------- H – O H

O – H ------ O – H ----H

(B)

Intramolecular hydrogen bonding:- In this type, hydrogen bonding occur within two atoms of the same molecule. This type of hydrogen bonding is commonly known as chelation.

O – H Example : 3

– Cl

(I)

o-Chlorophenol

H O

O

(II)

o-Nitrophenol

–N O H O H

(III)

O C

O

H O

2, 6-Dihydroxybenzoic acid

Page # 13

4.

Conclusion: (a) The chelation between the ortho substituted groups restricts the possibility of intermolecular hydrogen bonding. (b) Chelation does not take place in m – & p – isomers because the two groups far away from each other. (c) Ortho isomers of hydroxy, nitro, carbonyl compounds have low M.P. than their corresponding m – & p – isomers.

(B)

Dipole moment Due to difference in electronegativity polarity developes between two adjacent atoms in the molecule. The degree of polarity of a bond is called dipolemoment. (a) Dipole moment is represented by . =e×l  = Dipolemoment l = internuclear distance between two atoms, i.e., bond length in cm. e = magnitude of separated charge in e.s.u. e = 10–10 e.s.u. d = 10–8 cm So 1D = 1 × 10–18 esu.cm (b) The Debye (D) is the unit of dipole moment. (c) The dipole moment is represented by arrow head pointing towards the positive to the negative end. Cl C Cl

Cl Cl

(d) Dipole moment of the compound does not depend only on the polarity of the bond but also depends on the shape of the molecule. (e) Dipole moment of symmetrical compound is always zero. ( = 0) (f) Symmetrical compounds are those compounds which fulfill two conditions. (i) Central atom is bonded with the same atoms or groups. (ii) Central atom should have no lone pair of electrons.

Example : 4

CCl 4 , CH4 , BH3 , CO 2 Symmetrical molecules

O= C= O Resultant  = 0

F B Example : 5

F

F

Borontrifluoride  = 0.0 (g) Dipole moment of unsymmetrical compound is always greater than zero ( μ  0 ). .. .. .. Example : 6 CHCl3 H 2.O. H2 S NH3 ..

Page # 14

Unsymmetrical molecules:..

.. O

H

H

..

N H

H Water  = 1.84D

H Ammonia  = 1.46D

(h)   electronegativity of central atom or surrounding atoms present on the central atom of the molecule.

(i) Dipole moment of the trans derivative of the compound will only be zero if both the atoms attached to carbons are in the form a and b. C=C

C=C

μ0

 =0 Example : 7

H

Cl

H C=C

Cl

b

b

b

a

a

a

a

b

H C=C

Cl  = 1.85D (cis)

Cl

H  = 0D (trans)

(ii) If both will not be atoms then  trans may or may not be zero. (iii) If group have linear moments, then the dipolemoment of the trans isomer will be zero. H3C

Example : 8

CH3

H C=C

C=C H

H

CH3

 = 0.33D

H

H3C  = 0D

(iv) If group have non-linear moments, then the dipolemoment of the trans isomer will not be zero. H

Example : 9

COOH C=C

HOOC

H

 = 2.38D

(j) Dipolemoment of disubstituted benzene are :Case I:- When both groups x and y are electron donating or both groups are electron withdrawing then:μ  μ12  μ2 2  2μ1μ 2 cos θ

1 = dipolemoment of bond c – x 2 = dipolemoment of bond c – y θ = angle between x and y

Page # 15

 If value of  will be more, then cos  will be less.

o-derivative > m-derivative > p-derivative  in decreasing order (i) If x = y and both are atoms then dipole moment of para derivative will be zero. Cl

Example : 10 Cl

 = 0 (ii) If x and y are same groups and group have linear moments then the dipolemoment of para derivative will be zero. CH3

CN

CH3

CN

 = 0

 = 0

Example : 11

Cl

Cl Cl

Example : 12  = 2.54D

Cl  = 1.48D

(iii) If x and y are same groups and x = y and group have non-linear moments then the dipolemoment of para derivative will not be zero. OH

COOEt

OH

COOEt

Example : 13

μ0 μ0 Case-II:- When one group is electron withdrawing and the other group is electrondonating then:μ  μ12  μ2 2 – 2μ1μ 2 cos θ

So

p-derivative > m-derivative > o-derivative  in decreasing order CH3

CH3

CH3 CN

Example : 14 CN

CN

 in increasing order Page # 16

LECTURE # 6 (C) Criteria of Boiling Points : The boiling point of a liquid is the temperature where its kinetic energy is sufficient to overcome the intermolecular attractive forces. Boiling point  (a) Intermolecular hydrogen bonding (b) Molecular weight attraction (c) dipole-dipole attraction (d) Strength of vander waal's Conclusion : (a) Intermolecular hydrogen bonding increases the B.P. of the compound and also its solubility in water. (b) Intramolecular hydrogen bonding (chelation) decreases the B.P. of the compound and also its solubility in water. (a) Hydrogen Bonding (i) Water:- Water has the lowest molecular weight among hydrides of the VI group of periodic table, it has the highest boiling point. Water molecules associate through intermolecular hydrogen bonding and thus require more energy to separate the molecules for vaporization. H H H H

H–O Compound B.P.

H–O

H–O H2O 100ºC

H–O–

H2S –59.6

H2Se –42ºC

(ii) Alcohols : The successive replacement of hydrogen atom of the –OH group of alcohol by alkyl group to form ether blocks the probability of hydrogen bonding reduces and thus B.P. of alcohols are higher than ether.

B.P.

CH2 – OH

CH2 – OCH3

CH2 – OCH3

CH2 – OH 197ºC

CH2 – OH 125ºC

CH2 – OCH3 84ºC

(a) Surface area of the compound : Boiling point  Surface area of the compound More is the surface area of the compound more will be the B.P. of isomeric compounds. According to MW, B.P. of isomeric compounds will be the same but according to surface area, B.P. will be different. Example : 15

(I)

(II)

(III)

(b) Dipole-Dipole attraction : (i) Boiling point of the compound, which is polar in character but has no X – H group, depends on the magnitude of dipole-dipole attractions present in the compound. B.P.  Dipole-Dipole attraction Example : 16

Ethene Ethyne

(B.P. = – 102°C) (B.P. = – 75°C)

(ii) Alkynes has higher B.P. than corresponding alkanes / alkenes, due to greater dipolemoment in alkyne then in alkene.

Page # 17

(iii) Boiling point of cis isomer is always more than the trans isomer, dipole-dipole attraction in the cis isomer is more than the trans isomer. (c) As molecular weight increases B.P. of the homologue also increases Example : 17

o-Chlorophenol has lower boiling point in comparison to its p-isomer. O H

Cl o-Chlorophenol : (B.P. = 178ºC)

H–O

Cl ------ H – O

Cl --------- H – O

Cl -----

Intermolecular hydrogen bonding in p-Chlorophenol (B.P. = 217ºC)

Example : 18 CH2 OH

CH2OCH3

CH2OCH3

CH 2OH 197ºC

CH 2OH 125ºC

CH 2OCH 3 85ºC

OH CH3 – C – CH3

OH CH3 – CH – CH2 – CH3 99ºC

CH3 – CH2 – CH2OH 118ºC

CH3 83ºC

Boiling points of aldehydes and ketones are lower than alcohols. Examples - 19 Arrange following for decreasing order of boiling point 1.

CH3 – I

CH3 – Br

CH3 – Cl

2.

CH3 – CH2 – OH

CH3 – CH2 – F

CH3 – CH3

Ans.

(1) CH3 –I > CH3 – Br > CH3–Cl > CH3–F

CH3 – F

(2) CH3–CH2–OH > CH3–CH2–F > CH3–CH3

Examples - 20 Arrange the following compound in the order of boiling point.

(I)

(II)

OH

OH

OH

Cl

CHO

Ans.

(I)

>

(II)

>

>

CHO

Examples : 21 Ortho - hydroxy, nitro-, carbonyl, carboxylic or chloro compounds have lower melting and boiling points than the respective meta or para isomer due to interamolecular, H-bonding in ortho substituted compound.

Page # 18

Reason:- Due to intramolecular hydrogen bonding association of molecules does not take place. Chelation also explains the low boiling point of enolic form of acetoacetic ester than ketonic form. O

O – H ---- O CH3 – C = CH – C – OC2H5 (enol form) Chelation

CH3 – C – CH2 – COOC2H5 (Keto form)

(E)

Melting Point : (a) Melting point:- The M.P. of a liquid is the temperature where its kinetic energy is sufficient to overcome the intermolecular attractive forces.  The heavier the molecule and stronger the intermolecular forces, higher will be the M.P. of the compound.  M.P. generally increase with increase in number of carbon atoms in most of the homologues series. Examples : 24 The M.P. of covalent compounds increases with increase in molecular weight except water. Mol. wt. M.P. B.P. H2O 18 0ºC 100ºC H2S

34

–83ºC

–59.6ºC

H2Se

81

–64ºC

–42ºC

H2Te

130

–54ºC

–1.8ºC

Reason:- In water molecule, hydrogen bonding is present so become more compact and more energy is required to get them seperated during evaporation / melting. M.P. depends upon symmetry of compounds in isomers, more symmetric compound have higher M.P. C | i.e. C  C  C > C – C – C – C – C > C  C  C  C | | C C

LECTURE # 7 Relative strength of Organic Acids : (a) Defination : (1) Arrhenius Acid : The compounds which furnish H+ ion in aqueous solution are called Arrhenius acids. Ex. H2SO4, HNO3, HCl, HClO4 etc. (2) Bronsted Acids : The species which are H+ ion donors are called Bronsted acids. Ex. NH4+ , H3O+, etc. All Arrhenius acids are Bronsted acids. (3) Lewis Acids : The lone pair acceptors are known as lewis acids. They have vacant p or d orbitals. Ex. BX3, AlX3, ZnX2 etc.

Page # 19

(b) Scale for Measurement of Acid Strength : (Arrhenius or Bronsted acids)

RCOO H  

Ka 



RCOOH

Where Ka  acid dissociation constant.

A strong acid is defined as the acid which furnish more number of H+ ion in aqueous solution OR the acid which is more ionised in aqueous solution. So, a stronger acid has higher value of Ka , or it has lower value of pKa. pKa = –log Ka (c)

(i) Inorganic acids : The mineral acids are inorganic acids. These are considered as completely ionised in aqueous solution and are described as strong acids. [H2SO4 > HCl > HNO3].

(ii) Organic acids : They are R – SO3H > R – COOH > Ph – OH > ROH. They are weakly ionised in water, so these are weaker acids then mineral acids. (d) Prediction of Acid strength : (i)

Anion/conjugate base of HX

(ii) (Cb – conjugate base) A stronger acid has more stable anion, so a stronger acid forms a more stable conjugate base. Factors affecting stability of conjugate base/anion : (i) Presence of EWG in the alkyl (–R) part of the acid increase stability of anion, and hence increases acidic strength. 

–H  

Periodicity in Acid Strength of Hydrides : (1) Along the period from left to right : As electronegativity increase, Ka  (i) CH4 < NH3 < H2O < HF.

(Ka)

Conjugate base/Anion :

(stability)

E.N. dominates over size decrease. (2) Along the group from up to down : As size increases, Acid Strength  (i) HF < HCl < HBr < HI Page # 20

(3) Organic acids :

Explanation : >

>

in acidic character

Both of equal stability

Negative charge partially localised to oxygen. ROH

* In sulponic acid three equivelent resonating structure is formed. * In carboxylic acid, the carboxylate anion has more resonating stabilization of negative charge. The two resonating structure are equally stable and are equally contributing to R.Hybrid. * The phenoxide (C6H5O–) is a less stable anion, although it has five R.S. but three R.S. (II, III, IV) have negative charge at C-atom, so are less stable and less contributing. * The alkoxide anion does not have any resonance stabilization, so it is least acidic.

Page # 21

LECTURE # 8 4.

Experimental Order (Ka) :

(Ar – Aromatic)

(1) Aliphatic acid

Ex.1

Ans.

(11) > (2) > (7) > (3) > (6) > (10) > (1) > (4) > (5) > (8) > (9)

Ex.2

CH3COOH < ClCH2COOH < Cl2CHCOOH < Cl3CCOOH

Ex. 3

Ex.4

>

(a) H–COOH (I = 0)

>

>

CH3–COOH (+I)

Page # 22

(b) CH3 – COOH (+I)

>

CH3–CH2–COOH (+I)

(c) C–COOH > C–C–COOH > C–C–C–COOH Thus, as alkyl size increase, acid strength decreases. (e) C–C–C–C–COOH >

>

(f) F – CH2 – COOH > Cl – CH2 – COOH > Br – CH2COOH > I – CH2 – COOH (g) (h) CH3 – CH2 – COOH < CH2 = CH – COOH < CH  C – COOH sp3 sp2 sp

(2) Dicarboxylic acids :

1st one is stabilizing, 2nd one destabilizing. In case of Polybasic acids (compounds having more than one acidic H) , the successive acid dissociation constant always have order Ka1 > Ka2 > Ka3 > ............... In second dissociation

ion is taken out from a negatively charged species, so it is difficult.

(ii)

Ka1  I > II > III > IV pKa1  IV > III > II > I (iii)

Maleic acid is stronger acid than fumaric acid.

Page # 23

*

Ka1 > Ka1’ Ka2’ > Ka2

LECTURE # 9 Acidic strength in Aromatic acid : Acidic strength in aromatic acid depends upon following factors (i) Steric effect (steric inhibition of resonance) or Ortho effect (ii) Electronic effect of substituents (iii) Intermolecular and intramolecular hydrogen bonding. Steric effect (ortho effect) : In case of 1,2-disubstituted benzene if the substituents are bulky then due to steric repulsion (vanderwaal repulsion) the group go out of plane w.r.t. benzene ring. Due to this change in planarity the conjugation between the substituents on benzene is slightly diminished. This effect is called steric inhibition of resonance. (ortho effect) In case of ortho substituted benzoic acid due to steric inhibition of resonance the conjugation between benzene and carboxylic group is diminished and the +m effect benzene ring to the carboxylic is diminished and as the acidity increases w.r.t. benzoic acid and w.r.t. meta and para substituent benzoic acid inspite any electronic effect of substituent G may be  – NO2 , – R , – X , – COOR , – COOH except  – OH , – NH2, – C  N O

O

O–H

C

G

–

C

O

G



H  

(COOH group goes out of plane) Substituted Benzoic Acid : (1) (G = –m, –I) COOH

COOH

COOH

NO2 (c)

NO2 (b)

C

O

O

+

–H+

–H+

–H+ O

NO2

O

C

O

O

N

O

O N

–I, –m

C

O –I (m)

N O

O –I (w), –m Page # 24

Ka order = ortho > para > meta > benzoic acid (2) G = (–I) ................. CCl3

Ka order = ortho > meta > para > benzoic acid (3) G = (–I > +m) – Cl, Br, F, I

Ka order = ortho > meta > para > benzoic acid (4) G = (+m > –I) ..... OCH3

Ka order = meta  ortho > benzoic acid > para (5) G = (+I, H.C.) ........ R (Alkyl group)

Ka order = ortho > benzoic acid > meta > para

<

Ex.

Ex.

<

<

<

>

Electronic effect : In case of different O – susbstituted benzoic acids the acidity is not decided by the basis of ortho effect but it is decided by the. electronic effect of substituents.

Ex.1

I

+I II

+m>–I III

–I>+m II

–I,–m V

Ka order = V > VV > V > VV > VVV Page # 25

Ex.2 +I II

I

–I III

–I II

–I V

Ka order = V > VV > VVV > V > VV At meta positions only electronic effect is operative and m effect is not operative.

Ex.3 +I II

I 3.

–I III

–I II

–I V

Ka order = V > VV > VV >VVV > V Intermolecular& Intramolecular hydrogen bonding Intramolecular (6-membered) H-bond due to ortho is effective only in Phenols and not in –COOH (7-membered, insignificatnt).

COOH

Ex.1 COOH

–H +

–H+

OH – O C

O

OH

COO

Ortho anion stabilised by intramolecular H-bonding. Ka B.P. Solubility M.P. Separation

o>p o

Ex.2

Ka order : (i) > (ii) > (iii) > (iv). Note:

Ortho effect is operative only in o-substituted Benzoic acids and not in o-substituted phenols.

Page # 26

LECTURE # 10 (3) Acid strength of Phenols (effect of substituents) : There is no ortho effect in phenol

>

>

(1) EWG = (–m , –I) (–NO2, CHO, –COR, C  N, etc.)

Ka order = Para > Ortho > Meta > Phenol (2) EWG = (–I)

with no mesomeric effect

>

>

>

Ka order = Ortho > Meta > Para > Phenol (3) EWG = (–I > + m) X = F, Cl, Br, I

Ka order = Ortho > Meta > Para > Phenol

Page # 27

(4) G = R (Alkyl group) = + I, H.C.

Ka order : Phenol > Meta > Para > Ortho (5) G = (+m > –I) (–OH, –NH2, –OR, etc.)

Ka order : Meta > Phenol > Ortho > Pera

LECTURE # 11

Ex.1

>

(1)

>

>

(II)

(III)

>

>

(2)

(I)

(IV)

(V)

(VI)

only (–I) Page # 28

Ka order = I > II > III > VI > IV > V

(3)

(I)

(II)

(III)

(IV)

(V)

(VI)

Ka order = I > II > III > IV > V > VI

>

(4)

>

(5)

(No acidic H)

>

(6)

>

>

picric acid Reaction of Acids with salts : (1)

Ex.

NaX + Salt = NaOH + HX Weak acid

HY  Strong Acid

NaY + Salt = NaOH + H – Y (S.A.)

HX W.A.

2 NaCl + H2SO4  Na2SO4 + 2HCl Na2SO4 + 2HCl  No reaction

Remark : A stronger acid displaces the weaker acid from any metal salt. The weaker acid is released out as a gas or liquid or precipitates out as a solid. The weaker acid cannot displace the stronger acid from the salt. Ex.

CH3COONa + CH3SO3H  CH3COOH + CH3SO3Na (feasible) CH3COONa + PhOH  PhONa + CH3COOH (not feasible)

Q.

Which of the follwing reaction is possible ? (A) CH3COOH + HCOONa Not possible (reverse is possible)

ONa

OH (B)

+

 Not possible

NO2 Page # 29

SO3Na (C)

COOH +

 Not possible

(D*) HC  C – Na + H2O  HC  C – H + NaOH

LECTURE # 12 Basic strength (I) DEFINITION:

 

 









Ex:

(a) Arrhenius base: Those compound which furinishes OH - ions in aquous solutions are known as arrhenius base. NaOH, KOH, Ca(OH)2 etc. (b) Bronsted base:- e– pair donor or H ion acceptor. :NH3, R N H2, R2 N H, R3 N , H2 O, R O H, R – O – R (c) Lewis base:- e– pair donor to H ion.

 NH4

 

R – O – R + H+

+ R– O–R H –



NH 3 + H +

(II) BASICITY: It is the tendency to accept H ion, or it is the case of acceptance of H ion. H3N: + H __________easily..

 

Ex:

H – O – H + H __________less easily.. Thus, NH3 > H2O in basicity. Less electronegative atom (N) donates electron pair easily.

(III) SCALE OF BASICITY/BASIC STRENGTH:-



In aq. solution:R – NH 2 + H2O Base

+ R – NH3 + OH Conjugate Acid (C.A.)

 [RNH3 ][OH - ] Kb = – [where Kb = Base dissociation constant]. [RNH2 ]

pKb = –logKb Note : A stronger base always has a weaker C.A. and vice versa.

(IV) PERIODICITY IN Kb (BASIC STRENGTH):Factors:- (i) E.N. of element__________E.N.  , Kb  (ii) Size of element__________size  , Kb  (a) CH3 > NH2 > OH > F - -------- E.N. CH - is strongest base in periodic table. 3

(b) CH3 > P H2 > S H > Cl-------- E.N.

(c) F - > Cl - > Br - >  - -------- size Page # 30





(d) - O H > - SH-------- size (e) H2 O > H2 S (f) :NH3 > :PH3

(V) CARBANION BASES (:C - ) : (i) CH3 – CH2 > CH2 = CH > CH  (E.N.  , Kb  ) (ii) CH3 – CH2 – CH2 >> CH2 = CH – CH2 (delocalised lone pair) due to Resonance (iii) CH2 = CH – CH2 > CH3 – C – CH2

O

CH2

(better resonance due to –ve charge on ‘O’)

-

–

(iv)

< (Resonance Stabilisation)

(localised –ve charge)

Criteria for deciding basec strengths :1. Electronic effect : +I, +m group increases electron density at nitrogen atom and increases is lone pair donating ability therefore basicity increases. –I, –m group decreases electron density at nitrogen and hence decreases its lone pair donating ability, therefore basicity decreases. 2. Hybridisation of Nitrogen atom : The lone pair donating ability of nitrogen atom changes with hybridisation.

sp3(N) > sp2(N) > sp(N)

CH3 – CH2 – CH2 – N H2 > CH2 = CH – CH2 – N H2 > CH  C – CH2 – N H2 – –

Ex.



Kb  lone pair donating ability



sp(N) > sp2(N) > sp3(N)



Electrongeativity :







(VI) NITROGENOUS BASES –N: :-



Classification:- (i) Aliphatic amines (R NH2, R2 NH, R3 N)



(ii) Aromatic amines (Ph – NH2) or Anilines (iii) Amides R – C – NH2



O (iv) Amidines R – C – NH2 :NH Among these, amides are the weakest bases. (iv) > (i) > (ii) > (iii)

(i) Aliphatic amines : Consider the following molecules

,

,

,

By visiting + I effect of methyl in above example we may expect basic nature as NH3 < MeNH2 < Me2NH < Me3N Which is not true This is due the fact that basic strength of an amine in water is determined not only by ease of electron donation (protonation) of N atom but also by the extent to which cation so formed can undergo SOLVATION Page # 31

and become stabilised by H atom attatched to N atom greater is possibility of solvation via H bonding by water. Alkyl groups are hydrophobic and inhibits H bond. – –

H  H H – N:    + = O, solvation = max., Steric Hinderance = min. (1) H – –

H H (2) R – N:  +, solvation, Steric Hinderance H

– –

H H (3) R – N:  + , solvation, Steric Hinderance R – –

R H R – N:  +max., solvation min., Steric Hinderance max. (4) R In aq. solvation, the general order of basic strength is R2NH > R3N > RNH2 > NH3. (R = Ethyl)

or, in some cases, it is R2NH > RNH2 > R3 N > NH3 (R = Methyl) Explanation: It will be seen that the introduction of an alkyl group into ammonia increases the basic strength markedly as expected. The introduction of a second alkyl group further increases the basic strength, but the net effect of introducing the second alkyl group is very much less marked than with the first. The introduction of a third alkyl group to yield a tertiary amine, however, actually decreases the basic strength in both the series quoted. This is due to the fact that the basic strength of an amine in water is determined not only by electron-availability on the nitrogen atom, but also by the extent to which the cation, formed by uptake of a proton, can undergo solvation, and so become stabilised. The more hydrogen atoms attached to nitrogen in the cation, the greater the possibilities of powerful solvation via hydrogen bonding between these and water:

Thus on going along the series, NH3  RNH2  R2NH  R3N, the inductive effect will tend to increase the basicity, but progressively less stabilisation of the cation by hydrating will occur, which will tend to decrease the basicity. The net effect of introducing successive alkyl groups thus becomes progessively smaller, and an actual changover takes place on going from a secondary to a tertiary amine. Conclusion : (1) Secondary amines are stronger bases than tertiary amines. Reason:- Solvation is less in 3º amines and more steric hinderance to H ion. (2) All alkyl amines (1º, 2º, 3º) are stronger bases than ammonia (due to + effect of ‘R’ group). (3) In gaseous phase, the basic strength order is 3º > 2º > 1º > NH3 (+ effect of R group).

– CH2 – NH2 >

(4)







 

: NH3 > NH2 – NH2 > NH2 – O H Ammonia Hydrazine Hydroxylamine

 

(3)

– CH2 – NH2

 

(2)



(1)



Ex.

N H3 > R –– O –– R > R – O – H > H – O – H (+) (+ )

N:

>

N:

(More compact)  Cyclic Amine is more basic than acyclic amine (if degree of ‘N’ is same).

Page # 32

–

–

(5) :NH2

:NH2

(ii) > (iii) > (i) > (iv)

(6) Kb order : I > II > III

(7) I Sulphonamide

II Amide

III Ar. Amine

IV Aliphatic amine

Kb order : IV > III > II > I (8)

CH2 — CH2 — NH2 | Ph I

CH3 — CH — NH2 | Ph

CH3 — CH2 — NH | Ph

II

III

Kb order : I > II > III (9)

Ph—NH2 I

Ph2NH II

Ph3N III

Kb order : I > II > III

(10)

I

II

III

IV

Kb order : IV > II > III > I



LECTURE # 13 (ii) Aromatic amines (Ph – NH2) or Anilines :

When the lone pair lies in conjugation with a multiple bond, it resides in ‘2p’ atomic orbital, so that it can get resonance stabilisation. Aniline is a weaker base than NH3 because it has delocalised lone pair.

Page # 33

Ex.

Which of them is stronger base CH3 – NH2 Since ease of donation of lone pair of N is basicity, CH3 – NH2 is more basic due to + I effect of – CH3 group. Aryl amines aniline is very less basic since lone pair of N are involved in resonance.

Ex.

Which of them is strong base

In pyrole lone pairs are involved in resonacne therefore it is less basic. But in pyridine lone pairs are in perpendicular plane of  orbitals therefore not involved in resonance Substituted Anilines:(i) G = ERG (+m, HC, +)__________Kb  (ii) G = EWG (–m, –)__________Kb  Steric effect of ortho-substituted G (ortho effect) :

–

H2N:

G –

–

–

H + H–N–H G –

H 

(a) Ortho-substituted anilines are mostly weaker bases than aniline itself. (b) Ortho-substituent causes steric hinderance to solvation in the product (conjugate acid i.e. cation). (c) The small groups like –NH2 or –OH do not experience (SIR) due to small size. (1) G = (–m, –); NO2

–

NH2 (a)

(b)

(c)

(d)

Kb : dbc a

(2) G = (–); CCl3 –

–

–

NH2 (Aniline > p > m > o).

CCl3

–

CCl3

–

–

NH2

NH2

NH2 –

CCl3 (3) G = (– > +m); Cl –

–

–

–

NH2 –

NH2

NH2 Cl

aniline > p > m > o

Cl

–

NH2 –

Ex.

Cl

Page # 34

Only (–) decides the order. (4) G = (+, HC); R = –CH3 (Toluidines)

NH2 –

–

–

– O effect, + HC

+m

CH3

NH2

–

–

CH3

–

NH2

NH2

CH3 +w HC  more do min ating

(5) G = (+m > –);

Kb order : P > Aniline > O > M

LECTURE # 14 

(iii) Amides :

In amides lone pair donation atom is oxygen which is more electronegative. so it can hold negative charge more effectively, so it donation tendency decrease (kb decreases).

(iv) Amidines : (a)

Hybridisation of ‘N’ : sp3 > sp2 > sp

(b)

In case of amidines, the doubly-bonded ‘N’ is more basic in nature. Although, both the ‘N’ are sp2 hybridised. The lone pair of most basic ‘N’ lies in sp2 hybrid orbital (localised).

–

x R y H2N – C = NH –

R x y NH2 – C = NH

(1) sp2 sp2 2 (2) lone pair in sp H.O. lone pair in 2p A.O. (3) lone pair localised lp delocalised (4) Basicity y > x. Strongest organic Nitrogenous base:(+m)   (+m) NH 2 – C – NH 2

Guanidine

:NH2 Page # 35

+ :NH2 – C = NH2 NH2

 NH2 :NH Heterocyclic Compounds (Nitrogenous base) : (1) Pyridine (C5H5N:)

–

–

+ NH2 = C – NH2 NH2

 – –

H

S.F.:- H – H

H

6e – Aromatic Stronger  base

–

N:

–

H

 

 A.O. Diagram:-





N



  H NH2 – C – NH2  :NH2 – C – NH2

2

sp hybrid orbital (lone pair) localised

(2) Pyrrole (C4H5N:):6 e – Aromatic Very weak base Non  aromatic after it donates lone pair

2

:N – H (sp )

S.F.:-



 (lp. in 2p A.O.)



A.O. Diagram:-



N–H



Complete delocalisation of e–

>

 N H H

–

N 

>

H

–

:NH2 H

+ – 3 N ––– sp (4e ) – H H Non-aromatic

NH3

–

Aromatic

 

–

+ N H Aromatic

–

1.

2.

O

3.

a>c>b>d

 N H (c) sp3/2º (–) –

–

 N

(a) (b) sp3/2º sp2 localised    NH2 – C – NH2 ; R – C – NH2 :NH (1)

 N H (d) sp2 –

 N H

; R–C–R ;

:NH (2) sp2

NH (3) sp2

Kb order : (1) > (2) > (4) > (3) > (5) –

:NH2(x) 4.

N  (y)

 N(z) H

Basicity order:- y > x > z

–

Ex:-

Page # 36