Effects Strengths

8/10/2019 Effects Strengths

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ORG NIC LECTURE NOTES

GOC-II

Electronic effects (I, R, m, Hyperconjucation), Physical properties

Acidic strength & basic strength


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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 electronega-tive atom due to this shifiting of ea dipole is created between the two atoms.Due to this dipole the bonds ein the chain are also shifted and the chain becomes polarised this effect ofpolorisation in the chain due to a dipole is called inductive effect.

Diagram showing Ieffect

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 thechange(iii) Inductive effect is operative only through bond it does not affect or participate the bond electron(iv) The CH bond is the reference of inductive effect i.e. the electron movement or polarity in the CHbond is considered to be neglible and the I effect of Hydrogen is taken to be zero.

2. Types of inductive effect :

(a) IEffect :

The group which withdraws electron cloud is known as Igroup and its effect is called Ieffect. Variousgroups are listed in their decreasing Istrength as follows.

> > > NO2

> SO3H > CN > CHO > COOH > F > Cl > Br > I

> OR > OH > NH2 > C CH > C

6H

5 > CH = CH

2 > H.

(b) + IEffect :

The group which repel (or release electron) cloud is known as + Igroup & this effect is + Ieffect.

> > C(CH3)

3 > CH (CH

3)

2 > CH

2 CH

3 > CH

3 > D > H

Example-1 Since NO2is Igroup it pulls or withdraws e from cyclohexane ring making it e deficient

Example-

Example-

Due to e donating nature of carbon chain has become partially negative but COOH is Igrouptherefore carbon chain has become partially positive.Example -

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


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(a) CH3

CH2

CH2

(b)

(c) CH2

NC (d)

(e) (f)

(g) (h)

(i) (j) CH CH3 NO2+ CH

+

(k)

O

-- -

--

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 decid in g acid ic stre ngt h of

i ) Carbox yl ic acid

I effect Acidic strength (presence of Igroups increases acidic character)+ I effect Basic strength (presence of + Igroups increases basic character)

Example :

(I)

Cl|

COOHCHCHCHCH 223 (II)

Cl|

COOHCHCHCHCH 223

(III)

Cl|

COOHCHCHCHCH 223 (IV) CH3 CH2 CH2 CH2 COOH

IVIIIIII

orderstrengthAcid

Explanation :

We know that Ieffect increases acidic strength. Now morever it is distance based effect so where the Igroup is nearest to COOH, It enters strong effect and makes acid strong.

Example:(I) O

2N CH

2 COOH (II) F CH

2 COOH

(III) H3CO CH

2 COOH (IV) CH

3 CH

2 COOH

IVIIIIII

orderstrengthAcid

Since NO2has strong Ieffect its influence will make corresponding acid strongest ( Ieffect acid


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

Rule for writing resonating structure :(i) In resonating structure only p-orbitals eare shifted, bond eare not involved in resonance, therefore thebond skeleton well remain same in two resonating structures.(ii) The atoms are not moved(iii) The no. of paired p eare same and pair of unpaired p eare 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 hybrideis 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 = CH

2

2. CH2= CH CH = CH

2 CH

3 CH = C = CH

2

3. CH2= CH

4. CH2= CH CH = CH

2

5. CH2= CH CH = CH

2

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. CH

2= CH C CH (Show resonance)

2. CH2= CH CH

2 CH = CH

2(Not show resonance)

3. (Show resonance)

4. (Not show resonance)

5. CH2= C = CH

2(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 bond, positive charge, negative charge, odd electron or lone pair electron.

1. CH2= CH CH = CH

2(Conjugation between C = C and C = C)

2. CH = CH CH2 2+

(Conjugation between +ve charge and C = C)

3. CH = CH NH2 2. .

(Conjugation between lone pair and C = C)


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4. CH2= CH 2HC

(Conjugation between odd electron and C = C)

5. CH = CH CH = CH CH2 2

(Conjugation between negative charge and C = C)

Q. Write resonating structure for each of the following molecules :

(a) 33 CHCCHCHCH||O

(a)

(b) NCCHCH..

NH2 (b)

(c) (c)

(d) CH3 O CH = CH (d)

(e) (e)

(f) :..

OCCH3

(f)

(g) (g)

(h) (h)

(i) (i)

(j) (j)

Q. Draw resonating structure for following molecules

, CH2= C = O ,

Ans.+

+

+

CH C = O3. ..

.

OCCH3 CH = C = O2. ..

.

OCHC 2 N = N = N NNN2


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LECTURE # 3

Mesomeric Effect : When a group releases or withdraw p orbital electron in any conjugated system then it is calledM 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 , SO3

H , CHO, CH = NH, (G = H, R, OH, OR, NH2

, NHR, NR2

, X)

e.g.

1.

2.

O N = O:. . +

O N O:. .. . +

+

O N O:. .. . +

+

O N O:. .. . +

+O N = O:

. . +

3. H C = CH C N:2+

H C CH = C = N:2 . .

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

,


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Ans.

(a)

NHCCH3

O

NHCCH3

O

NHCCH3

O

NHCCH3

O

NHCCH3

O

(b)

CCH3

O

CCH3

O

(m effect group)

CCH3

O

CO CH3 C

O CH3

CO CH3

CO CH3

Note:

1. 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. +M group increases electron density in benzene ring while M group decreases electron density in thebenzene ring.

Q. Write electron density order in the following compound.

(a)

Ans. III> I> II> IV

(b)

Ans. I > II > III . IV

(c)

Ans. IV > I > III > II


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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 IIStability order I > 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 morestable.e.g.

1.

Stability order II > I.

2.

Stability order II > I.(4) 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 givento 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 ofbonding is same).

CH2= CH CH = CH

2CH

2= CH CH = CH CH = CH

2

I IIStability order II > I.


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

(cyclic and conjugate)

Stability order . II

(7) 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 otherthen the system is called cross conjugation.

CH2= CHCH = CH CH = CH2 >linear conjugation

LECTURE # 4(C) Hyper conjugation :

When a sigma CH bond of sp3 hybridised carbon is in conjugation with bond at p-orbital, half filledp-orbital or vacant p-orbital, then the bond pair eof sigma CH bond overlep with adjacent p-orbital. Thisphenomena is called hyperconjugation.

Like resonance hyperconjugation is also a stabilising effect but the effect of resonance is moredominating than hyperconjugation, since in resonance only p-orbital eoverlape 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 sp3hybridised -carbon and the hydrogen for hyperconjugation arecalled hyperconjugation hydrogen atom.

2 . H y p e r co n ju ga t io n in a lk e n e

3 . H yp erco nj u ga tio n in c a rb oca ti o n

4 . H y p e rco nju ga tio n in ra d ic a l

CH C2

H2

CH = C2

H2

HH

5 . H y p e rco nju ga tio n in t o lu e ne


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(i) Stability of alkenes no. of hyperconjugative structuresionHydrogenatH

1

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

Example :

C = C C = CH CH3 C = CH2

H C3 H C3 H C3

H C3 H C3 H C3

CH3

CH3

Stability in decreasing order

(b) Heat of hydrogenation : Greater the number of hydrogen results greater stability of alkene. Thusgreater extent of hyperconjugation results lower value of heat of hydrogenation

Example : CH2= CH

2< CH

3 CH = CH

2< CH

3 CH = CH CH

3(Heat of Hydrogenation)

(c) 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< (CH

3)

3

(b) CH3 > CH

3 CH

2 CH

2 > >

CHCH CCH

3

3

3

(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 < CH

3 CH

3< CH

3

(b) CH3 > CH

3 CH

2 > >

(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 completelytransfered from a double or a triple bond to one of the atoms joined 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 supported or opposed by other permanent effects in various conditions asfollows.

Case I: Electromeric effect may be supported or opposed by Ieffect as :

(A) CH3 CH3


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(B) CH3 CH3

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

(A) E CH = (B) CH

2 Br

E

Br

(A) is the condition of E effect supported by +m effect and opposed by Ieffect, (B) is the condition of Eeffect opposed by + m effect and supported by - Ieffect.Since + m> Ieffect, so the (A) is valid case.

Case III:If the multiple bond is present between two different atoms, the electromeric shift will take placein 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 thedisplacement of electrons is away from the atom or group, and E when displacement of electrons istowards the atom or group.

+ E effect + NC

E effect +

LECTURE # 5

Physical properties

(A) Hydrogen Bonding1. Definition :

The hydrogen bond is an electrostatic attractive force between covalently bonded hydrogen atom of onemolecule 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 inwhich hydrogen is attached to a strongly electronegative element.

F

H

The positive charge on hydrogen will be attracted electrostatically by the negative charge on F atom byanother molecule of HF.

H F ----- H F ------ H F

Covalentbond

hydrogenbond

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 electronegativ ity and smaller the size of the atom (F, O, N), the stronger is the hydrogen bond.

H F ----- H10 Kcal/mole

> H O ----- H7 Kcal/mole

> H N ----- H2 Kcal/mole


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3. Types of Hydrogen Bonding:-

(A) Intermolecular hydrogen bonding:- In such type of linkages the two or more than two molecules of thesame compound combine together to give a polymeric aggregate.This phenomena is also known as association.

Example : 2

(I) Hydrogen bonding in formic acid (Dimerisation)

H C C H

O H ----- O

O ----- H O

(II) In m - Chlorophenol

--Cl O H ------ Cl OH ------

(III) In water

- - O H ------- O H ----- O H

H H H

(IV) In p-Nitrophenol

N

O+

----- H O

O --------- H O

O

N

O -------+

(V) In p-Nitrophenol & water

N

O+

O H ------ O H -----

O --------- H O

H

H

(B) Intramolecular hydrogen bonding:- In this type, hydrogen bonding occur within two atoms of the samemolecule. This type of hydrogen bonding is commonly known as chelation.

Example : 3 (I)

O H

Cl o-Chlorophenol

(II)

O O

O

N

H

o-Nitrophenol

(III)

C

O

OOH

H

O

H

2, 6-Dihydroxybenzoic acid


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4. Conclusion:(a) The chelation between the ortho substituted groups restricts the possibility of intermolecular hydrogenbonding.(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 momentDue 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

= Dipolemomentl = internuclear distance between two atoms, i.e., bond length in cm.e = magnitude of separated charge in e.s.u.e = 1010e.s.u.d = 108cmSo 1D = 1 1018esu.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.

C

Cl

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 : 4moleculeslSymmetrica

CO,BH,CH,CCl 2344 O = C = OResultant = 0

Example : 5

F

B

F F

Borontrifluoride= 0.0

(g) Dipole moment of unsymmetrical compound is always greater than zero ( 0 ).

Example : 6 H O2. .. . H S2

. .

. . CHCl3 . .NH3


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Unsymmetrical molecules:-

N

H

H

H

Ammonia

= 1.46D

. .

O

H

HWater

= 1.84D

. .

..

(h)

electronegativity of central atomor 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 tocarbons are in the form a and b.

C = C

b

ba

a

= 0

C = C

a

bb

a

0

Example : 7

C = C

H H

Cl Cl

= 1.85D(cis)

C = C

H

HCl

Cl

= 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.

Example : 8 C = C

H H

CH3H3C

= 0.33D

C = CH

H

CH3

H3C

= 0D

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

Example : 9 C = C

H

H

COOH

HOOC

= 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:-

cos2 212

22

1

1= dipolemoment of bond c x

2= dipolemoment of bond c y

= angle between x and y


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

in decreasing order

o-derivative > m-derivative > p-derivative

(i) If x = y and both are atoms then dipole moment of para derivative will be zero.

Example : 10

Cl

Cl

= 0

(ii) If x and y are same groups and group have linear moments then the dipolemoment of para derivative willbe zero.

Example : 11

CH3

CH3

CN

CN

= 0 = 0

Example : 12

Cl

Cl

= 2.54D

Cl

Cl= 1.48D

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

Example : 13

OH

OH COOEt

COOEt

0 0

Case-II:- When one group is electron withdrawing and the other group is electrondonating then:-

cos2 212221

Soin decreasing order

p-derivative > m-derivative > o-derivative

Example : 14

CH3CN

CH3

CN

CH3

CN

in increasing order


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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'sConclusion :

(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 inwater.

(a) Hydrogen Bonding

(i) Water:- Water has the lowest molecular weight among hydrides of the VI group of periodic table, it hasthe highest boiling point.

Water molecules associate through intermolecular hydrogen bonding and thus require more energy to sepa-rate the molecules for vaporization.

H O H O H O H O

H H H H

Compound H2O H

2S H

2Se

B.P. 100C 59.6 42C

(ii) Alcohols :

The successive replacement of hydrogen atom of the OH group of alcohol by alkyl group to form etherblocks the probability of hydrogen bonding reduces and thus B.P. of alcohols are higher than ether.

CH OH2

CH OH2

CH OCH2 3

CH OH2

CH OCH2 3

CH OCH2 3

B.P. 197C 125C 84C

(a)Surface area of the compound :Boiling point Surface area of the compoundMore 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 bedifferent.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 themagnitude of dipole-dipole attractions present in the compound.

B.P. Dipole-Dipole attraction

Example : 16 Ethene (B.P. = 102C)Ethyne (B.P. = 75C)

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


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(iii) Boil ing point of cis isomer is always more than the trans isomer, dipole-dipole attraction in the cis isomeris 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.

Cl

OH

o-Chlorophenol : (B.P. = 178C)

H O Cl ------ H O

Cl --------- H O Cl -----

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

Example : 18

CH OH2

CH OH

197C2

CH OH2

CH OH

197C2

CH OH2

CH OH

197C2

CH OH2

CH OH

197C2

CH OH2

CH OH

197C2

CH OH2

CH OH

197C2

CH OH2

CH OH

197C2

CH OH2

CH OH

197C2

CH OH2

CH OH

197C2

CH OH2

CH OH

197C2

CH OH2

CH OH

197C2

CH OCH2 3

CH OH125C

2

CH OCH2 3

CH OH125C

2

CH OCH2 3

CH OH125C

2

CH OCH2 3

CH OH125C

2

CH OCH2 3

CH OH125C

2

CH OCH2 3

CH OH125C

2

CH OCH2 3

CH OH125C

2

CH OCH2 3

CH OH125C

2

CH OCH2 3

CH OH125C

2

CH OCH2 3

CH OH125C

2

CH OCH2 3

CH OH125C

2

CH OCH2 3

CH OCH85C

2 3

CH OCH2 3

CH OCH85C

2 3

CH OCH2 3

CH OCH85C

2 3

CH OCH2 3

CH OCH85C

2 3

CH OCH2 3

CH OCH85C

2 3

CH OCH2 3

CH OCH85C

2 3

CH OCH2 3

CH OCH85C

2 3

CH OCH2 3

CH OCH85C

2 3

CH OCH2 3

CH OCH85C

2 3

CH OCH2 3

CH OCH85C

2 3

CH OCH2 3

CH OCH85C

2 3

CH CH CH OH

118C3 2 2 CH CH CH CH

99C3 2 3

OH CH C CH3 3

OH

CH383C

Boiling points of aldehydes and ketones are lower than alcohols.

Examples - 19 Arrange following for decreasing order of boiling point

1. CH3 I CH

3 Br CH

3 Cl CH

3 F

2. CH3 CH

2 OH CH

3 CH

2 F CH

3 CH

3

Ans. (1) CH3I> CH

3 Br > CH

3Cl > CH

3F (2) CH

3CH

2OH > CH

3CH

2F > CH

3CH

3

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

(I) (II)

Ans. (I)

OH

CHO

>

OH

CHO(II)

OH

>

Cl

>

Examples : 21

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


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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.

CH C CH COOC H3 2 2 5

O

(Keto form)

CH C = CH C OC H3 2 5

O

(enol form) Chelation

O H ----

(E) Melting Point :

(a) Melting point:-The M.P. of a liquid is the temperature where its kinetic energy is sufficient to overcomethe 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 0C 100C

H2S 34 83C 59.6C

H2Se 81 64C 42C

H2Te 130 54C 1.8C

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

i.e.

C|

CCC|

C

> C C C C C >

C|

CCCC

LECTURE # 7Relative strength of Organic Acids :

(a) Defination :

(1) Arrhenius Acid : The compounds which furnish H+ion in aqueous solution are called Arrhenius acids.Ex. H

2SO

4, HNO

3, HCl, HClO

4etc.

(2) Bronsted Acids : The species which are H+ion donors are called Bronsted acids. Ex. NH4+, H

3O+,

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.BX

3, AlX

3, ZnX

2etc.


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(b) Scale for Measurement of Acid Strength :

(Arrhenius or Bronsted acids)

RCOOHHRCOO

Ka

Where Kaacid 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 K

a, or it has lower value of pK

a.

pKa= log K

a

(c) (i) Inorganic acids : The mineral acids are inorganic acids. These are considered as completely ionisedin aqueous solution and are described as strong acids. [H

2SO

4> HCl > HNO

3].

(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 increasesacidic strength.

H

Periodicity in Acid Strength of Hydrides :

(1) Along the period from left to right :As electronegativity increase, Ka

(i) CH4

< NH3 < H

2O < 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


21/36Page 21

(3) Organic acids :

Explanation :

> > in acidic character

Both of equal stability

Negative charge partiallylocalised 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 tworesonating structure are equally stable and are equally contributing to R.Hybrid.

*The phenoxide (C6H

5O) 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.


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LECTURE # 84. Experimental Order (K

a) : (Ar Aromatic)

(1) Aliphatic acid

Ex.1

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

Ex.2 CH3COOH < ClCH

2COOH < Cl

2CHCOOH < Cl

3CCOOH

Ex. 3 > >

Ex.4 (a) HCOOH > CH3COOH

(I= 0) (+I)


23/36Page 23

(b) CH3 COOH > CH

3CH

2COOH

(+I) (+I)

(c) CCOOH > CCCOOH > CCCCOOH

Thus, as alkyl size increase, acid strength decreases.

(e) CCCCCOOH > >

(f) F CH2 COOH > Cl CH

2 COOH > Br CH

2COOH > I CH

2 COOH

(g)

(h) CH3 CH

2 COOH < CH

2= CH COOH < CH C COOH

sp3 sp2 sp

(2) Dicarboxylic acids :

1stone is stabilizing, 2ndone destabilizing.

In case of Polybasic acids (compounds having more than one acidic H) , the successive acid dissociationconstant always have order

Ka1 > Ka

2 > Ka

3 > ...............

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

(ii)

Ka1 I > II > III > IV

pKa1IV > III > II > I

(iii) Maleic acid is stronger acid than fumaric acid.


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* Ka1 > Ka

1

Ka2 > Ka

2

LECTURE # 9Acidic 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 planaritythe conjugation between the substituents on benzene is slightly diminished. This effect is called stericinhibition of resonance. (ortho effect) In case of ortho substituted benzoic acid due to steric inhibition ofresonance the conjugation between benzene and carboxylic group is diminished and the +m effect benzenering to the carboxylic is diminished and as the acidity increases w.r.t. benzoic acid and w.r.t. meta and parasubstituent benzoic acid inspite any electronic effect of substituent

G may be NO2, R , X , COOR , COOH

except OH , NH2, C N

CO HO

G H

CO

O

G

(COOH group goes out of plane)

Substituted Benzoic Acid :

(1) (G = m, I)

COOH

NO2

C

N

OO

+

O+

H

, mI

COOH

NO2

C

N

OO

+

O

H

(b)

(m)IO

COOH

NO2

C

N

OO

O

(c)

(w), mI

O

H+


25/36Page 25

Kaorder = ortho > para > meta > benzoic acid

(2) G = (I) ................. CCl3

Kaorder = ortho > meta > para > benzoic acid

(3) G = (I> +m) Cl, Br, F, I

Kaorder = ortho > meta > para > benzoic acid

(4) G = (+m > I) ..... OCH3

Kaorder = meta ortho > benzoic acid > para

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

Kaorder = ortho > benzoic acid > meta > para

Ex. < < <

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

Ex.1

+ I + m > I I> + m I, mI II III II V

Kaorder =

V VV V VV VVV


26/36Page 26

Ex.2

+ I I I II II III II V

Kaorder =

V VV VVV V VV

At meta positions only electronic effect is operative and m effect is not operative.

Ex.3

+ I I I II II III II V

Kaorder =

V VV VV VVV V

3. Intermolecular& Intramolecular hydrogen bonding

Intramolecular (6-membered) H-bond due to ortho is effective only in Phenols and not in COOH (7-mem-bered, insignificatnt).

Ex.1

COOH

H+

C

OH O

O

H+

OH

COOH

COO

Ortho anion stabilised by intramolecular H-bonding.

Ka

o > pB.P. o < pSolubility o < pM.P. o < p

Separation Ortho is steam volatile.

Ex.2

Kaorder : (i) > (ii) > (iii) > (iv).

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


27/36Page 27

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


28/36Page 28

(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) > > > > >

(2)

(I) (II) (III) (IV) (V) (VI)

only(I)


29/36Page 29

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

(3)

(I) (II) (III) (IV) (V) (VI)

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

(4) > > (No acidic H)

(5) >

(6) > >

picric acidReaction of Acids with salts :

(1) NaX + HY NaY + HXSalt = NaOH + HX Strong Salt = NaOH W.A.

Weak acid Acid + H Y (S.A.)

Ex. 2 NaCl + H2SO

4Na

2SO

4+ 2HCl

Na2SO

4+ 2HClNo reaction

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

Ex. CH3COONa + CH3SO3HCH3COOH + CH3SO3Na (feasible)

CH3COONa + PhOHPhONa + CH

3COOH (not feasible)

Q. Which of the follwing reaction is possible ?(A) CH

3COOH + HCOONa Not possible (reverse is possible)

(B)

OH

+

ONa

NO2

Not possible


30/36Page 30

(C)

SO Na3

+

COOH

Not possible

(D*) HC C Na + H2O HC C H + NaOH

LECTURE # 12

Basic strength

(I) DEFINITION:

(a) Arrhenius base: Those compound which furinishes OH- ions in aquous solutions are known as arrheniusbase.

Ex: NaOH, KOH, Ca(OH)2etc.

(b) Bronsted base:- epair donor or H ion acceptor. :NH3, R

NH2, R

2

NH, R3

N, H2

O, R

O

H, R

O

R

(c) Lewis base:- epair donor to H ion.

+NH + H3 4NH

R +

O R + H

R +O

H R

(II) BASICITY:

It is the tendency to accept H ion, or it is the case of acceptance of H ion.

Ex: H3N: + H __________easily..

H

O

H + H __________less easily..Thus, NH

3> H

2O in basicity.

Less electronegative atom (N) donates electron pair easily.

(III) SCALE OF BASICITY/BASIC STRENGTH:-

In aq. solution:-

R NHBase

2+ H2O+ -R NH + OH

Conjugate Acid (C.A.)3

Kb=

-

]RNH[

]OH][RNH[

2

3

[where Kb= Base dissociation constant].

pKb= logK

b

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. , K

b

(ii) Size of element__________size , Kb

(a) CH3---------------

> NH2---------------

> OH---------------

> F --------------- -------- E.N.

CH3--------------- is strongest base in periodic table.

(b) CH3> P

---------------H

2> S

---------------H > Cl

----------------------- E.N.

(c) F --------------- > Cl--------------- > Br--------------- > --------------- -------- size


31/36Page 31

(d) O--------------- H > S--------------- H-------- size

(e) H2

O> H2

S

(f) :NH3> :PH

3

(V) CARBANION BASES (:C )--------------- :

(i) CH3 CH2

-> CH

2= CH

-> CH (E.N. , K

b )

(ii) CH3 CH2 CH2-

>> CH2= CH CH2-

(delocalised lone pair) due to Resonance

(iii) CH2= CH CH2

-> CH

3 C

O

CH2-

(better resonance due to ve charge on O)

(iv)

CH2-

(ResonanceStabilisation)

sp2(N) > sp3(N)

Kblone pair donating ability sp3(N) > sp2(N) > sp(N)

Ex. CH3 CH2 CH2

NH2> CH2= CH CH2

NH2> CH C CH2

NH2

(VI) NITROGENOUS BASES N:

:-

Classification:- (i) Aliphatic amines (R

NH2, R

2

NH, R3

N)

(ii) Aromatic amines (Ph

NH2) or Anilines

(iii) Amides

O

R C NH2

(iv) Amidines

:NH

R C NH2

Among these, amides are the weakest bases.(iv) > (i) > (ii) > (iii)

(i) Aliphatic amines :

Consider the following molecules , , ,

By visiting + Ieffect of methyl in above example we may expect basic nature asNH

3< MeNH

2< Me

2NH < Me

3N Which is not true

This is due the fact that basic strength of an amine in water is determined not only by ease of electrondonation (protonation) of N atom but also by the extent to which cation so formed can undergoSOLVATION


32/36Page 32

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

(1)

H

H N:

H

H

+= O, solvation = max., Steric Hinderance = min.

(2)

H

R N:

H

H

+, solvation, Steric Hinderance

(3)

H

R N:

R

H +, solvation, Steric Hinderance

(4)

R

R N:

R

H

+ max., solvation min., Steric Hinderance max.

In aq. solvation, the general order of basic strength is R2NH > R

3N > RNH

2> NH

3. (R = Ethyl)

or, in some cases, it is R NH > RNH > R N > NH2 32 3 (R = Methyl)

Explanation: It will be seen that the introduction of an alkyl group into ammonia increases the basic strengthmarkedly as expected. The introduction of a second alkyl group further increases the basic strength, but thenet effect of introducing the second alkyl group is very much less marked than with the first. The introductionof 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 byelectron-availabili ty on the nitrogen atom, but also by the extent to which the cation, formed by uptake of aproton, can undergo solvation, and so become stabilised. The more hydrogen atoms attached to nitrogen inthe cation, the greater the possibilities of powerful solvation via hydrogen bonding between these and water:

Thus on going along the series, NH3RNH2R2NH R3N, the inductive effect will tend to increase thebasicity, but progressively less stabilisation of the cation by hydrating will occur, which will tend to decreasethe basicity. The net effect of introducing successive alkyl groups thus becomes progessively smaller, andan 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 3 amines and more steric hinderance to H ion.(2) All alkyl amines (1, 2, 3) are stronger bases than ammonia (due to +effect of R group).(3) In gaseous phase, the basic strength order is 3 > 2 > 1 > NH

3(+effect of R group).

Ex. (1) CH NH2 2 > CH NH2 2

(2)Ammonia

NH: 3 > NH 2 NHHydrazine

2

> NH O2 HHydroxylamine

(3)

NH3>

R O R(+ ) (+ )

> R

O

H > H

O

H

(4) N:

(More compact)

> N:

Cyclic Amine is more basic than acyclic amine (if degree of N is same).


33/36Page 33

(5)

:NH2

:NH2

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

(6)

Kborder : I > II > III

(7)

I II III IVSulphonamide Amide Ar. Amine Aliphatic amine

Kborder : IV > III > II > I

(8)Ph|

NHCHCH 222

Ph|

NHCHCH 23

Ph|NHCHCH 23

I II III


(9) PhNH2

Ph2NH Ph

3N

I II III


(10)

I II III IV

Kborder : 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 getresonance stabilisation.

Aniline is a weaker base than NH3because it has delocalised lone pair.


34/36Page 34

Ex. Which of them is stronger base

CH3 NH

2

Since ease of donation of lone pair of N is basicity, CH3 NH

2is more basic due to + Ieffect of CH

3group.

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 inperpendicular 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) :

H N:2

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 NH2or OH do not experience (SIR) due to small size.

Ex. (1) G = (m, ); NO2

(a) (b) (c) (d)

NH2

acbd:Kb

(2) G = (); CCl3

CCl3

NH2

CCl3

NH2

CCl3

NH2

NH2

(Aniline > p > m > o).

(3) G = (> +m); Cl

ClNH2

Cl

NH2

Cl

NH2

NH2

aniline > p > m > o


35/36Page 35

Only () decides the order.

(4) G = (+, HC); R = CH3(Toluidines)

CH3

NH2

O effect,

+HC

CH3

NH2

+m CH3

NH2

+w

atingmindomoreHC

NH2

(5) G = (+m > );

Kborder : 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 chargemore effectively, so it donation tendency decrease (k

bdecreases).

(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 sp2hybridised.The lone pair of most basic N lies in sp2hybrid orbital (localised).

NH C = NH H N C = NH

(1) sp sp(2) lone pair in sp H.O. lone pair in 2p A.O.(3) lone pair localised lp delocalised(4) Basicity y > x.

2 2

2 2

2

x y x y

R R

Strongest organic Nitrogenous base:-

NH C NH2 2

:NH2

(+m) (+m)

NH C NH2 2

:NH2

NH C NH2 2

:NH2

NH C NH2 2

:NH2

NH C NH2 2

:NH2

NH C NH2 2

:NH2

NH C NH2 2

:NH2

NH C NH2 2

:NH2

NH C NH2 2

:NH2

NH C NH2 2

:NH2

NH C NH2 2

:NH2

NH C NH2 2

:NH2

NH C NH2 2

:NH2

NH C NH2 2

:NH2

NH C NH2 2

:NH2

Guanidine


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