LECTURE NOTES Session - 2012-13
ORGANIC CHEMISTRY TOPIC : GOC - II CONTENTS :
1 2. 3. 4. 5. 6. 7. 8. 9.
Inductive Effect Resonance Effect (mesomeric effect) Hyperconjugation Electromeric effect Aromaticity Directive Influence Reaction Intermediates (Free radical, Carbocation, Carbanion ) Tautomerism Acid & Base
Proposed ISEET/IIT-JEE Syllabus CONCEPTS : Inductive, Resonance and Hyperconjugation Effect, Applications of Electronic Effects, Aromaticity and Directive Influence, Reaction Intermediates, Tautomerism, Acid & Base
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 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) I effect is distance dependent and negligible after three carbon atom in the chain (iii) Inductive effect is operative only through bond it does not affect 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.
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- 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.
Page # 2
Example -
(a)
CH3
CH2
(c)
CH2
CH2
(b)
CN
(d)
(e)
(f)
(g)
(h)
(i)
(j)
+
CH 3
CH
+
CH
–
NO2
O -
(k)
- -
- -
Applications of Inductive effect : (i) Acidity (iii) Dipole moment (v) Reactivity of compound in chemical reaction (a)
(ii) Basicity (iv) Stability of intermediate
In deciding acidic strength of 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 H2 C H2 C CH COOH | Cl
(II)
CH3 CH2 CH CH2 COOH | Cl
(III)
CH3 CH CH2 CH2 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 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 character).
Page # 3
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 carbocation Example :
(c)
< CH3 CH2
< (CH3 )2 C H < (CH3 )3 C
Stability of carbon free radical : Carbon free radicals are stabilised by + I effect. Example :
(B)
Stability : CH3
Stability : CH3 < CH3
CH2 < CH3
< CH3
CH3
Resonance Effect The delocalisation of adjacent (parallel) p-orbital electron is called resonance. Through resonance the e– density in the molecule becomes uniform, due to uniformity 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. These lewis structures 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 hybrid. The resonance hybrid 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.
The most stable resonating structure contribute maximum to the resonance hybrid and less stable resonating structure contribute minimum to resonance hybrid.
Page # 4
Rules for writing resonating structure : (i) (ii) (iii) (iv) (v)
In resonating structure only p-orbitals electron are shifted, bond electron are not involved in resonance, therefore the bond skeleton will remain same in two resonating structures. The atoms are not moved The no. of paired e– are same and no. of unpaired electron are same in two resonating structure. The octet rule is not violated (for second period element) High energy structure are rejected as resonating structure or their contribution to the resonance hybrid is least. Opposite charge 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 have 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:1. 2.
CH2 = CH – CH = CH2 + CH 2 = CH – CH 2
(Conjugation between C = C and C = C) (Conjugation between +ve charge and C = C)
3.
.. CH 2 = CH – NH 2
4.
CH2 = CH – CH2
(Conjugation between odd electron and C = C)
5.
– CH2 = CH – CH = CH – CH2
(Conjugation between negative charge and C = C)
(Conjugation between lone pair and C = C)
Page # 5
Q.
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)
(k)
(k)
+
+ +
..
..
(l) CH2 = C = O
(l) CH 2 = C = O
(m)
(m)
C H2 – C O –2
N– N N Page # 6
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)
Structure with complete octet of each atom is more stable. e.g.
Stability order II > I. (4)
Negative charge on more electronegative atom and positive charge on less electronegative atom are more stable. e.g.
1. Stability order II > I.
2.
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 II > III > I . (5)
Two different compounds in which one compound has more conjugation is more stable (provided nature of bonding is same). CH2 = CH – CH = CH – CH2 – CH = CH2 CH2 = CH – CH = CH – CH = CH – CH3 I II Stability order II > I.
Page # 7
(6)
In two compounds if one is aromatic and another is non aromatic and conjugation is equal in both the compounds then 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
Mesomeric Effect : When a group releases or withdraw orbital electron in any conjugated system then it is called M effect group and effect is known as mesomeric effect.
(a) +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 > – F > –Cl > –Br > –I > –NO > –NC
e.g.
1.
2.
3.
4.
(b) – M group : Electron withdrawing group Relative order of –M group : – NO2 > –SO3H > –CN > –CHO >
> –COX >
>
> –COOH > –CONH2
Page # 8
e.g.
1.
– + . .– O – N – .O: .
– + . .– O – N – .O: .
– + O – N = .O: .
– + .–. O – N – .O: . +
+
2.
– + .. O – N = O:
+
Q
+ – H 2C – CH = C = .N:.
H 2C = CH – C N:
3.
By drawing resonating structures of following molecules, Judge whether the group attatched to ring exerts + m or – m effect.
,
Ans. (a)
O || – NH – C – CH3 (+ m group) NH–C–CH3 NH–C–CH3 O
(b)
NH–C–CH3
NH–C–CH3
O
NH–C–CH3
O
O
O
–C–CH3 (–m effect group) O O
Note: 1.
C
CH3
O
C
CH3
O
C
CH3
O 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 exists.
etc.
Page # 9
2.
+M group increases electron density in benzene ring while –M group decreases electron density in the benzene ring.
(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 overlap 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– overlap while in hyperconjugation molecule orbitals overlap with - molecule orbital. * Hyperconjugation is also called no bond resonance. Condition : There must be at least one hydrogen at sp3 hybridised -carbon.
Hyperconjugation in alkene
Hyperconjugation in carbocation
Hyperconjugation in radical H • CH 2 – C H2
• H CH 2 = CH2
Hyperconjugation in toluene
Page # 10
Applications of hyperconjugation : (a)
Stability of Alkenes:- Hyperconjugation explains the stability of certain alkenes over other alkenes. Stability of alkenes no. of hyperconjugative structures
1 HHydrogenat ion
Stability of alkenes Number of alpha hydrogens and Number of resonating structures CH3
H3C
H3C
H3C
Example : (b)
CH3
H3C
C = CH2 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)
Page # 11
(e)
Stability of free radical : Greater the number of -hydrogen results greater stability of carbon free radical Example :
(a)
< CH3 –
(b) CH3 –
< CH3 –
– CH3 < CH3 –
> CH3 – CH2 –
>
>
(due to resultant of inductive effect and hyperconjugation)
(D)
Electromeric effects
(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 joined by the bond at the requirement of attacking species.
– CH2 – E (ii)
If the attacking reagent or species is removed, charges disappear 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 as follows. Case I : Electromeric effect may be supported or opposed by I effect as : (A) CH3
CH3 –
(B) CH3 –
CH3 –
–
–
In the condition of (A) E effect is supported 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)
(B) CH2
– CH =
E
– Br
–
– Br
(A) is the condition of E effect supported by +m effect and opposed by – I effect, (B) is the condition of E effect opposed by + m effect and supported 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 electron 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
Page # 12
Lecture -33
(Aromatic - 1)
1. Aromatic Character : [The Huckel 4n + 2 rule] The following three rules are useful in predicting whether a particular compound is aromatic or non–aromatic. 1.
Aromatic compounds are cyclic and planar.
2.
Each atom in an aromatic ring is sp2 hybridised.
3.
The cyclic molecular orbital (formed by overlap of p-orbitals) must contain (4n + 2) electrons, i.e., 2, 6, 10, 14 ........ electrons. Where n = an integer 0, 1, 2, 3,..............
Comparision between Aromatic, Antiaromatic and Non-Aromatic :
Page # 13
Examples of Aromatic Compounds : n
Aromatic compound / ions
-electrons = (4n + 2)
(Table ) Salts
H
Ph
+
+
+
0
Examples of salts
OH Br
2
Ph
H Cyclopropenyl cation
Hydroxydiphenyl Cyclopropenyl bromide
H
SbCl6
+
Cyclopropenyl hexachloroantimonate +
Ph
Ph ++
0
2
2BF4 Ph
Ph
+ Cyclobutenyl di-cation
1
++
Cyclobutenium salt
tetraphenylcylobutenyl fluoroborate
-
-
6
Cyclopentadienyl anion
Cyclopentadienide salts
+
K
Potassium cyclopentadienide
+
1
+
6 Tropylium cation (Cycloheptatrienyl cation)
Tropylium salt
+
Br
-
Troplyium Bromide
, 2
10
3
14
Napthalene ,
azulene
Anthracene
Phenanthrene H
4
18
H
H
H
H H
[18] Annulene
Page # 14
3. Directive Influence of Substituents In Benzene The first substituent may occupy any position in benzene ring i.e. one and only one monosubstituted benzene is obtained. The next group may go to ortho, meta or para position, it is the group already present in the benzene nucleus that determines how readily the attack occurs and at which position, of the ring it occurs. In other words, the group attached to the ring not only affects the reactivity but also determines the orientation of substitution. This is called directive influence of substituents in bezene nucleus. The substituent group is able to activate or deactivate the ring due to a number of factors like inductive effect, electromeric effect, resonance effect and hyperconjugative effect.
Classification of Substituent Groups:E.E.
F.G.
Act./Deact. of Benzene Ring Strongly activating
(1)
+m, +I
O–
(2)
+m > –I
(3)
+I, H.C.
NH2 > NHR > NR2 > OH > OR > –NHAC > –OAC R (alkyl)
X = F, Cl, Br, I –N = O, –N
(5)
–I only
(6)
–m, –I
–
–I, +m
activating
moderately activating
Deactivating
– –
–
(4)
m-directing
C > –C–C C C
C
Me > Et > – C
o/p-directing
C
Strongly deactivating NR 3 , NH3 , CX3 – –NO2, CO , –CN, –COOH, Strongly deactivating – –COOR, –SO3H, COCl
(1)
(2)
Page # 15
(3)
ORTHO-PARA RATIO : Factors:(1) EE:- In case of +m, –I groups +m is same at o/p positions but –I is more at ortho position. So para position is more e– rich. % para is more than % ortho. (2) In case of alkyl group ortho position is more e– rich than para position. (3) – I > + m, halogen : In halobenzenes para position is more e¯ rich than ortho. (4) Steric factor : If size of substituent G increases or size of E+ increases than para products is formed as a major product. DIRECTIVE INFLUENCE IN DISUBSTITUTED RING : (1) If both groups shows +M effect then electrophile attack in the influence of more +M group. (2) If one group is + M effect and another is –M effect then electrophile attack in the influence of more +M group. (3) If both groups shows –M effect then electrophile attack in the influence of more –M group.
Br / Fe
1.
2
2.
(CH3)2C = CH2 / H
3.
+
CH3
CH3
NO2
HNO H SO
4 3 2
4.
Cl
Cl
CHO
CHO Br Br / FeBr
5.
3 2
OH
OH
Page # 16
FREE RADICAL Ionic reactions are those in which covalent bonds break heterolytically and in which ions are involved as reactants, intermediates, or products. Another broad category of reaction mechanisms that involve homolysis of covalent bonds with the production of intermediates possessing unpaired electrons called radicals (or free radicals):
Free radicals (i) Homolytic cleavage (ii) Condition for its generation Sun light, peroxide, High temperature (iii) Neutral species with odd e¯ (iv) Gaseous phase or non polar solvent (v) Paramagnetic
(vi) No rearrangement (generally) (vii) Hybridisation state – sp2 (generally) Inhibitors : like oxygen (decreases rate or reaction) Stability of free radical (i) Hyperconjugation (ii) + I, + M (iii) Decreasing order of stability is ...... > 3° > 2° > 1° >
Stability of radicals –
a
–
CH 3 – CH – CH 2 – CH 2 CH3 – C – CH2 – CH3 CH3 CH3
CH3 – CH – CH – CH3 CH3
–
1.
b
c
2.
d>b>a>c>e a
3.
c>a>b
b
c
d
e
CH2 = CH
a
c>b>d>a
b
c
d Page # 17
H
–
–
–
H
–
H
CH3
c>d>a>b –
4.
c •
d
CH 3 – CH – CH 2 – CH 3
CH3 – CH2 – CH2 – CH2
a
b
CH3 (CH3)2CCHCH3
–
•
•
5.
b
•
a
(CH3)3C
c
d>c>b>a
d
Carbocation
Carbocation
A carbon intermediate which contain three bond pair & a positive charge on it is called carbocation. , CH3 –
Methyl carbocation etc. Ethyl carbocation
Characteristic (i) sp2 or sp (ii) Classical & non classical (iii) Bond angle 120° (iv) Diamagnetic (6e¯ ) Hybridisation : Carbocation may be sp2 & sp hybridised Hybridisation
Example
sp2
Page # 18
sp
H2 C =
, HC
Structure of following carbocations (i) (ii) (iii) (iv)
(Allyl)
(v)
(Benzyl)
(vi)
(Phenyl carbocation)
Draw their orbital diagram and mention the nature of hybridisation in each case.
Stability of carbocations Factors affecting stability of carbocation (i) + I effect (ii) Hyperconjugation (iii) Resonance stabilization (iv) + m effect of substituent groups with lone pair (O, N)
General stability order :>
>
>
>
>
>
>
>
>
The carbocation is not possible at following bridge head positions I and II
,
(Non classical carbocation)
Exceptionally stable carbocations are
and
aromatic
Practice problem :
+ CH2
+ CH2
–
–
–
+ CH2
–
OMe Stability order : 1 > 3 > 2
–
(A)
–
OMe
OMe 3
+ CH2
+ CH2
+ CH2
–
–
2
–
1
(B)
CH – 3
Stability order : 3 > 1 > 2
–
–
CH3
CH3
Page # 19
1
2
+
+
3
+
+
+
–
OMe
(C)
–
–
CH3
1
2
3
+ CH2
CN
4
5
–
(D)
Stability order : 3 > 5 > 1 > 2 > 4
+ CH2
–
+ – CH2
CH – 3
–
– CH3
Stability order : 6 5 1 2 4 3
–
CH3
CH3 3
+ CH2
+ CH2
–
+ CH2
–
–
–
–
2
–
(E)
1
CH2D
NHCOCH3
OMe
4
5
6
CH3 – CH – CH3 CH3 – CH – NH2 2 3
CH3 1
.. CH3 – C O .. 4
Stability order : 3 > 4 > 2 > 1
+
+ CH3 –C CH 3
+ – CH – 1
1
+ 2
Stability order : 3 > 2 > 1
3
+ + CH3 – CH2 – CH2 – CH2 CH 3 – CH – CH 2 – CH 3 a b
–
+ (CH3)2C – CHCH3 CH3 c
Stability order : 3 > 1 > 2
3
+
(H)
(J)
–C+
2
+
(I)
Stability order : 1 > 3 > 2
.. + O .. 3
Stability order : d > c > b > a
+ (CH3)3C d
+ Draw all isomeric form of C5H11 carbocation, which one is the most stable. + C–C–C–C–C (a)
+ C–C–C–C–C (b)
+ C–C–C–C–C (c)
C + C–C–C–C (d)
–
(G)
+ .. O .. 2
.. O .. 1
– –
(F)
Page # 20
– –
C C – C – C+ C (g)
–
C C–C–C–C + (f)
–
C + C–C–C–C (e) + (f) C5H11 is most stable. (K)
+ .. CH3 – CH2 – NH – CH2 – CH – CH3 (b)
.. + CH3 – CH2 – NH – CH – CH2 – CH3 (a) CH3 – CH2 – NH . . – CH2 – CH2 – CH2 + (c) +
Stability order : a > b > c
+
+
+
(L)
Stability order : d > c > b > a b
+ CH2
+ CH2
–
–
+ CH2
d
–
+ CH2
c
–
a
(M)
Stability order : a > d > c > b a
b
c
d
+ +
(N)
Stability order : b > a > c
+ (a)
(O)
(b)
>
(c)
>
CH3 – CH – CH3 > CH3 – CH – CH2Cl > CH3 – CH – CH2NO 2
(P)
Stability order : a > c >b
Rearrangement : Whenever an Intermediate carbocation is formed in reaction it rearranges to a more stable one. Not all carbocations rearrange but only those which can produce more stable species can only rearrange. (i) The driving force responsible for carbocation rearrangement is formation of more stable carbocation. (ii) Shifting of H, alkyl, aryl, bond (1, 2) (iii) Ring expansion (more strained ring to less strain ring) (iv) Ring contraction Shifting of H, alkyl, aryl, bond (1, 2)
Page # 21
CH2 = CH – CH2 –
CH2 = CH –
– CH3
Carbocation rearrangement involving ring expansion.
One very stable carbocation reported is cyclopropylmethyl carbocation. This unique stabilisation is seen in this case of three member ring only. (more stable than Benzyl) cyclopropyl methyl carbocation
Ring contraction : -
Page # 22
+ +
Complete octet :
– –
CH3 + .. (i) CH3 – C – C = O ..
(III)
O
+
CH3 CH3 (IV) CH 3 – CH – CH – N – CH 3 +
–
CH3
.. (lI) CH3 – CH2 – O . . – CH – CH – CH3 + CH3
+ .. (VI) CH3 – CH – CH – Cl ..
..
+ .. (V) CH 3 – CH 2 – CH = O ..
CH3
Rearrange (if applicable) these carbocation into more stable form :-
(b)
+ CH – 2
–
CH3
(c) CH3 – C – CH – CH3 + CH3
(i)
+
(f)
+ – CH2
(g)
+ (h) CH2 – CH – CH – CH3
–
(e)
(d)
CH3 CH3 +
––
CH3
+
–
+ (a) CH3 – CH2 – CH2
CH3
+ – CH –
Bond dissociation
Carbanion
1.
Dissociation of covalent bond, (Homolysis and Heterolysis)
2.
Type of Reaction intermediates formed ( C+,
,
,:
)
Page # 23
Carbanion A carbon intermediate which contain three bond pair and a negative charge on it, is called carbanion. ,
Methyl carbanion
CH3 –
Ethyl carbanion
(CH3)2
Isopropyl carbanion
Hybridisation : Hybridisation of carbanion may be sp3, sp2 & sp. Hybridisation Example sp3
, CH3 –
sp2
H2 C =
sp Structure of following carbanions
HC
(a)
(b)
(e)
(f)
, CH2 = CH –
(c)
(d) CH2 = CH –
[Draw the orbital diagram and mention the type of hybridisation involved in each case] Stability of carbanion : Carbanions are stabilised by electron withdrawing effect as (i) – I effect (ii) – M effect (iii) Delocalisation of charge If -position of a carbanion has a functional group which contains multiple bond (C = C, C = O, C N, NO2 etc) or carries an electronegative atom, such carbanions are stablised by resonance hence more stable than simple aryl carbanion.
Example of stability order : (A)
CH C .. 1
(Stability order) : 1 > 2 > 3
.. CH2 (B)
.. CH C
1
..
2
..
.. (Stability order) : 2 > 1 > 3 > 4
3
4
CH3
–
..
(C)
.. CH3
1
2
3
–
..
(Stability order) : 2 > 1 > 3 > 4
4
Page # 24
..
..
..
(D)
(Stability order) : 3 > 4 > 2 > 1 Br
1
NO2
2
Cl
3
4
(E)
(Stability order) : 1 > 2 > 3
(F)
.. CCl3 1
.. CF3 2
(G)
.. Cl – CH2 – CH2 1
.. F – CH2 – CH2 – CH2 2
.. CH3 – CH CH3 3
.. CH3 – CH2 – CH2 4
–
(Stability order) : 1 > 2
CH C .. 3
.. CH2 – NO 2 1
(Stability order) :
–
–
–
CHO
NO2
2 .. CH –
–
– (J)
.. CH3 – CH2 – CH2 – CH2 a
.. (CH 3)2C – CH 2CH 3 c
(k)
O .. CH2 – C – CH3 a
.. CH –
+ NMe3
+ SMe2
4
4>2>6>3>5>1
5
–
–
.. CH –
3 –
1
–
(I)
.. CH
–
.. CH –
.. CH – 2
(Stability order) : 1 > 2 > 3
–
(H)
(Stability order) : 1 > 2 > 4 > 3
CN
6
.. CH3 – CH – CH2CH3 b .. (CH 3)3C d O .. CH2 – C – H b
O .. .. CH2 – C – OCH .. 3 c
a>b>d>c
e>b>a>c>d
O .. CH2 – C – NH2 d
Page # 25
(L)
>
>
(Stability order)
Reaction in which carbanion intermediates are formed Li / ether (a) R – X
Mg / ether (b) R – X
(c) R –
(Lot of carbanion like character in R)
Na / NH3 ( ) C – H
– R OH
(d) R – CH2 –
(e)
(f) CH3 – CH = O (g) Decarboxylation of alkanoic acid by using sodalime dereasing order of decarboxylation CH3 – CH2 – CH2 – COOH > CH3 CH COOH | CH3
SO3Na (C)
(dereasing order of decarboxylation)
COOH +
Not possible
(D*) HC C – Na + H2O HC C – H + NaOH (E) CH3COOH + NaCl Not possible (F) Ph–COOH + KBr Not possible (G) Picric acid + K2SO4 Not possible
(H*)
+
+
TAUTOMERISM
Page # 26
(1) Definition : Tautomers are two structural isomers (having different functional groups) which exist in dynamic equilibrium with each other. The isomers change into each other by shifting of H-atom from one position to another position and simultaneously the position of bond also changes. (2) General formula of compounds showing tautomerism | C X Y H
X=Y
can be : C = O
C = NH
(i)
N=O
NO O
r
(ii)
(iii)
(iv)
–C–C=O H Keto form
–
–
–
– –
(3) Keto-enol tautomerism General Expression:-
– C = C – OH enol form
Condition:- Presence of at least one – H wrt C = O group. Draw enol forms of following carbonyl compounds
O
CH3 – C = CH – CH3 and CH2 = C – CH2 – CH3 OH OH
–
(iii) CH3 – C – CH2 – CH3
CH3 CH2 = C – OH
–
CH3 (ii) CH3 – C = O
CH2 = CH – OH
–
(i) CH3 – CH = O
–
Ex.-1
Page # 27
(4) Comparison of stability of Keto-enol tautomers Keto form
Enol form
| CC | || H O
| C C | OH
The
group is more resonance stabilized
An alkene is less resonance stabilized
(Positive charge at ‘C’ and negative charge at ‘O’ atoms)
(Positive charge at ‘O’ and negative charge at ‘C’ atoms)
In case of monocarbonyl compounds, the compound mainly exists in Keto form (99%). The enol form is insignificant (less than 1%) As stability of alkene increases (due to increase in hyper-conjugation or resonance) % enol content (alkene content) increases. The -dicarbonyl (or 1, 3-dicarbonyl) compounds have significant enol contents due to resonance stabilization of enol form
O keto form (
O 25%)
H3C – C = CH – C – CH3 OH O enol form ( 75%)
–
H3C – C – CH2 – C – CH3
Compare % enol contents Ex. 1
Keto form
Enol form
I : CH3CH = O
CH2 = CH – OH
II : CH3 – CH2 – CH = O
Sol.
CH3 – CH = CH – OH
III : III > II > I
Page # 28
Ex. 2
CH3 C CH3 < CH3 CH2 C CH3 || || O O
Ex. 3
I :
CH3 C CH3 || O
II :
CH3 C C CH3 || || O O
III :
CH3 C CH2 C CH3 || || O O
Sol.
..... (% enol)
CH2 C CH3 | OH CH2 C C CH3 | || OH O
III > > II > I 76% 1.2% 0.01%
Ex. 4
CH3 | < CH3 CH C CH3 || O
CH3 C CH C CH3 | || OH O
(enol content) >
Ex. 5
Properties of Keto and enol forms 1.
Boiling point :
>
Reason : The keto form is more polar and the enol form has intramolecular H-bonding 2.
Solubility : In polar solvents (like H2O). The more polar keto forms is more soluble in water, than the less polar enol form. The enol form has intramolecular H-bonding
3.
Polarity
4.
Optical properties The carbonyl compounds which have alpha asymmetric carbon (C*) atom and have – H hydrogen atom, racemise due to tautomerisation on keeping in aqueous solution.
Keto form > enol form
Page # 29
5.
Chemical properties The -dicarbonyl compounds in aqueous solution show chemical properties of both the forms. The lab tests of following functional groups are positive. (i)
(ii) C = C
(iii) – OH
Keto-enol tautomesion in phenols and its derivatives
1.
In phenol (enol form is much more stable due to aromaticity)
+
2.
In catechol
3.
In Resorcinol :
4.
In Quinol (Hydroquinol)
5.
Trihydroxy benzene
(a)
(b)
(1, 2, 4)
Page # 30
(c)
(1, 3, 5) (% enol < % keto)
Q.1
Which of the following compounds will racemise on keeping in aqueous solution
(A*) CH3 CH CHO | D Q.2
(D) Ph CH CH2 C CH3 | || D O
O || (C*) Ph C CD 3
O || (D*) Me 3 C C CH3
(B*) Me2CD – CHO
(C*)
(D)
Which of the following will show tautomerism
(A*)
Q.4
Me | Ph C C CH3 (C) | || Et O
Which of the following will show tautomerism
(A) Me3C – CHO
Q.3
CH3 | C CH3 Ph C (B*) | || H O
(B)
Write the tautomer of
OH HO
OH
HO
OH
(A)
(B)
OH Q.5
‘X’ is the smallest dicarbonyl compound which has significant enol content. Predict S.F. of ‘X’.
Sol.
X = H C CH2 C H || || O O
Q.6
A (C6H10O2) is an optically active dicarbonyl compound which significantly exists in optically inactive enol form. Identify A.
Sol.
A = CH3 C CH C H || | || O C 2H 5 O
Q.7
An aromatic compound X of molecular formula C9H8O2 exists in keto form and predominantly in enolic form Y. Write all the possible structure of X and Y.
or
CH3 CH2 C CH C H || | || O CH3 O
Sol.
(X)
(Y)
Page # 31
Ph | X = OHC CH CHO
Q.8
Ph | Y = HO HC C CH O
Which of the following pairs are tautomers.
(A)
&
(B*)
and
Page # 32
Relative acidic strength of Organic Acids : (a)
Definition : (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.
(b)
Scale for Measurement of Acid Strength :
RCOO H
Ka
(c)
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 (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 > HNO3 > HCl]. (ii) Organic acids : These are weakly ionised in water, so these are weaker acids than mineral acids. R – SO3H > R – COOH > Ph – OH > ROH.
(d)
Prediction of Acid strength : (i)
(Anion/conjugate base of HX)
(ii)
(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 CH4 < NH3 < H2O < HF.
(Ka) Page # 33
Conjugate base/Anion :
(stability)
E.N. dominates over size decrease. (2) Along the group from up to down : As size increases, Acid Strength HF < HCl < HBr < HI (3) Organic acids : RSO3H > RCOOH > PhOH > R–OH in acidic character Explanation : O O O || | || H RSO3H R – S — O R – S O R – S O || || | O O O
(i)
(ii)
(both of equal stability)
(iii)
Negative charge partially localised to oxygen. (iv)
*
ROH
In sulponate anion three equivalent resonating structure is formed.
* In carboxylate anion 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. 4.
Experimental (Ka) Order :
Page # 34
(1) Aliphatic acid
Ex.1
Ans.
Ka order : (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)
(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.
Page # 35
(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 H ion is taken out from a negatively charged species, so it is difficult. (ii)
Ka1 I > II > III > IV (iii)
pKa1 IV > III > II > I
Maleic acid is stronger acid than fumaric acid.
Ka1 > Ka1'
(3)
;
;
Ka2' > Ka2
Acidic strength in Aromatic acid : Acidic strength in aromatic acid depends upon following factors Page # 36
(i) (ii) (iii)
Steric effect (steric inhibition of resonance) or Ortho effect Electronic effect of substituents 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 acidity increases w.r.t. benzoic acid inspite any electronic effect of substituent
O
O
O–H
C
G
H
–
C
O
G
G may be – NO2 , – R , – X , – COOR , – COOH except – OH , – NH2, – C N (COOH group goes out of plane) Substituted Benzoic Acid : (1) (G = –m, –I) COOH NO2
–H + O
C
O
+
O N
–I, –m
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
Page # 37
(5) G = (+I, H.C.) ........ R (Alkyl group)
Ka order = ortho > benzoic acid > meta > para
<
Ex.
<
Ex.
<
<
>
Note : In case of different ortho 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.
+I II
I
+m>–I III
–I>+m IV
–I,–m V
Ka order = V > IV > I > II > III
Ex.
I
+I II
–I III
–I
–I IV
V
Ka order = V > IV > III > I > II At meta positions only inductive effect is operative and mesomeric / hyperconguative effect is not operative in substituted benzoic acid.
Ex.
I
+I II
–I III
–I
–I IV
V
Ka order = V > IV > II >III > I Intramolecular (6-membered) H-bond due to ortho is effective only in Phenols and not in –COOH (7-membered, insignificatnt).
Page # 38
Ex. COOH –H+ OH
COO
Ortho anion stabilised by intramolecular H-bonding.
Ex.
Ka order : (i) > (ii) > (iii) > (iv).
(4)
Acid strength of Phenols (effect of substituents) : Note:
Ortho effect is operative only in o-substituted Benzoic acids and not in o-substituted phenols.
>
>
(1) EWG = (–m , –I) (–NO2, CHO, –COR, C N, etc.)
Ka order = Para > Ortho > Meta > Phenol (2) EWG = (–I)
with no mesomeric effect
Page # 39
>
>
>
Ka order = Ortho > Meta > Para > Phenol (3) EWG = (–I > + m) X = F, Cl, Br, I
Ka order = Ortho > Meta > Para > Phenol (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
Ex.
>
(1)
>
>
>
>
Ex.
(I)
(II)
(III)
(IV)
(V)
(VI)
only (–I)
Page # 40
Ka order = I > II > III > VI > IV > V
Ex.
(I)
(II)
(III)
(IV)
(V)
(VI)
Ka order = I > II > III > IV > V > VI
>
Ex.
Ex.
>
(No acidic H)
>
Ex.
(picric acid) >
>
Reaction of Acids with salts : (1)
NaX
+
Salt of Weak acid
HY Strong Acid
NaY + HX
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.
2 NaCl + H2SO4 Na2SO4 + 2HCl
Ex.
Na2SO4 + 2HCl No reaction
Ex.
CH3COONa + CH3SO3H CH3COOH + CH3SO3Na (feasible)
Ex.
CH3COONa + PhOH PhONa + CH3COOH (not feasible)
Q.
Which of the follwing reaction is possible ? (A) CH3COOH + HCOONa Not possible (reverse is possible)
Page # 41
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
(d) - O H > - S H-------- size (e) H2 O > H2 S (f) :NH3 > :PH3
Page # 42
(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
(better resonance due to –ve charge on ‘O’)
CH2
-
–
(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 N H2, R2 N H, R3 N)
(ii) Aromatic amines (Ph – N H2) 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 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 + = O, solvation = max., Steric Hinderance = min. (1) H – N: H
Page # 43
– –
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 >
– CH2 – NH2
(1)
N H3 > R –– O –– R > R – O – H > H – O – H (+ ) (+ )
(4)
N:
>
(3)
: NH3 > NH2 – NH2 > NH2 – O H Ammonia Hydrazine Hydroxylamine
(2)
N:
(More compact) Cyclic Amine is more basic than acyclic amine (if degree of ‘N’ is same).
:NH2
–
(5)
–
Ex.
:NH2
(ii) > (iii) > (i) > (iv)
Page # 44
(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
(ii) Aromatic amines (Ph – N H2) 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. 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.
Page # 45
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) :
G
–
–
H2N:
–
–
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 : dbc a
(2) G = (–); CCl3
–
–
–
NH2
–
CCl3
–
NH2
NH2
NH2
–
CCl3
–
(Aniline > p > m > o).
CCl3 (3) G = (– > +m); Cl
NH2
–
–
–
NH2
NH2
–
Cl
–
NH2
–
Cl
–
aniline > p > m > o
Cl
Only (–) decides the order. (4) G = (+, HC); R = –CH3 (Toluidines)
–
–
NH2
NH2
–
O effect, + HC
+m
CH3
–
–
CH3
–
NH2
NH2
–
Ex.
CH3 w + HC more do min ating Page # 46
(5) G = (+m > –);
Kb order : P > Aniline > O > M
(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 : Hybridisation of ‘N’ : sp3 > sp2 > sp
(a)
(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).
–
2
x R y H2N – C = NH
–
R x y NH2 – C = NH
2
(1) sp sp 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) NH2 – C – NH2
Guanidine
:NH2
+ :NH2 – C = NH2 NH2
+ NH2 = C – NH2 NH2
–
–
NH2 :NH Heterocyclic Compounds (Nitrogenous base) : (1) Pyridine (C5H5N:)
S.F.:- H – H–
–H N:
–
–
H
H
6 e – Aromatic Stronger base
A.O. Diagram:-
N
H NH2 – C – NH2 :NH2 – C – NH2
2
sp hybrid orbital (lone pair) localised Page # 47
(2) Pyrrole (C4H5N:):-
6e – 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
–
1.
N H
–
:NH2
H
+ – 3 N ––– sp (4e ) – H H Non-aromatic
–
+ N H Aromatic
–
–
NH3
Aromatic
O
2.
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)
;
:NH (2) 2 sp
N H (d) sp2
–
N H
R–C–R ; NH (3) 2 sp
Kb order : (1) > (2) > (4) > (3) > (5)
–
:NH2(x) 4.
N (y)
N(z) H
Basicity order:- y > x > z
–
Ex:-
Page # 48
