Chemistry. GENERAL ORGANIC CHEMISTRY - 4 Session.

Chemistry

GENERAL ORGANIC CHEMISTRY - 4

Session

Session Objectives

1. Resonance or Mesomeric effect

2. Inductive effect

3. Electromeric effect

4. Hydrogen bond

5. Hyper conjugation

6. Steric effect

Session Objectives

Resonance or Mesomerism

1. All the properties of a compoundcannot be explained by single structure.

2. Canonical structures or resonancecontributing structures–differ in positionof electrons.

3. Delocalisation of electrons leads to decrease in potential energy of molecule.

canonical structures of benezene

Resonance hybridstructure of benezene

Resonance or Mesomerism

4. Resonance hybrid is more stable thancanonical structures.

5. Resonance structures are imaginary.

6. Resonance energy = Actual energy of hybrid–energy of most stable contributing structure.

7. Resonance is measure of stability.

Rules for Drawing Resonance Structure

1. The molecule should be planar.

2. It contains an alternating system of single and double bonds(a conjugated system).

3. The relative positions of nuclei should remain unchanged (e.g. tautomerism).

4. The negative charge must preferably lie on the most electronegative atom.

5. The charge needs to be preserved in all the resonating structures.

6. The electrons always move away from a negative charge.

7. Arrows should be drawn to indicate the direction of the movement of electrons.

Types of Resonance

C C OH — C — C–

O H

(+M or +R effect of –OH group)

+

+R or +M effect

–R or –M effect

C C CH O — C — C CH — O–

(–R effect of –CHO group)

+

Types of Resonance

For substituted benzene

NH2 NH2 NH2

+

–

+

–

etc.

+R effect of –NH2 group.

N

O O

N

O O

N

O O–

etc.

+

+

–

+

–

+

+

– –

–R effect of –NO2 group.

Inductive Effect

1. Permanent effect in saturated carbon chain compounds.

2. Group attached to carbon chain shouldhave tendency to release or withdraw electrons.

Types of inductive effect

+ I effect effect –electron donating groupse.g., CH3 , C2H5

– I effect effect –electron withdrawing groupse.g., - NO2 , –CN

Features of Inductive Effect

1. Chloroacetic acid is a stronger acid than acetic acid because

Cl - CH2 C

OH

OCl-CH2 - C

O

O

H++

-I

CH3-C

OH

OCH3-C

O

O

H++

+I

Ka = 1.4 × 10-3

Ka = 1.75 × 10-5

Features of Inductive Effect

2. The larger is the electron-withdrawing effect of a group, the greater is the –I (inductive) effect.

F CH2 COOH Br CH2 COOH

-32.5×10 -31.3×10Ka

3. Inductive effect is additive

Cl3CCOOH Cl2CHCOOH

Ka-12.3×10 -25.4×10

Features of Inductive Effect

4. Since this effect is transmitted through a chain it becomes less effective with distance

ClCH2CH2COOH ClCH2CH2CH2COOH

Ka-48.32×10 -53.02×10

Electromeric Effect

Temporary effect which is observed in presence of reagents involving transfer of electrons in an unsaturated system.

X Y X Y

in presence of reagent + –

in absence of reagent

Electromeric Effect

R

C

H

CH

HC — C

R

H

H

H

R — CH — CH3

Br

C — CH3

R

H

HBr12

+ –

Br – + H+

Addition of HBr to an alkene

Hyperconjugation or no bond resonance

(a) Involves and bond orbitals

(b) More the number of hyperconjugative structures, more will be the stability of ion or molecule

H – C – C

H H

H H

+ H – C = C – H

H

H H

+

H C = C

H H

H H

+H H

H H+

H – C = C

Structure of ethyl carbonium ion

Hyperconjugation or no bond resonance

(c) The number of hyperconjugative structures in an alkene is obtained by the number of C — H bonds attached to the carbon bonded directly to the double bonded carbon atoms.

CH CH

H

H

CH2 CH CH

H

H

CH2

–

+

CH CH

H

H

CH2

–+

CH CH

H

H

CH2

–

+

Significance of Hyperconjugation

H3C — CH — CH CH2H3C — CH CH — CH2

H+

(2 hyperconjugative structures)1–butene

H3C — CH CH — CH3 H3C — CH — CH

H

CH2

–

+

(6 hyperconjugative structures)2–butene

More stable

Relative strength of organic acids

R - C

OH

O

R - CO

O

H+

R - C

O

O

Resonance structures

+

Class Exercise

Class exercise 1

The hybridization of carbon atoms C — C single bond in vinylacetylene is

(a) sp3 - sp3 (b) sp - sp2

(c) sp2 - sp (d) sp3 - sp

H2C CH — C CH1 2 3 4

Vinylacetylene

Solution :

Hence answer is (c).

Class exercise 2Allyl isocyanide has

(a) 9 bonds and 4 bonds

(b) 8 bonds and 5 bonds

(c) 8 bonds, 5 bonds and 4 non-bonding electrons

(d) 9 bonds, 2 bonds and 2 non-bonding electrons

H2C CH — CH2 — N C

Allyl isocyanide

+ –

The compound has 3 bonds and one lone pair, i.e. two non-bonding electrons. It also contains 9 -bonds. Hence answer is (d).

Solution:

Class exercise 3

Among the following which has the most acidic -hydrogen?

CH3CCH2CHO

O

CH3CCH2CCH3

O O

CH3CCH2COOCH3

O

(d) CH3CHO

(a) (b)

(c)

Solution

H3C — C — CH2 C — CH3

O O

H3C — C — CH2 C — OCH3

O O

H3C CH

O

H3C — C — CH2 — CH

O O

two e-withdrawing keto groups.

keto and ester group.

one aldehyde group.

one keto and one aldehydic carbonyl group.

Solution

Since e-withdrawing nature of

gas varies as aldehyde > keto > ester

Then most acidic a-H atom is present in

C O

H3C C — CH2 — CH

O O

Hence answer is (a).

Class exercise 4

The decreasing order of acidity among phenol, p-methylphenol, m-nitrophenol and p-nitrophenol is

(a) m-nitrophenol, p-nitrophenol, phenol, p-methylphenol

(b) p-nitrophenol, m-nitrophenol, phenol, p-methylphenol

(c) p-methylphenol, phenol, m-nitrophenol, p-nitrophenol

(d) phenol, p-methyl phenol, p-nitrophenol, m-nitrophenol

Solution

OH OH

CH3

OH

NO2

OH

NO2

+I –I –I, –R

Electron withdrawing groups increase acidic strength while electron donating group decreases the same.

So the proper decreasing order of acidic strength is

OHOH

NO2

OH

NO2

OH

CH3

> > >Hence answer is (b).

Class exercise 5

In the following compounds, the order of basicity is

H

NN

N

H H

N

O

(I) (II) (III) (IV)

(a) I > IV > II > I (b) II > I > IV > III

(c) III > I > IV > II (d) IV > I > III > II

Solution

N

H

N N

O

H

N

H

sp3

I

sp2

II

sp3

IV

sp2

III

Between I and IV, IV is less basic because of the –I effect of oxygen atom.

II is more basic than III as the lone pair on N-atom in III is not available for protonation as it is involved in resonance.

Therefore, the correct order is I > IV > II > III

Hence answer is (a).

Class exercise 6

Account for the order acidity in the following compounds.

CH3CCH2COOH CH3CH2COOH

O

HC CCH2COOH H2C = CHCH2COOH

(i)

(ii)

Solution

(i) H3C C CH2 COOH > CH3 CH2 COOH

O

(a) (b)

In compound (a), electron-withdrawing keto group increases the acidic strength by decreasing the O — H bond strength, while no such effect is there in compound (b).

Carbon atoms attached to triple bond is sp hybridised and more electron-withdrawing than sp2 hybridised carbon atom.

Hence, such order in acidic strength is observed.

HC C CH2 COOH > CH2CH CH2 COOH(ii)

Class exercise 7

Which of the following two amines is more basic and why?

CCl3CH2CH2CH2NH2 or CCl3CH2CH2NH2

Solution

Electron-withdrawing groups decrease the charge density on N-atom of organic amines and hence decrease the basic strength. In Cl3C CH2 CH2 CH2 NH2, the electron withdrawing — CCl3 is far apart from — NH2 group as compared to Cl3C CH2 CH2 NH2.

Hence, the former is more basic in nature.

Class exercise 8

Do you think

can show tautomerism?

CH3CH CH2

OH

Yes. The tautomeric form is CH3COCH3.

Solution:

Class exercise 9

Which one will be more acidic ?

OH

CH3NO2

CH3

OH

CH3C

CH3

N

I II

Solution

Because of steric inhibition of resonance conjugate base of I will not be stabilised by resonance. But for II there is no such steric inhibition of resonance.

Class exercise 10

Which hydrogen is maximum acidic in the following compound?

NO2

O — H

H — C C

COOH

Solution:

Carboxylic hydrogen is maximum acidic.

Thank you