Chemistry

General organic chemistry-I

Session Objectives

Session Objectives

1. Sigma and pi bonds

2. Tetravalency of carbon

3. sp3,sp2,sp hybridization in carbon

4. Structural representations

5. Wedge and dash representation

6. Fischer projections

7. Newmann and sawhorse projections.

Introduction

Why are there so many carbon compounds?

catenation.

CH3_ CH2 – CH2

_ CH2_ CH3 , and

Ability to form chains and rings type compounds

Ability to form multiple bonds.

3 3CH – CH CH – CH

Sigma and pi bonds

+

+ or

Interunclearaxis

p-orbitals Sidewayoverlapping

Final electroncloud picture

+ + + +

+ + + +

+

+ +

s-s overlap s-p overlap p-p overlap

sp3 hybridization in carbon

2p 2p

2p

2s 2sElectron

Prom otion

Electron

Hybridization sp hybridorbitals

3

1s 1s 1s

Energy

C(groundstate)

C(excitedstate)

C(excited andhybrid state)

Tetravalency of carbon

sp3

sp3

sp3sp3

H

H

HH

1s

1s

1s1s

H

H

H

H

C

109° - 28´

sp2 hybridization in carbon

unhybrid pz orbitals overlap to form pi bond, e.g. ethylene molecule.

2p 2p

2s 2sElectron

Prom otion

sp2

Hybridization sp hybridorbitals

2

1s 1s 1s

Energy

C(groundstate)

C(excitedstate)

C(excited andhybrid state)

Unhybridizedp-orbital

sp2 hybridization in carbon

sp2

sp2

sp2

Pz

Pz

120°

120°

120°

There are five sigma bonds and one pi bond in ethylene.

sp hybridization in carbon

2p 2p

2p

2s 2sElectron

Prom otion

sp

Hybridization sp hybridorbita ls

1s 1s

Energy

C(groundstate)

C(excitedstate)

unhybridizedorbita ls

sp hybridization in carbon

-bond

H H

-bond

2pz2py

2py2pz

sp sp sp sp

1s1s

HH

H — C — — C — H

— —— —

Structural representations

Here the bond C — H or C — C is represented by .. or xxe.g.

Four electrons of carbons (x)

One electron of four hydrogen atoms each (.)

(x.) represents a C — H bond.

Lewis structure:

CH HH

H

Complete structural formula

H — C — C — C — C — C — H

H

H

H

H

H

H

H

H

H

H

Condensed structural formulae

CH3 — CH2 — CH2 — CH2 — CH3 or CH3 (CH2)3CH3

Structural representations

Bond line structure

(Pentane) O

Ethylmethyl ether

Wedge and dash representation

H

H

HH

Ce.g., Methane

1. Shows three dimensional structure

Above the planeWedged bonds

Dashed bonds Below the plane

plain lines On the plane of the paper.

Fischer projections

Cl

H OH

Br

–––––––––––––

H OH

Br

C

––Cl

Vertical bonds indicate below the plane and horizontal bonds indicate above theplane of the paper

Newmann and sawhorse projections.Rotation around a bond in a linear structure

CH3

CH3

H H

HHCH3

H

H H

H

H3C

H H

H H

CH3

CH3 H H

H H

CH3

CH3

Class Test

Class exercise 1How many bonds are present in the following structures?

(i) Benzene (ii) Vinylacetylene

(iii) Ethylacetate (iv) Benzophenone

(v) Naphthalene (vi) Phenylacetylene

and

Solution

C

H

HC — C

H

C — H

(i) (ii)

H — C — C — O — C — H

H

H

O H

H

C

O(iii) (iv)

H

H

H

H

H

C C — H(v) (vi)

Class exercise 2

How many types of hybrid orbitals are present in the following structure?

(i) Allene

(ii) Conjugated diene with lowest molecular mass

(iii) Isolated diene with lowest molecular mass

(iv) Styrene

(v) Acetophenone

(vi) Aspirin

Solution

C

H

HC C

H

H

(a)

C

H

HC — C

H

HC

H H(b)

Solution

C

H

HC — C — C

H

HC

H H H

H

(c)

CH

HC

H(d)

Solution

C C

OH

H

H

H

H

H

H

H

(e)

C O

O

O — C — C — H

H

H

H

H

H

H

H

O H

H

(f)

Class exercise 3

How many types of c-atoms are present in the following structure?

3 4(i) CH C (ii)

(iii) 2 5 4(iv) C H C

3 2 32 2(v) CH CH CH CH

Solution

(i) Two types, i.e. 1° and 4°

(ii) Four types, i.e. 1°, 2°, 3°, 4°

(iii) Two types, 1° and 4°

(iv) Three types, 1°, 2° and 4°

(v) Three types, 1°, 2° and 3°

Class exercise 4

How many monochloro derivates are possible for the following compounds?

(i) Isopentane (ii) Iso-octane

(iii) Isobutane (iv) Neopentane

(v) n-pentane (vi) n-butane

Solution

(i) CH3 — CH — CH2 — CH3

CH3

as it

contains four types of replacable H-atoms. These are two types of 1°, one 2° and 3° carbon atoms.

CH3 — CH — CH2 — C — CH3

CH3 CH3

CH3(ii)

Solution

3 2 2 2 3(v) CH — CH — CH — CH — CH 3

3 2 2 3(vi)CH — CH — CH — CH 2

CH3 — CH — CH3

CH3

(iii)

CH3 — C — CH3

CH3

CH3(iv)

Class exercise 5

Structure of alkane (having molecular mass 86) which gives only four monochlorode-rivatives on monochlorination will be

CH3 — CH2 — CH2 — CH2 — CH2 — CH3(i)

CH3 — CH — CH2 — CH2 — CH3

CH3

(ii)

3 2 33(iii) CH C — CH CH

CH3 — CH — CH — CH3

CH3 CH3

(iv)

Solution

Thus, molecular formula of alkane

is .

Compound (b) will give four monochloro derivatives as it contains of 1° C-atom, 2° C-atom and 3° C-atom.

6 14C H

Hence answer is (b).

Class exercise 6

CH2 CH — C CH

C — C single bond in the given structure is formed by using orbitals

3 3(i) sp — sp 2(ii) sp — sp

2(iii) sp — sp 2 2(iv) sp — sp

Solution

H2C CH — C CH1 2 3 4

The hybridisation at C2 — C3 bond is sp2-sp type.

Hence answer is (c).

Class exercise 7

In which of the following molecules the highlighted carbon is hybridized?

(CH3)CH2 — C

H

CH2

*

(CH3)2 — C O

H

*

CH3 — CH2 — O — CH3

*CH3 — C — OH

NH

*

(a) (b)

(c) (d)

Solution

Only in (C) the carbon highlighted is forming four single bond. Hence, it is the sp3 hybridised carbon. Rest all carbons are forming double bonds. So they must be sp2 and not sp3 hybrid carbons.

Hence answer is (c).

Class exercise 8

Which of the following molecules has only sp3 hybridised carbon?

(a) CH3COOH (b) (CH3)3COH

(c) CH3CHO (d) (H2N)2CO

Solution

CH3 — C — OH

O

(sp3) (sp2)

(a)

CH3 — C — OH

CH3

CH3

(All sp3)

(b)

Solution

CH3 — C

H

O

(sp3) (sp2)

(c)

C

H2N

H2NO

sp2(d)

Hence answer is (d).

Class exercise 9

How many types of carbon atoms are present in the following?

CH3 3 2 32 3CH CH CH C CH

3 32 3CH CH CH CH C CH

CH3 — CH —

CH3

C

4

2CH CH C C H

(i) (ii)

(iii) (iv)

(v)

(vi)

Solution

(i) Three types: 1°, 2° and 3°

(ii) Four types: 1°, 2°, 3° and 4°

(iii) Three types: 1°, 2° and 3°

(iv) Four types: 1°, 2°, 4° and vinylic

(v) Three types: 1°, 3° and 4°

(vi) Two types: vinylic and acetylinic

Class exercise 10

How can you say that the product corresponding to primary carbon atom is negligible as compared to tertiary carbon atom during the monobromination of isobutane while gives relative rate of abstraction of H-atom from 1°, 2° and 3° carbon atoms are 1 : 82 : 1600?

Solution

CH3 — CH — CH3

CH3

Br2CH3 — CH — CH2 — Br

CH3

CH3 — C — CH3

Br

CH3

1° 9 × 1 = 9

3° 1 × 1600 = 1600

Total post = 1609

Solution

9×100=0.5%

1609Percentage of 1° =

Percentage of 3° = 1600

×100=99.5%1609

Thus, B is major product and A is negligible.

Thank you