Presiding Officer Training module 2024 lok sabha elections
Dr. R. S. Shinde B.Sc. I Sem-I Organic chemistry (1) (1).ppt
1. Ravindra S. Shinde
M.Sc., SET, GATE, B.Ed., Ph.D.
Assistant Professor ,
Department of Chemistry and Industrial Chemistry,
Dayanand Science College, Latur- 413512, Maharashtra, INDIA.
Chapter 2
Reaction Intermediate
2. Reaction Mechanism
• Detailed description of sequence
of steps involved in group from reactants
to products.
• Reactant intermediate product
3. Bond Cleavage
B
:
A
B
:
A A : B A– : B+
+vely charged ion – carbocation
-vely charged ion – carbaanion
Heterolytic Cleavage
Homolytic Cleavage
B
A
B
:
A
Free radicals.
4. Carbonium ion
• Planar – sp2 hybridised
bond angle 120o
• Has six electrons
• Stabilized by resonance or
inductive effort
or hyperconjugation
C
Empty unhybridised
p-orbital
sp2
Hybridisation of
carbon
Planar Strucutre of carbnion
+
5. Examples of
Carbonium ion
Benzyl cation
CH2 CH2 CH2
etc
+
+ +
CH2 CH CH2 CH2 CH CH2
+ +
Stabilised through
resonance
Allyl cation
H2C CH+
Vinyl cation
no resonance hence unstable.
6. Stability of
Cabocation
(i) By inductive effect
The resonance effect is always more
predominant than the inductive effect
in stabilizing an ion.
CH3
C
CH3
CH3
CH3
C
CH3
H
> > H C
CH3
H
> >
3° 2°
+ +
1°
+
7. Stability of Cabocation
(ii) By hyperconjugation
H3
C — C
CH3
CH2 — H
H3C — C
CH3
CH2H
H3C — C
CH2H
C
CH3
CH3
CH3
CH2
H
+
+
+
etc. +
Thus, tertiary carbocation is more stable than
secondary and so on.
8. Carbanion
• Pyramidal - sp3 hybridised
bond angle 109.28
• Has eight electrons
• Trigonal Pyramidal
• Stabilized by resonance or by inductive effect.
. . sp3 hybrid orbital
containing lone pair
Tetrahedral structure of carboanion
11. Stability of Carbanion
(iii) Electron-donating groups destabilize
a carbanion while electron-withdrawing
groups stabilize it.
NO2 3
OCH
>
2
CH
2
CH
12. Free Radical
• Trigonal Planer
• Has seven electrons
• Stabilized by resonance or by inductive effect.
• Order of stability of free radical 3o >2o> 1o
C
(
Unhybridised orbital
containing odd electron
120oC
sp2
hybridised carbon
Planar Sturcutre
+
13. Classification of
Reagents
Nucleophilic Reagents (Nucleophiles)
• Attacks the positive end of a polar
bond or nucleus-loving is known as
nucleophile.
• Generally, negatively charged or
electron rich species are nucleophilic.
3 3 2 3
e.g. OH , OCH , CN , I , CH COO , NH , CH
2 3 3 2
H O, NH , NH — NH
N
..
..
NH3,
CH3
— O — CH3
,
..
..
..
C2
H5
— OH,
..
..
H2O,
• All nucleophiles are in general Lewis bases.
14. Classification of
Reagents
Electrophilic Reagents (Electrophiles)
• Attacks a region of high electron
density or electron-loving is known as
electrophile.
• All positively charged or electron
deficient species are electrophilic.
3 2
H , CH , NO , Cl , Br , Ag
15. Classification of
Reagents
• Neutral reagents which contain an
electron-deficient atom are also
electrophiles.
AlCl3, SO3, BF3, SOCl2, POCl3, FeCl3, ZnCl2
• All electrophiles are in general Lewis acids.
16. Carbenes
• Divalent carbon compound.
• Carbon atom is linked to two adjacent
groups by covalent bonding.
• A carbene is neutral and possesses
two free electrons, i.e. a total of six
electrons.
• Electron deficient.
17. Carbenes
Carbene is of two types
(i) Singlet carbene:
(ii) Triplet carbene:
Triplet carbene is more stable than single carbene.
CH2 hybridisation sp2
it is v-shaped
CH2 hybridisation sp
it is linear shaped
23. Class Exercise - 3
Which of the following is the most
effective group in stabilizing a free
radical inductively?
(a) F (b) I
(c) Br (d) Cl
24. Solution
Since free radical is electron deficient,
any substituent with more electron
releasing and less electron
withdrawing ability will stabilize the
radical inductively.
The decreasing order of
electronegativity of halogens is: F >
Cl > Br > I
Hence answer is (b).
25. Class Exercise - 4
Which of the following is not a
nucleophile?
(a) CN– (b) BF3
(c) RNH2 (d) OH–
26. Solution
Among the following, BF3 is only
electron deficient. Hence, it will not
act as a nucleophile.
Hence answer is (b).
27. Class Exercise - 5
Which of the following is the correct
order regarding –I effect of the
substituents?
(a) –NR2 > –OR > –F
(b) –NR2 > –OR < –F
(c) –NR2 < –OR < –F
(d) –OR > –NR2 > –F
28. Solution
–I effect increases with electronegativity
of atom. The decreasing order of
electronegativity is
F > O > N
The correct order for –I effect is
–NR2 < –OR < –F
Hence answer is (c).
29. Class Exercise - 6
The least stable carbonium ion is
(a) (b)
(c) (d)
3 2
H CCH
6 5 2 2
C H — CH — CH
6 5 2
C H — CH
6 5 6 5
C H — CH — C H
30. Solution
Among the following, (a) is stabilised
through +I effect and (b) is
destabilized through –I effect of
phenyl ring. Other two are stabilised
through resonance.
Hence answer is (c).
31. Class Exercise - 7
Arrange the following ions in the
decreasing order of stability.
2
H
C
CH3
CH3
CH3
+
+
+
(a) (b) (c) (d)
32. Solution
CH2
+
. It is a primary cation.
Hence, minimum stability.
CH3
CH3
+
and
+
(c) (b)
are secondary cations.
Hence, stabilised through +I effect of –CH3 group which
decreases with distance. (c) is more stable as compared to (b).
(d) is most stable as it is tertiary cation and stabilised through +I
effect of –CH3 group and hyperconjugation.
The order is (d) > (c) > (b) > (a)
33. Class Exercise - 8
Arrange the following radicals in
order of their decreasing stability
3 2 3 3 6 5 2 2 2
CH CH , (CH ) C, C H CH , CH CH CH
34. Solution
Radicals are stabilised through
electron releasing resonance and
inductive effect.
CH2
CH2
etc.
More resonating structure
H2C CH — CH2 H2C — CH CH2
35. Solution
One resonating structure, although
both are primary radicals.
Among and , later is a tertiary radical.
Hence, more stable.
The decreasing order of stability is
H3C — CH2 (CH3)3C
C6H5CH2 > H2C CH — CH2 > (CH3)3C > H3CCH2
