1) The document discusses various organic chemistry reaction mechanisms including SN1, SN2, addition and substitution reactions. 2) It provides step-by-step explanations of the mechanisms of reactions such as hydrolysis of ethers with HI, addition of Grignard reagents to carbonyls, and acid-catalyzed dehydration of alcohols. 3) Limiting factors for different reaction mechanisms are discussed, such as steric hindrance preventing SN2 in tertiary halides and stability of carbocation intermediates determining SN1 rates.

1. ORGANIC CHEMISTRY
REACTION MECHANISMS
(1) Substitution Nucleophilic Bimolecular (SN 2 )
Proposed by Edward Davis Hughes and Sir Christopher Ingold
(1937).
The reaction between a primary halide and a nucleophile follows
second order Kinetics i.e., rate depends on the concentration of alkyl
halide as well as nucleophile.
e.g. rate µ [alkyl halide] [nucleophile]
HO +
Solid wedge
represents the bond coming out of the paper dashed
line represent bond going down the paper and a straight line represent
bond in the plane of the paper.
The incoming nucleophile interact with the alkyl halide causing the
C—X bond to break while forming a new C—OH bond. These two
processes takes place in a single step simultaneously and no intermediate
is formed. Inversion of configuration takes place during the process.
Carbon atom in the transition state is simultaneously bonded to five
atom, therefore unstable and cannot be isolated.
Tertiary halides donot undergo SN2 mechanism due to steric hinderance.

2. Since the nucleophile attacks from opposite side of halide atom, the
three alkyl groups do not permit the nucleophile to attack on carbon
atom, the order of reactivity followed is
Primary halide > Secondary halide > Tertiary halide
H
H
H
Nu:
X
H
Nu:
H
C
Nu:
X
H
H
Ethyl 1°
(1)
H
C
H
X
H
H
Methyl
(30)
H
H
H
H
Isopropyl 2°
(0.02)
C
H
Nu:
H
C
H
X
H
H C
H
C
H
H H
tert-butyl 3°
(0)
(2) Substitution Nucleophilic Unimolecular (SN 1 )
Tertiary halide proceeds via SN1 mechanism. Rate of reaction depends
only on the concentration of alkyl halide
rate µ [alkyl halide]
The reaction takes place in two steps
Step-I. The polarised C—X bond undergo slow cleavage to produce
a carbocation and a halide ion.
Br
Step-I is slowest and reversible.
Step-II. The carbocation thus formed is attacked by nucleophile to
complete the substitution reaction.

3. Effect of Solvent : SN1 reactions are favoured in protic solvent (a)
as step-I involves the C—Br breaking for which the energy is obtained
through solvation of halide ion with proton of protic solvent.
(b) Polar solvent promote ionization of halide ion.
Since the reaction proceeds through formation of carbocation, so
greater the stability of carbocation, faster will be the rate of reaction
therefore
⊕
⊕
b g b g
⊕
⊕
⊕
⊕
⊕
Ph3 C > Ph 2 CH > PhCH2 > allyl > CH3 3 C > CH3 2 CH > CH3 CH2 > CH 3
For this reason allylic and benzylic halides show high reactivity
toward SN1 mechanism. The carbocation thus formed get stabilised through
resonance.
⊕
⊕
H2 C = CH—CH 2 ¬ →

⊕
CH2
⊕
H2 C—CH = CH 2
CH2
CH 2
⊕
CH2
⊕
For alkyl group, the reactivity of halides R—X follow the same
order in both mechanisms.
R—I > R—Br > R—Cl > R—F
Vinyl halides neither undergo SN1 nor SN2 mechanism. SN 2 mechanism
is hindered by the fact that carbon atom attains a negative charge and
SN1 mechanism is hindered by resonance and no ionization possible.

4. SN2
SN1
1. SN2 reaction follow 2nd order
SN 1 reaction follow first order
kinetics.
kinetics.
2. Inversion of configuration takes Retention of configuration and also
place.
racemisation takes place.
3. No effect of solvent.
More polar sovlent more is the rate
or reaction.
(3) Acid Catalysed Hydration of Alkene
Alkene reacts with H O in presence of mineral acid as catalyst to
2
form alkenes. In unsymmetrical alkene the reaction proceed according
to Markovnikov rule.
> C = C <+ H2 O
H+
C —C
H OH
OH
H
C
+
H
O
3 CH = C —CH—CH
2
H
CH
3
3
Step-1. Protonation of alkene to form carbocation by electrophilic
attack of hydronium ion (H 3O+ )
+
HO
>C = C< + H—O—H
H
⊕
—C—C— +
H
: :
H
2

5. Step-2. Nucelophilic attack of water on carbocation.
Step-3. Deprotonation to form alcohol.
(4) Addition of Grignard Reagent on Carbonyl Compounds
Step-I. Nucleophilic addition of Grignard reagent to carbonyl group
to form an adduct.
δ+
δ−
δ+
δ ++ δ
> C = O+ R− Mg X
−
→

L
M —OM g—X O
P
—C
M
P
M
P
NR
Q
− 2+
−
Bond in Grignard Reagent is highly polar carbon being non-metal
and magnesium metal, So Mg reactes to oxygen to form adduct.
Step-II. Hydrolysis of adduct yield alcohol.
OH
+ –
–
++
–
—C—O Mg X
—C—OH + Mg
R
R
X
(5) Acid Catalysed Dehydration of Alcohol
Alcohols undergo dehydration by heating with concentrated H 2SO4.
H 2 SO4
CH 3 CH 2OH  → H2C = CH2 + H 2O
443 K

6. Secondary and tertiary alcohols are dehydrated under milder conditions.
OH
85% H 3PO4
CH3 —CH—CH3  → CH 3CH = CH2 + H2 O

443 K
CH
CH2
3
20% H 3P O 4
H3 C—C —OH   → CH 3 —C —CH 3 + H 2O

350 K
CH3
The reaction proceed in three steps—
Step-1. Formation of protonated alcohol.
H2 SO 4  → H⊕ + HSO4

Step-2. Formation of carbocation : — It is the slowest step and rate
determining step.
Step-3. Elimination of Proton.

7. The acid used in step-1 is released in step-3. To drive the equilibrium
to the right ethene is removed fast.
(6) Reaction of Ether with HI
Step-1. The reaction start with protonation of ether.
Step-2. Iodide is a good nucleophile. It attack the least substituted
carbon atom of the oxonium ion formed in step-1 and displaces an
alcohol molecule by SN 2 mechanism. Thus in the cleavage of mixed
ethers with two different alkyl group, the lower alkyl group forms alkyl
iodide and larger forms alcohol.
⊕
M
M
M
N
⊕
P
P
P
Q
I − + CH 3 —O—CH2 CH3 → I … CH 3 … O…CH 2CH3 → CH3 I + CH3 CH 2 OH
H
H
When HI is in excess and reaction is carried at high temperature
alcohol reacts with another molecule of HI to form another alkyl
iodide.

8. Step-3.
When one of the alkyl group is tertiary, the halide formed is a
tertiary halide.
CH
CH3
3

H3 C —C—O CH 3 + HI  → CH 3 OH + CH 3 —C —I
CH3
CH3
Due to formation of tertiary carbocation (stable).
CH
CH
3
CH3
3
⊕
Slow


H 3 C—C —O CH 3 + HI  → H3C —C—O—CH 3  → CH3OH + CH 3 —C⊕
CH3
CH3 H
CH
CH
3
CH3 —C ⊕ + I
CH3
CH3
→

3
CH 3 —C—I SN1
CH3

9. In anisole :
+
H — O— C H 3
O—CH 3
+ HI
+ I
The CH3 —O bond is weaker then C H5 —O bond because the carbon
6
atom of benzene ring is sp2 hybridised and there is a partial double
bond character. There attack of I
break the CH3—O bond from CH3I.
OH
↓
H —O—CH 3 + I
→

+ CH3 I
(7) Addition of HCN to >C = O
The reaction proceeds by attack of nucleophile.
OH
> C = O + HCN  →

C
CN
Step-1. Generation of nucleophile.
HO
HCN + OH
: CN +
2
Step-2. Nucleophilic attack of CN – on carbonyl group.

10. 
CH 3COOH + C2 H5OH H 2SO 4 → CH3COOC 2 H5 + H2 O
(8) Esterification
Step-1. Protonation of carbonyl oxygen activate the carbonyl group
towards nucleophilic addition of alcohol. Proton transfer in the tetrahedral
intermediate convert the OH– group into
FOH I
H K
−
+
2
Step-2. Transfer of Proton.
⊕
Step-3. OH 2 is a better leaving group and eliminated as H2 O.
Protonated ester so formed finally loses a H+ (Proton) to give ester.

11. (9) Mechanism
Addition of NH3 , NH2OH, NH2 NH2 , C6H5 NHNH2 or NH 2 CONHNH2
to >C = O.
Step-1. Addition of ammonia derivative to >C = O
Step-2. Elimination of
H
2
O to form product.
Where X = – H, —R, —OH, —CONHNH2 or —NHC6H5
The pH of the reaction is controlled at 3.5, in strongly acidic medium
proton is captured by amino grup to form salt
..
⊕

R NH 2 + H⊕  → R NH 3
In basic medium, OH – cannot attack to electro-negative oxygen
atom.
δ+
δ−
−
> C = O + OH  → No reaction

Hence no product is formed in strongly acidic or basic medium.
(10) Some Important Reactions
(A) Friedel Craft Reaction
Addition of alkyl (R) or aryl group (COR) to benzene nucleus in
presence of Anhydrous AlCl 3 (Lewis acid).

12. (a) Alkylation —
CH3
C
H
+l C
Anhy. AlCl
3
+ HCl
(b) Acetylation or Acylation —
COCH3
Anhy. AlCl
C
HC
+ OCl 3
+ HCl
COCH
3
H
CO3
Anhy. AlCl
+
+ HCl
O
H
CO3
(c) Benzoylation —
CH
CO 65
COCl
C
+ H6
+ HCl
5
Mechanism
Step-I. Generation of electrophile, AlCl 3 is Lewis acid and generate
electrophile.
⊕
CH 3Cl + AlCl 3  → AlCl 4 + C H3


13. Step-II. Formation of intermediate.
Step-III.
H
CH3
CH3
+
+ AlCl 4
HCl
l
++AlC 3
Characteristics :
(1) More stable carbocation will form the product. e.g.

14. (2) Phenol and aniline form addition product with AlCl 3 donot
form carbocation.
(3) Chlorobenzene and vinyl cloride do not form carbocation.
(4) Addition product will be determined by ortho, meta and para
directing group e.g.
CH3
CH3
C
H
+l C
AlCl
3
CH3
CH3 +
+ HCl
CH3

15. (B) Aldol Condensation
Aldehyde and ketones having one or more α –H-atom when warmed
with dilute base undergo self addition reaction known as aldol condensation.
OH
NaOH
CH 3 CH = O : + H CH 2 CHO  → CH 3 —CH—CH 2 CHO
Whatever be the size of aldehyde, attack comes from α –H-atom and
product is β .
Ketones also undergo self addition to form ketol.
(C) Cross Aldol Condensation Reaction
When two different aldehydes are condensed together 4 products are
formed.

16. OH
NaOH
CH 3 CH2CH = O + H CH2 —CHO  → CH 3CH2 —CH—CH 2 —CHO
Mechanism :
Step-I.Formation of carbanion.
−
−
CH3 CH = O + OH

CH 2 —CH = O ← → CH2 = CH —O + H2O
α–H-atom is removed by base as HO.
2
Step-II.
O–
CH3 CH = O + CH 2CHO  → CH 3 —CH —CH2 —CHO

Step-III.
–
O
OH
H+ OH −
−
CH3 —CH—CH 2 CHO   → CH 3 CH—CH 2 CHO + OH

Cross Aldol Involving Aldehyde and Ketone
O
OH
O
−
OH

CH 3CH = O + H CH 2 —C—CH3   → CH3 —CH—CH 2 —C —CH3
(D) Cannizzaro Reaction
Disproportionation of an aldehyde lacking α –H-atom like HCHO,
CH CHO, R C—CHO to salt of an acid and a primary alcohol is known as
6
5
3
Cannizzaro Reaction.
2 HCHO + NaOH  → CH 3OH + HCOONa


17. CHO
2
CHOH
2
+ NaOH
COONa
+
CH3
CH3
CH3
–

2 CH3 —C —CHO + NaOH  → CH 3 —C—CH2OH + CH3 —C—COONa+
CH3
CH3
CH3
Mechanism :
Step-I.Reversible addition of OH– to >C = O.
OH
C6H5 —CH = O + OH−  → C 6H 5 —C —O

H
Step-II.Transfer of hydride ion (H– ) to another aldehyde.
OH
O
OH
–
O
Slow
C 6H5 —C—O + C —C6H5   → C 6H5 —C = O + H —C —C6 H5

H
H
Step-III
H
O
C 6H5 —C —O + C 6H 5 —CH 2OH

18. Step-III. The acid and alkoxide ion so obtained involve in proton
exchange to yield more stable product of salt and alcohol. Similarly
O
O–
O–
OH −
−
H—C —H + OH  → H—C—H   → H —C—H + H2 O


–
OH
O–
O
H
O
H—C—H + C = O  → H—C

O–
H
+ CH3OH
O–
Markownikov’s Rule (MKR)
“In addition of HX molecule to an unsymmetrical alkene, H-atom goes
to the C-atom which has already larger no of H-atoms attached to it.”
1. The molecule to be added is known as addendum e.g.
+
−
+
−
+ −
+ −
+ −
+ −
+ −
−
+
H Cl , HB r , H I, HF , HCN , H NO3 , HH SO4 , H OC l
2. The positive part of adendum goes to the carbon atom which has
already larger number of H-atoms attached to it.
3. The negative part of addendum goes to the C-atom which has lesser
no of H-atoms attached to it.
Br

CH 3 —CH = CH 2 + H Br  → CH3 —CH CH3

19. 4. When the reaction is carried in presence of some peroxide, the
product is reverse of MKR, called Kharasch or peroxide effect.
Benzoyl

CH 3CH = CH2 + HBr   → CH 3CH2 CH 2 Br
Peroxide
Mechanism—MKR proceed via carbocation formation—
1. Larger alkyl group polarise theπ -bond
Br
 →

δ+
δ−
⊕
Br


CH 3 CH = CH + H⊕  → CH 3 CH —CH 3  → CH 3 —CHCH 3
⊕
2 . H attack first and generate secondary carbocation.
3. Br– attack to carbocation to form product.
Another example is
Br
δδ +
δδ −
+
Br−
⊕
CH3 CH2 CH = CHCH 3 + H → CH3 CH3 CHCH2 CH3  → CH3 —CH2CHCH2

CH3
Peroxide effect proceed via free radical mechanism
O
O
O
•
hv
•
C 6H5 —C—O—O—C—C 6H5  → 2C6H5 —C —O  → C 6 H 5 + CO2


One free radical always generate another free radical
•
•

C 6 H 5 + HBr  → C 6H 5 + B r
•
•
HBr
•


CH 3 CH = CH 2 + Br  → CH 3 C HCH2Br  → CH 3 CH2 CH 2Br + B r
l l l