Basic Organic Chemistry-merged.pdf

Basic
Organic
Chemistry
KEEP
CALM
÷
☒¥

Organic compounds:- The compounds c were obtained from living
organisms.
Vital Force theory (VFT) :- No organic compound can be prepared in a
lab as it requires a mysterious force called
vital force c exists only in living organisms.
Urea was first artificially prepared organic compound in lab by
Wohler.
[Berzelius]
Characteristics of c-atom:-
Tetravalency:- Valency = 4
At. no :- 6
Electronic configuration:- 2,4
i
#

Tendency to form multiple bonds
Double bond Triple bond
Eg:- CH2 — CH2
HCHO — O CH3 — C — N
Catenation:- Self linking property of an atom.
Representation of organic compounds:-
1) Complete structural formula:- All covalent bonds are shown
in this method.
H — C —H H — C — O — H
H
H
H—C—C—O—H
H O
H
H
H
on
-

2.) Condensed structure :- Atleast one covalent bond is hidden.
Eg:- CH3—CH3 CH3—CH—CH2 CH3—COOH
3.) Bond-line formula :- Straight lines are used to represent a
Intersection pts Molecule/compound
—> C atoms & attached H-atoms are not written.
3.3
E-
qÉ
W
r
E
÷
i.
.
•
•

Determine total no. of sigma and pie bonds:-
Hybridisation and shape:-
No of Bond Hyb Pair orbitals Shape Bond angle
4
3
2
SP3
SP2
SP
0
1
2
Tetrahedral
Trigonal planar
Linear
109
120
180
0
0
0
T
o
⇐
É
9
?
EEE
i
s
¥É
6
-0
④

Eg:- Alkene will be a planar molecule.
In cumulated polyene/Allene
No. of double bond and ring in Allene
If ODD If EVEN
A B A OR B
Parallel
A B
÷
-
-
-
Ei
%
!
E-
T
.
U
•
a-
⇐
¥
1-
-
-
-
-

Applications of hybridisation:-
Electronegativity
Bond length
C > C > C
SP SP2
SP3
[50%] [33.3%]
[25%]
B.L. SP3 > SP2 > SP
Degree of unsaturation / Index of hydrogen Deficiency:-
—> No. of moles of H2 molecules c are required to remove bonds &
rings from it. D.U = sum of rings + pie bonds

E.g :- CH3—CH = CH2 nH2 CH3—CH2—CH3
nH2
E
E
E
÷
É
or
n
m
o
v
E
11
N
÷÷÷¥
E
En
É
É
o
n
n
É
,
É
z
N
-0
E
É
É
E
u

Classification of organic compounds:-
Based on structure
Open chain/Acyclic/Aliphatic Closed / Cyclic chain
Saturated Unsaturated
C—C
(Single bond)
C C/ C C
Homocyclic/
Carbocyclic
Heterocyclic
Eg Benzene
Pyranose/
Furanose ring
Alicyclic Aromatic
Have properties similar
to aliphatic compounds
Alicyclic Aromatic
(Satisfy hackal’s
rule)
:

Based on functional groups
—COOH
—SO3H
C2O3
H/R—C—O—R
R—C—R
R—SH
R—O—R
Carboxylic acid
Sulphonic acid
Acid anhydride
Ester
Ketone
Thioalcohol
Ether

R/H—C—X
R/H—C—N
R—CN
R—NC
R/H—C—H
R—OH
R—N
R/H
R/H
R—S—R
Acid halide
Acid Amide/ Amide
Cyanide
Isocyanide
Aldehyde
Alcohol
Amine
Thioether

Homologous series:-
A series/group of compounds having same functional group but
different molecular formula.
Have same general formula
Have same chemical properties
Successive members of a homologous series differ by CH2 unit.
No homologous series can have more than one different general
formula.
Types of C-atoms
CH3
1
CH2 C—H C
2 3 4
:
:
rarer

Types of Alkyl halides & Alcohol:-
C—OH C—Cl
1 —> 1
2 —> 2
3 —> 3
1 —> 1
2 —> 2
3 —> 3
Types of Amines:-
NH2 NH N N
1 2 3 4
%
0
at
o
o
ii.
0m
0
:
:
:
.

IUPAC
NOMENCLATURE
TAKE IT EASY
¥
☒:¥

Word root:- Represent total no. of C-atoms in longest / principal /
main C - chain.
n n
Word root Word root
1
2
3
4
5
6
7
8
9
10
Meth
Eth
Prop
But
Pent
Hex
Hept
Oct
Non
Dec

1 Suffix / Primary Suffix:-
Represents Saturation / Unsaturation present in C-chain .
— ane , —> ene , —> yne
1 Prefix / Primary prefix:-
Represents nature of parent C-chain.
Open chain Cyclo

2 Prefix / Secondary Prefix :-
Represents side chains / substituents attached with main C-chain.
X
F
Cl
Br
I
R
CH3
NO2
NO
OR
OCH3
0C2H5
C — C
O
Halo
Fluro
Chloro
Bromo
Iodo
Alkyl
Methyl
Nitro
Nitroso
Alkoxy
Methoxy
Ethoxy
Epoxy
1

2 Suffix / Secondary Suffix:-
Represents principal functional group present.
Functional group 2 Suffix 2 Prefix
COOH Oic acid
Carboxylic acid
Carboxy
SO3H Sulphonic acid Sulpho
C — O — C
O O
Oic Anhydride
:

C — X Carbonyl halide
C — NC
O
Amide
Carboxamide
Carbamoyl
CN
Nitrile
Carbonitrile
Cyano
NC
CHO
Isonitrile Isocyano
al
Carbaldehyde
Oxo
Formyl
Oyo halide Halo carbonyl
O
:

C
O
one Keto
Oxo
OH
SH
N
C O
O
ol
thiol
amine
oate
carboxylate
Hydroxy
mercepto
amino
alkonyl oxy
alkoxy carbonal
2 P + 1 P + W.R + 1 S + 2 S
0
:
0
:

How to select Principal C - chain?
Multiple bonds
No. of C-atoms
No. of substituents
Numbering:- P.C.C > Multiple bond > lowest set of locents >
Alphabetical order
rrr

(1)
(2)
4-Ethyl,3 methylheptane
3-Ethyl,2 methylhexane

(3)
(4)
4-Ethyl 5 methyloctane
3,4 dimethylheptane

(5)
(6)
2 Chloro 4 methylpentane
6-Bromo 3 methylhex-2ene
8

(7)
(8)
2-Bromo 5 Chloro 3-Ethylhex-3-ene
5-Bromo 4 methylhex-2-yne
or
at

(9)
(10)
3[1-Bromo ethyl]2,4 dimethyl
hexa 1,4 diene
3-Chloro 4 ethyl
5-methyl
hex-4en-1-yne
3
oh

(11)
(12)
3-ethyl pent-3en
-2one
4 Ethyl 3,5 dimethyloctane

(13)
(14)
2,2 Dibromo 4-Chloro pentane
3 Ethyl 4 nitro hex-2-ene
8
&
%

(15)
(16)
4-Ethyl 5 methyl oct-ane
3[2-Bromoethyl] penta 1-3
diene
oh

(17)
(18)
2-Ethyl 3 methyl hex-1-en-4yne
4-Ethyl hept-2ene-6yne
Z

(19)
Complex substituent :- Substituted substient is called complex
substituent.
Always written in Brackets
3[1-Bromomethyl]
pent-4ene-2-ol
r
E

(1)
(2)
3ethyl 4methoxy
Pent-4-en-2 amine
3-Chloro 2 ethenyl
Pent-4-en-1amine
:
El

(3)
(4)
Firstly always write
substituent attaches
to ester.
Methyl 4,5 Bis[1-bromoethyl]
hex-5-enoate
4[1,1 dimethylethyl]
2-ethyl hex-5-enoic acid
gg
°
£

(5)
(6)
3-ethyl 4 methyl pent-4-en
2-isonitrile
3-Bromo-N methyl N-methyl
But-3-en-amide
£
:>

(7)
(8)
4 Amino-3-ethyl-5methyl
hept-5-en-2-ol
3-Bromo 2 methyl but-3ene-
oic acid
É
E
O
E

(9)
(10)
2-ethyldiene 3 nitrohex-5-en-1ol
2-Ethyl 3[1-methylpropyl]
pent-4-en-1-nitrile
§
3
0
I

(11)
(12)
3-ethyl-N-methyl pent-4
en-2-amine
3-ethyl 4hydroxy 5 methyl
hex-5-ene-2one
€
:

(13)
(14)
4-keto 2,5 dimethyl
heptanal
3-ethoxy-2-methyl 5-oxo
pentanoic acid
If carbon aldehyde part of PCC, then
oxo.
÷
I

(15)
(16)
Ethyl 2-ethoxy -4hydroxy
5-methyl-6-oxo-hexonate
5 Cyano-4ethoxy
2methylpent-3-en-1 oic acid
If terminal grp consider as
substituent,then except C- of
aldehyde no other C is count.
3
•
:
I
0
0

(17)
(18)
4-Ethanoyloxy 3-nitrobut-2en-oic acid
3ethyl hex-2-4diol
C of ester not attached with
PCC Alkanoyloxy
I
°
O
E
O
O
E

(19)
(20)
3-Amino 2 ethyl pentane,
1,5 dioic acid
Butane 1,2,4 tricarboxylic
acid
E
E
☐
①
I
É
.
E-

(21)
(22)
3-ethyl-5-formyl 2nitrosohex-5-en-1amide
2-ethyl 3keto 4methyl
hex-4-enoyl chloride
0
E
o
8
0
§
°

(23)
(24)
4-methoxy carbonyl-2-methyl
butanoic acid
2-ethyl 3methyl pentane 1,4diol
0
E
E
O
E

(25)
(26)
4[1-hydroxymethyl]
3-methyloctane 2,6diol
2,3diethylpent 1,5dinitrile
E
E
E
E
0

(27)
(28)
Butane 1,2,4
tricarbaldehyde
Heptane 1,4,6 tricarbonitrile
§
°
;
-
}

(29)
3ethyl but-3-en-2-sulphonic
acid
É

IUPAC Naming of Cyclic compounds:-
Priority order :- Principal functional grp > Multiple bond > No. of carbon ring
(1)
(2)
2Cyclopropyl butane
1[1-methylethyl]
cyclobutane

(3)
(4)
2-cyclopentyl but-2-ene
3-ethyl cyclopent-1-ene

(5)
(6)
6-ethyl 1-methyl
cyclohexane-ene
3[3-hydroxycyclobutyl]
butane-2-one
O
E

(7)
(8)
5-hydroxy cyclopent-2-en
carboxylic acid
2-Bromo 5-formyl
cyclopent-3-ene
carboxamide
0
EN
É
O
E
at

(9)
(10)
4-formyl cyclopent-2-ene
1-carbonitrile
Cyclohexylcyclobutane-1-
carboxylate
E
o
o
:

(11)
(12)
3[1-cyclopent-2-enyl]butane-2-amine
2[3-ketocyclohexyl] ethan-1-al
EN
O
O

(13)
(14)
Ethyl-6-ketocyclohex-3-
en-carboxylate
5-carboamyl cyclo hex-2-
ene carbonyl bromide.
i.
£
.
÷

(15)
(16)
Cyclopentyl 3ethyl but-3-1-oate
2[2-hydroxy cyclohexyl]
cyclobutan-one
Koo
E

(17)
(18)
4-amino cyclohexyl 1,3 diol
6-hydroxy cyclohex-4-en 1,3 dicarboxylic acid
E
E
§
8
E
Et
E

IUPAC Naming Anhydride
C —O —C
O O
Propanoic anhydride
If different , then write in alphabetical order.
O
OH OH
O
Butanoic ethanoic anhydride
(1)
Ethanoic Anhydride
i.

(2)
(3)
Pentanoic-propanoic anhydride
Butane-dioc-anhydride
÷
°
o
•
:*
.

IUPAC Naming of aromatic compounds:-
S.P + P.P + W.R + P.S + S.S
Benzene X
(1) (2)
Bromobenzebe
2-Bromo-4-Chloro-1
nitrobenzene
i.
⑨

(3)
(4)
4-Bromo-2-Chloro
1-nitrobenzene
1-Bromo 2Chlorobenzene
or
•
i.
B

(5)
(6)
1-Bromo-3ethoxy-5-Fluro
benzene
1-Bromo-3-ethyl benzene
0
§
.
0
*
B

# If functional group attached to benzene ring.
S.P + P.P + W.R + P.S + S.S
Single word :- Depends upon functional group
(1) (2)
Phenol Benzoic acid
¥
E

(3)
(4)
Aniline
Benzamide
§
É
:

(5)
(6)
Benzaldehyde
Benzoyl Chloride
E
o
?

(7)
(8)
Benzonitrile
Alkyl benzoate
:
E
o
v

(11)
(12)
3-Amino 4-bromophenol
4-Amino benzaldehyde
Er
E
B
É
É

(13)
(14)
2-formyl 5-hydroxy benzonitrile
2-Cyanobenzoic acid
E
É
É
88

(15)
(16)
2-ethoxybenzoic acid
4-cyano 3 hydroxy benzamide
Bo
E
É
É
o
v
8

(17)
(18)
4Amino 2-nitrobenzoyl
chloride
Ethyl 3-Bromocarbonyl benzoate
8
EN
O
OU
OU
Er
0
↳

(19)
(20)
4-nitrosotoluene
3-Nitro anisole
E
Es
E
o

If more than 1 functional group:-
S.P + P.P + W.R + P.S + S.S
Benzene
(1) (2)
Benzene 1,2 diol 5-nitrobenzene 1,3 dicarboxylic
acid
§
E
E
E
E

(3)
(4)
4-Aminobenzene 1,2 dicarbaldehyde
2-nitrosobenzene 1,4 dicarboxamide
É
É
É
É
E
EE
E

(5)
When open chain is principal carbon chain??
If functional group, multiple bond , more than 3C-atoms,
substituents .
Use Phenyl for benzene ring.
(1)
2-hydroxy benzene 1,4 dicarbonitrile
1-Phenylpropan-1-ol
E
:
E
E

(2)
(3)
3-Phenylbut-1-en
2-Phenylpropanal

(4)
(5)
2-Phenyl ethanol-1-amine
2-nitrophenol
É
E
E

(6)
(7)
O-Methyl benzaldehyde
M-Ethylbenzoic acid
§
É
É

Bicyclo compounds :- Have atleast 2C-atoms.
(1)
Bicyclo (X,Y,Z) alkane X>Y>Z
X+Y+Z+2
Bicyclo (4,1,0) heptane

(2)
(3)
Bicyclo (3,3,0) octane
Bicyclo (2,2,1) heptane
•

(4)
(5)
Bicyclo (1,1,0) butane
Bicyclo (2,2,2) octane
iii.
i.

Spirometer compounds:- Have 1 common atom.
(1) (2)
(X,Y) alkane
Increasing order (X+Y+1)
Spiro (3,4) octane Spiro (2,2) pentane
:

(3)
(4)
Spiro (2,5) octane
Spiro (3,3) heptane

Common ‘s name
Radical:-
CH4 CH3
CH3—CH3 CH2—CH3
If more than one different radicals are present, then different prefix radicals are
present, then different prefix will be n , iso , neo , sec , tert etc.
n (Normal):- Radical is a straight chain with free valency .
CH3 — CH2 — CH2 — Cl n-propyl chloride
CH3 — CH2 —CH2 —NH2 n-butyl amine
I
I

Iso:- Second last C-have methyl group attached to it.
CH3 — CH — CH2 —Br
CH3 Isobutylbromide
Naming always start from free valency.
Neo:- Second last C- is attached with 2 methyl groups.
CH3 — C —CH2 —NH2
CH3
CH3
neo-pentyl amine
r

Sec / S :- Free valency is present at 2nd carbon.
CH3 — CH —CH2 —CH3
Sec-butyl
tert / ter / t :- Free valency is present at 3** Carbon
CH3 — C
CH3
tert-butyl

Vinyl :- It is removed from Sp2 carbon.
CH2 — CH — H CH2 — CH —Cl
Vinyl chloride
Alyl :- H is removed from Sp3 carbon.
CH2 — CH —CH2 —H
CH2 — CH —CH2 — OH Allyl alcohol
Benzyl :- when H -removed from Sp3 hybridised C of toluene.
CH2 —H CH2 — OH
Benzyl alcohol
Ph—CH2—H
C6H5—CH2—H
—CH2—H
I
9
D-

Acrylic :- C— C — G
CH2 — C — COOH —> Acrylic acid
Total C- included
CH2 — CH — CN —> Acryonitrile
Croton :- Same but C - atoms becomes 4.
C — C — C — C CH3 — CH — CH — COOH
Crotonic acid
Acet :- CH3 — C — CH3—C—H CH3—C—NH2
O O
Acetaldehyde Acetamide
→
Er
→
¥

CH3 — C —OH
O
CH3 — C — NH2
O
Acetic acid Acetamide

General organic chemistry
Inductive effect/ Transmission effect
Induction of polarity in the non-polar bond due to presence of a polar bond is
known as inductive effect.
Eg:- CH3—CH2—Cl
Is a permanent effect
Is weak and poor effect
Distance dependent effect i.e
Effects only in Sigma bond crowding
Invalid in - bond crowding
(Acc. To above example)
is
is
Es
Es
E
E
r
r
r
a
4

TYPES
+I effect/+group -I effect/+Igroup
e- donating group
Has tendency to increase e-
density in the C-chain attached
with it.
e- withdrawing grouop
Decreases e- density in the C-chain
attached with it.
Eg:- CH3—CH2—O Eg:- CH3—CH2—OR2
MT
:
:
rn

CH2 > NH > O > COO > —C—CH3 —CH3 > CH > CH2–CH3 >
CD3 > CH3 > D > H
CH3
CH3
CH3
CH2 , नह , ओ , carboxylate , 3 > 2> 1 >CD3 > CH3
-I effect :- Shown by all neutral except saturated alkyles.
HYDROGEN T
aken as reference
Inductive effect considered to be zero
CH3

—OR2 > —NF3 > NR3 > NH3 > NO2 > SO3H > CN > —C—R
> —F > —Cl > Br > I > OR > OH > C—CH > NH2 > Ph > CH = CH2 > H
ओ र 2 >, ना फरहान >, ना र
ि त
ि क >, ना हीरा लाल >, ना ओमपुरी >,
SO3H >, Cyanide >, Acid >, Father >, Collector >, Beta >, Inspector >, Aur >,
OH >, Alkyne >, Amine >, Benz , ene
O
-

Resonance
Conjugation:- If in a molecule /ion , a orbital can overlap with more than one
p-orbital, conjugation is said to be there.
Eg:-
Conditions for a compound to have conjugation :->
P-orbitals must be present at adjacent atoms.
P-orbitals must be parallel.
€7
:

Units having conjugation
1.) Conjugation
2.) (+ve charge ) or - vacant orbital conjugation
3.) (-ve charge)
Sp2 hybridised with 2 electrons
E
f
E-
¥
±

4.) Lone pair conjugation
5.) Umpaired e- conjugation
6.) Lone pair - vacant orbital conjugation
Sp2 hybridised + p-orbital having 1
electron
CH3—O—CH2—CH3
I
:
:
É
É

In which of the following compounds conjugation is absent ??
:
:
ñ
I
S
N
=
>
i
✗

CH3—C—O
O
a-
>
s
✗
ñ
F
-

•
•
•
•
>
E
S
E
E
>
Er
>

CH3—C
O
O
In this , N-don’t have vacant p-orbital
✗
✗
it
E
E
'
g-
✗

Delocalisation
T
ake place in a conjugated system.
Increases stability of conjugated system.
Resonance
If a molecule / ion can’t be represented by single
structure ,then 2 or more hypothetical structures are
used to represent .
These structures are called resonating structure/ Canonical structures.
"
÷
.
r
r
r

Rules to write valid resonating structures:-
(1) Relative position of atoms must be same .
i.e Sigma bond must not be broken .
(2) No. of unpaired electrons must remain same.
(3) 2nd period elements must not have more than 8electrons in their outermost
shell.
Resonance hybrid:-
Actual structure of a conjugated system and its
contribution of all valid resonating structures.
Contribution Stability of resonating structures
r

Relative stability of different resonating structures:-
Resonating structure having more no. of atoms with complete octet will be
more stable.
Eg:- (1) CH3—CH—CH=CH—O—CH3
(2) CH3—CH=CH—CH=O—CH3
Stability 2>1
More no. of covalent bonds = more no. of atoms whose octet is complete
Bonds will always be same.
Non-polar resonating structures will be more stable than resonating structures.
CH3—C—OH CH3—C—OH
O
O
1 > 2
:
T
9

Resonating structures having -ve charge present at more electronegative atom
will be more stable.
CH2=C—H CH2—C—H
O O
1 > 2
Resonating structure having +ve charge present at most electropositive atom
will be more stable.
Resonating structure having opposite charge close to each other will be
more stable.
Resonating structure having like charges close to each other will be less
stable.
r
r
r
r

Compare stability:-
(A)
(1) CH3—C=O—H
O
(2) CH3—C—OH
O
(3) CH3—C—OH
O
(4) CH3—C—OH
O

Solution:- 4 > 1 > 3 > 2
(B)
(1)
(2)
(3)

Solution :- 3 > 1 > 2
(C)
Solution :- 2 > 3 > 1
(1) (2)
(3)

Resonance energy:- Defined as the difference between energy of most
stable resonating structure and that of resonance
hybrid.
Draw resonating structures:-
CH2 CH CH CH CH2
CH2 CH CH CH CH2
CH2 CH CH CH CH2
CH2 CH CH CH CH2
Es
i
:
÷
Is
÷
÷
→
-
n

CH2 CH CH CH CH2
CH2 CH CH CH CH2
CH2 CH CH CH CH2 CH2 CH CH CH CH2
CH2 CH CH CH O CH3
CH2 CH CH CH O CH3
CH2 CH CH CH O CH3
CH2 CH CH CH O CH3
÷
÷
Csi
:
Es
'
÷
i
÷
is
do
÷
÷
is
:
:
]
>
it
>
it

Mesomeric effect:-
Permanent polarity induced/ produced in a conjugated system due to
delocalization of p-electrons.
Will be a permanent effect.
Strong effect.
Distance independent effect.
(Only effects ortho and para positions, meta-unaffected)
Two types
+M -M
Increases e- density in attached
C-chain
Decreases e- density in attached
C-chain
If lone pair is attached on 1st atom —> +M effect
If lone pair is absent and atom is attached to most electronegative atom by
multiple bond —> -M effect
n
^^
9
99

Check +M or -M effect :-
(1) NH2
(2) —C—CH3
O
(3) —O—Eth
(4) CN
+M
-M
+M
-M

(5) CH=NH
(6) NO2
(7) N
Me
Ethyl
(8) NO2
(9) —CH2
-M
-M
+M
-M
-M

(10) COOH
(11) —O
(12) —C—O—Me
0
(13) —NH
(14) —NH—C—Me
O
-M
+M
-M
+M
+M

(15) —C—NH—Eth
O
(16) F
(17) NO
(18) —Ph
(19) —CH=CH2
-M
+M
+M/-M
+M/-M
+M/-M

(20) CH3
(21) CH2—OH
APPLICATIONS OF RESONANCE
Complete conjugation (if all atoms are involved in conjugation)
No Yes
Non-aromatic compound Count e- involved
[4n+2] e- [4n] e-
n=0,1,2,3…… n=0,1,2,3,4
Satisfy Haeckal ‘s rule
Aromatic
Anti-aromatic
No conjugation
⇐
⇐
÷

Stability :- Aromatic > Non - Aromatic > Anti- Aromatic
= —> 1
—> 1
- —> 1
—> 0
+/ v.o
Identify the following ; whether Aromatic, Non- Aromatic and
Anti- Aromatic:-
(1) (2) (3)
N A AA
÷
@

(4) (5) (6)
(7)
(8) (9)
A
(Because vacant p-orbital)
N AA
A A N
as

(10) (11) (12)
(13) (14) (15)
AA
A AA
A A
A
-
•
•
-m

(16) (17) (18)
(19) (20) (21)
N A A
A N
N Because no
p-orbital
present.
•
•
••
:
:

(22) (23) (24)
(25) (26)
(27)
A N A
N
A A
(Due to angle strain, it
becomes non-polar & no
overlapping &no
conjugation.)
(Because stability can
bear angle strain.)
_m

Quasi-Aromatic
Structure in which we have to draw resonating structure to check if it is
Aromatic , Non-Aromatic or Anti-Aromatic.
(1)
(2)
←
€

(6)
(7)
Bond length:-
A > B > A — B > A — B
A — B > A B > A B
€
I
÷

Compare C — C bond length:-
(A) (1)
(2)
(3)

Solution : 1 > 2 > 3
(B) (1) (2)
(3)

Solution :- 3 > 2 > 1
(C) (1)
(2)
(3) (4)

Solution : 2 > 3 > 4 > 1
(D) (1) (2)
(3) (4)
:

Solution : 2 > 4 > 3 > 1
# While comparing bond length within a molecule , check by
resonating structures.
(1)
CH3 C
O
O
a
b
→
←

Solution:-
CH3 C
O
O
R.S 1
a
b
R.S 2
2
1
1
2
1.5
1.5
Bond length

solution:-
CH3 C
O
OH
R.S 1 R.S 2
X
Y
2
1
1
2
Stability R.S 1 > 2
So, X > 1.5
Y < 1.5

Hyperconjugation Effect:-
Nathen and Baker effect / No - Bond Resonance
Here, bond overlaps with Sigma bond.
Delocalization of and electrons.
Permanent effect.
Stronger than inductive but weaker than mesomeric.
- hydrogen must be present for hyperconjugation and p-orbital must not be
completely filled.
C C
— carbon —> Sp3 hybridised
geo
-
⇐
ji
→
s
b
⑤
⇐
^^
^^
8

Units which can show hyperconjugation :-
(1)
(2)
(3)
C C
C C
C C C
C- anion can’t participate in hyperconjugation.
In hyperconjugation , when C-H bond is involved , it will be an e- donating
effect. Hence called +H effect.
•
⑦
⑦
⑤
a

Calculate total no. of Hydrogen which are involved in
hyperconjugation:-
(1)
(2)
i
•

Solution:- 1.) 4 H
2.) 3 H
(3)
(4)
88

Solution :- 3.) 8 H
4.) 6 H
(5)
(6)
88
•

Solution :- 5.) 6 H 6.) 3 H
If hyperconjugation , effect of a group attached with Benzene ring increases
e- density at ortho & para positions & meta position remain unaffected.
Applications of Hyperconjugation effect
(1) Heat of hydrogenation / Heat of hydrogenation of Alkene
C C
H2/Pt
C + Heat
(Always exothermic)
HOH No. of double bonds/ pie bonds
HOH 1
Stability
C
If bond is involved in resonance , then stability will be more.
r
r
88

Compare HOH
(A) (1)
(2)
(3)
E
É
E-
*

Solution :- 2 > 1 > 3
(B) (1)
(2)
(3)
É
I
¥

Solution :- 3 > 2 > 1
(C) (1)
(2)
(3) (4)
ET
É
I
E-

Solution :- 1 > 4 > 3 > 2
(D)(1)
(2)
(3)
(4)
(5)
É
FEE
j
É
OÉ

Solution :- 2 > 4 > 5 > 3 > 1
Electronic effect
Complete transfer of electrons towards one of the two atoms on demand
of external attacking reagent.
Temporary effect
Strong effect
Case 1
C C
H
C C
H
Case 2
C O
CN
C O
CN
E
ra
n

Two types
+E effect -E effect
If attacking reagent and electrons
are on same atom.
Eg:- Case 1
On different atom.
Eg:- Case 2
In which of the following Electronic effect is possible:-
(1) (2)
r
^
f.
it
.
.

Solution :- On all except 3rd.
For electronic effect to take place , presence of - bonds is
must.
Bond breaking / Bond fission
Homolytic
Heterolytic
Homolytic
A — B A + B
A B A — B
Free radicals are formed.
:
:
K
§
~

Factors favouring :- High temperature.
Non- polar solvent.
Presence of free - radical attacking reagent.
Presence of peroxides.
Presence of light.
Heteroytic:-
A — B —> A + B
Leads to formation of cation and anion.
Factors favouring :- Low temperature
Presence of polar solvents.
Presence of ion attacking reagents.
:
r

Reaction intermediate
R —> P
Single step reaction
R —> A —> B —> C —> P
Multistep reaction
A & B short lived,unstable, reactive
quantity formed during reaction
Reaction intermediates
No. of intermediates :- n-1 ( n = no. of steps)

Minima represents Intermediate
whereas maxima represents no.
of steps.
§
¥÷

ORGANIC
CHEMISTRY
Believe
YOURSELF
F-ion

Carbo-cation / Carbonium ion intermediate
Act as electrophile.
Used in E1 , SN1 and Addition electrophile.
Has 3 bond pairs , lone pair = 0
6 e- in its outermost shell.
Zero unpaired e- [Diamagnetic]
Stability
Stability
Electron donating group
Electron withdrawing group
+M , +H , +I
-M , -H , -I
Resonance
At meta-position , only Inductive effect is considered.
a
b—c—X
d
b—c
d
+ X
a
✗ ✗ ✗

Compare stability of carbo-cations
1.) CH2-CH2 2.) CH3-C
CH3
CH3
3.) CH3
CH3
CH 4.) CH3
A.)

Solution
1.) 2.)
3.) 4.)
2 > 4 > 3 > 1
1 H 7 H
3 H
4 H
✗-
4-
✗-
✗-

Solution
1.) 2.)
3.) Ph-CH2 4.)
3 H
3 > 1 > 4 > 2
24th
[R
]
[ R]
✗-

Solution
1.) 2.)
3.)
4.)
3 > 1 > 4 > 2
[R] [I]
[R] + I
4 ✗H

Solution
Less electronegative atom donates lone pair easily.
2 > 1 > 4 > 3

F.)
1.) 2.)
3.) Ph-CH2 4.)
5.)

Solution
1.) 2.)
3.) Ph-CH2 4.)
5.)
5 > 3 > 4 > 2 > 1
C-Cation involved in resonance with lone pair will be more
stable than involved in resonance with benzene.

G.)
1.) 2.)
3.) 4.)
5.)
U
O
e
O
Eth

Solution
1.)
2.)
3.) 4.)
5.)
1 > 5 > 3 > 2 > 4
+ I U - I
0 + I
e
-
M
O
Eth
+±

Solution
1.) 2.)
3.) 4.)
1 -I
+M
+I
3 H
3 > 4 > 1 > 2
✗
a
✗
0 Me

Solution
1.) 2.)
3.) 4.)
3 H 2 H
1 H
0 H
1 > 2 > 3 > 4
•
: :

Solution
4 > 3 > 2 > 1
At meta position only inductive effect

Solution
1.) 2.)
3.)
4.)
+M , -I +M , -I
-I
2 > 1 > 3 > 4

M.)
1.) 2.)
3.)
4.)
2
0
V02
02

N.) 1.) 2.)
3.)
4.)
CH3 CD3
CH3 CD3
O

Solution
1 > 2 > 4 > 3
+I : CD3 > CH3
+H : CH3 > CD3

Solution
1.)
2.)
3.)
4.)
4 H 6 H
Aromatic
Resonance
3 > 4 > 1 > 2
CH2
✗
✗

Conditions in which sp2 hybridised carbon can be positive:-
According to Bredt ‘s rule , sp2 hy C- can’t be positive.
Ist case-> x,y,z -> Bicyclo compound , z = 0
2nd case-> If atleast one ring has 8 atoms
-
" "
g.

Cyclopropane
Dancing resonance
Most stable carbocation
Q.) In which of the following molecules resonance is absent
1.)
2.)
3.)
4.)
C

Solution :-
In 2 and 4
In 1 and 3

Carbo Anion / Carbanion
It has 8 e- in its outermost shell.
3 bond pairs and 1 lone pair.
Zero vacant p-orbital.
Diamagnetic in nature.
Sp3 hybridised {if not involved in resonance } and Sp2 {if involved in resonance}
Stability e- WG
EDG
-M , -H , -I
+M , +H , +I
Resonance
b—C
a
d
—z b—C
a
d
+ Z
→
→
→
→
→
✗ ✗ ✗

Compare stability:-
A.) 1.) 2.) CH3
3.) CH3-CH2
4.)

Solution
1.) 2.)
3.) 4.)
7 H 1 H
3 H
4 H
2 > 4 > 3 > 1
: :

Solution
4 > 3 > 1 > 2
C- Anion involved in resonance with C = 0 will be more
stable than involved in resonance with benzene.

F.)
1.) 2.)
3.) 4.)
Solution :- 4 > 3 > 2 > 1

G.)
1.) 2.)
3.) 4.)
Solution:- 1 > 2 > 3 > 4
i.

H.)
1.) 2.)
3.)
4.)
Solution :- 1 > 3 > 2 > 4
2
Eth

I.)
1.) Ph-CH2 2.)
3.) 4.)
Solution:- 2 > 1 > 3 > 4

J.)
1.)
2.)
3.)
Solution:- 1 > 3 > 2
Eth Eth
Eth Eth
Eth
Eth

FREE RADICAL
7 e- in its outermost shell.
3bp electrons.
Zero lone pair e- s.
1 unpaired e-.
Sp2 hybridised and p-orbital has 1 e-
Stability
Electron donating group
Electron withdrawing group
Resonance
b—C
d
a
—e b—C
d
a
+ e :
→
→
✗ ✗

Compare stability
A.)
1.) CH3 2.) CH3-C
CH3
CH3
3.) CH3-CH2 4.)CH3-CH-CH3

Solution:- 2 > 4 > 3 > 1
B.)
1.) 2.)
3.) 4.)

Solution:- 3 > 4 > 2 > 1
C.)
1.)
2.)
3.) 4.)

Solution:- 2 > 3 > 1 > 4
D.)
1.) Ph-CH2 2.) CH2 CH-CH2
3.) CH2-CH2-CH2 4.) CH3-CH-CH3
O

Acidic strength
H A
H A
H A
H A
Stability :- A > A
Acidic character :- HA > HA
Acidic strength Stability of conjugate base (anion)
Acid - H+
E W G
E D G
÷

Compare Acidic Strength:-
(A)
(1) (2)
(3) (4)
CH4 NH3
H2O HF

Solution : 4 > 3 > 2 > 1
If -ve charge is present at those atoms which lie in the same
period , then deciding factor will be electronegativity.
(B) (1) (2)
(3) (4)
HF HCl
HBr HI

Solution :- 4 > 3 > 2 > 1
While going down a group ; size will be deciding factor.
(C) (1) (2)
(3)
CH3 — CH3 CH2 — CH2
CH CH

Solution :- 3 > 2 > 1
Acidic strength % of S-character.
(D)
E
EE
r
¥
ÉE

Solution :- c > b > d > a
(E)
:

(F)
Solution :- b > a > c > d
(a) (b)
(c)
OH
Cl
OH
Br
OH
F

Solution :- b > a > c
(7) (A) (B)
(C)
Cl
COOH
Cl
COOH
Cl
COOH

Solution:- a > b> c
(8) (a) (b)
(c) (d)
OH
F
OH
Cl Cl
OH
Cl
OH
Br
Br

Solution :- d > c > a > b
Priority order :- Distance > No. > Strength
(9)
(a) (b)
(c) (d)
Ph — OH CH2 — COOH
H2O CH3 CH2 OH

Solution:- b > a > c > d
(10)
(a) (b)
(c) (d) (e)
OH
Me
OH
OH OH
OH
Eth
O CHO F

Solution:- d > e > b > a > c
On basis of H
(11) (a) (b)
(c) (d)
OH
OH
OH OH
Cl
i

Solution:- d > c > b > a
(12) (a) (b)
(c) (d)
OH OH
OH OH
Me
Me
Me

Solution:- d > c > a > b
(13)(a) (b)
(c) (d)
OH OH
OH
OH

Solution:- a > d > c > b
(14) (a) (b)
(c) (d)
OH
NO2
OH
OH OH
NO2
NO2

Solution:- b > a > c > d
(15) (a) (b)
(c) (d)
OH
NO2
NO2
OH
OH OH
NO2
NO2 NO2
NO2
NO2

Solution:- d > a > c > b
Because at meta position, only inductive effect is present.
(16)(a) (b)
(c) (d)
COOH
Eth
COOH
COOH COOH
CHO F
Me—N—Eth

Solution:- c > d > a > b
(17)(a) (b)
(c) (d)
COOH COOH
Cl
COOH
NO2
COOH

Solution:- c > b > d > a
(18)
(a) (b)
(c) (d)
COOH
Eth
COOH
Eth
COOH
Eth
COOH

Solution:- a > d > c > b (Due to ortho effect.)
(19 (1) (2)

Solution:- 2 > 1
(20)(1) (2)
NO2 NO2

Solution :- 1 > 2
(21)
(1) (2)
OH
NO2
OH
NO2

Solution:- 1 > 2
(22) (1) (2)
OH
CN
OH
CN

Solution:- 2 > 1
(23)(1) (2)
(3) (4)
COOH
NO2
COOH
NO2
COOH
NO2
COOH

Solution:- 2 > 1 > 3 > 4
(24) (1) (2)
(3)
CH3 — C — OH
O
Ph — OH

(25) (1) (2)
(3)
HCOOH
CH3COOH
Ph—COOH

(26)(1) (2)
(3) (4) (5)
CH3 — CH3 CH3 — NH2
CH3 — OH CH2 — CH2 CH — CH

Solution:- 3 > 5 > 2 > 4 > 1
(Terminal alkyne stable than Amino and OH)
(27)
(1) (2)
:

Solution:- 2 > 1
(28)
(1) (2)
Solution:- 1 > 2
CHCl3 CHF3

Acidic strength order :-
R —SO3H > H3O / ROH2 > Picric acid > Squaric acid > R —COOH >
NH4 > R—NH3 > Ph—OH > H2O > ROH
Acidic strength Ka 1
PKa
Base strength
Base :- Proton Covalent bonds —> Base
—> atom attached should have lone pair / -ve charge
क
े साथ
it

B + H —> B—H
B + H —> B—H
Stability B > B
Basic strength :- BH > BH
Basic strength 1
Stability of anion
1
Resonance
E D G
E W G
:

Compare basic strength :-
(1)(1) (2)
(3) (4)
CH3 NH2
OH
F

Solution:- 1 > 2 > 3 > 4
(2)
(1) (2)
(3) (4)
Cl
Br I
F

Solution:- 1 > 2 > 3 > 4
(3)
(1) (2)
(3)
(4)
CH3COO Ph — O
CH3 — SO3 CH3 — CH2 — O

Solution:- 4 > 2 > 3 > 1
Comparing Neutral atoms:-
B H
B H
B — H
B — H
Jitna easily lone pair donate
Basic strength
(1)
(1) (2)
(3)
CH3 — O H CH3 — NH2
CH3 — F
:
IÉ
.
/
KE
-
:
•
•
••
:
i
@

(2)
(1) (2)
(3)
CH3 — CH2 — NH2 CH3 — CH — NH
CH3 — C — N
•
•
•
•
•

(3)
(1) (2)
NH2 NH2
:

Solution:- 1 > 2
(4)
(1) (2)
(3) (4)
-
.

Solution:- 1 > 2 > 3 > 4
(5)
(1) (2)
(3)
(4)
N
H
NH2
N
N
H

Solution:- 4 > 3 > 2 > 1
Lone pair involved in maintaining aromaticity will be less stable
then involved resonance.
(6)
(1) (2)
(3) (4)
NH2
NH2
NH2
Ph—NH2

Solution:- 2>1>4>3
(7) (1) (2)
(3) (4)
N
H
N —H
N — H
N
H
i

Solution:- 2>3>4>1
(8)(1) (2)
(3) (4)
NH2
OMe
NH2
CHO
NH2
F
NH2
Eth

Solution:- 1>4>3>2
(9)
(1)
(2)
(3) (4)
NH2
NO2
NH2
NO2
NH2
NO2
NH2

Solution:- 4>2>3>1
(10)
(1) (2)
(3) (4)
NH2
Eth
NH2
Eth
NH2
Eth
NH2

Solution:- 3>2>4>1 (Due to ortho effect)
(11)
(1) (2)
(3)
NH2
NH
NH2
NH
NH2
NH2
Amidine unit
Guanidine unit
:

Solution:- 3>2>1
(12) (1) (2)
(3)
(4)
N
N
H
— N
H N
NH2
N — H
N
H

Solution:- 2>1>4>3
(13)
(1)
(2)
(3) (4)
CH3 — NH2 CH3
CH3
NH
CH3 — N
CH3
CH3
NH3

Solution:- 3>2>1>4 2>1>3>4
In gas phase / non-polar In polar medium
(14)
(1) (2)
(3) (4)
C2H5 — NH2
C2H5 — NH
C2H5
C2H5 — N
C2H5
C2H5
NH3

Solution:- 3>2>1>4 2>3>1>4
Gas phase Polar medium
Basic strength Kb 1
PKb
i

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