Chapter 16 General Organic Chemistry Part 2 by TEACHING CARE Online coaching and tuition classes

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It is convenient to classify the numerous reactions of the various classes of organic compound into four types,

Substitution reactions, Addition reaction,
Elimination reactions, Rearranged reactions,

Substitution reactions

Replacement of an atom or group of the substrate by any other atom or group is known as substitution

reactions.

Leaving group

Substituting or attacking group

Examples : CH3 – CH2 – Br + NaOH ¾¾¾® CH3 – CH2OH+ NaBr

Ethyl bromide Ethyl alcohol

(Bromine atom is replaced by hydroxyl group)

CH₃OH + H – Br ¾¾¾® CH₃ – Br + H₂O C6 H6 + HNO3 ¾¾^H2S¾^O4 ¾⁽^con¾¾^c^.) ® C6 H5 NO2 + H2O

Methyl alcohol

Methyl bromide

Benzene

(Conc.)

Nitrobenzene

(Hydroxyl group is replaced by bromine) (Hydrogen is replaced by

CH3 – CH = CH2 + Cl2 ¾¾⁵⁰⁰¾^°^C ® Cl – CH2 – CH = CH2 + H – Cl

NO2

group)

Propylene Allyl chloride

(Hydrogen is replaced by chlorine)

C6 H6 + Cl2 ¾¾^F^e^C¾^l3 ® C6 H5 – Cl+ H – Cl

Benzene Chlorobenzene

(Hydrogen is replaced by chlorine)

Types of substitution reactions : On the basis of the nature of attacking species substitution reactions are classified into following three categories,

(1) Nucleophilic substitution reactions, (2) Electrophilic substitution reactions,

(3) Free radical substitution reactions

(1) Nucleophilic substitution reactions

Many substitution reactions, especially at the saturated carbon atom in aliphatic compounds such as alkyl halides, are brought about by nucleophilic reagents or

R – X +

Substrate

OH ^

Nucleophile

¾¾¾® R – OH +

X 

Leaving group

Such substitution reactions are called nucleophilic substitution reactions, i.e., SN

substitution and N for nucleophile).

reactions (S stands for

The weaker the basicity of a group of the substrate, the better is its leaving

Leaving power of the group µ 1

Basicity of thegroup

   

Example :

¾¾HI¾>¾H¾Br¾> ¾HC¾l ¾> H¾¾F ®

Decreasing acidity

¾¾¾I B¾r ¾C¾l F¾¾®

Increasing basicity

Decreasing leaving ability

The leaving power of some nucleophilic groups are given below in decreasing order,

|| 

CF₃ – S– O > Br

S– O > CH₃

S– O > C6 H5 – S– O > CH3 – S– O

  || 

Å  

|| 

|| ||

O O

I > Br > CF3 – C– O > H3 O > Cl > F > CH3 – C– O
- In these reactions leaving group of the substrate is replaced by another nucleophile. If reagent is neutral than leaving group is replaced by negative part of the reagent. Negative part of the reagent is always nucleophilic in

+d –d   

R – L ¾¾^E ^–¾^N¾^u ® R – Nu + L ; R – L + Nu ¾¾® R – Nu + L

In SN

reactions basicity of leaving group should be less than the basicity of incoming nucleophilic group.

Thus strongly basic nucleophilic group replaces weakly basic nucleophilic group of the substrate.

Example :

R – Cl ¾¾^O^H¾ ® R – OH + Cl



( NaOH)

Basicity of OH ^



is more than Cl

Q Q



hence OH



replaces Cl as Cl

R – OH ¾¾^C¾^l ´® R – Cl + OH

(HCl)

   

Basicity of Cl is less than OH hence Cl will not replace OH as OH .

Unlike aliphatic compounds having nucleophilic group as leaving group, aromatic compounds having same group bonded directly with aromatic ring do not undergo nucleophilic substitution reaction under ordinary

The reason for this unusual reactivity is the presence of lone pair or p bond on key atom of the functional group. Another factor for the low reactivity is nucleophilic character of aromatic ring.

The SN

reactions are divided into two classes, SN 2

and SN1

reactions.

Distinction between S_N² and S_N¹ reactions

Factors

S_N² Reactions

S_N¹ Reactions

Number of steps

Reaction rate and order

Molecularity TS of slow step

Reacting nucleophile

Stereochemistry

Q Q

One: R : L+ : Nu ® R : Nu+ : L

Second order:

Rate µ [Substrate] [Nucleophile] or Rate = K ₂[RL][: Nu^ ]

Bimolecular

d – d –

: Nu– – – C – — : L

The nucleophile attacks the carbon of the substrate exclusively from the back side.

Complete inversion of configuration takes place.

Two: (i) R : L ¾¾^S^l^o¾^w ® R⁺ + : L^

(ii) R⁺ + : Nu^– ¾¾^F^a¾^st ® R : Nu

First order:

Rate µ [Substrate] or Rate = K₁[RL]

Unimolecular

d + d –

: Nu – – – C – – – L – – – Nu :

The nucleophile can attack the carbon of the substrate both on the back and front sides although the back side attack predominates.

Inversion and retention takes place.

Electrophilic substitutions reactions : Electrophilic substitution involves the attack by an

It is represented as SE

(S stands for substitution and E stands for elctrophile). If the order of reaction is 1, it is

written as SE1 (unimolecular)and if the order is 2, it is SE2 (Bimolecular).

Free radical substitution reactions : Free radical substitution reactions involves the attack by an free These reactions occurs by free radical mechanism which involves : Initiation, Propagation and Termination steps. Examples,
- Chlorination of methane : The chlorination of methane in the presence of ultraviolet light is an examples of free radical substitution (HOMOLYSIS).

CH4 + Cl2 ¾¾^U¾^V ®

CH3Cl

HCl

Methane

light

Methyl chloride

The reaction does not stop with the formation of methyl chloride (CH₃Cl) but the remaining hydrogen atoms are replaced one by one with chlorine atoms to give rise chain reaction.

CH3 Cl + Cl2 ® CH 2 Cl2 + HCl CH 2 Cl2 + Cl2 ® CHCl3 + HCl CHCl₃ + Cl₂ ® CCl₄ + HCl

Reactivity of the halogens for free radical substitution is in the order :

F2 > Cl2 > Br2 > I 2

Arylation of aromatic compounds (Gomberg reaction) : The reaction of benzene diazonium halide with benzene gives diphenyl by a free radical substitution

C6 H5 – H +

C6 H5 N2 X

Benzene diazonium halide

¾¾^Alk¾^ali ® C6 H5 – C6 H5 + N2 + HX

Diphenyl

Wurtz reaction : Ethyl bromide on treatment with metallic sodium forms butane, ethane and ethylene by involving free radical

C2 H5 – Br + 2Na + Br – C2 H5 ¾¾^Eth¾^er ® C2 H5 – C2 H5 + 2NaBr

C2 H5 – Br + Na ® C2 H5 + NaBr

. .

C2 H5 + C2 H5 ® C2 H5 – C2 H5 (Butane)

2CH3

C H2

¾¾Disp¾rop¾ortio¾nat¾i¾on ® CH3

= CH2

Ethane Ethylene

Allylic bromination by NBS (N-Bromosuccinimide) : NBS is a selective brominating agent and it

normally brominates the ethylenic compounds in the allylic

(CH₂ = CH – CH₂ -)

position. This type of reaction

involving substitution at the alpha carbon atom with respect to the double bond is termed Allylic substitution. It is also used for benzylic bromination. Some examples are:

(a) CH₃ – CH = CH₂ +

CH2 – CO

N – Br ¾¾^C^C¾^l4 ® Br – CH2 – CH = CH2 +

CH2 – CO

Propene

CH2 – CO

Allyl bromide

CH2 – CO

NBS Succinimide

Note : ® Halogenation of alkanes occurs by a free radical mechanism and is catalysed by radical initiators

like (C6 H5COO)2 , Pb(CH3 )4 , Pb(C2 H5 )4

etc.

Addition reactions

These reactions are given by those compounds which have at least one p bond,

i.e., (> C = C <,-C º C-,- C -, C º N). In this reaction there is loss of one p bond and gain of two s bonds.

Thus product of the reaction is generally more stable than the reactant. The reaction is a spontaneous reaction.

Types of addition reactions : Addition reactions can be classified into three categories on the basis of the nature of initiaing species.

(1) Electrophilic additions (2) Nucleophilic additions and (3) Free radical additions

(1) Electrophilic addition reactions

This reaction is mainly given by alkenes and
Electrophilic addition reactions of alkenes and alkynes are generally two step
Alkenes and alkynes give electrophilic addition with those reagents which on dissociation gives electrophile as well as

If the reagent is a weak acid then electrophilic additon is catalysed by strong acids (Generally

H2 SO4 ).

Unsymmetrical alkenes and alkynes give addition reactions with unsymmetrical reagents according to Markownikoff’s

The negative part of the reagent adds on that doubly bonded carbon of the alkene which has least number of hydrogen (s).

This rule can be used only in those alkenes which fulfil the following conditions:

Alkene should be
Substituent/substituents present on doublly bonded carbon/(s) should only be +I
If phenyl group is present on doublly bonded carbon, then both doublly bonded carbons should be substituted by phenyl

For example, the following alkenes will give addition according to the Markownikoff’s rule.

CH₃ – CH = CH ₂ ,

CH3

C = CH2 ,

CH3

C = CH – CH₃

C6 H5

C = CH – C6 H5,

C6 H5 – CH = CH2

Following alkenes will not give addition reaction according to Markownikoff’s rule.

CH 2 = CH 2 ,

R – CH = CH – R,

C = C

R C6 H5

C = C

C6 H5

Unsymmetrical alkenes having the following general structure give addition according to anti Markownikoff’s

CH ₂ = CH – G

where G is a strong –I group such as – CX3 ,- NO2, – CN,-CHO,-COR,-COOH,- C– Z (Z = Cl, OH, OR, NH 2 )

Example: CH ₂

CH 2

= CH – CHO + HCl ¾¾Ant¾i-Ma¾rko¾vnik¾ov a¾ddi¾tio¾n ® |

CH 2

CHO

Mechanism of electrophilic addition reactions is as followes,

C = C

+ ¾¾^Slo¾^w ®

+ |

– E

Olefin

E C C

Electrophile |

Å | Q

Fast |

Carbonium ion

C– C– E + X ¾¾¾® C– C– E

| Nucleophile | |

Addition product

Nucleophilic addition reactions : When the addition reaction occurs on account of the initial attack of nucleophile, the reaction is said to be a nucleophilic addition Due to presence of strongly electronegative

oxygen atom, the p -electrons of the carbon-oxygen double bond in carbonyl group ( C = O ) get shifted towards

the oxygen atom and thereby such bond is highly polarised. This makes carbon atom of the carbonyl group electron deficient.

Å  d ⁺ d –

« C– O º C– O

Example : The addition of HCN to acetone is an example of nucleophilic addtion.

CH3

C = O+ HCN ®

CH3 OH C

CH3 CN

Acetone Acetone cyanohydrin

The mechanism of the reaction involves the following steps:

Å 

Step 1. HCN gives a proton (H) and nucleophile, a cyanide ion (CN) .

HCN ® H ^Å + CN ^

Step 2. The nucleophile

(CN ^ )

attacks the positively charged carbon as to form an anion [ H ^Å

does not

attack the negatively charged oxygen as anion is more stable than cation].



 CH3 CN

CH3

C = O ® CN

CH3



C – O or

CH3 O

CH3 CN

Step 3. The proton (H ⁺ ) combines with anion to form the addition product.

CH3

CH₃  Å |

CH3 OH

CN CH3

C – O+ H ® NC – C – OH

CH3

or C

CH3 CN

In C = O

compounds, the addition of liquid HCN gives cyanohydrin and the addendum is

CN ^–

ion

(addition is catalysed by bases or salts of weak acids and retarded by acids or unaffected by neutral compounds) and not HCN directly.

d + d –



Å   O Å OH

C = O ¾¾+ H¾^–C¾N¾®

(A.R.)

C– O ¾¾CN¾® C ¾¾H¾® C

CN CN

Nucleophilic addition (AN ) reactions on carbonyl compounds will be in order:

C = O >

H3 C

C = O >

H3C

C = O

Note : ® Decreasing order of nucleophilic addition in some species.

O O

|| ||

C6 H5 CH2COCH3 > CH3COCH3 > C6 H5 – C– CH3 > C6 H5 – C– C6 H5

CHO > –COCH3 > –COCl > –COOCH3 > –CONH 2 > –COOH
Free radical addition reactions : Those reactions which involve the initial attack by a free radical are known as free radical reactions. Addition of hydrogen bromide to alkenes (say, propylene) in the presence of peroxide (radical initiator) follows free radical Free radical reactions generally take place in non-polar

solvents such as

CCl 4 ,

high temperature, in presence of light or a free radical producing substance like O₂

and

peroxides. The essential steps of the reactions are as follows.

Initiation (Formation of free radical)

. . .

(a)

RO – OR ®

Peroxide

2RO

Free radicals

(b)

HBr + RO ® ROH + Br

. . | .

Propagation. CH ₃

CH = CH ₂ + Br ®

CH₃ – C H – CH ₂ – Br or

Secondary (2°) free radical (more stable)

CH ₃ – C H – C H ₂

Primary (1°) free radical (less stable)

CH3 – CH– CH2 – Br + HBr ® CH3 – CH2 – CH2 – Br + Br

and so on

. .

Termination : Br+ Br ® Br – Br

Polymerisation of ethylene to polyethylene and vinyl monomers to polyvinyl polymers are free radical addition reactions.

nCH2 = CH2 ¾¾^Cata¾^ly¾^st ®(- CH2 – CH2 -)n

Ethylene

Polyethylene

Elimination reactions

Elimination reactions are formally the reverse of addition reactions and consist in removing the two groups (Generally, one being a proton) from one or two carbon atoms of a molecule to form an unsaturated linkage or centre.

Elimination reaction is given by those compounds which have a nucleophilic group as leaving group,

Å Å Å

R Å R

i.e., X, OH, OR, N 2 , N ₃ , H₃O, N

R , S

R R

Elimination reactions are generally endothermic and take place on heating. Elimination reactions are classified into two general types,

a– elimination reactions or 1, 1-elimination
b– elimination reaction or 1, 2-elimination
a– elimination reactions or 1, 1-elimination reactions : A reaction in which both the groups or atoms are removed from the same carbon of the molecule is called a– elimination This reaction is mainly given by gem dihalides and gem trihalides having at least one a– hydrogen.

..  Å

CHX3 ¾¾^Alc.¾^KO¾^H¾^/^D ® CX2 + X+ H

..  Å

CH 2 X 2 ¾¾^Alc,¾^KO¾^H¾^/^D ® CHX+ X+ H

CH3 CHI 2 + Zn ® CH3 CH+ ZnI 2

Product of the reaction is halocarbenes or dihalocarbenes. Carbenes are key intermediates in a wide variety of chemical and photochemical reactions.

b– elimination reactions or 1, 2-elimination reactions : Consider the following reactions,

Å 

CH3 – CH 2 – CH 2 – L ® CH3 – CH = CH 2 + H+ L

b a

A reaction in which functional group (i.e., leaving group) is removed from a– carbon and other group (Generally hydrogen atom) from the b– carbon is called b– elimination reaction. In this reaction there is loss of two

s bonds and gain of one p bond. Product of the reaction is generally less stable than the reactant.

Types of b– elimination reactions : In analogy with substitution reactions, b– elimination reactions are divided into three types:

(i) E₁ (Elimination unimolecular) reaction, (ii) E₂ (Elimination bimolecular) reaction and (iii) E_1cb (Elimination unimolecular conjugate base) reaction

E₁ (Elimination unimolecular) reaction : Consider the following reaction,

CH3



C H O/ D

CH3 

CH3 – C– Cl ¾¾2 ¾5 ¾¾® CH2 = C

CH3

C₂H₅OH + Cl

Reaction velocity depends only on the concentration of the substrate; thus reaction is unimolecular

Rate µ [Substrate]

Product formation takes place by formation of carbocation as reaction intermediate (RI).
Since reaction intermediate is carbocation, rearrangement is possible in E₁
Reaction is carried out in the presence of polar protic
The E₁ reaction occurs in two steps,

CH3

|d ⁺ ^–d Å 

Step 1.

CH3 – C– Cl ® CH ^+d

C – – – – – Cl ¾¾^Slow¾^st¾^ep ® CH – C– CH + Cl

CH3

3 |

CH3 (TS1 )

Å ê

3 3

CH3

(RI )

ú Å

CH3

Step 2.

B + H – CH ₂ –

CH3

– CH3 ® êB – – – –H – – – –CH 2 – – – –

CH3

CH3ú ¾¾^f^a¾^s^t ® B H + CH 2 = C

CH 3

Energy profile diagram for E₁ reaction is,

TS2 û

E₂ (Elimination bimolecular) reaction : Consider the following reaction,

CH3

CH 2

CH 2

Br ¾¾^B^a^s¾^e⁽¾^B⁾ ® CH3

CH = CH ₂

Å 

H+ Br

Reaction velocity depends only on the concentration of the substrate and the base used; thus reaction is bimolecular Rate µ[Substrate] [Base]
Since the reaction is a bimolecular reaction, the product formation will take place by formation of transition state (TS).
Rearrangement is not take place in E₂ reaction but in case of allylic compound rearrangement is
Reaction is carried out in the presence of polar aprotic
The E₁ reaction occurs in one step,

B H H

éB^–d – – – – H^+d H ù

| | ê ⁝ | ú 

CH₃ – C – C – H ¾¾^Slow¾^st¾^ep ® ê CH₃ – C C – Hú ¾¾^f^a¾^s^t ® CH₃ – CH = CH₂ + BH = Br

| ⁝

êë H TS Br úû

Energy profile diagram for E₂ reaction is,

E_1cb (Elimination unimolecular conjugate base) reaction

E_1cb mechanism is limited to substrates with substituents which can stabilise the carbanion as reaction

Thus b– carbon should contain strong –I group, e.g.,

group.

– C -, – NO2,

C º N

or other carbanion stabilising

This reaction is given by those compounds, which have poor leaving group, otherwise carbanion will not be
b– hydrogen should be highly acidic so that it can easily be removed as proton to give



CF3 – CHCl2 ¾¾^C2 H¾5¾^O ® CF2 = CCl2 ;

The E_1cb reaction occurs in two step,



C6 H5 – CH2 – CH2 – F ¾¾^C2 H¾5¾^O ® C6 H5 – CH = CH2



Step 1. CF3 – CHCl2 ¾¾^C2 H¾5¾^O ®

Fast step

CF3 – CCl2

Conjugate base of sustrate

which is stabilised by CF₃ and Cl

Step 2.

F – CF2



CCl2

¾¾Slow¾st¾ep ® CF2



= CCl₂ + F

Orientation in b – elimination reactions : If substrate is unsymmetrical, then this will give more than one Major product of the reaction can be known by two emperical rules.
- Saytzeff rule : According to this rule, major product is the most substituted alkene e., major product is

obtained by elimination of H from that b– carbon which has the least number of hydrogen. Product of the reaction

in this case is known as Saytzeff product.

CH3

b ₂ |

CH – CH

| a

b₁

¾¾Alc.¾KO¾H /¾D ® CH3

HCl

C = CH– CH₃

CH3

Saytzeff product

Hofmann rule : According to this rule, major product is always least substituted alkene e., major product is formed from b– carbon which has maximum number of hydrogen. Product of the reaction in this case is known as Hofmann product.

CH3

C – CH2

¾¾Alc.¾KO¾H ¾/ D ® CH

CH3

₃– C– CH2

CH = CH₂

| b 2

CH3

a b1

CH3

Hofmann product

Note :® In E₁ reactions, product formation always takes place by Saytzeff rule.

In E_1cb reactions, product formation always takes place by Hofmann
In E₂ reactions, product formation takes place by Saytzeff as well as Hofmann In almost all E₂

reactions product formation take place by Saytzeff rule.

(3) Examples of b – elimination reactions

Dehydrohalogenation is removal of HX from alkyl halides with alcoholic KOH or (Potassium tertiary butoxide) and an example of a–b elimination,

KNH 2

or OK – ter-Bu

e.g.,

CH3 – CH 2 X ¾¾^Alc.¾^K^O¾^H ® H 2 C = CH 2 ;

CH3 – CH– CH3 ¾¾^Alc.¾^K^O¾^H ® CH3CH = CH2

(- HX)

Ethene

| (- HX)

Propene

CH3 – CH2 – CH– CH3 ¾¾^Alc.¾^K^O¾^H ® CH3 – CH = CH – CH3 + CH3 – CH2 – CH = CH2

| (- HX)

2- Butene (Major)

1- Butene (Minor)

Dehydration of alcohol is another example of elimination When acids like conc.

H 2 SO4 or

H₃ PO₄ are used as dehydrating agents, the mechanism is E₁ . The proton given by acid is taken up by alcohol.

Dehydration is removal of

H ₂O from alcohols,

e.g.,

CH3 – CH 2 – OH ¾¾^Con¾^c. ^H¾2SO¾4 ,1¾⁷⁰^°¾^C ® H 2 C = CH 2

(- H2O)

CH3 – CH2 – CH2 – OH ¾¾^Con¾^c. ^H¾2 SO¾4 ,1¾⁷⁰^°¾^C ® CH3 – CH = CH2

(- H 2 O)

Note :® Dehydration of alcohols is in the order:

Tertiary > Secondary > Primary

(3°)

(2°)

(1°)

2° and 3° alcohol by E₁ process and 1° alcohol by E₂ Alcohols leading to conjugated alkenes are more easily dehydrated than the alcohols leading to non-conjugated alkenes.

CH ₂ = CH – CH – CH₃

is easily dehydrated than CH3 – CH 2 – CH – CH3 and so

OH OH OH

> >

Dehalogenation : It is removal of halogens, g.,

CH 2 – CH 2 + Zn dust ¾¾ⁱⁿ ^C¾^H3O¾^H^,¾^he¾^at ® H 2 C = CH 2

| |

Br Br

(-ZnBr₂ )

Ethylene

Ethylene bromide

Dehydrogenation : It is removal of hydrogen, e.g., CH3 – CH – CH3 ¾¾^Cu^,¾³⁰⁰¾^°^C ® CH3 – C– CH3

Isopropyl alcohol

Rearrangement reactions

(- H2 )

Acetone

The reactions, which involve the migration of an atom or group from one site to another within the molecule (nothing is added from outside and nothing is eliminated) resulting in a new molecular structure, are known as rearrangement reactions. The new compound is actually the structural isomer of the original one.

It is convenient to divide rearrangement reactions into following types:

Rearrangement or migration to electron deficient atoms (Nucleophilic rearrangement) : Those rearrangement reactions in which migrating group is nucleophilic and thus migrates to electron deficient centre which may be carbon, nitrogen and

Y  B: B

– | | – Y | | | |

C – C– ¾¾¾® – C– C– ® – C– C–

| | Å | |

X : X X

Bridged or

non-classical carbocation

X= Nucleophilic species, Y = Electronegative group, B = Another nucleophile.

Examples:

Pinacol – pinacolone rearrangement

CH3 CH3

O CH3

| |

CH –

¾¾^D¾¾® CH

|| |

– –

₃ C —

C — CH3

30% H2SO4

₃ C C — CH3

CH3

Pinacol

Wagner – Meerwein rearrangement

CH3

Pinacolone

CH3

C – CH

₂OH ¾¾^H2S¾^O¾4 ® CH3 –

C = CH – CH₃

= C – CH2

– CH3

CH3

Neopentyl alcohol

Benzilic acid rearrangement

2-Methyl-butene – 2

2-Methyl-butene -1

O O

|| ||

C6 H5 OH

C6 H5 C– C– C6 H5 ¾¾^N^a^O¾^H ^/¾^K^O¾^H ® C

Benzil

6 5



COO

Wolf rearrangement : CH₃ – COOH

Bezilic acid anion

CH2 N 2 ® CH3 – CH2 – COOH

Allylic rearrangement

CH3

Acetic acid

diazo methan

CH3

Propinoic acid

CH3

C – CH = CH ₂

¾¾^H¾+ ® CH

C = CH– CH

₂OH

3- Methyl-but-1-ene-3-ol

3- Methyl-but-2-ene-1-ol

Sommelet – Hauser rearrangement

Å 

CH₂NMe₃X

¾¾^N^a^N¾^H¾2 ®

CH₃

CH₂Nme₂

Benzyl quaternary ammonium salt O-Substituted benzyl tertiary amine

Hofmann rearrangement :

RCONH2 + Br2 + 4 KOH ® RNH2 + 2KBr + K2CO3 + 2H2O

Curtius rearrangement :

RCON₃ ¾¾^D ® RNCO+ N₂

Acid azide

Isocyanate

Schimdt rearrangement :

RCOOH + HN3 ¾¾^H2 S¾^O¾4 ® R – NH2 + CO2 + N2

Baeyer Villiger reaction : CH3- CO – CH3 + CH3 – COOOH ¾¾^H2S¾^O¾4 ® CH3 – COO – CH3 + H 3 CCOOH
- Rearrangement or migration to electron rich atoms (Electrophilic rearrangement) : Those rearrangement reactions in which migrating group is electrophile and thus migrates to electron rich

Examples :

Stevens rearrangement :

Å 

(CH3 )2 N – CH2 – COC6 H5 ¾¾OH¾®(CH3 )2 N – CH – COC6 H5

CH2C6 H5

Keto-quaternary ammonium salt

(- H2O)

C6 H5

CH2 –C6 H5

Amino keto

C6 H5

Witting rearrangement :

C6 H5

C = O + H2C = P(C6 H5 )3 ®

C = CH2 +

C6 H5

(C6 H5 )3 P = O

Triphenyl phosphine oxide

1,1- Diphenyl ethylene

 

Favorskii rearrangement : CH3 – CO – CH2 – Cl + OC2H5 ® CH3 – CH 2 – COO – C2 H5 + Cl

O COOC H

Cl 2 5

¾¾^C2 H¾5O¾^N¾^a ®

Rearrangement or migration to free radical species (Free radical rearrangement) : Those rearrangement reactions in which the migrating group moves to a free radical centre. Free radical rearrangements are comparatively
Aromatic rearrangement : Those rearrangement reactions in which the migrating group moves to aromatic Aromatic compounds of the type (I) undergo rearrangements in the manner mentioned below,

X–Y X–H

® +

X–H

(I)

The element X from which group Y migrater may be nitrogen or oxygen.

Examples :

(i) Orton rearrangement (ii) Claisen rearrangement

Cl COCH₃

H COCH₃

H COCH₃ N

O.CH₂–CH= CH₂

CH –CH= CH

¾¾dil. H¾Cl ® +

Phenyl allyl ether

¾¾200¾°¾C ®

2 2

o- Allyl phenol

N– Chloroacetanilide o– Chloroacetanilide p– Chloroacetanilide

Fries rearrangement

O–C–CH₃ OH

¾¾Anh¾y. A¾lC¾l3 ®

COCH₃

Phenylacetate (Ester) o- and p– hydoxy acetophenone

Organic compounds having same molecular formula but differing from each other at least in same physical or chemical properties or both are known as isomers (Berzelius) and the phenomenon is known as isomerism.

For example,

*Ethyl alcohol (C₂H₆O)*	*Dimethyl ether (C₂H₆O)*
CH₃ – CH₂ – OH	CH3 – O– CH3
ß	ß
It is liquid.	It is a gas.
Its boiling point is 78°C	Its boiling point is – 24°C.
It reacts vigorously with sodium and evolves hydrogen.	It does not react with sodium.
It reacts with HI and forms ethyl iodide, C₂ H₅ I .	It reacts with HI and forms methyl iodide, CH₃I .

The difference in properties of isomers is due to the difference in the relative arrangements of various atoms or groups present in their molecules. Isomerism can be classified as follows:

Isomerism

Constitutional or structural isomerism Without referring to space, the isomers differ in the arrangement of atoms within the molecule is called structural isomerism. Thus structural isomers have:

Same molecular formula
Same empirical formula
Same molecular weight
Different properties
Different structural formula

Configurational or stereo isomerism

The isomerism arises due to different arrangement of atoms or groups in space. It deals with the structure of molecules in three dimensions. Thus stereoisomers have:

Same molecular formula
Same empirical formula
Same molecular weight
Different properties
Different orientation of atoms or molecules in space

Chain	Position	Ring chain	Functional	Meta	Tauto –	Geometrical	Optical	Conformational
isomerism	isomerism	isomerism	isomerism	merism	merism	isomerism	isomerism	isomerism

Chain, nuclear or skeleton isomerism : This type of isomerism is arises due to the difference in the nature of the carbon chain (e., straight or branched) which forms the nucleus of the molecule.

Examples : (i) C₄H₁₀ : CH3 – CH 2 – CH 2 – CH3 ,

n-Butane

CH₃ – CH – CH₃

CH3

Isobutane

CH3

C₅H₁₂ : (Three) CH3 – CH2 – CH 2 – CH 2 – CH3 , CH3 – C H – CH 2 – CH3 , CH₃ – C– CH₃

n-Pentane

C₄H₈ : CH3 – CH2 – CH = CH2 , CH3 –

a -butylene

C =

CH3

CH 2

CH3

Isopentane

CH3

Neopen tan e

C₅H₈ :

HC º C – CH 2 – CH 2 – CH3 ,

1-Pentyne

Isobutylene

HC º C – CH ₂ – CH₃

CH3

2-Methyl – 1 – butyne

4 3 2 1 2 3

C₄H₇N : C H3- C H 2 – C H 2 – C N

Butane nitrile

CH ₃ – C H – C H₃

| 1CN

2-Methyl propane nitrile

Note : ® Except alkynes chain isomerism is observed when the number of carbon atoms is four or more than four.

Chain isomers differ in the nature of carbon chain, e., in the length of carbon chain.
The isomers showing chain isomerism belong to the same homologous series, e., functional group, class of the compound (Cyclic or open) remains unchanged.
Chain and position isomerism cannot be possible together between two isomeric If two compounds are chain isomers then these two will not be positional isomers.
Position isomerism : It is due to the difference in the position of the substiuent atom or group or an unsaturated linkage in the same carbon

Examples :

C₄ H₁₀ O : CH3 – C H – CH 2 – OH , CH3 – C – CH3

CH3

Isobutylalcohol

CH3

t -Butyl alcohol

C₃H₆ Cl₂ : CH₃ – CCl₂ – CH₃ , CH₃ – CH₂ – CH – Cl₂ , CH3 – C H– C H 2 , C H ₂ – CH ₂ – C H ₂

2,2-Dichloro propane,

1,1-Dichloro propane

| | | |

Cl Cl Cl Cl

C₄ H₈ : CH3 – CH 2 – CH = CH 2 , CH3 – CH = CH – CH3

1,2-Dichloro propane

1,3-Dichloro propane

[Gem-two, vic-one and a , w -one]

1-Butene 2-Butene

C₄ H₆ : CH3 – CH 2 – C º CH , CH3 – C º C – CH3 , CH 2 = C = CH – CH3 , CH2 = CH – CH = CH2

1-Butyne

2-Butyne

1,2-Butadiene

1,3-Butadiene

C₃ H₆ O₂ : CH3 – C H– CHO , C H 2 – CH 2 – CHO

2-Hydroxy propanal

3-Hydroxy propanal

C₇ H₇ NO₂ (Three aromatic) :

CH₃

o – Nitrotoluene

NO₂

m – Nitrotoluene

NO₂

p – Nitrotoluene

C₈ H₁₀ (Three aromatic) :

CH₃ CH₃

CH₃

o – Xylene

CH₃

m – Xylene

CH₃

p – Xylene

C₆ H₃ (OH)₃ (Three aromatic) :

OH OH OH OH OH

OH OH OH

Paragallol

1,2,4 – Trihydroxybenzene

Phloroglucinol

C₆ H₃ X₂Y (Six aromatic) :

X X X X X X

X X Y Y

Y X X Y X

Y X

C₆ H₃ XYZ (Ten aromatic) :

X X X

Y Y Z Y Z
Z Y Y Z Z

X X X X Z Z

Z Y Y Z

Y Y

C₆ H₁₄ : CH3 – C H – CH2 – CH2 – CH3 , CH3 – CH 2 – C H – CH 2 – CH3

CH3

2-Methylpentane

C₄ H₁₁ N : CH3 – NH – CH2 – CH2 – CH3

N -Methyl -1- propaneamine

CH3

3- Methylpentane

CH₃ – NH – C H – CH₃

CH3

N -Methyl-2-propaneamine

Note : ® Aldehydes, carboxylic acids (and their derivatives) and cyanides do not show position isomerism.

Monosubstituted alicylic compounds and aromatic compounds do not show position
Structural isomers which differ in the position of the functional group are called For

example, (i) CH 3 – CH 2 – CH 2 – OH

(ii) CH ₃ – CH – CH ₃

Functional isomerism : This type of isomerism is due to difference in the nature of functional group present in the The following pairs of compounds always form functional isomers with each other.

Examples :

Alcohols and ethers (C_n H_2n+2O)

C₂H₆O :

CH₃ – CH ₂ – OH ;

Ethyle alcohol

H3 C – O – CH3

Dimethyl ether

C₃H₈O : CH3 – CH 2 – CH 2 – OH ;

n – propyl alcohol

C2 H5 – O – CH3

Ethyl methyl ether

C₄H₁₀O : CH3 – CH2 – CH2 – CH2 – OH ;

n-Butyl alcohol

C2 H5 – O – C2 H5

Diethyl ether

Aldehydes, ketones and unsaturated alcohols …etc. (C_n H_2nO)

C₃H₆O : CH3 – CH 2 – CHO

Propionaldehyde

; CH₃ – C– CH₃ ;

Acetone

CH₂ – CH– CH₃

1,2-Epoxy propane

; CH₂ = CH – CH₂OH

Allyl alcohol

Acids, esters and hydroxy carbonyl compounds …etc. (C_n H_2nO₂)

C₂H₄O₂ : CH3COOH

Acetic acid

; HCOOCH₃

Methyl formate

C₃H₆O₂ : CH3 – CH2 – COOH

Propionic acid

; CH₃COOCH₃

Methyl acetate

; CH3 CHCHO

2- Hydroxy propanal

; CH₃ – C– CH₂ – OH

1- Hydroxy propan- 2-one

Alkynes and alkadienes (C_n H_2n-2)

C₄H₆ : CH3 – CH2 – C º CH

1- Butyne

; H₂C = CH – CH = CH₂

1,3- Butadiene

; CH₃ – C º C – CH₃ ;

2-Butyne

H₂C = C = CH – CH₃

1,2- Butadiene

Cyanides and isocyanides ( – CN and – NC )

C₂H₃N :

CH3CN ;

Methyl cyanide

CH3 NC

Methyle isocyanide

Nitro alkanes and alkyl nitrites ( – NO₂ and

O – N = O )

C₂H₅NO₂ : C2 H5 – N

Nitro ethane

; C₂ H₅ – O – N = O

O Ethyl nitrite

Amines (Primary, secondary and tertiary)

C₃H₉N : CH3 – CH 2 – CH 2 – NH 2

Propan -1- amine

; CH₃ – CH₂ – N

CH3

CH₃ – CH – CH₃ ;

NH2

N -Methyl ethanamine

CH3

CH3 – N

CH3

Propan – 2- amine

N,N –Dimethyl methanamine

Alcohols and phenols

C₇ H₈ O ;

CH₂OH

CH₃

Benzyl alcohol o-Cresol

Oximes and amides

C₂H₅NO : CH₃ – CH = NOH

Acetaldoxime

Thio alcohols and thio ethers

; CH₃ – C– NH₂

Acetamide

C₂H₆S :

C2 H5SH

Ethyl thioalcohol

; CH₃ – S – CH₃

Dimethyl thioether

Ring-chain isomerism : This type of isomerism is due to different modes of linking of carbon atoms, e.,

the isomers possess either open chain or closed chain sturctures.

Examples :

C₃H₆ : CH3 – CH = CH2

Propene

CH2

; H2C – CH2

Cyclopropane

H 2 C

— CH 2

CH3

C₄H₈ :

CH3 – CH 2 – CH = CH 2

1- Butene

; |

H 2 C

| ;

— CH 2

H 2 C – CH 2

Methyl cyclopropane

H2C

H2 C

CH2

CH2 –

CH2

Cyclobutane

CHCH3

C₆H₁₂ :

H2C

| ;

CH2

C H2

C H 2 – CH 2

Methyle cyclopentane

; CH3 CH2CH2CH2CH = CH2

1-Hexene

Cyclohexane

Note : ® Ring – chain isomers are always functional isomers.

Metamerism : This type of isomerism is due to the difference in the nature of alkyl groups attached to the polyvalent atoms or functional Metamers belong to the same homologous series. Compounds like ethers, thio-ethers ketones, secondary amines, etc. show metamerism.

Examples :

C₄H₁₀O :

C2 H5 – O – CH3

Diethyl ether

; C3 H7 – O – CH3

Methyl propyl ethers

C₅H₁₀O : C2 H5 – CO – C2 H5 ; C3 H7 – CO – CH3

Diethyl ketone

C₄H₁₁N : C2 H5 – NH – C2 H5

Diethyl amine

Methyl propyl ketone

; C3 H7 – NH – CH3

Methyl propyl amine

C H N : C H – N

CH3 ; C H – N

C2 H5

5 13 3 7

CH3 2 5

CH3

Dimethyl propyl amine

C₆H₁₅N : C3 H7 – NH – C3 H7

Dipropyl amine

Diethyl methyl amine

; C2 H5 – NH – C4 H9

Ethyl butyl amine

Note : ® If same polyvalent functional group is there in two or more organic compounds, then never write chain or position isomerism, it will be metamerism e.g.,

CH3 – C– CH2 – CH2 – CH3

(Pentan- 2-one)

; CH3CH2 – C– CH2CH3

(Pentan-3-one)

are metamers and not position isomers.

CH3 – C– CH2CH2CH3 ; CH3 – C– C H – CH3

are metamers and not chain isomers.

|| ||

O O

CH3

(Pentan- 2-one)

(3-Methylbutan- 2-one)

Alkenes does not show

(6) Tautomerism

The type of isomerism in which a substance exist in two readily interconvertible different structures leading to dynamic equilibrium is known as tautomerism and the different forms are called tautomers (or tautomerides).

The term tautomerism (Greek: tauto = same; meros = parts) was used by Laar in 1885 to describe the phenomenon of a substance reacting chemically according to two possible structures.

It is caused by the wandering of hydrogen atom between two polyvalent atoms. It is also known as Desmotropism (Desmos = bond and tropos = turn). If the hydrogen atom oscillates between two polyvalent atoms linked together, the system is a dyad and if the hydrogen atom travels from first to third in a chain, the system is a triad.

Dyad system : Hydrocyanic acid is an example of dyad system in which hydrogen atom oscillates between carbon and nitrogen atoms. H – C º N ⇌ C =→N – H
Triad system

Keto-enol system : Polyvalent atoms are oxygen and two carbon atoms.

Examples :

O H OH

|| | |

C– C ⇌

(Keto)

C = C–

(Enol)

O OH

|| |

Acetoacetic ester (Ethyl acetoacetate) :

CH3 – C– CH 2 COOC2 H5

Keto form

⇌ CH₃ – C = CHCOOC₂H₅

Enol form

Acetoacetic ester gives certain reactions showing the presence of keto group (Reactions with

HCN, H 2 NOH, H 2 NNHC6 H5 ,

etc.) and certain reactions showing the presence of enolic group (Reactions with

Na, CH3 COCl, NH 3 , PCl5 , Br2 water and colour with neutral FeCl3 , etc.).

Acetyl acetone : CH3 – C– CH 2 COCH3

Keto form

CH₃ – C = CHCOCH₃

Enol form

C– CH 3

= C– CH 2

Keto form Enol form

OH O

Enol form Keto form

O OH

O O HO OH

Keto form

Enolisation is in order

Enol form

CH3COCH3 < CH3COCH2COOC2 H5 < C6 H5COCH2COOC2 H5 < CH3COCH2COCH3 < CH3COCH2CHO

O OH OH

|| H^Å | Å |

Acid catalysed conversion CH3 – C– CH 2 – R

CH3

– C–

C H – R ¾¾^–^H¾® CH3 – C = CH – R

Base catalysed conversion

Keto

Å | (Enol)

 OH

|| OH ||

| H O |

CH₃ – C– CH ₂ – R

– H O

CH3 – C– C H – R ¾¾® CH3 – C = CH – R ¾¾2¾® CH3 – C = CH – R

Keto ² 

–OH ^

(Enol)

Triad system containing nitrogen

Examples :

Nitrous acid :

H – O – N = O

Nitrite form

H – N

Nitro form

Nitroethane : CH3 – CH2 N

Nitro form

CH₃CH = N

Acid form

Characteristics of tautomerism

Tautomerism (cationotropy) is caused by the oscillation of hydrogen atom between two polyvalent atoms present in the The change is accompanied by the necessary rearrangement of single and double bonds.
It is a reversible intramolecular
The tautomeric forms remain in dynamic Hence, their separation is a bit difficult. Although their separation can be done by special methods, yet they form a separate series of stable derivatives.
The two tautomeric forms differ in their stability. The less stable form is called the labile form. The relative proportion of two forms varies from compound to compound and also with temperature, solvent etc. The change of one form into another is also catalysed by acids and
Tautomers are in dynamic equilibrium with each other and interconvertible (⇌).
Two tautomers have different functional
Tautomerism has no effect on bond
Tautomerism has no contribution in stabilising the molecule and does not lower its
Tautomerism may occur in planar or nonplanar

Note : ® Keto=enol tautomerism is exhibited only by such aldehydes and ketones which contain at least one a -hydrogen. For example CH3CHO,CH3CH2CHO,CH3COCH2COCH3 .

Tautomerism is not possible in benzaldehyde

(C6 H 5 CHO), benzophenone

(C6 H 5 COC6 H 5 ) , tri

methyl acetaldehyde, (CH₃ )₃ C – CHO and chloral CCl₃ – CHO as they do not carry a – H .

Number of structural isomers

Molecular formula	*Number* of *isomers*
Alkanes
C4 H10	Two
C5 H12	Three
C6 H14	Five
C7 H16	Nine
C8 H18	Eighteen
C9 H20	Thirty five
C10 H22	Seventy five
Alkenes and cycloalkanes
C3 H6	Two (One alkene + one cycloalkane)
C4 H8	Six (Four alkene + 2 – cycloalkane)
C5 H10	Nine (Five alkenes + 4 – cycloalkanes)
Alkynes
C3 H4	Two

C4 H6

Monohalides

C3 H7 X C4 H9 X C5 H11 X

Dihalides

C2 H4 X2

C3 H6 X2

C4 H8 X2

C5 H10 X 2

Alcohols and ethers

C2 H6 O C3 H8 O C4 H10 O C5 H12O

Aldehydes and ketones

C3 H6 O C4 H8 O C5 H10 O

Monocarboxylic acids and esters

C2 H4 O2 C3 H6 O2 C4 H8 O2 C5 H10 O2

Aliphatic amines

C2 H7 N C3 H9 N C4 H11 N

Aromatic compounds

C8 H10 C9 H12 C7 H8 O

Six

Two Four Eight

Two Four Nine

Twenty one

Two (One alcohol and one ether) Three (Tow alcohols and one ether) Seven (Four alcohols and three ethers)

Fourteen (Eight alcohols and six ethers)

Two (One aldehyde and one ketone) Three (Two aldehydes and one ketone) Three (Four aldehydes and three ketone)

Two (One acid and one ester) Three (One acid and two esters) Six (Two acids and four esters)

Thirteen (Four acids and nine esters)

Two (One 1°-amine and one 2°-amine)

Four (Two 1°-amines, one 2°-amine and one 3°-amine)

Eight (Four 1°-amines, three 2°-amines and one 3°- amines)

Four Nine Five

The compounds which have same molecular formula but differ in properties due to different spatial arrangement of atoms or groups in space are known as geometrical isomers and the phenomenon is known as geometrical isomerism. The isomer in which same groups or atoms are on the same side is known as cis form and the isomer in which same groups or atoms are on the opposite side is called trans-isomer.

Examples :

H – C– COOH

H – C – COOH

Maleic acid (cis)

H – C– COOH

HOOC–C–H

Fumaric acid (trans)

H₃C – C– COOH

H – C – COOH

Citraconic acid (cis-isomer)

H₃C – C– COOH

HOOC – C – H

Mesaconic acid (trnas -isomer)

H – C– Cl

H – C – Cl

cis-1,2-Dichloroethylene

H – C– Cl

Cl – C – H

trans-1,2-Dichloroethylene

H₃C – C– H

H₃C – C – H

cis-Butene- 2

H₃C – C– H

H – C – CH₃

trans-Butene- 2

Conditions for geometrical isomerism : Compound will show geometrical iosomerism if it fulfils the following two conditions
There should be frozen rotation about two adjacent atoms in the molecule. frozen rotation about carbon, carbon double bond in

ozen rotation about carbon, carbon single bond in cycloalkanes.

Both substituents on each carbon should be different about which rotation is frozen. If these two conditions are fulfilled, then compound will show geometrical

Note : ® The compounds of the following type will not show geometrical isomerism.

a – C– a x – C– a a – C– a

x–C–y

a–C–a

x–C–x

Note the similar atoms (Groups) on one or both of the carbon

(2) Distinction between cis– and trans– isomers

By cyclization method : Generally, the cis-isomer (g. maleic acid) cyclises on heating to from the corresponding anhydride while the trans-isomer does not form its anhydride at all.

H – C– COOH ¾¾He¾at ® H – C– CO

H – C – COOH

Maleic acid(cis)

|| O

H – C – CO

Maleic anhydride

Note : ® Note that the two reacting groups (-COOH) are near to each other.

H – C– COOH ¾¾^He¾^at ® No anhydride

HOOC – C – H

Fumaric acid (trans)

Note : ® Note that the two reacting groups (-COOH) are quite apart from each other, hence cyclisation is not possible.

By hydroxylation (Oxidation by means of

KMnO₄ , OsO₄ or

H2O2

in presence of

OsO₄ ) : Oxidation

(Hydroxylation) of alkenes by means of these reagents proceeds in the cis-manner. Thus the two geometrical isomers of an alkene leads to different products by these reagents. For example,

H – C – COOH

H – C– COOH

Maleic acid (cis)

¾¾^K^M¾ⁿ^O¾4 ®

H OH COOH

meso -Tartaric acid

H – C – COOH

¾¾^K^M¾ⁿ^O¾4 ®

H OH COOH

HOOC

OH H

HOOC – C– H

Fumaric acid (trans)

HOOC

OH H

H OH

COOH

(+)-Tartaric acid (-)-Tartaric acid

By studying their dipole moments : The cis-isomer of a symmetrical alkene (Alkenes in which both the carbon atoms have similar groups) has a definite dipole moment, while the trans-isomer has either zero dipole moment or less dipole moment than the cis- For example, 1,2-dichloroethylene and butene-2.

H – C– Cl

H –C–Cl

cis – Dichloroethylene

(m =1.9 D)

H – C– Cl

Cl–C–H

trans– Dichloroethylene

(m =0.0 D)

H – C– CH₃

H–C–CH₃

cis– Butane-2

H – C– CH₃

CH₃ –C–H

trans– Butane-2

In trans-isomer of the symmetrical alkenes, the effect produced in one half of the molecule is cancelled by that in the other half of the molecule.

In case of unsymmetrical alkenes, the cis-isomer has higher dipole moment than the corresponding trans-

isomer.

For Example,

H₃C – C– Cl

CH₃CH₂–C–Cl

cis-2,3-Dichloropentene-2(High dipole moment)

CH₃ – C– Cl

Cl–C–CH₂CH₃

trans-2,3-Dichloropentene-2(Less dipole moment )

Similar is the case with hexene-2.

H3C H

C = C

CH2CH2 – CH3

H CH3

C = C

CH2CH2 – CH3

cis -Hexene – 2(more polar) trans -Hexene – 2(Less polar)

Note : ® Note that the

CH2CH2 – CH3

has more +I effect than the

group, hence dipole

moment of the two polar bonds do not cancel each other in the trans isomer. Thus trans-isomer is also polar, but less than the corresponding cis-isomer.

By studying other physical properties: (a) The cis-isomer of a compound has higher boiling point due to higher polarity, higher density and higher refractive index than the corresponding trans-isomer (Auwers-skita rule).

b. p
m. p

CH₃ – C– H

CH₃ – C–H

cis-2-Butene 4°C

-139°C

CH₃ – C– H

H–C–CH₃

trans-2Butene 1°C

-106°C

H – C– Cl

H – C – Cl

cis-1,2-Dichloroethene

60°C

-80°C

H – C– Cl

Cl – C – H

trans-1,2-Dichloroethene

48°C

-50°C

(b) The trans-isomer has higher melting point than the cis-isomer due to symmetrical nature and more close packing of the trans-isomer.

Stability : Trans-isomer is more stable than cis-isomer due to steric

Note : ® Terminal alkenes such as propene, 1-butene and 2-methyl propene do not show geometrical isomerism.

Cis-trans isomers are configurational isomers but not mirror images, hence cis and trans isomers are always diastereomers.
Non-terminal alkenes with the same atoms or groups either on one or both the carbon atoms of the double bond such as 2-methyl-2-butene, 2,3-dimethyl –2 – butene etc. do not show geometrical

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Substitution reactions

(1) Nucleophilic substitution reactions

Addition reactions

(1) Electrophilic addition reactions

Elimination reactions

(3) Examples of b – elimination reactions

Rearrangement reactions

(6) Tautomerism

Nitrous acid :

Number of structural isomers

(2) Distinction between cis– and trans– isomers

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