1. General Organic Chemistry 
  2. Hydrocarbons 
  3. Electrophilic Aromatic Substitution (General Mechanisms Only) 
  4. Alkyl Halides, Alcohols and Ethers 
  5. Qualitative Analysis 






In general organic chemistry, we would emphasise on the basics of organic chemistry, which are very important to you to understand organic chemistry in a more lucid way.


In hydrocarbons, we will see about the various preparation methods and properties of hydrocarbons and last but not the least we will discuss about stereo specific reactions.


In Electrophilic aromatic substitutions, we will see the general mechanisms for the reactions, and also about orientation of electrophiles in a monosubstituted benzene ring 


In alkyl halides, alcohols and ethers our main emphasis will be on substitution, elimination and we will also see the other properties of these compounds.


In qualitative analysis, we will basically see the reagents that can separate cations and anions and also about the reactions of cations and anions.




The chemistry dealing with study of carbon and its compounds can be termed as organic chemistry.


If we observe carefully, we find that in organic chemistry, carbon is involved in covalent bonding with other elements, and concepts of covalent bonding are expressed in terms of hybridisation.



The first question which arises in our mind is how to find the hybridisation by looking at structure of molecule, then the answer is quite simple  i.e., 

(i) If a carbon is bonded to four atoms then it involves sp3 hybridised orbitals for formation of that compound.

(ii) If a carbon is bonded to three atoms then hybridisation is sp2 

(iii) If a carbon is bonded only to two atoms then hybridisation is sp


Illustration 1: Identify the hybridisation of each carbon atom in the following compounds 

  (a) (b)

(c) Allene  (carbon) (d) Allylene (carbon) 

(e) Acetyl Acetone (f) Acetonyl Acetone 


(a) , here carbon is attached to 3 other atoms i.e., 1 oxygen and 2 hydrogens ∴ it is sp2 hybridised

(b) , sp3 hybridised 


(c) Allene is propadiene i.e., H2C = C = CH2, here central carbon is sp hybridised while terminal carbons are sp2 hybridised. 

(d) Allylene is propyne i.e.,   

CH3 carbon is sp3 hybridised C – 2, C – 3 carbons are sp hybridised 

(e) 1,3,5 carbons are sp3 hybridised 

2,4 are sp2 hybridised 



C –1, C – 3, C – 4, C – 6 are sp3 hybridised whereas C –2, C –4 are sp2 hybridised 

Fission of Covalent Bond  

After knowing  the concept of hybridisation, let us now try to know how many ways, various organic compounds undergo reactions. If we consider an organic reaction A + R – X ⎯⎯→ A – R + X, in this reaction, the bond between R – X is broken and a new bond is formed  between A – R, the mechanism of reaction depends on how the bonds were broken, the breaking of bonds can take place in two ways.

(a)  ————→

If the bond is broken such that each atom is taking one electron, then this cleavage is called as homolytic fission.


 (b) R – X ⎯⎯→ R+ + :X 

R — X⎯⎯→:R + X+ 


If the bond breaks in a way such that both the electrons of bond are taken by the same atom, then this is called heterolytic fission.


Depending on the above two processes, organic reactions are mainly classified as 


Apart from the above  the two other important types of organic reactions are molecular rearrangement and pericylic reactions.


Reactions which involve homolytic fission, proceed via free radical intermediates, some examples of this type will be seen in free radical substitution, free radical addition reactions. On the other hand, reactions involving heterolytic fission may lead to formation of either an electron deficient carbon or electron excess carbon, which are usually called as carbocations and carbanions respectively. 

In if A is more electronegative than carbon, then a carbocation —A → or carbonium ion is formed, so now any negatively charged species or species containing lone pair of electrons can attack this resulting in the formation of products. Such type of reactions ⎯⎯→ are said to be involving carbocation intermediates In ——A if A is less electronegative than carbon then a carbanion can be generated  ——— A ⎯⎯→ + A+


 If any positively charged species or electron deficient species can attack and result in the formation of products. This reaction involves a carbanion intermediate


+ B+ ⎯⎯→


Depending upon their nature the attacking species in organic reactions are classified as 

  1. Free radicals 
  2. Electrophiles 
  3. Nucleophiles 


Free radicals are the species containing only one unpaired electron. They are very reactive and their life times are extremely short in solution state. Free radicals are usually obtained by thermal or photochemical cleavage of molecules. 



Electrophiles are species which contain positive charge or electron deficient compounds without any charge on them 


Ex: H+, Br+, Cl+, , , BF3,AlCl3,ZnCl2  carbenes, FeCl3 


Nucleophiles are the species which contain negative charge or the molecules where one of the atom contains lone pair of electrons. 


CH ≡ C etc.


After knowing about the types of species that can attack a carbon, now let us see the mechanistic part of reactions.


Mechanism is the pathway of the conversion of reactants to products. For studying the organic reactions we have to go through the field effects which are responsible for these reactions, depending on these effects we can explain why certain substrates undergo reactions by only one particular type of mechanism.


Effect of structure on reactivity of organic compounds 

The effect of structure on reactivity can be divided into three major types: field, resonance and steric. First let us see about the field effects.

Inductive effect: The effect because of which there is permanent displacement of electrons towards the more electronegative atom is termed as inductive effect. 


Inductive effect is always transmitted along a chain of carbon atoms 

Now since chlorine has greater electronegativity than carbon, the electron pair between C1  and Cl will be displaced more towards the chlorine atom because of which C1 acquires a small positive charge which will be less than that on C1 and in the same way C3 also acquires  a small positive charge, as length of carbon chain goes in increasing, this effect starts diminishing, i.e., after C3 carbon inductive effect will be negligible. It is represented  by symbol  →⎯, where the arrow points towards more electronegative atom 

  →⎯ →⎯ →⎯ →⎯


Any atom or group which attracts electrons more strongly than hydrogen in
R3 C-H is said to have negative inductive  effect. In the groups are also said to be electron attracting or electron withdrawing groups.


Any atom or group which attracts electrons less strongly than hydrogen in R3 C-H is  said to have positive inductive effect. The groups are also said to be electron repelling or electron releasing.


Some of the examples of +I and –I effect causing groups 

– I effect + I effect
NO2, -CN, -COOH, F,Cl, Br, OAr, COOR,OR,OH,Ph2, = CH2, , -CR3, -CHR2, -CH2R,-CH3, , COO


Illustration 2: Assign the following groups to be either exhibiting +I or –I effect in the following molecules ?


Solution : (i)  is an –I effect causing group 

(ii) -s   – I effect causing group 

(iii)  –OC6H5 is –I effect causing group 

(iv) is  +I effect causing group 


Applications of Inductive effect 

Increase in –I effect decreases the bond length 

C-F, C – Cl, C – Br, C – I 

C-F bond has shortest bond length 


Stability of Carbocations: 

Positive inductive effect causing group increases the stability of carbocations

, , ,

As number  of positive inductive effect causing groups increases, they will release electrons and hence the electron deficiency on the carbon atom containing positive charge will decrease and hence as an uncharged or less charged structure is more stable than the charged ones, its stability will be greatly increased.


Negative inductive effect increases the acidity of carboxylic acids. As the distance between negative inductive effect causing group and –COOH group increases, the acidic character will decrease.

Ex:  , ,


In the above example  the decreasing order of acidity is as follows 

III > IV > V > VI > I > II 


Basic Character of amines: 

The basic character of amines is due to the presence of unshared electron pair on nitrogen atom which accepts proton.Due to +I effect of alkyl groups, nitrogen atom becomes richer in electron because of which the lone pair of electrons on nitrogen atom is more easily available in amines than in ammonia. Trifluoro amine has no basic property due to negative inductive effect of fluorine.

Therefore the relative basic character of amines on the basis of inductive effect alone should be 3° amine > 2° amine > 1°amine 

However the above order was not found to be true instead the order is 2° amine > 1° amine > 3° amine.

We say that  steric hindrance is the only reason why 3°amine is least basic, because is amines, the three bonding pairs and the lone pair occupy sp3  orbitals, and a lone pair causes crowding of bonding pairs, protonation will relieve crowding by transforming the lone pair to a bond pair. In going from Me2NH to Me3N, the ion becomes less solvated and so the stabilisation is lost and as decrease in solvation is more pronounced than the lone pair availability due to inductive effect, hence the basic character decreases. 


Mesomeric Effect

The phenomenon in which electrons are displaced from a double bond to adjacent single covalent bond, or from a double bond to an atom or from an atom containing unshared pair of electrons to the adjacent single covalent bond, is termed as mesomeric effect or resonance effect.


In mesomeric effect or resonance effect if electrons are transferred away from an atom or group then it is said to be causing –M effect.

Here Y is said to be –M effect causing group

If the electron pair is transferred towards the aotm, then it is said to be causing +M effect.


+M effect causing groups  -M effect causing groups 
, etc. N, -SO3H

-COOH, -COOR etc


Applications of Mesomeric effect:

Basic character of amines: 

Aromatic amines are weakly basic when compared aliphatic amines because in aromatic amines the lone pair of electrons are not readily available due to resonance.




Bond length in Allylic system


In the case of CH2 = CH – CH2Cl, there should be two bond lengths i.e, C – C and C  = C but only one bond length is found out whose value is intermediate value of C = C and C – C because due to resonance  C – C be changed to C = C.


Here, the atoms covered  the localised electron by delocalised electrons must lie in a plane. In picryl iodide, bondlengths, for ortho and para   nitro groups are different. Oxygens of  p-introgroup are in the plane of the ring and b has partial double character,  while the oxygen of o – nitrogroups are forced out of plane by large iodine atom 

Hyper conjugation : Delocalisation which takes place involving ‘σ’ electrons is normally termed as hyperconjugation. When a carbon attached to atleast one hydrogen is attacked to an unsaturated atom or one with an unshared orbital, canonical structures can be written which are termed as hyper conjugative structures.


In the above example, overlap of the σ orbital of C – H bond and π orbital of –C bond is resulting in hyperconjugation.


Till now we have gone through all the fundamental aspects required for knowing about an organic reaction, now we shall look into some structural features of organic compounds.


Isomerism: Isomers are the compounds containing same molecular formula but different properties. It is divided into following types. 


Of all the above types we will discuss about functional group; tautomerism  and stereo isomerism.


Functional Group isomerism:

Class of organic compound Gen. Formula  Functional Gp isomer
Alkanes  CnH2n+2
Alkenes  CnH2n Cycloalkanes 
Alkynes  CnH2n-2 Cycloalkenes, alkadienes
Alkyl halides  CnH2n+1X
Alcohols  CnH2n+1 OH Ethers 
Aldehydes  CnH2nO Ketones, Cyclic ethers 
Carboxylic acids  CnH2nO2 Esters, hydroxyaldehydes 



This is a phenomenon which involves rapid shifts back and forth among the molecules 

Keto – enol Tautomerism

R = H, alkyl , if then the equilibrium lies more towards the left, keto form has C – H, C – C,
C = O bonds where as enol form has C – O, C = C, O – H,. The bond energies prove that keto form is thermodynamically more stable  than enol by about 12 Kcal / mol.




The isomers which differ only in the orientation of atoms in space are known as stereoisomers.


Optical activity: 

Any material that rotates the plane polarised light is said to be optically active. The property of non superimposability of an object on its mirror image is called as chirality. One of the main preliminary tests for knowing whether a compound is optically active or inactive is presence of symmetry.


The symmetry elements can be further classified into (I) Axis of symmetry (2) Plane of symmetry (3) Centre of symmetry.

If a molecule is rotated through an axis , where ‘n’  is even no after one followed by reflection in a plane perpendicular to the original axis and results in an identical molecule, then it is said to posses n – fold alternating axis of symmetry. 


Example : 

Althoguh the molecule is resulting in an identical by rotation about 360, but because of plane of symmetry’s presence it is a chiral.

I and III are identical hence here C2 axis of symmetry is present 

Meso tartaric acid


Different arrangement of atoms (groups) in space in a molecule that can be readily converted into one another is known as conformation. For this we have to know a bit about various projection formulae.

C2H6 (ethane)

—– Indicates that atoms are behind the plane (α

Projecting atoms in front of the plane (β

 __ In the plane 


Conformations of ethane : – X3C – CX3 (X = H) 


In case of ethane two conformations are possible I indicates eclipsed conformation where the energy is high and the other one is the more stable staggered conformation. 


Eclipsed is less stable than staggered by 3Kcal / mole due to presence of bond eclipsing (or) bond opposing or Torsional  or Pfitzer’s strain.


Comparison of stability of most preferred Meso  and Active compounds 


This is known by finding out the number of interactions in the system

Meso Active
2L-s 2L-s
2L-M 2L-M
2M-S 1M-M


AS (EM-M + ES-S) > E2 (m-s) 

The most preferred meso is more stable than most preferred active when gauche interactions are repulsive.


Illustration 3: Meso 2,3 butane diol reacts slowly with a carboxyl compound than active 2,3 butane diol, explain the reason.



After seeing the conformational isomers, let us now begin our exploration into compounds containing asymmetric carbons.


An organic compound containing ‘n’ number of asymmetric carbons should have 2n number of optical isomers. Let us consider the example of tartaric acid. 


The molecule can be represented in plane formula as below.


Formula  (IV)  on rotation through 180° in plane of paper becomes identical with III. Now the number of isomer is 3. If the force which rotates the plane of polarised light is from H to OH then. 

(i) Will rotate plane of polarised light to the right 

(ii) Will rotate plane of polarised light to the left 

(iii) Upper half  will rotate to the right while the lower half will rotate to the left and hence the compound will be optically inactive 

 (I) and (ii) are enantiomers 




Hydrocarbons are the compounds containing carbon and hydrogen. They are divided into two types.


Hydrocarbons containing only single bonds are known as saturated hydrocarbons, alkanes are examples of this type of compounds. Let us discuss first about alkanes and then we will continue our discussion with the other hydrocarbons. 

Alkanes: These are the compounds containing C – H, C – C bonds. All the bonds present are ‘σ’ bonds. Their general formula is CnH2n+2.


Methods of Preparation: 

(i) From Alkyl halides 

(i) Reduction : – Alkyl halides are reduced in presence of LiAlH4, LiEt3BH, NaBH4 etc.
R – X

LiAlH4 can reduces Alkyl halides of all types i.e., vinylic, bridgehead and cyclopropyl halides also. 

LiEt3BH reduces primary, secondary, allylic, benzylic and neopentyl substrates but not tertiary or aryl halides.

NaBH4 can be used for reduction in the presence of –COOH, -COOR, -COO groups without  affecting them.

One halogen of a gem dihalide can only be reduced by using BH4SnH.


(b) Wurtz reaction: Two moles of Alkyl halide undergo reaction with sodium metal to give an alkane.

R – X + Na + Na  + X – R ⎯⎯→ R – R

The basic point which we have to rember here is that an alkane containing double the number of carbon atoms as that of alkyl halide is obtained.


Mechanism: Two types of mechanism have been proposed for this reaction 

(a) C2H5 – Br + 2Na ⎯⎯→ C2H5Na+ + NaBr 

C2H5 – Na+ + C2H5Br ⎯⎯→ C2H5– C2H5 + NaBr 


(b) C2H5 – Br ⎯⎯→ +  

 ⎯⎯→ C2H5 – C2H5


Best yields of alkanes are obtained by using single  alkyl halide i.e,  if we take one mole of an alkyl halide R – X and second mole of  other halide R – X mixture of products is obtained. 


(2) Reaction with organometallic compounds 

Alkyl bromides, chlorides and iodides react with dialkyl copper in ether or THF to give alkanes 

R – X + R2CuLi ⎯⎯→ R – R


R May be  Primary alkyl , allylic, benylic, aryl, vinylic, or allenic and may contain Keto, COOH, COOR and CONR2 groups. R in R2 CuLi may be primary allyl, vinylic allylic or aryl. R2 does not react with ketones.


(3) From Carboxylic acids 


(a) Sodium salt of a carboxylic acid on treatment with soda lime gives alkane containing one carbon less than the original carboxylic acid 

R — — ONa + NaOH R — H + Na2CO3 


(v) Kolbe’s electrolysis: – 

When sodium salt of a carboxylic acid is subjected to electrolysis an alkane is obtained.

2CH3COONa ⎯⎯→ CH3 — CH3 + NaOH + CO2 


Properties of alkanes 

Halogenation: Alkanes undergo halogenation by free radical mechanism


X2    ⎯⎯→ (1) Chain initiating

+ RH ⎯⎯→ R – X + ⎯⎯→ (2) Chain Propagating 

+ ⎯⎯→ R – R  } Chain terminating 


The various halogenation agents are sulfuryl chloride (SO2Cl2), mixture of Br2 and HgO, NBS, CCl4, Cl2O, COCl2. In all these cases chain initiating catalyst is required.


A tertiary hydrogen is easily abstracted  when compared to secondary hydrogen which is easily abstracted when compared to primary hydrogen. The ratio of their reactivity towards  chlorination is 5.0 : 3-8:1.0. In bromination ratios are 1600:82:1 respectively. 


Nitration: They undergo vapour phase nitration to yield nitroalkanes 

CH3 – CH3 CH3 – CH2 NO2


Pyrolysis: Alkanes undergo decomposition by the action of heat to yield smaller alkanes and alkenes.



Compounds containing C = C functional group are termed as alkenes. Their general formula is CnH2n

Alkenes can exhibit geometrical isomerism. Alkenes having structure Cab = Ccd, Cab = Cab can exhibit geometrical isomerism



(1) Dehydrohalogenation of alkyl halides 

 + KOH

Dehydrohalogenation is an example of β – elimination reaction. It was found that the dehydrohalogenation reaction when carried out in presence of a strong base follows second order kinetics.

Rate = K (Rx] [ B]


E2 Mechanism


(1) This reaction involves single step

(2) Favoured in polar aprotic solvent 

Base pulls a proton away from carbon and simultaneously a halide ion departs and a double bond is formed.



Saytzeff’s rule: In dehydrohalogenation the preferred product is the alkene that has greater number of alkyl groups 

Reacting of R – X towards E2 = 3° >>


Illustraion 4:

Although R4N+ undergo this reaction why do not undergo this reaction 

Solution:  Above reaction involves a base promoted elimination where a moderately basic R3N is the leaving group. But when RNH3 is used, once the abstraction of proton takes place we have an amine but not ion, and amines do not undergo this type of reaction.


2-methyl –3- pentyl  tosylate was heated in n-butanol with no added base, what are the products obtained.

As there is not strong base and reaction conditions favour unimolecular reaction.


(a) Dehydration of alcohols:-

Dehydration of alcohols is usually carried out by heating alcohol with sulfuric or phosphoric acid or by passing the  alcohol vapour over a catalyst 

Ease of dehydration of alcohols : 3° >>


Illustration 5: Neohexyl alcohol A + B + C + D

Identify products A, B,C and D

Steroelectronic factors for elminations 

Two leaving groups must have anti  conformation 


Transition state has less energy 




Reaction of Alkenes 


  1. Addition of Hydrogen 

Quantity of heat evolved when one mole of an unsaturated compound is hydrogenated is called as heat of hydrogenation. Based upon heats of hydrogenation we can know stability of alkenes. Based on heats of hydrogenation stability order is as follows:

R2C = CR2 > R2C = CHR > R2C = CH2, RCH = CHR > RCH = CH2 > CH2 = CH2 


Addition of hydrogen halides

Addition involves two steps 

⎯⎯→ :X (slow)

⎯⎯→ (Fast )


In the slow step attack on loosely held  Pie electron of alkene is taking place and the attacking reagent is an electron deficient species, hence such reactions are classified as electrophilic additions.

Since this reaction is involving  the formation of a carbocation intermediates, rearrangement are bound to occur and rearrranged products are usuallly obtained. Generally in electrophilic addition Markonkoff’s rule is  followed. The rule says that negative part of addendum is attached to carbon atom containing least number of hydrogens. 




Addition of Halogens 

In  the addition of bromine, the intermediate involved is cyclic bromonium ion.


Addition of water

(a) Oxymercuration – demercuration 

Oxymercuration demercuration gives markonikov’s product without any rearranged product.


(b) Hydroboration – oxidation 

With diborane alkenes undergo hydroboration to yield alkyl boranes which on  oxidation gives alcohols.

We get Antimarkonikov’s product by following Markonikov’s rule.


Addition of HBr in the presence of Peroxide: – 

Presence of peroxide initiates a free radical mechanism

(1) H — O — O — H — 2

(2)  + HBr — + H2





As 2° free radical is more stable  is formed as major product 


Alkenes are hydroxylated by using certain reagents like KMnO4, OsO4 and peroxy acids to give glycols.

Addition of alkenes with KMnO4 is cis addition while with Peroxyacids it is a trans addition 




Ozonolysis: It is one of the important reactions by which structural elucidation of alkenes is performed.


Hydrocarbon containing C C bond are termed as alkynes 




(1) From Alkyldihalides  —CC—


(2) ⎯⎯→ —C C—R + LiX


Reactions of Alkynes 


(1) Reduction 


(2) Hydration: Alkynes undergo addition of H2O in presence of H2SO4, HgSO4  to give carboxyl compounds.

Except  acetylene all alkynes will give ketones as major products 


  1. Formation of acetylides: – 

Hydrogen attached to triply bonded carbon at end of chain shows appreciable acidity, they react with sodium metal to give acetylides 

HC CH + Na ⎯⎯→ HC C Na+ H2

Because of this property terminal alkynes can be distinguished from non terminal alkynes as well as from alkenes 



Electrophilic Aromatic Substitutions proceed by only one mechanism in the substrate.

Above mechanism is known as arenium ion mechanism. The attacking species, Electrophile may be positive ion or a dipole. In these cations the intermediate formed is not aromatic, but by loss of proton in the last step it is reverting back to its aromatic character.

The important reactions of this type are 

  1. Nitration 2. Sulfonation 
  2. Halogenation 4. Friedel – Craft’s alkylation 
  3. Friedel – Craft’s acylation 


Nitration: – It is usually carried out by using a mixture of concentrated nitric acid and sulphuric acid. The other nitrating agents which can be used are.

N2O5 in CCl4 is used in presence of P2O5 when anhydrous conditions are required 

(b) Ethylnitrate is used to carry out nitration in alkaline medium


HO – NO2 + H2SO4 ⎯⎯→ + HSO4 

  ⎯⎯→ + H2

H2O + H2SO4 ⎯⎯→ H3O+ + HSO4 


HNO3 + 2H2SO4  ⎯⎯→ NO2+ + 2HSO4 + H3O+


Once an electrophile is generated put NO2+ instead of E in general mechanism and it gives mechanism of nitration.


Sulfonation: It is usually affected by using Con H2SO4 or with fuming H2SO4 or chlorosuforic acid.


2H2SO4 SO3 + H3O+ + HSO4 



Aromatic compounds can be brominated or chlorinated by treatment with bromine or chlorine in the presence of a catalyst. When thallium (III) acetate is catalyst many substrates react with high regioselectivity at para to an O-P directing group. Iodine is least reactive among halogens. Except for active substrates, an oxidizing agent should be normally present to oxidise I2 to a better electrophile.

Ex: HNO3, HIO3, SO3, CH3CO3H, H2O2.



X – X + FeX3 ⎯⎯→ X+ + X – FeX3


Fridel – Craft’s alkylation:

The electrophile involved in this reaction is R+ (Alkyl carbo cation)

It is usually generated from Alkyl halides and even from alcohols.

Ex: RCl + AlCl3  ⎯⎯→ R+ + AlCl4 

ROH + AlCl3 ⎯⎯→ ROACl2 ⎯⎯→R+ + OAlCl2

ROH + H+  ⎯⎯→ RO+H2 ⎯⎯→ R+ + H2

In the case of  olefins proton donating acid is definitely required without it reaction does not take place.


Reactivity of Catalysts: 

AlBr3 > AlCl3 > GaCl3 > FeCl3 > SbCl5 > BCl3. As the reaction involves formation of a carbocation intermediate, rearrangements are bound to occur and this is serious limitation of this method.


Illustration 6:


Now 3° butyl group (cation) is electrophile end product obtained is 3° butyl benzene.


Identify X



In the above example ring closure takes place, the product obtained will be 

Friedel – Crafts Acylation:  

The electrophile involved in this reaction is  (Acyl carbocation). Apart from acylhalides carboxylic acid, anhydrides and ketenes can also be used 


Formylation cannot be done

Because formyl chloride or formic anhydride are not stable at ordinary temperatures. When non polar solvents a free ion is not the attacking species but the attack is mainly by complex.


Orientation in mono substituted benzenes:

When a group is attached to bezene ring and increases the rate of electrophilic substation reactions, it is termed as an activating gp, whereas the gp which decreases the rate of electrophilic substitution reactions is a deactivating group.


Activating groups:  –R, etc.

Activating groups: -NO2, -SO3H, etc.

The rate of  electrophilic aromatic substitution depends on tendency of substituent to release or withdraw electrons. The classification of substituents into activating and deactivating can be perfectly understood by knowing the resonance structures.


Resonance structures for electrophilic substitution:-

If X is an electron withdrawing group, it destabilises the resonance structure III and hence it acts as a deactivating group by making the carbon still more positive on the other hand if X is electron releasing then it stabilises the carbocation by donating electrons and hence increases the rate of such reactions.


If a group already present on the benzene ring orients the incoming group to ortho, para positions it is called as an o,p directing group while if it directs to meta it is termed as a meta directing group.


Resonance structures for ortho para and meta attacks 


If x is an electron releasing group, the structures III, V are stabilised hence ortho para attack is more favoured than meta attack.

If X is an electron withdrawing group then structures III, V will be highly destabilised and as a result of this ortho para attack is not possible but in meta attack, all the structures will be more stabilised when compared to that of ortho para attack hence m – attack will take place.


In the case of halogens their reactivity is governed  by –I effect but orientation is governed by +m effect hence although they are deactivating groups they are o,p directors.





Organic compounds containing carbon – halogen  bond are termed as alkyl halides. Gen formula  of Alkyl halides is CnH2n+1X.




  1. From alcohols:

Alcohols when treated with HX or PX3 will give halides .

The above reaction is valid for preparation of primary, secondary or tertiary halides but isobutyl alcohol or neopentyl alcohol will give rearranged products. For reaction of primary and secondary alcohols with HCl catalyst like ZnCl2 is required.


ROH + Ph3P + CCl4 ⎯⎯→ R — Cl + Ph3PO + HCCl3


  1. From ethers 

Ethers can be cleaved by heating with conc HI or HBr. HBr react more slowly than HI. Dialkyl ethers and aryl alkyl ethers can be cleaved. Usually methyl ethers are cleaved so that methyl iodide or bromide is formed as product.


  1. From silver salt of carboxylic acids 


Silver salt of carboxylic acid on heating with Br2 gives alkyl halide containing one carbon less than original carboxylic acid.



Chemical properties: 

(1) Alkyl halides mainly undergo substitute and elimination reactions. Substitution reactions usually take place by SN2 and SN1 mechanism. 


SN2 mechanism :-


The reaction takes place in a single step through the formation of transition state intermediate 2 stands for bimolecular this is always not the same as 2nd order. If a large excess of nucleophile is present the mechanism may be bimolecular even experimentally determined kinetics will be first order. When the leaving group is present at bridge head carbon such compounds cannot react by SN2 mechanism since the nucleophilecannot approach from the rear side.


Branching at either α or β carbons decreases the rate of SN2 reactions. Tertiary and neopentyl systems react by SN2 mechanism very slowly and synthetically are useless.


SN1 mechanism:

R – X R+ + X 

R+ + Y R – Y


These reactions proceed by carbocation intermediates as the intermediate carbocation is planar, the incoming nucelophile can attack either from the front side or from the rear side, this results in an  inversion of configuration as well as retention of configuration if the original alkyl halide is optically active.


For SN1 reactions branching at α carbon increases the rate. The rate of reactions are dependent on difference in free energy between starting compounds and transition states. Secondary alkylhalide can react by SN1 mechanism when solvent of high polarity is use. Rate of SN1 reactions are increased when there is a double bond at the position, because the intermediate carbocations are stabilised  by resonance.


Effect of Nuelcophiles on substitution reactions:

Whenever we go down a group the nucelophilicity increases while basic character decreases. Smaller negatively charged nucleophiles are more solvated by usual polar protic solvents, because negative  charge of Cl is more concentrated than the charge of I. ∴ many nucleophilic substitutions with small negatively charged nucleophilie are more rapid in aprotic solvents 


Order of nucleophilicity

RS > ArS > I > CN > OH > > Br > ArO > Cl > Py > Ac 


Effect of the reaction medium:-

The effect of solvent polarity on the rate of SN1 reactions depend on whether the substrate is neutral or positively charged. For neutral substrates more polar the solvent, faster will be the rate of reaction. SN1 reactions of unionized substrates are more rapid in protic solvents which can form hydrogen bonds with the leaving group.


  1. Alkyl halides undergo elimination reactions by two mechanisms E2 mechanisms have already been discussed in alkenes. Let us now mainly study about E1 eliminations.

E1 mechanism is a two step process in which rate determining step is ionisation of substrate to give a carbocation. 


In E1 mechanism solvent itself acts as the base and when external bases are added the mechanism shifts to E2.



Organic compounds containing C – OH bonds are termed as alcohols. If only one –OH group is present they are called as monohydric alcohols.


(1) Reduction of carbonyl compounds :-

containing compounds are reduced to by reagents for LiAlH4

-H ⎯⎯→ CH3CH2OH 


  1. Carbonyl compounds react with Grignard reagents to give Alcohols. Formaldehyde gives 1° alcohol. All other aldehyde give 2° alcohols, while ketones give 3° alcohols.


  1. Alcohols are also obtained by hydroboration – oxidation of alkenes 

RCH = CH2 + (BH3)2 ⎯⎯→  RCH2CH2OH 


Reaction of Alcohols:-

  1. H of O-H in alcohols is weakly acidic. The order of decreasing acidity is 

H2O > R–OH(1°) > ROH (2°) > ROH(3°) > RC ≡ C-H >RCH3

  1. Alcohols undergo dehydration at 170°C to give alkenes, while at 140°C they give ehters.


C2H5OH + H2SO4 +

⎯⎯→ CH3CH2OC2H5 + H2SO4 


  1. Alcohols  react with hydrohalides to give corresponding alkyl halides .



Organic compounds containing R–O–R functional group are called as ethers.




Williamson synthesis 

R—X + R′ONa+ ⎯⎯→ R—O—R′

Or +

ArO Na+ R–OAr

Alkylhalide should be 1° or 2° 3° usually leads to elimination reaction.


Reactions of Ethers 

Clevage  by acids 

R–O–R′ + HX ⎯⎯→ R–X+R′–OH

Reactivity of HX: HI > HBr > HCl




The branch of chemical analysis which aims to find out the constituents of a mixture or compound is known as qualitative analysis.


The detection of cations (Basic radicals) and anion (Acidic radicals in a salt in a mixture is known as inroganic qualitative analysis.

Acidic radicals have been roughly divided into following types

Reagent  Observation Inference 
Salt + dil HCl or Cold H2SO4 Colourless gas, which turns lime water milky and puts off burning splinter 
Colorless gas, with odour and burning sulphur
Evolution of brown funes with pungent odour 
Colorless gas evolution  with rotten egg smell  S–2
Salt conc H2SO4 and heat  Evolution of colourless pungent gas a glass rod dipped in NH4OH gives dense white fumes  Cl
Colorless purgent gas is evolved .  yellow ppt with a glas rod dipped in AgNO3 Br
Pungent colorless or light brown gas is evolved 


Analysis of Basic or cationic radicals 

Reagent  Color ppt  Radicals 
Ist group dil  HCl White  AgCl, PbCl2, Hg2Cl2
IInd group dil HCl + H2S gas  Black or brown or yellow  CuS, Bi2S3, PbS SnS 

CdS, Sb2S3,  As2S3

IIIrd group NH4Cl + NH4OH White 


Blue or green 




IVth group NH4OH+ H2S White chocolate pink black ppt. Zn, Ni or Co Flesh colored   
Vth group NH4Cl + NH4OH + (NH4)2CO3 White  Ba2+ , Sr2+, Ca2+, Mn+2
VI No special reagent  Mg2+,






  1. Write  the following alcohols in the increasing order of their reactivity towards gaseous HBr. 2-butanol, 2-methyl-1-propanol, 2-methyl-2-propanol.


  1. Racemic 1,2-dibromo-1,2-diphenyl loses HBr on treatment with alc. KOH (dehydrohalogenation) to yield only trans-1-bromo-1,2-diphenylethane. Give reason.


  1. Calculate the total number of steroisomers in the following compounds :


  1. Addition of HCl on 1,3 butadiene gives two products. Explain


  1. An alkane A (C5H12) on chlorination at 300o gives a mixture four different mono chlorinated derivatives B, C, D and E. Two of these derivatives give the same stable alkene F on dehydrohalogenation on oxidation with hot alkaline KMnO4 followed by acidification of F gives two products G and H give structures of A to H with proper reasoning.


  1. 0.37gm of ROH was added to CH3MgI and the gas evolved measured 11.2c at STP. What is the molecular wt of ROH? On dehydration ROH gives an alkene which on ozonolysis gives acetone as one of the products. ROH on oxidation easily gives an acid containing the same number of carbon atoms. Gives the structures or ROH and the acid with proper reasoning.


  1. (C7H14 (A) decolorises Br2 in CCl4 and reacts with Hg(OAC)2 in THF. H2O followed by reduction with NaBH4 to produced a resolvable compound B. (A) undergoes reductive ozonolysis to give the same compound (C) obtained by oxidation of 3-hexanol with KMnO4. Identity A, B and C compound D, and isomers of A reacts with BH3. THF and then H2O2/OH to give chiral E. Oxidation of E with KMnO4 or acid dichromate affords a chiral carboxylic acid F. Ozonolyses of D, gives after reduction with Z the same compound G obtained by oxidation of 2-methyl-3-pentanol with KMnO4 Identify D, E, F and G.


  1. A compound A with molecular formula C5H10 reacts with alkaline potassium permanganate to form compound B (C5H12O2). A on reduction yields 2-methylbutane and on ozonolysis forms aetone and acetaldcehyde. Deduce the structural formula of A and B write equations for the above reacions.


  1. An alkene (A) on ozonolysis yields acetone and an aldehyde. The aldehyde is easily oxidised to an acid (B). When (B) is treated with bromine in presence of phosphorous if yields compound (C) which on hydrolysis gives a hydroxy acid (D). This acid can also be obtained from acetone by the reaction with hydrogen cyanide followed by hydrloysis. Identify the compounds A, B, C and D.


  1. (a) Explain why a nonsymmetrical ether is not usually prepared by heating a mixture of ROH and R′OH in acid. 

(b) Why is it possible to prepare t-butyl ethyl ether by heating a mixture of t-butanol and ethanol? 

(c) Would you get any di-t-butyl ether from this reaction? Explain. 

(d) Can t-butyl ethyl ether be made by heating H2C = CH(CH3)2 and ethanol?


  1. An organic compound (A) gives positive Liebermann reaction and on treatment with CHCl3/KOH followed by hydrolysis gives (B) and (C). Compound (B) gives colour with Schiff’s reagent, but not  (C) which is steam volatile. (C) on treatment with LiAlH4 gives (D), C7H8O2 which on oxidation gives (E). Compound (E) reacts with (CH3CO)2O in CH3COOH solvent to give a pain reliever (F). Gives structures of (A) to (F).


  1. A white precipitate was formed slowly when AgNO3 was added to a compound A with molecular formula C6H13Cl. Compound A on treatment with hot alcoholic KOH gave a mixture of two isomeric alkenes B and C having the formula C6H12. The mixture of B and C on ozonolysis furnished the four compounds,




What are the structures of A, B and C?


  1. An open chain compound (A), C5H8O, is optically active. When (A) is hydrogenated is presence of palladium as catalyst, it absorbs two moles of hydrogen per mole of (A) to produce compound (B), C5H12O, which is optically inactive. However when (A) is warmed with dilute H2SO4 in presence of HgSO4 it gives compound (C), C5H10O2 which is still optically active. (C) responds to the iodoform test. On oxidation with HNO3 (C) gives acetic acid and propanoic acid. Write the structures of (A), (B) and (C) and explain the above reactions.


  1. A mixture of two aromatic compounds A and B was separated by dissolving in chloroform followed by extraction with aqueous KOH solution. The organic layer containing compound A, when heated with alcoholic solution of KOH produced a compound C (C7H5N) associated with an unpleasant odour. The alkaline aqueous layer on the other hand, when heated with chloroform and then acidified gave a mixture of two isomeric compounds D and E of molecular formula C7H6O2. Identify the compounds A,B,C,D and E and give their structures.


  1. A primary alkyl halide (A), C4H9Br, reacted with alcoholic KOH to give compound (B). Compound (B) reacted with hydrogen bromide to give (C), an isomer of (A). When (A) was treated with sodium, it gave a compound (D), C8H18, which was different from the compound produced when n-butyl bromide was reacted with sodium. Draw the structural formula of A and write equations for all reactions. 
  2. On electrolysis, an alcoholic solution of sodium chloride gives a sweet smelling liquid (A) which gives carbylamine reaction and condenses with acetone to form hypnotic. What is (A)? Give reactions of its formation.


  1. A compound (X) on heating with an excess of NaOH solution gives a gas (Y) which gives white fumes on exposure to HCl. Heating is continued to expel the gas completely. The resultant alkaline solution again liberates the same gas (Y) when heated with Zn powder. However, when the compound (X) heated alone does not give nitrogen. Identify (X) and (Y).
  2. When 20.02 g of a white solid (X) is heated, 4.4 g of an acid gas (A) and 1.8 g of a neutral gas (B) are evolved leaving behind a solid residue (Y) of weight 13.8 g. (A) turns lime water milky and (B) condenses into a liquid which changes anhydrous copper sulphate blue. The aqueous solution of (Y) is alkaline to litmus and gives 19.7 g of white ppt. (Z) with barium chloride solution. (Z) gives carbon dioxide with an acid. Identify (A), (B), (X), (Y) and (Z).


  1. A certain inorganic compound (X) shows the  following reactions :

(i) On passing H2S through an acidified solution of (X), a brown ppt. is obtained.

(ii) The ppt. obtained in first step dissolves in excess of yellow ammonium sulphide.

(iii) On adding an aq. solution of NaOH to solution of (X), first a white ppt. is obtained which  dissolves in excess of NaOH.

(iv) The aq. solution of (X) reduces ferric chloride

Identify the cation of X and give chemical equations for the steps (i), (ii), (iii) and (iv).


  1. Classify the following substituents as activating or deactivating groups and write the structural formula of non-substitution products.
  2. a) C6H5CF3, Mono bromination. (B) C6H5CH(CH3)CH2CH3, mono sulfonation (C) Mononitration, C6H5COOCH3


  1. One mole of the compound (A) molecular formula (C8H12) incapable of showing stereoisomerism, reacts with only one mole of hydrogen on hydrogenation over pd. (A) undergoes ozonolysis to give a symmetrical diketone (B). C8H12O2. What are structures of A & B.


  1. Compound A (C8H9Cl) is hydrolysed by dilute acids to give compound B (C8H10O) compound B can be oxidised under mild conditions to compound C (C8H8O). Compound (C) upon reduction with Zn/Hg and concentrated HCl gives Ethyl benzene. Identify structures from (A) to (C).
  2. A metallic chloride A does not respond to chromyl chloride test, but it gives a white precipitate with limited amount of another metal chloride B and grey precipitate with excess amount of B. A when treated with KI gives a scarlet precipitate which dissolves in excess of KI forming an important reagent C used in qualitative analysis. Identify A,B,C and give various reactions involved.
  3. A white powder (A) on heating gives a colourless gas (B) and a  solid (C). The compound (C) assume a yellow colour on heating and changes to white on cooling. If  dissolves in dilute acids and the resulting solution gives a white precipitate with K4Fe(CN)6. Further (A) dissolves in dil HCl with the evolution of a gas, which is the same as B, which turns lime water milky. The solution so obtained gives a white precipitate  (D) with H2S in the presence of NH4OH. Another portion of the solution gives initially a precipitate (E) with NaOH solution, which dissolves in excess of the base. Identify the compounds (A) to (E)
  4. A certain alloy is dissolved in hot concentrated nitric acid and the solution evaporated till the residue is just moist water and HCl are then added the solution thus obtained is treated with H2S. A black precipitate is obtained which filtered off, dissolves in hot dilute HNO3 to  yield a blue solution which becomes deep blue on adding excess of NH4OH. The filterate, after treatment with H2S yield a black precipitate with ammonia. But if the filtrate is throughly boiled and then treated with dimethyl glyoxime in ammonical medium, a rose red precipitate is obtained. What were the metals present in the alloy.








  1. (I) Number of optical isomers = 22 = 4 (2dl pairs)

Number of geometrical isomers = 0

(Since, compound does not contain a ring or a double bond )

total number of isomers = 4 + 0 = 4 


  1. Mechanism of addition of HCl on 1,3-butadiene is as follows 

Step :1






In step 1 proton adds to one of the terminal cabons of 1,3 butadiene to form, as usual, the more stable carbonium ion, in this case a resonance  stabilized allylic cation. Addition of one of the inner carbon atoms would have produced a much less stable primary cation that could not be stabilized by resonance;

In step (2), a chloride ion forms a bond to one of the carbon atoms of the allylic cation that bears a partial positive charge. Reaction at one carbon atom results in the 1,2 – addition product, reaction at the other gives that 1,4 addition product.


5. A) CH3



B) CH3


HC3 — CH — CH2CH2Cl

C)       CH3


H3C — CH — CH — CH3






CH3 — C — CH2CH3




E)     CH3


ClCH2 — CH — CH2CH3




H3C — C = CH — CH3




  1. H3C

      CH — CH2OH



7. A) CH3 — CH2 — C — CH2 — CH2CH3









C) CH3CH2 — C — CH2CH3




D) CH3


CH3 — CH — C — CH2CH3



E)     CH3






F) CH3


CH3 — CH — CH — CH2 — CH3



G) CH3 — CH — C — CH2 — CH3

| ||








(a) A mixture of three ethers, R—O—, R—O—R, and R—O—R is obtained. 

(b) When alcohol is 3°, its onium ion easily loses water to form a carbocation, which is solvated by the other 2° or 1° alcohol to give the mixed ether preferentially. This is an example of an SN1 mechanism.


(c) No.t-Butanol does not solvate the 3° carbocation readily because of steric hindrance. 

(d) Yes. The addition of H+ to the alkene gives the same Me3C+ intermediate.


  1. (A) CH3CH2—CHCl— CH(CH3)2


(C) CH3—CH2—CH(CH3)2




15. (A)








  1. CHCl3


  1. X – NH4NO3

Y – NH3


  1. A – CO2 B – H2O X – KHCO3 Y – K2CO3

Z – BaCO3


  1. Cation of  X     – Sn+2

Reactions – (i) Sn+2 + H2S ⎯→ SnS ↓ + 2H=

brown ppt.


(ii) Sn+2 + 2NaOH ⎯→ 2Na+ +  Sn(OH)2

        white ppt.

Sn(OH)2  + 2 NaOH  ⎯→ Na2SnO2 + 2H2O


(iii) Sn+2 + 2Fe+3 ⎯⎯→ Sn+4 + 2Fe+2








  1. A metallic chloride which does not respond test for Cl may be HgCl2

2HgCl2 + SnCl2  ⎯⎯→ SnCl4 + Hg2Cl2

(A) (B) (white)

Hg2Cl2 + SnCl2   ⎯⎯→  SnCl4 + 2Hg 


Hg2Cl2 + 2KI ⎯⎯→ HgI2 + 2KCl

KI + HgI2 ⎯⎯→ K2HgI4

Excess) (C) Scar let red [Nessler’s reagent)


  1.               ZnCO3(s) ZnO(s) + CO2 (g) 

(A) (C) (B)

ZnO + 2HCl ZnCl2 Zn2[Fe(CN)6

C (White ppt)

ZnCO3 + 2HCl  ⎯⎯→ ZnCl2 + CO2 + H2O

(A)   (B)

ZnCl2 ZnS


(White ppt) 

ZnCl2 Zn(OH)2    Na2ZnO2

(E)   (Sodium zincate soluble)


  1.             The alloy containing Cu and Ni form nitrates of these metals by dissolving in nitric acid Cu(NO3)2 gives black CuS with H2S in acid medium which dissolves in hot dilute HNO3 to form blue copper nitrate which with excess of NH3 gives deep blue cuprammonium complex ion [Cu(NH3)4]2+. Filtrate still contains dissolved H2S. when ammonia is added, NiS is formed (black). But if the H2S is boiled off then Ni2+ ion gives a rose red precipitate with  DMG in ammonical solution.