b) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.
b) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.
b) Wurtz reaction
b) Wurtz reaction
a) sp 3 – s and sp 3 – sp 3
a) sp 3 – s and sp 3 – sp 3
a) 2 – methyl pentane
a) 2 – methyl pentane
c) C 2 H 6 and CO 2
c) C 2 H 6 and CO 2
b) C n H 2n
b) C n H 2n
a) C 3 H 6
a) C 3 H 6
c) H 2 C = C = O
c) H 2 C = C = O
d) 2, 3 – Dimethyl but – 2 – ene
d) 2, 3 – Dimethyl but – 2 – ene
a) 2 – methylbut – 2- ene
a) 2 – methylbut – 2- ene
a) 2 – chloro – 1 – iode – 2 – methyl propane
a) 2 – chloro – 1 – iode – 2 – methyl propane
a) trans – 2- chloro – 3- iodo – 2- pentane
a) trans – 2- chloro – 3- iodo – 2- pentane
c) configurational isomers
c) configurational isomers
c) alc. KOH
c) alc. KOH
d) slower
d) slower
d) both (a) and (b)
d) both (a) and (b)
a) Nitro benzene
a) Nitro benzene
b) – NO 2
b) – NO 2
d) Isopropyl chloride
d) Isopropyl chloride
b) action of sodium metal on iodomethane
b) action of sodium metal on iodomethane
a) C 8 H 18
a) C 8 H 18
a) Formaldehyde
a) Formaldehyde
c) pent – 1 – ene
c) pent – 1 – ene
d) 2, 2, 3, 3 – tetra chloro butane
II. Write brief answer to the following questions:
d) 2, 2, 3, 3 – tetra chloro butane
II. Write brief answer to the following questions:
1)
2)
3)
4) ethyl isopropyl acetylene
5)
1)
2)
3)
4) ethyl isopropyl acetylene
5)
All the activating groups are ‘ortho-para’ directors.
Example: – OH, – NH 2, -NHR, -NHCOCH 3, -OCH 3 -CH 3 – C 2 H 5 etc.
Let us consider the directive influences of phenolic (-OH) group. Phenol is the resonance hybrid of following structures.
In these resonance structures, the (-) charge residue is present on ortho and para position of ring structure. It is quite evident that the lone pair of electron on the atom which is attached to the ring involves in resonance and makes the ring more electron rich than benzene. The electron density at ortho and para positions increases as compared to the meta position. Therefore phenolic group activates the benzene ring for electrophilic attack at ‘ortho’ and ‘para positions and hence -OH group is an ortho-para director and activator.
In aryl halides, the strong -I effect of the halogens (electron withdrawing tendency) decreases the electron density of benzene ring, thereby deactivating for electrophilic attack. However the presence of lone pair on halogens involved in the resonance with pi electrons of benzene ring, increases electron density at ortho and para position. Hence the halogen group is an ortho-para director and deactivator.
All the activating groups are ‘ortho-para’ directors.
Example: – OH, – NH 2, -NHR, -NHCOCH 3, -OCH 3 -CH 3 – C 2 H 5 etc.
Let us consider the directive influences of phenolic (-OH) group. Phenol is the resonance hybrid of following structures.
In these resonance structures, the (-) charge residue is present on ortho and para position of ring structure. It is quite evident that the lone pair of electron on the atom which is attached to the ring involves in resonance and makes the ring more electron rich than benzene. The electron density at ortho and para positions increases as compared to the meta position. Therefore phenolic group activates the benzene ring for electrophilic attack at ‘ortho’ and ‘para positions and hence -OH group is an ortho-para director and activator.
In aryl halides, the strong -I effect of the halogens (electron withdrawing tendency) decreases the electron density of benzene ring, thereby deactivating for electrophilic attack. However the presence of lone pair on halogens involved in the resonance with pi electrons of benzene ring, increases electron density at ortho and para position. Hence the halogen group is an ortho-para director and deactivator.
It is prepared by the action of a mixture of con. HNO 3 and con. H 2 SO 4 (nitrating mixture) on benzene maintaining the temperature below 333 K.
Nitration:
Sulphuric acid generates the electrophile – NO 2 +, nitronium ion-from nitric acid. This is an example of aromatic electrophilic substitution reaction.
The generation of nitronium ion
H 2 SO 4 + HONO 2 →
+ HSO 4 –
To the nitronium ion (being an electron deficient species) the π bond of benzene, donates a pair of electrons forming a 6-bond. A species with a + ve charge is formed as an intermediate. This is called ‘arenium ion’ and is stabilised by Resonance.
In the last step, the hydrogen atom attached to the carbon carrying the nitro group is pulled out as a proton, by the Lewis base HSO 4 –, so that stable aromatic system is formed.
It is prepared by the action of a mixture of con. HNO 3 and con. H 2 SO 4 (nitrating mixture) on benzene maintaining the temperature below 333 K.
Nitration:
Sulphuric acid generates the electrophile – NO 2 +, nitronium ion-from nitric acid. This is an example of aromatic electrophilic substitution reaction.
The generation of nitronium ion
H 2 SO 4 + HONO 2 →
+ HSO 4 –
To the nitronium ion (being an electron deficient species) the π bond of benzene, donates a pair of electrons forming a 6-bond. A species with a + ve charge is formed as an intermediate. This is called ‘arenium ion’ and is stabilised by Resonance.
In the last step, the hydrogen atom attached to the carbon carrying the nitro group is pulled out as a proton, by the Lewis base HSO 4 –, so that stable aromatic system is formed.
In 1865, August Kekule suggested that benzene consists of a cyclic planar structure of six carbon with alternate single and double bonds.
There were two objections:
Benzene forms only one ortho disubstituted products whereas the Kekule’s structure predicts two o-di substituted products as shown below.
Kekule’s structure failed to explain why benzene with three double bonds did not give addition reactions like other alkenes.To overcome this objection, Kekule suggested that benzene was mixture of two forms (1 and 2) which are in rapid equilibrium.
Resonance description of benzene:
The phenomenon in which two or more structures can be written for a substance which has identical position of atoms is called resonance. The actual structure of the molecule is said to be resonance hybrid of various possible alternative structures. In benzene, Kekule’s structures I & II represented the resonance structure, and structure III is the resonance hybrid of structure I & II.
The structures 1 and 2 exist only in theory. The actual structure of benzene is the hybrid of two hypothetical resonance structures.
Spectroscopic measurements:
Spectroscopic measurements show that benzene is planar and all of its carbon- carbon bonds are of equal length 1.40 Å. This value lies between carbon-carbon single bond length 1.54 Å and carbon- carbon double bond length 1.34 Å.
Molecular orbital structure:
The structure of benzene is best de¬scribed in terms of the molecular orbital theory. All the six carbon atoms of benzene are sp 2 hybridized. Six sp 2 hybrid orbitals of carbon linearly overlap with six 1 s orbitals of hydrogen atoms to form six C – H sigma bonds. Overlap between the remaining sp 2 hybrid orbitals of carbon forms six C – C sigma bonds.
Formation of Sigma bond in benzene
All the σ bonds in benzene lie in one plane with bond angle 120°. Each carbon atom in benzene possess an un hybridized p-orbital containing one electron. The lateral overlap of their p-orbital produces 3 π- bond. The six electrons of the p-orbitals cover all the six carbon atoms and are said to be delocalised. Due to delocalization, strong π-bond is formed which makes the molecule stable. Hence unlike alkenes and alkynes benzene undergoes substitution reactions rather addition reactions under normal conditions.
All carbon atoms have the delocalized π MO is formed by p-orbitais the overlap of six p-orbitals.
Representation of benzene:
Hence, there are three ways in which benzene can be represented.
In 1865, August Kekule suggested that benzene consists of a cyclic planar structure of six carbon with alternate single and double bonds.
There were two objections:
Benzene forms only one ortho disubstituted products whereas the Kekule’s structure predicts two o-di substituted products as shown below.
Kekule’s structure failed to explain why benzene with three double bonds did not give addition reactions like other alkenes.To overcome this objection, Kekule suggested that benzene was mixture of two forms (1 and 2) which are in rapid equilibrium.
Resonance description of benzene:
The phenomenon in which two or more structures can be written for a substance which has identical position of atoms is called resonance. The actual structure of the molecule is said to be resonance hybrid of various possible alternative structures. In benzene, Kekule’s structures I & II represented the resonance structure, and structure III is the resonance hybrid of structure I & II.
The structures 1 and 2 exist only in theory. The actual structure of benzene is the hybrid of two hypothetical resonance structures.
Spectroscopic measurements:
Spectroscopic measurements show that benzene is planar and all of its carbon- carbon bonds are of equal length 1.40 Å. This value lies between carbon-carbon single bond length 1.54 Å and carbon- carbon double bond length 1.34 Å.
Molecular orbital structure:
The structure of benzene is best de¬scribed in terms of the molecular orbital theory. All the six carbon atoms of benzene are sp 2 hybridized. Six sp 2 hybrid orbitals of carbon linearly overlap with six 1 s orbitals of hydrogen atoms to form six C – H sigma bonds. Overlap between the remaining sp 2 hybrid orbitals of carbon forms six C – C sigma bonds.
Formation of Sigma bond in benzene
All the σ bonds in benzene lie in one plane with bond angle 120°. Each carbon atom in benzene possess an un hybridized p-orbital containing one electron. The lateral overlap of their p-orbital produces 3 π- bond. The six electrons of the p-orbitals cover all the six carbon atoms and are said to be delocalised. Due to delocalization, strong π-bond is formed which makes the molecule stable. Hence unlike alkenes and alkynes benzene undergoes substitution reactions rather addition reactions under normal conditions.
All carbon atoms have the delocalized π MO is formed by p-orbitais the overlap of six p-orbitals.
Representation of benzene:
Hence, there are three ways in which benzene can be represented.
1) 3 – chloro nitrobenzene
2) 4 – chlorotoluene
3) Bromo benzene
4) m – dinitro benzene
1) 3 – chloro nitrobenzene
2) 4 – chlorotoluene
3) Bromo benzene
4) m – dinitro benzene
Propene decolourises Br 2 /H 2 O it forms dibromo compound but propane does not react with Br 2 / H 2 O.
Propene decolourises Br 2 /H 2 O it forms dibromo compound but propane does not react with Br 2 / H 2 O.
(i) ethane
(ii) n – butane
(i) ethane
(ii) n – butane
Conformations of n-Butane:
n-Butane may be considered as a derivative of ethane, as one hydrogen on each carbon is replaced by a methyl group
Eclipsed conformation:
In this conformation, the distance between the two methyl group is minimum. So there is maximum repulsion between them and it is the least stable conformer.
Anti or staggered form:
In this conformation, the distance between the two methyl groups is maximum and so there is minimum repulsion between them. And it is the most stable conformer.
The following potentially energy diagram shows the relative stabilities of various conformers of n-butane.
Conformations of n-Butane:
n-Butane may be considered as a derivative of ethane, as one hydrogen on each carbon is replaced by a methyl group
Eclipsed conformation:
In this conformation, the distance between the two methyl group is minimum. So there is maximum repulsion between them and it is the least stable conformer.
Anti or staggered form:
In this conformation, the distance between the two methyl groups is maximum and so there is minimum repulsion between them. And it is the most stable conformer.
The following potentially energy diagram shows the relative stabilities of various conformers of n-butane.
C 3 H 8 (g) + 5O 2 (g) ) → 3CO 2 (g) + 4H 2 O(l)
propane
C 3 H 8 (g) + 5O 2 (g) ) → 3CO 2 (g) + 4H 2 O(l)
propane
Markovnikoff ‘s rule:
“When an unsymmetrical alkene reacts with hydrogen halide, the hydrogen adds to the carbon that has more number of hydrogen and halogen add to the carbon having fewer hydrogen”. This rule can also be stated as in the addition reaction of alkene/alkyne, the most electro negative part of the reagent adds on to the least hydrogen attached doubly bonded carbon.
Addition of water: (Hydration of alkenes)
Normally, water does not react with alkenes. In the presence of concentrated sulphuric acid, alkenes react with water to form alcohols. This reaction follows carbocation mechanism and Markovnikoff’s rule.
Markovnikoff ‘s rule:
“When an unsymmetrical alkene reacts with hydrogen halide, the hydrogen adds to the carbon that has more number of hydrogen and halogen add to the carbon having fewer hydrogen”. This rule can also be stated as in the addition reaction of alkene/alkyne, the most electro negative part of the reagent adds on to the least hydrogen attached doubly bonded carbon.
Addition of water: (Hydration of alkenes)
Normally, water does not react with alkenes. In the presence of concentrated sulphuric acid, alkenes react with water to form alcohols. This reaction follows carbocation mechanism and Markovnikoff’s rule.
CH 2 = CH + H 2 O + (O)
CH 2 OH – CH 2 OH
Ethylene Ethylene glycol
CH 2 = CH + H 2 O + (O)
CH 2 OH – CH 2 OH
Ethylene Ethylene glycol
1) 2, 3 – Dimethyl – 6 – (2 – methyl propyl) decane
2) 5 – (2 – Ethyl butyl ) – 3, 3, – dimethyldecane
3) 5 – (1, 2 – Dimethyl propyl) – 2 – methylnonane
1) 2, 3 – Dimethyl – 6 – (2 – methyl propyl) decane
2) 5 – (2 – Ethyl butyl ) – 3, 3, – dimethyldecane
3) 5 – (1, 2 – Dimethyl propyl) – 2 – methylnonane
i) 2 – butyne
ii) CH 2 = CH 2
iii)
iv) CaC 2
CaC 2 + 2H 2 O → Ca(OH) 2 + C 2 H 2 ↑
Calcium Carbide Acetylene
i) 2 – butyne
ii) CH 2 = CH 2
iii)
iv) CaC 2
CaC 2 + 2H 2 O → Ca(OH) 2 + C 2 H 2 ↑
Calcium Carbide Acetylene
1 – butyne reacts with ammoniacal AgNO 3 solution it forms white precipitate of silver acetylide but, 2 – butyne doesnot reacts with ammoniacal AgNO 3 solution.
1 – butyne reacts with ammoniacal AgNO 3 solution it forms white precipitate of silver acetylide but, 2 – butyne doesnot reacts with ammoniacal AgNO 3 solution.