Q.1. Write structures of different chain isomers of alkanes corresponding to the molecular
formula C4H10 and C5 H12. Also write their IUPAC names.
Q.2. Write IUPAC names of the following compounds :
(i) (CH3)3CCH2C(CH3) 3
(ii) (CH3) 2 C(C2H5) 2
(iii) tetra – tert-butylmethane
Q.3. Write structural formulas of the following compounds:
(i) 3-Methylhexane
(ii) 3-ethyl-2, 2–dimethylpentane
Q.4. Define conformers/rotamers.
Ans.The spatial arrangements of atoms which can be converted into one another by
rotation around a C-C single bond are called conformations or conformers or rotamers
Q.5. Write IUPAC names of the following compounds:
(i) CH3 – CH = CH2
(ii) CH3 – CH2 – CH = CH2
(iii) CH3 – CH = CH–CH3
Q.6. Out of pentane, 2-methylbutane and 2,2-dimethylpropane which has the highest
boiling point and why?
Ans. Pentane having a continuous chain of five carbon atoms has the highest boiling point
(309.1K) whereas 2,2 – dimethylpropane boils at 282.5K. With increase in number of
branched chains, the molecule attains the shape of a sphere. This results in smaller area of
contact and therefore weak intermolecular forces between spherical molecules, which are
overcome at relatively lower temperatures.
Q.7. Calculate number of sigma (σ) and pi (π) bonds in the following
(i) CH2 = C (CH2CH2CH3) 2
(ii) CH2 = CH – CH – CH3
|
CH3
(iii) CH3 – CH = CH–CH3
Q.8. Write structures and IUPAC names of different structural isomers of alkenes
corresponding to C5H10
Ans.pent-1-ene
pent-2-ene
2-methylbut-2-ene
3-methylbut-1-ene
2-methyl-but-1-ene
Q.9. Discuss about structural isomerism and stereoisomerism.
Ans. Structural Isomerism: Compounds having the same molecular formula but different
structure i.e. different arrangement of atoms within the molecule are called structural
isomers and the phenomenon is called as structural isomerism.
Types:
• Chain isomerism
• Position isomerism
• Functional isomerism
• Metamerism
• Tautomerism
Stereoisomerism: Isomers which have the same structural formula but have
different relative arrangement of atoms or groups in space are called stereoisomers and the
phenomenon is called as stereoisomerism.
Q.10. Draw cis and trans isomers of the following compounds. Also write their IUPAC names
:(i) CHCl = CHCl
(ii) C2H5CCH3 = CCH3C2H5
Q.11. Write IUPAC names of the products obtained by addition reactions of HBr to hex-1-ene
(i) in the absence of peroxide and
(ii) in the presence of peroxide.
Q.12. Define Hückel rule.
Ans. Huckel Rule gives information about the aromaticity. According to this rule, the
necessary and sufficient conditions for a molecule to be aromatic are:
Planarity
Complete decolisation of the pi electrons in the ring.
Presence of (4n+2) pi electrons in the ring where n is an integer (n= 0,1,2…)
Q.13How would you convert the following into benzene?
(i)Ethyne (ii) Ethene (iii) Hexane
Q.14. Out of benzene, m–dinitrobenzene and toluene which will undergo
nitration most easily and why?
Ans. -CH3 group is electron donating while –NO2 group is electron withdrawing. Therefore,
maximum electron density will be in toluene , followed by benzene and at least in m-dinitrobenzene. Therefore, the ease of nitration decreases in the order:
Toluene> benzene> m-dinitrobenzene
Q.15. Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also
give reason for this behaviour.
Benzene Hexane Ethyne
sp
3
sp
2
sp
Since s-electrons are close to the nucleus, therefore as the s-character of the orbital making
the C-H bond increases, the electrons of C-H bond lie closer and closer to the carbon atom.
In other words, the partial positive charge on the H-atom and hence the acidic character
increases as the s character of the orbital increases.
Thus the acidic character decreases in the order:
Ethyne> Benzene> Hexane
Q.16.Why does benzene undergo electrophilic substitution reactions easily and nucleophilic
substitutions with difficulty?
Ans. Due to the presence of an electron cloud containing 6 pi electrons above and below the
plane of the ring, benzene is a source of electrons. Consequently, it attracts the
electrophiles towards it and repels nucleophiles. As a result, benzene undergoes
electrophilic substitution reactions easily and nucleophilic substitution with difficulty.
Q.17. Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and
why?
Ans. cis-Hex-2-ene & Trans-Hex-2-ene
The dipole moment of a molecule depends upon dipole-dipole interactions. Since cis-isomer
has higher dipole moment, therefore it has higher boiling point.
Q.18. Explain the extra ordinarily stability of benzene though it contains three double
bonds?
Ans. Resonance or delocalisation of electrons usually leads to stability. Since in benzene all
the six pi electrons of the three double bonds are completely delocalised to form one lowest
molecular orbital which surrounds all the carbon atoms of the ring, therefore, it is extra-ordinarily stable.
Q.19. What are the necessary conditions for any system to be aromatic?
Ans. The necessary conditions for a molecule to be aromatic are:
(i)It should have a single cyclic cloud of delocalised pi electrons above and below the plane of
the molecule.
(ii)It should be planar. This is because complete delocalisation of pi electrons is possible only if
the ring is planar to allow cyclic overlap of p-orbitals.
(iii)It should contain Huckel number of electrons, i.e. (4n+2) pi electrons where n=0,1,2,3….etc.
formula C4H10 and C5 H12. Also write their IUPAC names.
Q.2. Write IUPAC names of the following compounds :
(i) (CH3)3CCH2C(CH3) 3
(ii) (CH3) 2 C(C2H5) 2
(iii) tetra – tert-butylmethane
Q.3. Write structural formulas of the following compounds:
(i) 3-Methylhexane
(ii) 3-ethyl-2, 2–dimethylpentane
Q.4. Define conformers/rotamers.
Ans.The spatial arrangements of atoms which can be converted into one another by
rotation around a C-C single bond are called conformations or conformers or rotamers
Q.5. Write IUPAC names of the following compounds:
(i) CH3 – CH = CH2
(ii) CH3 – CH2 – CH = CH2
(iii) CH3 – CH = CH–CH3
Q.6. Out of pentane, 2-methylbutane and 2,2-dimethylpropane which has the highest
boiling point and why?
Ans. Pentane having a continuous chain of five carbon atoms has the highest boiling point
(309.1K) whereas 2,2 – dimethylpropane boils at 282.5K. With increase in number of
branched chains, the molecule attains the shape of a sphere. This results in smaller area of
contact and therefore weak intermolecular forces between spherical molecules, which are
overcome at relatively lower temperatures.
Q.7. Calculate number of sigma (σ) and pi (π) bonds in the following
(i) CH2 = C (CH2CH2CH3) 2
(ii) CH2 = CH – CH – CH3
|
CH3
(iii) CH3 – CH = CH–CH3
Q.8. Write structures and IUPAC names of different structural isomers of alkenes
corresponding to C5H10
Ans.pent-1-ene
pent-2-ene
2-methylbut-2-ene
3-methylbut-1-ene
2-methyl-but-1-ene
Q.9. Discuss about structural isomerism and stereoisomerism.
Ans. Structural Isomerism: Compounds having the same molecular formula but different
structure i.e. different arrangement of atoms within the molecule are called structural
isomers and the phenomenon is called as structural isomerism.
Types:
• Chain isomerism
• Position isomerism
• Functional isomerism
• Metamerism
• Tautomerism
Stereoisomerism: Isomers which have the same structural formula but have
different relative arrangement of atoms or groups in space are called stereoisomers and the
phenomenon is called as stereoisomerism.
Q.10. Draw cis and trans isomers of the following compounds. Also write their IUPAC names
:(i) CHCl = CHCl
(ii) C2H5CCH3 = CCH3C2H5
Q.11. Write IUPAC names of the products obtained by addition reactions of HBr to hex-1-ene
(i) in the absence of peroxide and
(ii) in the presence of peroxide.
Q.12. Define Hückel rule.
Ans. Huckel Rule gives information about the aromaticity. According to this rule, the
necessary and sufficient conditions for a molecule to be aromatic are:
Planarity
Complete decolisation of the pi electrons in the ring.
Presence of (4n+2) pi electrons in the ring where n is an integer (n= 0,1,2…)
Q.13How would you convert the following into benzene?
(i)Ethyne (ii) Ethene (iii) Hexane
Q.14. Out of benzene, m–dinitrobenzene and toluene which will undergo
nitration most easily and why?
Ans. -CH3 group is electron donating while –NO2 group is electron withdrawing. Therefore,
maximum electron density will be in toluene , followed by benzene and at least in m-dinitrobenzene. Therefore, the ease of nitration decreases in the order:
Toluene> benzene> m-dinitrobenzene
Q.15. Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also
give reason for this behaviour.
Benzene Hexane Ethyne
sp
3
sp
2
sp
Since s-electrons are close to the nucleus, therefore as the s-character of the orbital making
the C-H bond increases, the electrons of C-H bond lie closer and closer to the carbon atom.
In other words, the partial positive charge on the H-atom and hence the acidic character
increases as the s character of the orbital increases.
Thus the acidic character decreases in the order:
Ethyne> Benzene> Hexane
Q.16.Why does benzene undergo electrophilic substitution reactions easily and nucleophilic
substitutions with difficulty?
Ans. Due to the presence of an electron cloud containing 6 pi electrons above and below the
plane of the ring, benzene is a source of electrons. Consequently, it attracts the
electrophiles towards it and repels nucleophiles. As a result, benzene undergoes
electrophilic substitution reactions easily and nucleophilic substitution with difficulty.
Q.17. Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and
why?
Ans. cis-Hex-2-ene & Trans-Hex-2-ene
The dipole moment of a molecule depends upon dipole-dipole interactions. Since cis-isomer
has higher dipole moment, therefore it has higher boiling point.
Q.18. Explain the extra ordinarily stability of benzene though it contains three double
bonds?
Ans. Resonance or delocalisation of electrons usually leads to stability. Since in benzene all
the six pi electrons of the three double bonds are completely delocalised to form one lowest
molecular orbital which surrounds all the carbon atoms of the ring, therefore, it is extra-ordinarily stable.
Q.19. What are the necessary conditions for any system to be aromatic?
Ans. The necessary conditions for a molecule to be aromatic are:
(i)It should have a single cyclic cloud of delocalised pi electrons above and below the plane of
the molecule.
(ii)It should be planar. This is because complete delocalisation of pi electrons is possible only if
the ring is planar to allow cyclic overlap of p-orbitals.
(iii)It should contain Huckel number of electrons, i.e. (4n+2) pi electrons where n=0,1,2,3….etc.
No comments:
Post a Comment