BADHANEDUCATION D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
1. How do you account for the formation of ethane during chlorination of methane?
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Solution:
The formation of ethane is due to the side reaction in termination step by the combination of two
CH3 free radicals.
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2. Write IUPAC names of the following compounds:
(e) 2-Methyl phenol
(f) 5-(2-Methyl propyl)-decane
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(d) 4-Phenyl-but-1-ene
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(c) Buta 1, 3-diene.
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(b) Pent-1-ene-3-yne
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Solution:
(a) 2-Methyl - but-2-ene
(g) 4-Ethyl-deca-1, 5, 8-triene.
3. For the following compounds, write structural formulas and IUPAC names for all
possible isomers having the number of double or triple bond as indicated :
(a) C4H8 (one double bond)
(b) C5H8 (one triple bond)
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CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in DELHI AND OTHER MAJOR CITIES
EMAIL:1 [email protected] web site www.badhaneducation.in
BADHANEDUCATION D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
Solution:
(a) (i) CH2 = CH - CH2 - CH3 But - 1-ene
(ii) CH3 - CH2 = CH - CH3 But - 2-ene
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2-Methyl propene
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(iii)
(b) (i) HC
(ii) CH3 - C
C - CH2 - CH2 - CH2 Pent - 1-yne
C - CH2 - CH2 Pent - 2-yne
(iii)
3-Methyl-but- 1-yne.
4. Write IUPAC names of the products obtained by the ozonolysis of the following
compounds :
(i) Pent-2-ene
(ii) 3,4-Dimethyl-hept-3-ene
(iii) 2-Ethylbut-1-ene
(iv) 1-Phenylbut-1-ene
(ii) Butan-2-one and pentan-2-one.
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(iv) Propanal and benzaldehyde.
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(iii) Methanal and pentan-3-one.
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Solution:
(i) Ethanal and propanal.
5. An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan-3-one. Write the
and IUPAC name of ‘A’.
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Solution:
The IUPAC name of A is 3-Ethyl-pent-2-ene.
6. An alkene ‘A’ contains three C – C, eight C – H bonds and one C – C p bond. ‘A’ on
ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of ‘A’.
Solution:
The IUPAC name of A is But – 2 –ene
7. Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the
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CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in DELHI AND OTHER MAJOR CITIES
EMAIL:2 [email protected] web site www.badhaneducation.in
BADHANEDUCATION D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
structural formula of the alkene?
Solution:
4-Ethyl-hex-3-ene
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8. Write chemical equations for combustion reaction of the following hydrocarbons:
(i) Butane
(ii) Pentene
(iii) Hexyne
(iv) Toluene
Solution:
(a) C4H10(g) +
4CO2(g) +5H2O(g)
(b) C5H10(g) +
5CO2(g) +5H2O(g)
(c) C6H10(g) +
6CO2(g) +5H2O(g)
(d) C7H8(g) +5O2(g)
7CO2(g) +4H2O(g).
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Solution:
The cis and trans structures of hex-2-ene
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9. Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p.
and why?
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The cis form will have higher boiling point . This is due to more polar nature leading to strong
intermolecular dipole-dipole interaction, thus requiring more heat energy to separate them.
10. Why is benzene extra ordinarily stable though it contains three double bonds?
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CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in DELHI AND OTHER MAJOR CITIES
EMAIL:3 [email protected] web site www.badhaneducation.in
BADHANEDUCATION D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
Solution:
Benzene is extra ordinarily stable due to resonance.
11. What are the necessary conditions for any system to be aromatic?
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Solution:
The essential conditions for any system to be aromatic are Planar, conjugated ring system with
delocalisation of (4π + 2) p electrons, where n is an integer.
12. Explain why the following systems are not aromatic?
Solution:
The compounds are not aromatic due to lack of cyclic cloud p-electrons having (4n + 2) p electrons.
13. How will you convert benzene into
(i) p-nitrobromobenzene
(ii) m- nitrochlorobenzene
(iii) p - nitrotoluene
(iv) acetophenone?
Solution:
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CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in DELHI AND OTHER MAJOR CITIES
EMAIL:4 [email protected] web site www.badhaneducation.in
BADHANEDUCATION D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
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14. In the alkane H3C – CH2 – C(CH3)2 – CH2 – CH(CH3)2, identify 1°,2°,3° carbon atoms
and give the number of H atoms bonded to each one of these.
Solution:
6H attached to 10 carbons
4H attached to 20 carbons
1H attached to 30 carbons
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15. What effect does branching of an alkane chain has on its boiling point?
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Solution:
The boiling point will be lower if there are more branching in alkane.
Solution:
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16. Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does
the result support Kekulé structure for benzene?
The result tells us that all the three products cannot be obtained by any one of the Kekule’s
structures showing that benzene is a resonance hybrid of the two resonating structure.
17. Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also
give reason for this behaviour.
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CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in DELHI AND OTHER MAJOR CITIES
EMAIL:5 [email protected] web site www.badhaneducation.in
BADHANEDUCATION D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
Solution:
H – C – C – H > C6H6 > C6H14
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The acidic property is maximum in ethyne due to maximum s orbital character in ethyne (50
percent) as compared to 33 percent in benzene and 25 percent in n-hexane.
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18. Why does benzene undergo electrophilic substitution reactions easily and
nucleophilic substitutions with difficulty?
Solution:
Due to the presence of p -electrons in benzene it can undergo electrophilic substitution reactions
easily. In Nucleophilic substitutions the nucleophile will be repelled by p -electrons. So benzene
undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty.
19. How would you convert the following compounds into benzene?
(i) Ethyne (ii) Ethene (iii) Hexane
Solution:
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Solution:
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20. Write structures of all the alkenes which on hydrogenation give 2-methylbutane.
21. Out of benzene, m–dinitrobenzene and toluene which will undergo nitration most
easily and why?
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CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in DELHI AND OTHER MAJOR CITIES
EMAIL:6 [email protected] web site www.badhaneducation.in
BADHANEDUCATION D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
CONTACT FOR ADMISSION GUIDANCE B.TECH BBA BCA, MCA MBA DIPLOMA AND OTHER COURSES 09810144315
Solution:
Due to electron releasing nature of the methyl group toluene undergoes nitration most easily.
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22. Suggest the name of a Lewis acid other than anhydrous aluminium chloride which
can be used during ethylation of benzene.
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Solution:
Another Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of
benzene is FeCl3 .
23. Why is Wurtz reaction not preferred for the preparation of alkanes containing odd
number of carbon atoms? Illustrate your answer bytaking one example.
Solution:
Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms
because of the formation of side products. For example, by starting with 1-bromo propane and 1bromo butane, hexane and octane are the side products besides heptane.
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CONTACT FOR MATHEMATICS GROUP/HOME COACHING FOR CLASS 11TH 12TH BBA,BCA,
DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in DELHI AND OTHER MAJOR CITIES
EMAIL:7 [email protected] web site www.badhaneducation.in