Some Basic Principles & Techniques Practice Questions - DPP for JEE

 Page 1


1. (a)
2. (b) The chemistry of the Lassaigne’s test of nitrogen is
3. (c) The atom or group which has more power to attract electrons in
comparision to hydrogen is said to have -I effect. Thus higher the
electronegativity of atom stronger will be the -I effect. As
electronegativity of N, O and F follow the order N < O < F hence 
based upon electronegative character order of-I effect is – NR
2
 < –
OR < – F.
4. (a) Carbinol is methyl  alcohol (CH
3
OH), hence vinylcarbinol should
be  (prop-2-enol)
5. (a)
Page 2


1. (a)
2. (b) The chemistry of the Lassaigne’s test of nitrogen is
3. (c) The atom or group which has more power to attract electrons in
comparision to hydrogen is said to have -I effect. Thus higher the
electronegativity of atom stronger will be the -I effect. As
electronegativity of N, O and F follow the order N < O < F hence 
based upon electronegative character order of-I effect is – NR
2
 < –
OR < – F.
4. (a) Carbinol is methyl  alcohol (CH
3
OH), hence vinylcarbinol should
be  (prop-2-enol)
5. (a)
6. (a) It is a bridge compound
1, 7, 7-trimethyl bicyclo [2. 2. 1] heptan-2-one
7. (a) Cyclohexane (iv) is non-planar and has chair conformation. In this
conformation, the bond angle is the normal tetrahedral angle (109°,
28’) and thus has no angle strain and hence is most stable. The rest
of the molecules are nearly planar and hence their stability depends
upon the angle strain in accordance with Baeyer’s strain theory.
Since cyclopropane has higher angle strain 
 than cyclopentane 
. Therefore cyclopentane  (iii) is more
stable than cyclopropane (i). Further, because of the presence of a
double bond in a three membered ring, cyclopropene  (ii) is the
least stable. Thus the order of stability is (iv) > (iii) > (i) >(ii).
8. (a) 
9. (b)
Page 3


1. (a)
2. (b) The chemistry of the Lassaigne’s test of nitrogen is
3. (c) The atom or group which has more power to attract electrons in
comparision to hydrogen is said to have -I effect. Thus higher the
electronegativity of atom stronger will be the -I effect. As
electronegativity of N, O and F follow the order N < O < F hence 
based upon electronegative character order of-I effect is – NR
2
 < –
OR < – F.
4. (a) Carbinol is methyl  alcohol (CH
3
OH), hence vinylcarbinol should
be  (prop-2-enol)
5. (a)
6. (a) It is a bridge compound
1, 7, 7-trimethyl bicyclo [2. 2. 1] heptan-2-one
7. (a) Cyclohexane (iv) is non-planar and has chair conformation. In this
conformation, the bond angle is the normal tetrahedral angle (109°,
28’) and thus has no angle strain and hence is most stable. The rest
of the molecules are nearly planar and hence their stability depends
upon the angle strain in accordance with Baeyer’s strain theory.
Since cyclopropane has higher angle strain 
 than cyclopentane 
. Therefore cyclopentane  (iii) is more
stable than cyclopropane (i). Further, because of the presence of a
double bond in a three membered ring, cyclopropene  (ii) is the
least stable. Thus the order of stability is (iv) > (iii) > (i) >(ii).
8. (a) 
9. (b)
Due to cis-addition of H
2
 to the triple bond, the reduced product
has a plane of symmetry and hence is optically inactive.
10. (d) The amount of s-character in various hybrid orbitals is as follows.
sp = 50%, sp
2
 = 33% and sp
3
 = 25%
Therefore s character of the C – H bond in acetylene (sp) is greater
than that of the C – H bond in alkene (sp
2
 hybridized) which in
turn has greater s character of the C – H bond than in alkanes.
Thus owing to a high s character of the C – H bond in alkynes, the
electrons constituting this bond are more strongly held by the
carbon nucleus with the result the hydrogen present on such a
carbon atom can be easily removed as proton. The acidic nature of
three types of C – H bonds follows the following order
Further, as we know that conjugate base of a strong acid is a weak
base, hence the correct order of basicity is
11. (b) is optically active
due to absence of plane of symmetry and center of symmetry.
12. (b) Hydrazine (NH
2
NH
2
) does not contain carbon and hence on fusion
Page 4


1. (a)
2. (b) The chemistry of the Lassaigne’s test of nitrogen is
3. (c) The atom or group which has more power to attract electrons in
comparision to hydrogen is said to have -I effect. Thus higher the
electronegativity of atom stronger will be the -I effect. As
electronegativity of N, O and F follow the order N < O < F hence 
based upon electronegative character order of-I effect is – NR
2
 < –
OR < – F.
4. (a) Carbinol is methyl  alcohol (CH
3
OH), hence vinylcarbinol should
be  (prop-2-enol)
5. (a)
6. (a) It is a bridge compound
1, 7, 7-trimethyl bicyclo [2. 2. 1] heptan-2-one
7. (a) Cyclohexane (iv) is non-planar and has chair conformation. In this
conformation, the bond angle is the normal tetrahedral angle (109°,
28’) and thus has no angle strain and hence is most stable. The rest
of the molecules are nearly planar and hence their stability depends
upon the angle strain in accordance with Baeyer’s strain theory.
Since cyclopropane has higher angle strain 
 than cyclopentane 
. Therefore cyclopentane  (iii) is more
stable than cyclopropane (i). Further, because of the presence of a
double bond in a three membered ring, cyclopropene  (ii) is the
least stable. Thus the order of stability is (iv) > (iii) > (i) >(ii).
8. (a) 
9. (b)
Due to cis-addition of H
2
 to the triple bond, the reduced product
has a plane of symmetry and hence is optically inactive.
10. (d) The amount of s-character in various hybrid orbitals is as follows.
sp = 50%, sp
2
 = 33% and sp
3
 = 25%
Therefore s character of the C – H bond in acetylene (sp) is greater
than that of the C – H bond in alkene (sp
2
 hybridized) which in
turn has greater s character of the C – H bond than in alkanes.
Thus owing to a high s character of the C – H bond in alkynes, the
electrons constituting this bond are more strongly held by the
carbon nucleus with the result the hydrogen present on such a
carbon atom can be easily removed as proton. The acidic nature of
three types of C – H bonds follows the following order
Further, as we know that conjugate base of a strong acid is a weak
base, hence the correct order of basicity is
11. (b) is optically active
due to absence of plane of symmetry and center of symmetry.
12. (b) Hydrazine (NH
2
NH
2
) does not contain carbon and hence on fusion
with Na metal, it cannot form NaCN; consequently hydrazine does
not show Lassaigne’s test for nitrogen.
13. (d) In staggered conformation any two hydrogen atoms on adjacent
carbon atoms are as far apart as possible there by minimising
repulsion between the electron clouds of s-bonds of two non-
bonded H-atomic (torsional strain)
Staggered form Eclipsed form
No torsional strain
14. (c) Steam distillation is the most suitable method of separation of 1 : 1
mixture of ortho and para nitrophenols as there is intramolecular
hydrogen bonding in o-nitrophenol.         
15. (c)  a double bond is formed
between C and Cl. Hence it is less reactive due to resonance    
16. (b) The resonance structures have same positions of nuclei and same
number of unpaired electrons.
17. (a) NO
2
 is activating group and CH
3
 and OCH
3
 are deactiving group.
Hence, the correct order of nucleophilic substitution reactions
18. (d) The stability of carbanions is affected due to resonance, inductive
Page 5


1. (a)
2. (b) The chemistry of the Lassaigne’s test of nitrogen is
3. (c) The atom or group which has more power to attract electrons in
comparision to hydrogen is said to have -I effect. Thus higher the
electronegativity of atom stronger will be the -I effect. As
electronegativity of N, O and F follow the order N < O < F hence 
based upon electronegative character order of-I effect is – NR
2
 < –
OR < – F.
4. (a) Carbinol is methyl  alcohol (CH
3
OH), hence vinylcarbinol should
be  (prop-2-enol)
5. (a)
6. (a) It is a bridge compound
1, 7, 7-trimethyl bicyclo [2. 2. 1] heptan-2-one
7. (a) Cyclohexane (iv) is non-planar and has chair conformation. In this
conformation, the bond angle is the normal tetrahedral angle (109°,
28’) and thus has no angle strain and hence is most stable. The rest
of the molecules are nearly planar and hence their stability depends
upon the angle strain in accordance with Baeyer’s strain theory.
Since cyclopropane has higher angle strain 
 than cyclopentane 
. Therefore cyclopentane  (iii) is more
stable than cyclopropane (i). Further, because of the presence of a
double bond in a three membered ring, cyclopropene  (ii) is the
least stable. Thus the order of stability is (iv) > (iii) > (i) >(ii).
8. (a) 
9. (b)
Due to cis-addition of H
2
 to the triple bond, the reduced product
has a plane of symmetry and hence is optically inactive.
10. (d) The amount of s-character in various hybrid orbitals is as follows.
sp = 50%, sp
2
 = 33% and sp
3
 = 25%
Therefore s character of the C – H bond in acetylene (sp) is greater
than that of the C – H bond in alkene (sp
2
 hybridized) which in
turn has greater s character of the C – H bond than in alkanes.
Thus owing to a high s character of the C – H bond in alkynes, the
electrons constituting this bond are more strongly held by the
carbon nucleus with the result the hydrogen present on such a
carbon atom can be easily removed as proton. The acidic nature of
three types of C – H bonds follows the following order
Further, as we know that conjugate base of a strong acid is a weak
base, hence the correct order of basicity is
11. (b) is optically active
due to absence of plane of symmetry and center of symmetry.
12. (b) Hydrazine (NH
2
NH
2
) does not contain carbon and hence on fusion
with Na metal, it cannot form NaCN; consequently hydrazine does
not show Lassaigne’s test for nitrogen.
13. (d) In staggered conformation any two hydrogen atoms on adjacent
carbon atoms are as far apart as possible there by minimising
repulsion between the electron clouds of s-bonds of two non-
bonded H-atomic (torsional strain)
Staggered form Eclipsed form
No torsional strain
14. (c) Steam distillation is the most suitable method of separation of 1 : 1
mixture of ortho and para nitrophenols as there is intramolecular
hydrogen bonding in o-nitrophenol.         
15. (c)  a double bond is formed
between C and Cl. Hence it is less reactive due to resonance    
16. (b) The resonance structures have same positions of nuclei and same
number of unpaired electrons.
17. (a) NO
2
 is activating group and CH
3
 and OCH
3
 are deactiving group.
Hence, the correct order of nucleophilic substitution reactions
18. (d) The stability of carbanions is affected due to resonance, inductive
effect and s - character of orbitals. Greater the number  of groups
having + I group (alkyl group) lesser stable would be the carbanion.
Further stability of carbanion decreases with decrease in s-character.
Benzene carbanions are stablized due to resonance, hence the
correct order is
= 
The correct order of stability of given carbanion is in the order I > III > II
> IV.
19. (c) Sterioisomerism involve those isomers which contain same ligands
in their co-ordination spheres but differ in the arrangement of these
ligands in space. Stereo-isomerism is of two type geomerical
isomerism and optical isomerism. In geomerical isomerism ligands
occupy different positions around the central metal atom or ion.
In optical isomerism isomers have same formula but differ in their ability
to rotate directions of the plane of polarised light.
20. (b)
21. (16.76) Wt. of organic substance = 0.25 g
V
1
 = 40 mL, T
1
 = 300 K
P
1
 = 725 – 25 = 700 mm of Hg
P
2
 = 760 mm of Hg (at STP)
T
2
 = 273 K