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Compound(s) that on hydrogenation produce(s) opticallyinactive compound(s) is (are)
JEE ADV. 2015
A B
C D
Compound(s) that on hydrogenation produce(s) opticallyinactive compound(s) is (are)
JEE ADV. 2015
A B
C D
Solution:
It involves nucleophilic substitution reaction (SN2 ) followed by
oxidation with oxidising agent and removal of water molecule.
A gaseous hydrocarbon gives upon combustion 0.72 g of waterand 3.08 g of CO2 . The empirical formula of the hydrocarbon is :
C2H4
C3H4
C6H5
C7H8
A
B
C
D
A gaseous hydrocarbon gives upon combustion 0.72 g of waterand 3.08 g of CO2 . The empirical formula of the hydrocarbon is :
C2H4
C3H4
C6H5
C7H8
A
B
C
D
25 g of an unknown hydrocarbon upon burning produces 88 g ofCO2 and 9 g of H2O. This unknown hydrocarbon contains :
20 g of carbon and 5 g of hydrogen
22 g of carbon and 3 g of hydrogen
24 g of carbon and 1 g of hydrogen
18 g of carbon and 7 g of hydrogen
A
B
C
D
JEE MAIN 2019
25 g of an unknown hydrocarbon upon burning produces 88 g ofCO2 and 9 g of H2O. This unknown hydrocarbon contains :
20 g of carbon and 5 g of hydrogen
22 g of carbon and 3 g of hydrogen
24 g of carbon and 1 g of hydrogen
18 g of carbon and 7 g of hydrogen
A
B
C
D
JEE MAIN 2019
Solution:
● Carboxylic acid is stronger acid than ammonium ion,
hence —COOH(X) is most acidic.
● Z (NH3+) is more acidic than Y (NH3
+) due to – I effect of
—COOH on Z.
Hence, overall acid strength order is X > Z > Y
Solution:
We know, a stronger acid produces its stable or weaker conjugate base. Here, CH(CN)3 produces the most stable conjugate base (NC)3C- . Stronger -R and -I effects of the CN- group, make the carbanion (conjugate base) very stable. The resonance hybrid structure of [(NC)3C]- is as follows:
However, halogen (X = Cl, Br, I) show -I effect but + R effect of halogens, destabilises the carbanion, X3C– (conjugate base of the haloform, HCX3 ).
Thus, CH(CN)3 is the strongest acid among the given options.
Solution:
Lone pair is not taking part in resonance, most basic. In
other cases, lone pair of nitrogen is part of delocalisation
which decreases Lewis base strength.
C2 is rotated anti-clockwise 120° about C2- C3 bond.The resulting conformer is :
A B
C Dgauche
partially eclipsed
staggered
eclipsed
2004
C2 is rotated anti-clockwise 120° about C2- C3 bond.The resulting conformer is :
A B
C Dgauche
partially eclipsed
staggered
eclipsed
2004
Out of the following the alkene that exhibits optical isomerism is :
3-methyl-2-pentene
4-methyl-1-pentene
3-methyl-1-pentene
2-methyl-2-pentene
A
B
C
D
Out of the following the alkene that exhibits optical isomerism is :
3-methyl-2-pentene
4-methyl-1-pentene
3-methyl-1-pentene
2-methyl-2-pentene
A
B
C
D
Which of the following will have least hindered rotation aboutcarbon-carbon bond ?
Ethane
Ethylene
Acetylene
Hexachloroethane
A
B
C
D
Which of the following will have least hindered rotation aboutcarbon-carbon bond ?
Ethane
Ethylene
Acetylene
Hexachloroethane
A
B
C
D
Solution:
Ethane has the smallest sized group (H) bonded
to carbons, hence there will be least hindered
rotation about C—C bond.
Solution:
are aromatic and stabilised by resonance. They follow Huckel’s rule.
Aromatic compounds are stable due to resonance while non-aromatics are not. According to Huckel’s rule (or 4n + 2 rule), “For a planar, cyclic compound to be aromatic, its π cloud must contain (4n + 2)π electrons, where, n is any whole number.” Thus,
is non-aromatic, hence, least stabilised by resonance.
The correct stability order for the following species is :
A B
C DII > I > IV > III
II > IV > I > III
I > III > II > IV
I > II > III > IV
The correct stability order for the following species is :
A B
C DII > I > IV > III
II > IV > I > III
I > III > II > IV
I > II > III > IV
Among the following four structure I and IV
It is true that :
A B
C Donly III is a chiralcompound
all four are chiralcompounds
only II and IV arechiral compounds
only I and II are chiralcompounds
Among the following four structure I and IV
It is true that :
A B
C Donly III is a chiralcompound
all four are chiralcompounds
only II and IV arechiral compounds
only I and II are chiralcompounds
Taking into account of hybridization and resonance effect, rank the following bonds in order of decreasing bond length.
A B
C DI > III > II
I > II = III
II = III = I
II > III > I
Taking into account of hybridization and resonance effect, rank the following bonds in order of decreasing bond length.
A B
C DI > III > II
I > II = III
II = III = I
II > III > I
Solution:
Taking resonance into consideration, all the bonds have partial double bond character. As seen by resonance, all the C−O bonds are identical and have the same bond length.∴ II = III = I [ order of bond length ]
Which of the following orders is not correct regarding −I -effect ofthe substituents?
-I < -Cl < -Br < -F
-NR3+ < -OR2
+
-NR3+ < -OR < -F
-SR < -OR < -OR2+
A
B
C
D
Which of the following orders is not correct regarding −I -effect ofthe substituents?
-I < -Cl < -Br < -F
-NR3+ < -OR2
+
-NR3+ < -OR < -F
-SR < -OR < -OR2+
A
B
C
D
Solution:
The correct order regarding —I effect of the
substituents is :
-I < -Cl < -Br < -F
These groups have the ability to withdraw electron
density toward themselves.
The correct order of acidity for the following compounds is :
A B
C DIII > IV > II > I
I > II > III > IV
I > III > IV > II
III > I > II > IV
JEE ADV. 2016
The correct order of acidity for the following compounds is :
A B
C DIII > IV > II > I
I > II > III > IV
I > III > IV > II
III > I > II > IV
JEE ADV. 2016
Solution:
● -OH group displays both kinds of effect; an electron withdrawing acid-strengthening
inductive effect from the meta-position and an electron-releasing acid weakening
resonance effect from the para-position (at this position, resonance effect overweighs
the inductive effect). Thus, III > IV.
● o-hydroxybenzoic acid (II) is far stronger than the corresponding meta and para
isomers as the carboxylate ion is stabilised by intramolecular H-bonding.
● 2,6-dihydroxybenzoic acid (I) forms carboxylate ion which is further stabilised by
intramolecular H-bonding, Thus, correct order is I > II > III > IV
The total number of contributing structures showinghyperconjugation (involving C — H bonds)for the following carbocation is :
2011
The total number of contributing structures showinghyperconjugation (involving C — H bonds)for the following carbocation is :
Ans : 6
2011
Solution:
Although the compound has two chiral carbons (indicated by stars), it does
not has four optically active isomers as expected. It is due to its existence in
cis-form only.
The above shown transformation does not exist due to restricted rotation
about the bridge head carbons, hence only cis-form and its mirror image
exist.
Solution:
It involves Electrophilic addition of alkenes, followed by Nucleophilic substitution mechanism.
The correct IUPAC name of the following compound is
2-methyl-5-nitro-1-chlorobenzene
3-chloro-1-methyl-1-nitrobenzene
2-chloro-1-methyl 1-4-nitrobenzene
5-chloro-4-methyl 1-1-nitrobenzene
A
B
C
D
JEE MAIN 2019
Solution:
The IUPAC name of the given compound is
2- chloro-1-methyl- 4-nitrobenzene
Here, the given compound contains two or more functional groups. So, the numbering is done in such a way that the sum of the locants is the lowest.
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How to Avail Discount ?
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Special Discount for this class
Link in Description
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