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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
The maximum number of isomer for an alkene with the molecularformula C4H8 is :
two
three
four
five
A
B
C
D
The maximum number of isomer for an alkene with the molecularformula C4H8 is :
two
three
four
five
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
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
Solution:
Only (I) and (II) are chiral compounds as all the four
attached substituents are different.
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 ?
Apply Coupon Code: HMCC
₹6,000/-(incl. of all taxes)
Visit: https://vdnt.in/JEECCE
Special Discount for this class
Link in Description
₹10,000