Q. Which of the following groups is a strong +M (mesomeric) director?
A.
–NO2
B.
–OH
C.
–COOH
D.
–CN
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Solution
The –OH group is a strong +M director as it donates electron density through resonance.
Correct Answer: B — –OH
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Q. Which of the following groups is a strong -I group?
A.
-F
B.
-CH3
C.
-OCH3
D.
-Br
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Solution
The -F group is a strong -I group due to its high electronegativity.
Correct Answer: A — -F
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Q. Which of the following groups is a strong -M group?
A.
-OH
B.
-OCH3
C.
-NO2
D.
-NH2
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Solution
-NO2 is a strong -M group as it withdraws electron density through resonance.
Correct Answer: C — -NO2
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Q. Which of the following groups is a strong electron-donating group?
A.
–NO2
B.
–OH
C.
–CN
D.
–COOH
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Solution
The –OH group is a strong electron-donating group through resonance (+M effect).
Correct Answer: B — –OH
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Q. Which of the following groups is a weak +M director?
A.
-OH
B.
-OCH3
C.
-NH2
D.
-CH3
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Solution
-CH3 is a weak +M director as it can donate some electron density but is less effective than -OH or -OCH3.
Correct Answer: D — -CH3
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Q. Which of the following groups is a weak -I effect group?
A.
–F
B.
–Cl
C.
–Br
D.
–I
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Solution
The –I effect decreases in strength down the group in the halogens, making –I the weakest -I group.
Correct Answer: D — –I
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Q. Which of the following groups is a weak -I effect?
A.
-CH3
B.
-Cl
C.
-Br
D.
-I
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Solution
The -CH3 group exhibits a weak -I effect as it is a weak electron-donating group.
Correct Answer: A — -CH3
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Q. Which of the following groups is a weak -M group?
A.
-CHO
B.
-COOH
C.
-NO2
D.
-F
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Solution
-F is a weak -M group as it has a slight electron-withdrawing effect but is not as strong as -CHO or -COOH.
Correct Answer: D — -F
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Q. Which of the following groups is considered a weak -I group?
A.
-NO2
B.
-CN
C.
-F
D.
-COOH
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Solution
-F is considered a weak -I group compared to -NO2 and -CN.
Correct Answer: C — -F
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Q. Which of the following groups is known for its +I effect?
A.
-CH3
B.
-OH
C.
-NO2
D.
-COOH
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Solution
-CH3 exhibits a +I effect as it donates electron density through inductive effect.
Correct Answer: A — -CH3
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Q. Which of the following groups shows +M effect?
A.
-NO2
B.
-CN
C.
-OCH3
D.
-COOH
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Solution
The -OCH3 group shows +M effect as it donates electrons through resonance.
Correct Answer: C — -OCH3
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Q. Which of the following groups shows a +M effect?
A.
–NO2
B.
–F
C.
–OH
D.
–CN
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Solution
The –OH group donates electron density through resonance, exhibiting a +M effect.
Correct Answer: C — –OH
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Q. Which of the following groups would decrease the electron density on a benzene ring?
A.
-OCH3
B.
-NO2
C.
-CH3
D.
-Cl
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Solution
-NO2 is a strong electron-withdrawing group that decreases electron density on the benzene ring.
Correct Answer: B — -NO2
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Q. Which of the following haloalkanes can be converted to an alcohol by hydrolysis?
A.
Bromobenzene
B.
Chloroethane
C.
Iodobutane
D.
All of the above
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Solution
Iodobutane can be converted to an alcohol by hydrolysis, while bromobenzene requires a different mechanism due to its aromatic nature.
Correct Answer: C — Iodobutane
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Q. Which of the following haloalkanes is expected to have the highest boiling point?
A.
Chloroethane
B.
Bromoethane
C.
Iodoethane
D.
Fluoroethane
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Solution
Iodoethane has the highest boiling point due to the larger size and polarizability of iodine.
Correct Answer: C — Iodoethane
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Q. Which of the following haloalkanes is least soluble in water?
A.
Chloroethane
B.
Bromoethane
C.
Iodoethane
D.
Fluoroethane
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Solution
Iodoethane is least soluble in water due to its larger size and lower polarity compared to the others.
Correct Answer: C — Iodoethane
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Q. Which of the following haloalkanes is most reactive towards nucleophilic substitution?
A.
1-bromopropane
B.
2-bromopropane
C.
3-bromopropane
D.
Bromobenzene
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Solution
3-bromopropane is more reactive due to the stability of the tertiary carbocation formed during the reaction.
Correct Answer: C — 3-bromopropane
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Q. Which of the following haloalkanes is the most reactive towards nucleophilic substitution?
A.
1-bromopropane
B.
2-bromopropane
C.
3-bromopropane
D.
bromobenzene
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Solution
3-bromopropane is a tertiary haloalkane, which is more reactive towards nucleophilic substitution due to steric hindrance and stability of the carbocation formed.
Correct Answer: C — 3-bromopropane
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Q. Which of the following haloalkanes will undergo elimination reaction more readily?
A.
1-chloropropane
B.
2-chloropropane
C.
3-chloropropane
D.
chlorobenzene
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Solution
3-chloropropane is a tertiary haloalkane and will undergo elimination more readily due to the stability of the resulting alkene.
Correct Answer: C — 3-chloropropane
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Q. Which of the following haloalkanes will undergo elimination reaction most readily?
A.
1-bromopropane
B.
2-bromopropane
C.
3-bromopropane
D.
bromobenzene
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Solution
2-bromopropane undergoes elimination readily due to the formation of a stable alkene.
Correct Answer: C — 3-bromopropane
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Q. Which of the following haloarenes can undergo nucleophilic substitution?
A.
Chlorobenzene
B.
Bromobenzene
C.
Iodobenzene
D.
All of the above
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Solution
All of the listed haloarenes can undergo nucleophilic substitution under suitable conditions.
Correct Answer: D — All of the above
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Q. Which of the following haloarenes is most reactive in nucleophilic aromatic substitution?
A.
Chlorobenzene
B.
Bromobenzene
C.
Nitrochlorobenzene
D.
Fluorobenzene
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Solution
Nitrochlorobenzene is most reactive due to the electron-withdrawing nitro group, which stabilizes the negative charge in the intermediate.
Correct Answer: C — Nitrochlorobenzene
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Q. Which of the following haloarenes is most reactive towards nucleophilic substitution?
A.
Chlorobenzene
B.
Bromobenzene
C.
Iodobenzene
D.
Fluorobenzene
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Solution
Iodobenzene is the most reactive towards nucleophilic substitution due to the weaker C-I bond compared to the other haloarenes.
Correct Answer: C — Iodobenzene
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Q. Which of the following has a bent molecular geometry?
A.
CO2
B.
H2O
C.
CH4
D.
BF3
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Solution
H2O has two lone pairs and two bonding pairs, resulting in a bent geometry.
Correct Answer: B — H2O
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Q. Which of the following has a bond angle of approximately 120 degrees?
A.
H2O
B.
NH3
C.
BF3
D.
CH4
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Solution
BF3 has a trigonal planar geometry with bond angles of approximately 120 degrees.
Correct Answer: C — BF3
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Q. Which of the following has a bond order of 0?
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Solution
He2 has a bond order of 0, calculated as (2 bonding electrons - 2 antibonding electrons)/2 = 0.
Correct Answer: A — He2
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Q. Which of the following has a bond order of 1?
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Solution
H2 has a bond order of 1, calculated as (2 bonding electrons - 0 antibonding electrons)/2 = 1.
Correct Answer: C — H2
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Q. Which of the following has a central atom with dsp3 hybridization?
A.
PCl5
B.
XeF4
C.
NH4+
D.
C2H2
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Solution
In PCl5, the phosphorus atom is dsp3 hybridized, resulting in a trigonal bipyramidal geometry.
Correct Answer: A — PCl5
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Q. Which of the following has a central atom with sp3d hybridization?
A.
XeF4
B.
SF4
C.
PCl5
D.
BrF5
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Solution
In BrF5, bromine undergoes sp3d hybridization to form five bonds with fluorine and has one lone pair.
Correct Answer: D — BrF5
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Q. Which of the following has a central atom with sp3d2 hybridization?
A.
XeF2
B.
SF6
C.
BrF5
D.
PCl5
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Solution
In BrF5, the bromine atom is sp3d2 hybridized, allowing it to form five bonds with fluorine.
Correct Answer: C — BrF5
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