Chemistry (School & UG)
Q. What is the stereochemical outcome of the electrophilic substitution of a chiral aromatic compound?
A.
Retention of configuration
B.
Inversion of configuration
C.
Racemization
D.
No stereochemical outcome
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Solution
Electrophilic substitution on a chiral aromatic compound typically leads to racemization due to the formation of a planar sigma complex.
Correct Answer: C — Racemization
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Q. What is the stereochemical outcome of the electrophilic substitution of a chiral benzene derivative?
A.
Racemic mixture
B.
Enantiomerically pure product
C.
Diastereomers
D.
No stereochemical change
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Solution
Electrophilic substitution on a chiral benzene derivative typically leads to a racemic mixture due to the planar nature of the benzene ring.
Correct Answer: A — Racemic mixture
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Q. What is the stereochemical outcome of the electrophilic substitution of toluene with chlorine?
A.
Racemic mixture
B.
Stereospecific
C.
No stereochemistry
D.
Enantiomers
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Solution
Electrophilic substitution on toluene does not lead to stereochemistry as the reaction does not create a new stereocenter.
Correct Answer: C — No stereochemistry
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Q. What is the stereochemical outcome of the reaction of (R)-2-bromobutane with NaOH in a polar aprotic solvent?
A.
Retention of configuration
B.
Inversion of configuration
C.
Racemization
D.
No reaction
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Solution
The reaction proceeds via an SN2 mechanism, leading to inversion of configuration at the chiral center.
Correct Answer: B — Inversion of configuration
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Q. What is the stereochemical outcome of the SN2 reaction of (R)-2-bromobutane with sodium iodide?
A.
(R)-2-iodobutane
B.
(S)-2-iodobutane
C.
(R)-2-bromobutane
D.
(S)-2-bromobutane
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Solution
The SN2 mechanism inverts the configuration at the chiral center, resulting in (S)-2-iodobutane.
Correct Answer: B — (S)-2-iodobutane
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Q. What is the stereochemical outcome of the S_N2 reaction?
A.
Retention of configuration
B.
Inversion of configuration
C.
Racemization
D.
No stereochemical change
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Solution
S_N2 reactions result in inversion of configuration at the carbon center where substitution occurs.
Correct Answer: B — Inversion of configuration
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Q. What is the stereochemistry of 2-butanol?
A.
R
B.
S
C.
R and S
D.
Neither
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Solution
2-butanol has a chiral center and can exist as both R and S enantiomers.
Correct Answer: C — R and S
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Q. What is the stereochemistry of 2-butanone?
A.
Achiral
B.
Chiral
C.
R/S configuration
D.
E/Z configuration
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Solution
2-Butanone is achiral because it does not have a chiral center.
Correct Answer: A — Achiral
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Q. What is the stereochemistry of 2-butene?
A.
Cis
B.
Trans
C.
Both cis and trans
D.
Neither
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Solution
2-butene can exist as both cis and trans isomers due to the presence of a double bond between the second and third carbon atoms.
Correct Answer: C — Both cis and trans
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Q. What is the stereochemistry of the product formed from the electrophilic substitution of 1,2-dimethylbenzene (o-xylene)?
A.
R
B.
S
C.
R and S
D.
None
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Solution
The product can have both R and S configurations due to the presence of two methyl groups that can influence the orientation of the incoming electrophile.
Correct Answer: C — R and S
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Q. What is the stereochemistry of the product formed from the reaction of (R)-2-bromobutane with sodium iodide in acetone?
A.
R
B.
S
C.
R and S
D.
No stereochemistry change
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Solution
The reaction leads to inversion of configuration, resulting in the formation of (S)-2-iodobutane.
Correct Answer: B — S
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Q. What is the stereochemistry of the product formed when 1,2-dibromobenzene undergoes electrophilic substitution?
A.
R
B.
S
C.
R/S mixture
D.
No stereochemistry
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Solution
The product does not have stereochemistry because the electrophilic substitution occurs at a position that does not create a chiral center.
Correct Answer: D — No stereochemistry
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Q. What is the stereochemistry of the product formed when 1,2-dimethylbenzene undergoes electrophilic substitution?
A.
Only one stereoisomer
B.
Two stereoisomers
C.
Three stereoisomers
D.
No stereoisomers
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Solution
Electrophilic substitution on 1,2-dimethylbenzene can lead to two different stereoisomers due to the presence of two methyl groups.
Correct Answer: B — Two stereoisomers
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Q. What is the term for the minimum energy required for a reaction to occur?
A.
Enthalpy
B.
Entropy
C.
Activation energy
D.
Gibbs free energy
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Solution
The minimum energy required for a reaction to occur is known as the activation energy.
Correct Answer: C — Activation energy
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Q. What is the term for the phenomenon where the adsorption of one species affects the adsorption of another?
A.
Competitive adsorption
B.
Cooperative adsorption
C.
Desorption
D.
Sorption
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Solution
Competitive adsorption occurs when the presence of one adsorbate affects the adsorption of another species on the surface.
Correct Answer: A — Competitive adsorption
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Q. What is the trend in atomic size as you move down a group in the periodic table?
A.
It increases
B.
It decreases
C.
It remains constant
D.
It fluctuates
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Solution
Atomic size increases down a group due to the addition of electron shells.
Correct Answer: A — It increases
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Q. What is the trend in electronegativity across a period in the periodic table?
A.
Increases
B.
Decreases
C.
Remains constant
D.
Fluctuates
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Solution
Electronegativity increases across a period due to increasing nuclear charge, which attracts electrons more strongly.
Correct Answer: A — Increases
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Q. What is the trend in electronegativity as you move down a group in the periodic table?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Fluctuates
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Solution
Electronegativity decreases down a group due to increased atomic size and shielding effect.
Correct Answer: B — Decreases
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Q. What is the trend in first ionization energy as you move down a group in the periodic table?
A.
It increases
B.
It decreases
C.
It remains constant
D.
It fluctuates
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Solution
First ionization energy decreases down a group due to increased distance from the nucleus and increased electron shielding.
Correct Answer: B — It decreases
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Q. What is the trend in ionization energy across a period?
A.
Increases
B.
Decreases
C.
Remains constant
D.
Increases then decreases
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Solution
Ionization energy generally increases across a period due to increasing nuclear charge.
Correct Answer: A — Increases
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Q. What is the trend in ionization energy as you move across a period from left to right?
A.
Increases
B.
Decreases
C.
Remains the same
D.
Varies unpredictably
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Solution
Ionization energy generally increases across a period due to increasing nuclear charge.
Correct Answer: A — Increases
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Q. What is the trend in metallic character as you move across a period from left to right?
A.
It increases
B.
It decreases
C.
It remains constant
D.
It first increases then decreases
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Solution
Metallic character decreases as you move from left to right across a period.
Correct Answer: B — It decreases
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Q. What is the typical pH range for phenolphthalein as an indicator?
A.
0-4
B.
4-10
C.
8-10
D.
10-14
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Solution
Phenolphthalein changes color in the pH range of approximately 4 to 10, making it suitable for many acid-base titrations.
Correct Answer: B — 4-10
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Q. What is the typical range of wavelengths for infrared (IR) spectroscopy?
A.
400-700 nm
B.
700-1400 nm
C.
1400-4000 nm
D.
4000-10000 nm
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Solution
Infrared spectroscopy typically operates in the range of 1400 to 4000 nm.
Correct Answer: C — 1400-4000 nm
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Q. What is the typical range of wavelengths for UV-Vis spectroscopy?
A.
100-400 nm
B.
400-800 nm
C.
200-800 nm
D.
800-2500 nm
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Solution
The typical range for UV-Vis spectroscopy is 200-800 nm, covering both ultraviolet and visible light.
Correct Answer: C — 200-800 nm
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Q. What is the typical range of wavelengths measured in UV-Vis spectroscopy?
A.
200-400 nm
B.
400-800 nm
C.
200-800 nm
D.
800-1200 nm
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Solution
UV-Vis spectroscopy typically measures wavelengths in the range of 200-800 nm, covering both ultraviolet and visible light.
Correct Answer: C — 200-800 nm
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Q. What is the unit of Faraday's constant?
A.
C/mol
B.
J/mol
C.
V
D.
A
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Solution
Faraday's constant is expressed in coulombs per mole (C/mol).
Correct Answer: A — C/mol
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Q. What is the unit of the rate constant k for a first-order reaction?
A.
mol/L·s
B.
s^-1
C.
mol/L
D.
L^2/(mol·s)
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Solution
For a first-order reaction, the unit of the rate constant k is s^-1.
Correct Answer: B — s^-1
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Q. What is the unit of the rate constant k for a second-order reaction?
A.
M
B.
M/s
C.
1/M·s
D.
M^2/s
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Solution
For a second-order reaction, the rate law is Rate = k[A]^2, and the units of k must be 1/(M·s) to ensure that the rate has units of M/s.
Correct Answer: C — 1/M·s
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Q. What is the value of the azimuthal quantum number (l) for a 3d orbital?
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Solution
The azimuthal quantum number (l) for a d orbital is 2.
Correct Answer: C — 2
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