A) Use a large excess of alkyl halide relative to the aromatic compound.
B) Use a large excess of benzene relative to the alkyl halide.
C) Use an alkyl halide without a Lewis acid catalyst.
D) Use a large excess of the Lewis acid catalyst.
Correct Answer
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Multiple Choice
A) Addition-substitution and substitution-addition
B) Addition-elimination and elimination-addition
C) Addition-addition and elimination-elimination
D) Elimination-substitution and substitution-elimination
Correct Answer
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Multiple Choice
A) I
B) II
C) III
D) IV
Correct Answer
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Multiple Choice
A) I
B) II
C) III
D) IV
Correct Answer
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Multiple Choice
A) Only I
B) Only II
C) Only I and II
D) Only III
Correct Answer
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Multiple Choice
A) I
B) II
C) III
D) IV
Correct Answer
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Multiple Choice
A) Because it is sterically very large.
B) Because it adds electron density to the meta position, thus activating it.
C) Because it stabilizes the intermediate cation.
D) Because it removes more electron density from the ortho and para positions than the meta position, thus deactivating the meta position less.
Correct Answer
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Multiple Choice
A) I
B) II
C) III
D) IV
Correct Answer
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Multiple Choice
A) I
B) II
C) III
D) IV
Correct Answer
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Multiple Choice
A) NO+
B) NO2+
C) NO3+
D) NO2H
Correct Answer
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Multiple Choice
A) Protonation of the aromatic ring
B) Deprotonation of the aromatic ring
C) Addition of the electrophile to the aromatic ring
D) Loss of the electrophile from the aromatic ring
Correct Answer
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Multiple Choice
A) Increasing the electronegativity of the halogen increases the reactivity of the aryl halide.
B) Increasing the number of electron-withdrawing groups increases the reactivity of the aryl halide.
C) Electron-withdrawing groups stabilize the intermediate carbanion, and lower the energy of the transition state.
D) When a nitro group is located meta to the halogen, the negative charge of the intermediate carbanion can be delocalized onto the NO2 group, thus stabilizing it.
Correct Answer
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Multiple Choice
A) I
B) II
C) III
D) IV
Correct Answer
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Multiple Choice
A) Br2 and FeBr3
B) NBS and light
C) Br2 in CCl4
D) NaBr and H2O
Correct Answer
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Multiple Choice
A) I
B) II
C) III
D) None of the above
Correct Answer
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Multiple Choice
A) CH3O
B) Cl
C) NO2
D) HSO3
Correct Answer
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Multiple Choice
A) [1] LiAlH4; [2]H2O
B) Zn (Hg) , HCl
C) NH3, NaOH
D) H2, Pd-C
Correct Answer
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Multiple Choice
A) Only I
B) Only II
C) Only III
D) Only I and II
Correct Answer
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Multiple Choice
A) I
B) II
C) III
D) IV
Correct Answer
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Multiple Choice
A) II > IV > I > III
B) II > III > IV > I
C) IV > II > I > III
D) IV > I > II > III
Correct Answer
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