A) 4
B) 5
C) 6
D) 7
E) 8
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A) I
B) II
C) III
D) IV
E) V
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A) Bromine radical selectively abstracts the tertiary hydrogen.
B) The transition state of the rate determining step is product-like.
C) The major product formed from this reaction is 1-bromo-2-methylpropane.
D) Bromine radical selectively abstracts the tertiary hydrogen and the transition state of the rate determining step is product-like.
E) Bromine radical selectively abstracts the tertiary hydrogen,the transition state of the rate determining step is product-like,and the major product formed from this reaction is 1-bromo-2-methylpropane.
Correct Answer
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A) I
B) II
C) III
D) IV
E) V
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A) 2-Methyl-1-bromopropane (almost exclusively)
B) 2-Methyl-2-bromopropane (almost exclusively)
C) A mixture of 50% (A) and 50% (B)
D) A mixture of 90% (A) and 10% (B)
E) Butyl bromide
Correct Answer
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Correct Answer
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A) I
B) II
C) III
D) IV
E) V
Correct Answer
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A) I
B) II
C) III
D) IV
E) V
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A) 3
B) 4
C) 5
D) 7
E) 9
Correct Answer
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A) There would be only 4 aromatic protons at low field.
B) The signal for the protons on the benzylic carbon would be a doublet.
C) The signal for the methyl protons would be a triplet.
D) The signal for the methyl protons would be a doublet.
E) The signal for the methyl protons would integrate for only 2 hydrogens.
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A) CCl4
B) HCCl3
C) CH2Cl2
D) CH3CH3
E) CH2=CH2
Correct Answer
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A) attack on the alkene by a Br+ ion.
B) attack on the alkene by a H+ ion.
C) attack on the alkene by a bromine atom,Br·.
D) attack on the alkene by a hydrogen atom,H·.
E) isomerization of the 2-bromobutane produced initially.
Correct Answer
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A) I
B) II
C) III
D) IV
E) V
Correct Answer
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View Answer
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A) I
B) II
C) III
D) IV
E) V
Correct Answer
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A) I
B) II
C) III
D) IV
E) V
Correct Answer
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