A) I
B) II
C) III
D) IV
Correct Answer
verified
Multiple Choice
A) A flat line (carboxylic acids are not IR active)
B) A sharp line at 2250 cm-1
C) A broad peak from 3800-2800 cm-1
D) A broad peak from 800-600 cm-1
Correct Answer
verified
Multiple Choice
A) Heat with hydrochloric acid
B) React with thionyl chloride (SOCl2)
C) React with sodium chloride
D) React with Cl2 and FeCl3
Correct Answer
verified
Multiple Choice
A) ethyl acetate
B) 2-propanone
C) 3,3-dimethyl-2-pentanone
D) ethyl 2,2-dimethylpropionate
Correct Answer
verified
Multiple Choice
A) I
B) II
C) III
D) IV
Correct Answer
verified
Multiple Choice
A) an alkane
B) an aldehyde
C) an alkene
D) a nitrile
Correct Answer
verified
Multiple Choice
A) 4-fluoro-3-methylbenzoate
B) 4-fluoro-3-methylbenzoic acid
C) 3-methyl-4-fluorobenzoic acid
D) 4-fluoro-5-methylbenzoic acid
Correct Answer
verified
Multiple Choice
A) 2.5 ppm
B) 2250 cm-1
C) 3800 cm-1
D) 1740 cm-1
Correct Answer
verified
Multiple Choice
A) I
B) II
C) III
D) IV
Correct Answer
verified
Multiple Choice
A) Heat with an alcohol and catalytic acid
B) Deprotonate with a base and react with an alcohol
C) Deprotonate with a base and react with an alkyl halide
D) Both (a) heat with an alcohol and catalytic acid and (c) deprotonate with a base and react with an alkyl halide
E) Both (a) heat with an alcohol and catalytic acid and (b) deprotonate with a base and react with an alcohol
Correct Answer
verified
Multiple Choice
A) The nucleophile is too basic.
B) Reforming the carbonyl is energetically favorable.
C) The leaving group is unstable and wants to be negatively charged.
D) There is no tetrahedral intermediate.
Correct Answer
verified
Multiple Choice
A) I
B) II
C) III
D) IV
Correct Answer
verified
Multiple Choice
A) I
B) II
C) III
D) IV
Correct Answer
verified
Multiple Choice
A) LiAlH4
B) DIBAL-H
C) NaBH4
D) H2, Pd/C
Correct Answer
verified
Multiple Choice
A) A ketone
B) An aldehyde
C) An ester
D) A nitrile
Correct Answer
verified
Multiple Choice
A) Nitrogen is a better leaving group.
B) Chloride is a better leaving group.
C) Nitrogen donates more electron density into the carbonyl.
D) The amide anion is less basic.
Correct Answer
verified
Multiple Choice
A) I
B) II
C) III
D) IV
Correct Answer
verified
Multiple Choice
A) I
B) II
C) III
D) IV
Correct Answer
verified
Multiple Choice
A) 2-fluorobutanonitrile
B) 2-fluoropentanonitrile
C) 3-fluoropentanonitrile
D) 2-fluorobutylcyanide
Correct Answer
verified
Multiple Choice
A) Yes
B) No
Correct Answer
verified
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