Aspirin can be made in the laboratory by reacting acetic anhydride C 4 H 6 O 3 with salicylic acid C 7 H 6 O 3 to form aspirin C 9 H 8 O 4 and acetic acid C 2 H 4 O 2 . The balanced equation is: C 4 H 6 O 3 + C 7 H 6 O 3 → C 9 H 8 O 4 + C 2 H 4 O 2 In a laboratory synthesis, a student begins with 5.00 mL of acetic anhydride (density = 1.08g/mL) and 2.08 g of salicylic acid. Once the reaction is complete, student collects 2.01 g of aspirin. Determine the limiting reactant, theoretical yield of aspirin, and percent yield for the reaction
Aspirin can be made in the laboratory by reacting acetic anhydride C 4 H 6 O 3 with salicylic acid C 7 H 6 O 3 to form aspirin C 9 H 8 O 4 and acetic acid C 2 H 4 O 2 . The balanced equation is: C 4 H 6 O 3 + C 7 H 6 O 3 → C 9 H 8 O 4 + C 2 H 4 O 2 In a laboratory synthesis, a student begins with 5.00 mL of acetic anhydride (density = 1.08g/mL) and 2.08 g of salicylic acid. Once the reaction is complete, student collects 2.01 g of aspirin. Determine the limiting reactant, theoretical yield of aspirin, and percent yield for the reaction
Solution Summary: The author explains the limiting reactant, theoretical yield, and actual yield of the reaction between salicylic acid and acetic anhydride.
Aspirin can be made in the laboratory by reacting acetic anhydride
C
4
H
6
O
3
with salicylic acid
C
7
H
6
O
3
to form aspirin
C
9
H
8
O
4
and acetic acid
C
2
H
4
O
2
. The balanced equation is:
C
4
H
6
O
3
+
C
7
H
6
O
3
→
C
9
H
8
O
4
+
C
2
H
4
O
2
In a laboratory synthesis, a student begins with 5.00 mL of acetic anhydride (density = 1.08g/mL) and 2.08 g of salicylic acid. Once the reaction is complete, student collects 2.01 g of aspirin. Determine the limiting reactant, theoretical yield of aspirin, and percent yield for the reaction
3. Predict the major product and give a mechanism for the following reactions:
(CH3)3COH/H₂SO4
a)
b)
NC
CH₂O
c)
LOCH,
(CH3)3COH/H2SO4
H,SO
-OH
Indicate if the aldehyde shown reacts with the provided nucleophiles in acid or base
conditions.
a NaBH4
be Li
eli
-NH2
P(Ph3)
f
KCN
g
OH excess
h CH3OH
i NaCHCCH3
Chapter 8 Solutions
Mastering Chemistry with Pearson eText -- Standalone Access Card -- for Introductory Chemistry (6th Edition)
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