Experiment 13 Oxidation reactions
Syntheses of cyclohexanone and adipic acid.
Purpose: one or two sentences states the objective of the experiment
Procedure: Please refer to Chemistry 230/231 lab manual, Camosun College, revised
July 2011, pages xxxx1 {note: use of superscript to give source of information]
Data and Results:
Give chemical equations using molecular structures (either full structures or skeletal structures – you may
use drawing programs such as ISIS Draw or Chem Draw) – Make sure that you give clear structures (no
cuts and paste from websites, please).
Eg.
OH
OH
+
+
HOCl
HCl + H 2O
Give Tables with Descriptive headings for List of reagents, hazard information (don’t forget to reference
your sources (use superscript)1
Give Table with Descriptive heading for physical data (melting point – actual, literature), colour of
product and a brief description, actually yield and percent yield.
Give calculation for theoretical and percent yield of products.
Eg. of calculations
Calculation of theoretical yield of cyclohexanone in the synthesis of cyclohexanol (note the title)
1. Determining the limiting reagent
Mole of cyclohexanol: Mole of HOCl = 1:1
1
molar mass
g
1 mole
Mole cyclohexanol = 6.5 mL x 0.962
x
= 0.062 mole
mL 100.2 g
g
1 mole
Mole HOCl = 90 mL x 0.06 wt/wt (HOCl in bleach) x 1.08
x
= 0.11 mole
mL 52.45 g
Mole cyclohexanol = volume x density x
HOCl is in excess and cyclohexanol is limiting reagent.
2. Determining the theoretical yield
Mole of cyclohexanol = mole of cyclohexanone
Gram of cyclohexanone =
3. Percent yield =
0.062 mole cyclohexanone x
isolate yield
x 100
theoretical yield
98 grams
= 6.08 grams
1 mole cyclohexanone
Discussion.
Avoid the use of personal pronouns (I, we, they, us, you, etc… - half a mark deduction for each usage to
a maximum of 3 marks). The discussion in scientific reports should be written in passive past tense.
For example: The cyclohexanone is collected in a beaker and it is a colourless liquid. (not passive past)
A colourless liquid was collected in a beaker, and after several analytical tests it was confirmed to be
cyclohexanone.
Refer to your tables (makes a good starting point). What was observed during the reaction? Any colour
change, any precipitate (why?), is it a homogeneous mixture, bubbling of gas? Was there any odour?
What was the appearance of the product after recrystallization. Can you explain why these changes are
occurring? Compare actual melting point to literature. Is it pretty accurate or any discrepancies? Use
curved arrows to propose a detailed mechanism (if known). For this lab on oxidation reduction reactions,
you don’t need a detailed mechanism, but you should refer to your textbook (at least one textbook
reference), or the internet (note: be selective on the site) for the theory.
What is the oxidizing agent and
what is the reducing age? What characteristics of a compound will not oxidize. For example, primary
alcohols can be oxidized to aldehydes which can further oxidize to carboxylic acids. How about
secondary alcohols.. Can they further oxidize to an ester or what will happen to further oxidize secondary
alcohols? Are there anything unusual about the reactions that are performed.
Give source of errors (two errors – refrain from suggesting transferring between glassware or compounds
spilling to reduce the yield).
Conclusion – one or two sentences
References - See sample for format
Questions – Answer questions 1-7 (You may need to look up the answer in your textbook).
Some hint to balancing half-reactions page 50 questions.
Cylohexanol + HOCl → Cyclohexanone + Cl −
HOCl + I−→ Cl− + I2
HOCl + HSO3−→ Cl− + SO42Cylohexanone + MnO4− → adipic acid (give molecular structure) + MnO2
MnO4− + I− → I2 + MnO2 (alkaline solution)
MnO4− + HSO3− → MnO2 + SO42- (alkaline solution)
MnO4− + HSO3− → Mn2+ + SO42- (acidic solution)
See next page for a review on how to balance redox reactions by half reaction method.
Balancing redox reactions by the method of half reactions.
In absence of half-reaction data sheet Balance the following redox reaction in acidic solution.
MnO4−
(aq)
+ C2O42- (aq)
→
Mn2+ (aq) + CO2(aq)
How to balance this reaction – in acidic solution
Step 1: Split the reaction into two half reactions and determine its half-reaction
MnO4− → Mn2+
C2O42- → CO2
Step 2: Balance the non-hydrogens and non-oxygens.
C2O42- → 2CO2
Step 3: Balance the oxygens (on side of equation) with H2O (on other side)
MnO4− → Mn2+ + 4H2O
C2O42- → 2CO2 (good no further atomic balancing required)
Step 4: Balance the hydrogens from water with H+ (on other side)
8H+ + MnO4− → Mn2+ + 4H2O
Step 5: Balance the charges of the half reaction using electrons.
5e− + 8H+ + MnO4− → Mn2+ + 4H2O
C2O42- → 2CO2 + 2e−
Step 6: Determine the common number of electrons required in the two half reactions (multiple each
reagent and electron by the common factor – if necessary).
(5e− + 8H+ + MnO4− → Mn2+ + 4H2O) multiply everything by a factor of 2
(C2O42- → 2CO2 + 2e− ) multiply everything by a factor of 5
Step 7: Sum – common number of reagents and electrons on left side and right side of equation cancel
How to balance this reaction – in basic solution?
Repeat Step 1-4 then after step 4 – For every H+
Add the same number of OH−
add the same number of OH− (same side)
on the other side of the equation.
The next steps are the same as Step 5, 6, and 7 are the same as in acidic solution.