Organic Chemistry Lab – Student Questions about Single and Double Unknown Analyses (Extra Credit)
Preparation for the Experiments


During these single unknown and double experiments, are we allowed to ask questions for further
clarification?
Of course! We just might not give you the answer, but ask you another question. 
Will we be allowed to bring the handouts to lab particularly the “classification (ID) tests for organic
chemicals and the “procedures for solid derivatives”
Yes, you may bring all handouts to lab. At the very least, bring the “Functional Group Tests to
Classify Organic Chemicals” and “Solid Derivative Synthesis Procedures” handouts, but be sure to
write the procedures you perform into your lab notebooks and take lots of observations.
You will need to write the following items into your notebook:
For the Single Unknown:
 Normal pre-lab notebook page components (for Table of Properties/Materials Chart, include at
least the acids and bases you will be using in the solubility tests)
 Protocol components (with room for data/observations, as always):
 “Make initial observations” section (pg 1 of Single Unknown handout)
 “Obtain a melting or boiling range” section (pg 1 of handout)
 “Limit potentials classes using solubility tests (acid-base chemistry)” section (pg 2 of handout)
 Solubility tests flowchart (pg 3 of handout) – draw entire flowchart on one page
 LEAVE SPACE (approximately 1 full notebook page) for “Functional Group Tests to Classify Organic
Chemicals” document
 Once you have narrowed your potential unknown’s class using the solubility tests, you will
choose appropriate functional group tests from this document and you will need to write in
the protocols for each before you perform them.
 Page 1 (including header) of the “Solid Derivative Synthesis Procedures” handout**
 “Recrystallization Technique” section from page 2 of the “Solid Derivative Synthesis Procedures”
handout**
 “Hot Filtration” section from page 2 of the “Solid Derivative Synthesis Procedures” handout
(including drawing)**
 LEAVE SPACE (approximately ½ notebook page - 1 full notebook page) for the specific solid
derivative synthesis procedure you will choose once you have an educated guess (post-functional
group testing) about the identity of your unknown.
For the Double Unknown:
 Normal pre-lab notebook page components (for Table of Properties/Materials Chart, include at
least ether and the acids and bases you will be using in the separation scheme)
 Protocol components (with room for data/observations, as always):
 “Test for water solubility” section (pg 1 of the Double Unknown handout)
 “Separation technique” section (pg 2 of the handout)
 Separation scheme (pg 3 of the handout) – draw entire separation scheme and its notes / ether
contents info
 “Recovery of separated compounds” section (pg 2 of the handout)
 “Characterization of separated compounds” section (pg 2 of the handout)
 LEAVE SPACE for any functional group characterization tests you might have to perform
 “Boiling/melting ranges” section (pg 1 of the handout)
 LEAVE SPACE for the specific solid derivative synthesis procedures (one for EACH of the two
unknowns) you will choose once you have an educated guess (post-characterization) about the
identity of each of your unknowns
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Physical Properties

Is every unknown going to initially be a solid or will some of the unknowns be liquid?
Some will be liquids, some will be solids. We prepare them such that you should have enough of whatever it
is to complete the experiment.

What is a split stopper, and how does it aid in chemical analyses?
Used for measuring a boiling point of a liquid without needing to set up a distillation.
A boiling point test tube with the appropriate volume of the liquid (and a boiling chip) will be sealed with
one of these, and a thermometer can be inserted through the opening.
A second opening or slit (see above photos) provides a vent through which vapors may escape, so that we
aren’t heating a closed system. We will show you how to do this.

For obtaining a boiling range, how common are fires? Do we need to consult our neighbors nearby to
avoid this, or is it more of being aware of our surroundings?
When the boiling range test is performed correctly, a fire is unlikely, but don’t put the flame at the top of
the test tube where the vapors are escaping. We will be sure to remind students in the lab to be aware of
their surroundings and where in the lab the boiling point tests should be performed.
Solubility Test Flowchart: These concepts are useful for both single and double unknown analyses.

How do you use the solubility test flowchart?
Start from the left side of the page (unknown) and work through the pathway. Once your unknown has
reacted with one of the reagents, you don’t need to continue performing the following tests.
EX 1.
Unknown + H2O = soluble (obvious mixing): Move up the soluble path and test the pH of the
unknown. At this point, you’re done. You likely have a class that matches pH of your unknown.
EX 2.
Unknown + H2O = insoluble (layers/not obvious mixing): Move down the insoluble path.
Unknown + NaHCO3 = soluble (indicated by bubbling/CO2 formation)
You’re done. You likely have a higher MW carboxylic acid or an acidic phenol.
EX 3. Unknown + H2O = insoluble (layers/not obvious mixing): Move down the insoluble path.
Unknown + 5% NaHCO3 = insoluble (no bubbles): Move down the next insoluble path.
Unknown + 3M NaOH = insoluble (no reaction observed): Move down to 3M HCl.
Unknown + 3M HCl = insoluble (no reaction observed): Move down to Conc H2SO4.
Unknown + H2SO4 = reaction observed
You’re done. You likely have a neutral oxygen-containing compounds, alkene, or nitrile.
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
When reviewing the flowchart provided in “Analysis of a Single Unknown” I was confused as to the amounts
of acid or base used for dissolving the solid derivative. (Linfield College Chemistry Webpage, Accessed 310-15).
Page 2 of the Single Unknown handout instructs to add ~0.5 mL of a reagent to your unknown.

In the single unknown handout, why is it okay to keep drawing off and putting on reagents when testing
for solubility? Will the reagents potentially mix together and give an incorrect result?
As long as you draw off most of the previous reagent it will be fine for the purposes of this step. All of the
solubility tests are aqueous (water, NaHCO3 in water, NaOH in water, HCl in water, until conc. H2SO4). If you
are concerned you can start with fresh unknown, but remember you have a limited supply.
Double Unknown – Separation Scheme according to Solubility Properties

Why is a solution solute soluble at one pH but insoluble in others (assuming the solvents are the same)?
Be careful with word choice. A solid molecule or compound (solute) will be soluble in a solvent if they are
“like” each other due to polarity or charge. Remember, “Like dissolves like.” (Two liquids that mix are
said to be miscible instead of soluble.) By manipulating the pH of the solution, you can protonate or
deprotonate the compound, altering its charge (i.e. to either an ion or neutral). Neutral molecules will
generally be more soluble in neutral solvents (organic, like ether); charged molecules will generally be
more aqueous-soluble. The ultimate purpose of manipulating the pH this way is to isolate your unknowns
from other another.

In the double unknown handout, the separation technique is described. In step two, it discusses testing
the lower layer’s pH to make sure to get the desired pH. Why is the lower layer after each test kept out of
the mixture and then all put together when the lower layer has to correct pH instead of putting it right back
in?
Review the purpose of manipulating the pH: Imagine you are adding acid to an ether layer to reach pH <2.
Acid is in an aqueous solution and that layer will be on the bottom due to its density. You are performing
this step in order to protonate the compound of interest in the organic layer – let’s imagine a basic such as a
neutral amine-in order to isolate it from the other unknown. As long as your solution still contains some of
the neutral, basic amine that can be protonated, there won’t be free protons in solution. Once you’ve
protonated all the available amine molecules, the pH will lower to <2, indicating free H+ floating around in
solution with nothing to protonate.
So: You extract the ether layer with a portion of aqueous HCl. H+ in the aqueous will be “used up” to
protonate the basic, neutral amine (RNH2 + H+  RNH3+). You check the pH of the aqueous layer and it’s not
low enough – signifying you have amine left to protonate, but not as much as you had before you began
adding acid. Thus, you can imagine that the next round of extracting with HCl, there will be even less need
for the H+, so you are closer to the goal of having free H+.
Bottom line: If you combined a new aqueous HCl wash with the original aqueous HCl layer, you
might observe an overestimated pH compared with the actual concentration of the neutral amine
that’s left, so you would likely end up adding more acid than you need to compensate for your
observed, but overestimated, pH. If you reused the original aqueous layer for subsequent
extractions, you might not have enough HCl available to keep protonating the amine. You don’t want
to add more aqueous than you need in case your unknown is slightly soluble in water.
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
For the double unknown separation scheme, if we have “ether z,” the procedure states not to dry over
CaCl2, do we dry it over anything? How do we recover the separated compounds?
Ether Z is not a final ether layer (see the separation scheme in the double unknown handout). You will
continue performing separation techniques on ether Z, so you are not done yet and thus it would be
redundant to dry this one. Once you have reached the points of Ether A, B, C, or D, you can assume you have
separated out the compounds and are ready to decide what to distill first.

How do you know the solubility of a molecule at a given pH?
Review the “Rationale for the separation scheme” section of pg 1 of the Double Unknowns handout.
If a molecule is an ion at a given pH, it will be soluble in the aqueous layer. If it is neutral, it will be soluble
in the organic layer. But if a molecule is really small it may be soluble in water. If that is the case, take the
pH of the aqueous layer and it will indicate if it is an acid, base or neutral compound.
The concept of pKa (measure of acid dissociation) is very important for this experiment. A compound with
a low pKa will be “willing” to lose a proton even with lots of free H+ in aqueous solution (i.e. in an
environment with a low pH). A compound with a high pKa will only be “willing” to lose a proton when there
is a low concentration of free H+ in aqueous solution (i.e. in an environment with a high pH).
Example 1: Let M-H represent a compound with a pKa of 10:
Imagine M-H is in an aqueous solution whose pH is 7. You would expect it to stay protonated (M-H) in this
environment because the concentration of free H+ in aqueous solution is not low enough for it to be
“willing” to donate a proton. (What would happen if the pH of the aqueous solution was 14 for this pKa=10
molecule?)
Example 2: Let M-H represent a compound with a pKa of 3:
Imagine M-H is in an aqueous solution whose pH is 7. You would expect it to give up its proton and be
deprotonated (M-) in this environment because it is “willing” to (or “prefers” to) give up a proton even
when free H+ concentration is high.
You can use this rationale for predicting the solubility of a molecule in its environment at a given pH.
Additionally, you can visually observe whether a compound is soluble in its solvent (i.e. aqueous/organic)
at a given pH based on whether the compound appears dissolved or if it crashes out as a precipitate. If both
aqueous and organic phases are present, the compound will migrate to the solvent it is more soluble in. If
only the aqueous phase is present but the compound is neutral (and thus not aqueous-soluble), it will likely
crash out as a solid (and vice-versa). If both phases are present and the compound is not enough “like”
either, it may still crash out as a solid or third layer.
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Characterization of Unknowns

In the Double Unknown, when you characterize the separated compounds, if you have a liquid, but not
the 5 mL needed for fractional distillation, can you perform a simple distillation? What would you want us
to do?
Remember, you may end up with 4 ether layers (A, B, C, D) but only have 2 unknowns so you will need to
strategize to determine which ether layers to try distilling first. Before you distill an ether layer, place a
drop or two on a clean watch glass and allow the solvent to evaporate. If there is an obvious residue left
behind this indicates you have something in this ether layer. If you still think you have something in your 5
mL of solution we will go from there.

Tips on performing the classification tests?
Add reagents in the specific order indicated in the procedure.
Note the initial and final conditions (clarity and color). Write down plenty of observations.
Compare your unknown against all appropriate controls as well as the compound you think your unknown
might be, if you have an idea.
EX. Think your unknown might be 2-propanol? Compare your unknown against primary, secondary,
and tertiary controls for the alcohol tests, using 2-propanol as your secondary alcohol control.

Why can you only trust the Lucas test for only 6C or smaller?
Your alcohol needs to be soluble in the Lucas Reagent or it won’t react, potentially giving a false negative.
More than 6C increases the likelihood of solubility problems.

If the unknown is a solid and the suspected pure compound is in solution, would many of the identifiers be
different from one another, or does the state of the compound have little effect on the melting and boiling
range, along with other properties?
Before characterizing anything, you will either recrystallize your solid unknown from its solution to purify
it; or you will distill your liquid unknown from its solution to purify it. If your unknown has a melting point
around room temperature, it might be in liquid state until after the distillation and then, as it cools, it might
crystallize into a solid.
Recrystallization and Hot Filtration

During recrystallization, is there any time you would use a solvent besides ethanol? If so, how do you
make that decision?
Yes, this is possible. We would help you with this. You ultimately want a solvent for recrystallization in
which:
When hot: desired compound is soluble + contaminants are soluble
When room temp – cold: desired compound is less soluble + contaminants remain soluble
You can also use the CRC handbook to look up solubility properties of the compound. We can help with this.

In hot filtration for solid derivative procedures, are we covering just the funnel with a watch glass or are
we covering the entire beaker and funnel?
The picture is not to scale, but the funnel will sit in a beaker with a similar diameter. The watch glass will
cover both.
5
Synthesis of Solid Derivatives

What is the specific purpose of the solid derivative procedures? Is it another confidence measure?
Review the first section of the “Solid Derivative Synthesis Procedures” handout. You may narrow down the
identification of your unknown to two isomers that have similar boiling points, similar solubility
properties, and similar IR spectra, etc. (i.e. you don’t have enough information to really choose between one
or the other). Changing one functional group on the molecule might produce a derivative compound that is
different enough from the derivative of the isomer to provide contrasting melting points, enabling you to
identify your unknown with greater confidence.

For solid derivative synthesis do you make a derivative of your unknown and of the authentic compound
that you suspect it to be?
Yes, you always want to compare your unknown to a control. The authentic compound serves as your
control for the solid derivative synthesis. When you characterize the solids you derive, you will take 3
melting ranges: One with just the derivative of your unknown; one with just the derivative of your
authentic control, and one with a 50/50 mixture of both. It is possible for two different compounds to have
the same melting point. However, if your 50/50 mixture has a depressed/wide melting range in
comparison to the pure samples of each, this indicates they are different compounds (and thus your
unknown is not what you thought it was).

Would making a solid derivative be the first thing to do after receiving the unknown and testing for water
solubility, or is synthesizing a solid derivative something that comes toward the end of the analysis, or
somewhere in the middle?
You want to have a pretty well-educated guess about the identity of your unknown before starting to
synthesize a solid derivative. Therefore, you won’t perform the solid derivative synthesis until after you
have performed characterization tests and narrow the possibilities down.

How certain should you be of your compound’s identity by the time you make a solid derivative?
Once you have obtained characterization data and have narrowed your unknown down to a class and a
boiling/melting point, you will look at a list of compounds with that similar information, and you can also
check our supply to see what we actually have available. Obviously, if we don’t have the compound, it is
unlikely you were given that compound.

How will I know what solid derivative to make if the identity of the unknown is still relatively unknown?
Once you have obtained characterization data and have narrowed your unknown down to a class and a
boiling/melting point, you will look at a list of compounds with that similar information, and you can also
check our supply to see what we actually have available. Obviously, if we don’t have the compound, you
probably wouldn’t have gotten that as your unknown. It would be ideal to have narrowed it down to one or
two possible identities. Once you believe you have identified your unknown (“parent” compound), we will
provide you with a list of feasible solid derivatives you can synthesize for that compound.

When producing a solid derivative of an aldehyde, would poor mixture of the solution with NaHCO3 result
in a lower yield of the crystals or some other result? What do you mean by “good crystals”? (P. 3)
The bicarb purifies your crystals (“good” = “pure”).
6

Would solid derivatives not be used for alkanes and alkyl halides? What other tools would be
useful in replacing the use of solid derivatives to give equally relevant data?
Review the first section of the “Solid Derivative Synthesis Procedures” handout. Making a derivative
molecule of a “parent” compound (i.e. your unknown) occurs by transforming a functional group into a new
functional group. Alkanes and alkyl halides will not make for suitable “parent” compounds since they don’t
have functional groups of much interest. You will perform additional qualitative tests on your compound in
this case, in lieu of making a solid derivative.

After you make the solid derivatives, if they exhibit similar properties, is that what indicates that they are
possibly the same compound?
See a couple possible situations:
Table A: Example of two different derivative compounds with similar melting ranges
Sample
Observed Melting Range
Derivative of Unknown
Accurate and narrow compared
with authentic
Derivative of Authentic Control
Accurate and narrow compared
with unknown
50/50 mixture
Broad and depressed compared
with above ranges
Interpretation
The derivatives have the same
melting range but are different
compounds, indicated by the wide
and low melting range of the
50/50 mixture (they are each
contaminating the other’s lattice
structure); Therefore, the parent
compounds are likely different
from one another.
Table B: Example of identical derivative compounds
Sample
Observed Melting Range
Derivative of Unknown
Accurate and narrow compared
with authentic
Derivative of Authentic Control
Accurate and narrow compared
with unknown
50/50 mixture
Accurate and narrow compared
with above ranges
Interpretation
The derivatives have the same
melting range and the 50/50
mixture is pure, indicated by the
accurate and narrow range
compared with the individual
samples (there is little to no
contamination); Therefore, the
parent compounds likely have the
same identity.
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