Revision sheet: laboratory techniques in organic chemistry (202-SNU-RE)
This sheet does not repeat the procedures: your course notes already give them, and a procedure learned by heart is useless in front of a mixture you have never seen. It gathers what really costs points in the fourth element of the organic chemistry competency, the one that asks you to carry out the synthesis, purification and characterization of a compound.
Everything rests on a single move, which comes BEFORE the choice of technique: naming the property that differs between the product and what you want to separate it from. Solubility between two phases, extraction. Acidity, acid-base extraction. Boiling point, distillation. Solubility as a function of temperature, recrystallization. Polarity, chromatography. The rest of the sheet unfolds this move, then puts a figure on what each slip costs: 0.65 g of crystals lost to a hot wash, 4.00 g of inseparable mixture from sodium hydroxide taken too early, and a theoretical mass 14 percent too high, hence a yield of 63.0 instead of 72.0 percent, on a limiting reagent identified wrongly.
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The thread of the chapter
A purification technique is not recited, it is DEDUCED: name the property that differs between the product and what you want to separate it from, and the technique is already chosen.
•A purification technique never exploits the product itself, it exploits a GAP between the product and what it is separated from. Name the gap, and the technique follows by itself.
•Solubility between two immiscible phases: liquid-liquid extraction.
•Acidity or basicity: acid-base extraction, which CREATES the gap by ionizing what you want to move.
•Boiling point: distillation, simple apparatus above a gap of about twenty degrees, column below.
•Solubility as a function of temperature: recrystallization.
•Polarity: chromatography, on a plate to analyze, on a column to separate.
The marker expects this sentence BEFORE any procedure: “the product and the impurity differ in their ..., so we purify by ...”. A procedure recited without it loses the justification point, which is often worth as much as the procedure itself.
The separatory funnel: density decides, not habit
•The DENSEST layer is at the bottom, without exception. Ether 0.713; methanol 0.792; toluene 0.867; ethyl acetate 0.902; water 1.000; saturated brine 1.20; dichloromethane 1.326; chloroform 1.489.
•Chlorinated solvents are heavier than water, almost all the others are lighter: the organic layer therefore changes sides from one procedure to the next.
•An extraction solvent must be IMMISCIBLE with water. Methanol, ethanol and acetone are out of the game whatever their density: they never form a second layer.
•KEEP ALL the layers until the final weighing: this is what makes a stopcock mistake repairable.
On the left the ether floats on the water, on the right the dichloromethane passes underneath: the organic layer changes sides, and only the density says so.
The field test, when in doubt: a drop of water placed at the top of the funnel. If it passes through the upper layer, that layer is organic.
What each final measurement proves, and what it does not
•MELTING POINT. A range below one degree proves PURITY. An impurity LOWERS the melting point and BROADENS the range.
•IDENTITY is never read from a melting point alone: it is proved by a MIXED melting point with an authentic sample, which stays sharp if it is the same compound and collapses otherwise.
•YIELD =mtheoreticalmobtained, the theoretical mass coming from the LIMITING reagent, identified in MOLES and never in mass.
•CONSTANT MASS. You weigh, put it back to dry, weigh again: two consecutive weighings within 0.01 g say that the solvent has gone.
In green, the pure substance melts sharply, within less than one degree. In red, the impurity LOWERS the melting point by 3.4 °C and BROADENS the range to 5.0 °C.
A yield above 100 percent is not a calculation error in nine cases out of ten: it is solvent, water or an impurity that the balance counts as product.
The rules in table form
Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.
Which technique, for which gap
Read this table from LEFT to right: you start from the property that differs, never from the name of the technique. The last row is the only one where no technique works, and it can be recognized in advance.
What differs
Clue in the statement
Technique
Solubility
2 phases
Extraction
Example: 75 mL of dichloromethane remove 2.40 g of product from the aqueous phase.
Acidity
pKa
Acid-base extraction
Example: pKa=4.20 against 10.20: bicarbonate takes only the first.
Tbp
gap >20 °C
Simple distillation
Example: Acetone and toluene: 111−56=55 °C, a simple apparatus is enough.
Tbp
gap <20 °C
Column distillation
Example: Propanols: 97−82=15 °C, and 5 plates to exceed 98.0 percent.
Hot solubility
a solid
Recrystallization
Example: 0.17 g per 100 mL at 0 °C against 6.80 at 95 °C: a ratio of 40.
Polarity
Rf
Chromatography
Example: Two esters at Rf=0.35 and 0.45: a gap of 0.10, hence a column and a lot of silica.
Nothing
azeotrope
no separationno technique separates
Example: Hydrochloric acid and water: distillation stalls at 20.2 percent by mass, at 108.6 °C.
What to do: Change the property: dry chemically, add a third substance, or separate on something other than boiling. Lengthening the column gains nothing, since α=1 implies y=x.
The red row is the only one a student cannot guess in the exam room: an azeotrope cannot be seen in the flask, it is read in a table before the apparatus is set up.
The washing reagent decides what leaves
Four compounds in ether, an amine, a carboxylic acid, a phenol and a hydrocarbon. Each wash must remove ONE only, and it is the strength of the reagent, not its availability, that decides.
Washing reagent
Target compound
What goes into the water
Hydrochloric acid
amine
ammonium ion
Example: pKa=5.10: at pH=1.0, the amine is more than 99.9 percent protonated.
Bicarbonate
carboxylic acid
carboxylate ion
Example: logK=6.35−4.20=2.15, so K=141: the reaction is quantitative.
Bicarbonate
phenol
nothingno reaction takes place
Example: logK=6.35−10.20=−3.85, so K=1.4×10−4.
What to do: Correct nothing: this is the goal. The phenol STAYS in the ether, and that is exactly what separates it from the carboxylic acid.
Hydroxide
phenol
phenolate ion
Example: logK=14.0−10.20=3.80, so K=6.3×103: quantitative.
Hydroxide
acid AND phenol
both togetherdoes not distinguish
Example: logK=9.80 and 3.80: both reactions are complete and 4.00 g come out mixed.
What to do: Run the bicarbonate first, which takes only the acid, and keep the hydroxide for the NEXT wash.
A general rule, and it goes beyond this chapter: to separate two species of the same kind, choose the reagent whose strength falls BETWEEN the two, never the most powerful one available.
The mistakes that cost marks
These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.
1.Assuming the organic layer is on top
the whole product, hence the whole procedure and the lab report mark
What not to write
“The organic layer is on top, so I open the stopcock and throw away what comes out.”
What to write
“The extraction solvent is dichloromethane, ρ=1.326 g/mL against 1.000 for water: the ORGANIC layer is the one at the BOTTOM, I collect it and keep the other.”
Why: Nothing says the organic layer floats: chlorinated solvents are denser than water. Density does the sorting in the funnel, and it is read from a table before opening the stopcock.
2.Washing the crystals with hot solvent after a recrystallization
13 yield points on a 5.00 g recrystallization, and the next question becomes impossible to calculate
What not to write
“I rinse the crystals on the Büchner funnel with the boiling water that was left, to remove the mother liquor thoroughly.”
What to write
“I wash with 10.0 mL of ICE-COLD water: at 20 °C the loss is 0.029 g, at 95 °C it would be 0.680 g, that is 23 times more.”
The curve rises from 0.29 to 6.80 g per 100 mL between 20 and 95 °C: it is the same gap that makes the product crystallize on cooling and redissolves it in a hot wash.
Why: The wash is there to remove the mother liquor left between the crystals, not to redissolve them. With hot solvent, you undo exactly what the cooling had just built.
3.Cooling quickly to save time
2 points, plus a recrystallization to redo when the melting point comes out too low
What not to write
“I put the flask in ice as soon as everything has dissolved, the crystals appear right away.”
What to write
“I let it cool SLOWLY in air to room temperature, and only then in ice: large crystals form, and the impurities stay in the mother liquor.”
On the left, slow cooling grows large crystals and leaves the mother liquor outside. On the right, the fine powder traps the mother liquor and its impurities.
Why: Abrupt cooling gives a fine powder each grain of which traps mother liquor. The product looks abundant, but its melting point is depressed and broadened: the impurity is inside the solid, and a wash no longer reaches it.
4.Starting the acid-base extraction with sodium hydroxide
the whole separation, that is 4.00 g of two compounds mixed and inseparable
What not to write
“I wash with sodium hydroxide, the strongest base I have: it will take away everything acidic.”
What to write
“I wash first with bicarbonate: logK=2.15 with the carboxylic acid but −3.85 with the phenol, a ratio of 106. Only the acid leaves.”
Bicarbonate deprotonates only acids with a pKa below 6.35: the acid at 4.20 leaves, the phenol at 10.20 stays. Sodium hydroxide takes both.
Why: Sodium hydroxide deprotonates the acid AND the phenol: reacidifying the extract makes them reprecipitate together. Bicarbonate is not a bargain-bin reagent, it is the DISCRIMINATING reagent of the chapter.
5.Measuring the retention factor from the bottom of the plate
1 point per spot, and a polarity ranking that shifts across the whole plate
What not to write
“The spot is at 1.86 cm and the front at 7.30 cm, so Rf=7.301.86=0.255.”
What to write
“Both distances are measured from the BASELINE, placed at 0.50 cm: Rf=7.30−0.501.86−0.50=6.801.36=0.200.”
In red, distances measured from the bottom of the plate give 0.255. In green, measured from the baseline, they give the true retention factor, 0.200.
Why: The retention factor compares two MIGRATIONS, not two positions. The error is systematic and always inflates the small retention factors, precisely those that decide the choice of eluent.
6.Reading a high retention factor as strong polarity
1.5 points, and the order of elution from the column stated backwards
What not to write
“Compound C climbs almost to the front, Rf=0.80: so it is the most polar of the three.”
What to write
“Silica is POLAR and holds back whatever resembles it: Rf=0.80 signals the LEAST polar compound, and it will come out FIRST from the column.”
The compound that stays near the baseline is the MOST polar, the one that climbs toward the front is the least polar: on silica, a high retention factor reads as low polarity.
Why: The stationary phase decides the direction of reading. On silica, climbing means escaping the polar phase; on a reversed phase, the same value would mean the opposite.
7.Concluding the identity of the product from a sharp melting point
2 points, and a report conclusion that claims more than the measurement
What not to write
“My solid melts sharply at 122 °C, like the benzoic acid in the table: so it really is benzoic acid.”
What to write
“The solid melts sharply, so it is PURE. To identify it, I run a MIXED melting point with an authentic sample: if it stays sharp, it is the same compound; if it is depressed and broadened, it is not.”
A sharp melting point says PURE, not WHO. Mixed with the authentic sample, the solid stays sharp if it is the same compound; otherwise it is depressed and broadened.
Why: Thousands of compounds melt within the same degree. A melting point measures PURITY; identity requires a second piece of information, here the absence of depression on mixing.
8.Calculating the yield in mass, or on the wrong reagent
3 points, and the whole yield of the next question
What not to write
“We use 12.588 g of acid against 16.18 g of alcohol: the acid is the one in short supply, so it is the limiting one.”
What to write
“n(alcohol)=88.1516.18=0.1836 mol and n(acid)=60.0512.588=0.2096 mol: the ALCOHOL is limiting even though it is the heavier.”
On the left, the alcohol weighs more than the acid; on the right, it supplies fewer moles, 0.1836 against 0.2096. The right-hand column designates the limiting reagent.
Why: An equation counts MOLES, not grams. A reagent of low molar mass supplies more moles at equal mass, and it is almost always the one that reverses the comparison.
9.Reporting a yield above one hundred percent without comment
2 points, and doubt cast on every other measurement in the report
What not to write
“I get 25.8 g for 23.89 g expected, that is 108 percent: the reaction worked very well.”
What to write
“108 percent is impossible: the excess mass, at least 1.91 g, is solvent, water or an unremoved reagent. I dry, I distil, then I weigh to constant mass.”
The theoretical mass is a ceiling: the 1.91 g that stick out are not product, but solvent, water or a reagent still to be removed.
Why: Conservation of matter forbids exceeding the theoretical mass. Such a figure is therefore not a result, it is the diagnosis of a badly purified product, and it is dealt with before being written down.
10.Forgetting to vent the funnel after a bicarbonate wash
the stopper pops, half the crude product ends up in the fume hood, and the procedure is lost
What not to write
“I stopper the funnel, shake vigorously for thirty seconds, then put the whole thing on the stand.”
What to write
“I shake for a few seconds, I INVERT the funnel and open the stopcock toward the back of the fume hood to vent: the excess acid releases 638 mL of carbon dioxide here.”
Why: Bicarbonate turns the acid into carbon dioxide: the gas forms in a closed vessel and the pressure rises within seconds. Venting is not a comfort precaution, it is the condition of the wash.
Which method to choose
Faced with a mixture: which technique, on which clue
We look neither at the names of the compounds nor at the chapter: we look at the physical state and the numerical data given in the statement.
If two immiscible LIQUIDS, or a product in aqueous solution → liquid-liquid extraction, then a brine wash
Example: 2.40 g of product taken up by 3×25 mL of dichloromethane
brine raises the density difference from 0.098 to 0.298 g/mL with ethyl acetate, that is 3 times more, and settling becomes clean
If the statement gives pKa values, or mentions an acid, a phenol or an amine → acid-base extraction: we ionize what we want to move
Example: pKa=4.20 and 10.20 in the same ether, separated by bicarbonate
If two miscible LIQUIDS with two given boiling points → simple distillation if the gap exceeds about twenty degrees, column below that
Example: 111−56=55 °C, simple apparatus; 97−82=15 °C, column
with α=2.50 and x0=0.500, one plate gives y1=0.714 and five plates exceed 98.0 percent
If the product is a heat-sensitive LIQUID that boils very high → steam distillation, which makes the mixture boil below 100 °C
Example: eugenol, Tbp=254 °C, carried over at 99.5 °C at the rate of 0.159 g per gram of water
If a SOLID accompanied by an impurity, both soluble when hot → recrystallization from the minimum volume of boiling solvent
Example: 5.00 g of benzoic acid in 73.5 mL of boiling water, 4.875 g recovered
the solvent is chosen on the RATIO of solubilities, here 0.176.80=40, never on the hot solubility alone
If two NONVOLATILE compounds with different retention factors → column chromatography, eluent chosen for Rf≈0.30
Example: Rf=0.35 and 0.45: a gap of 0.10, hence 40 g of silica per gram of mixture
If no branch applies, the property that differs has not yet been named: go back to the statement and look for the numerical datum that has not been used. And if the mixture boils at a constant temperature as though it were a pure compound, it is an azeotrope: no column separates it.
Which reagent for which acid-base wash
We sweep the acidity scale through increasing values of pKa, and each wash must remove only ONE species.
Each wash removes ONE species and only one: hydrochloric acid takes the amine, bicarbonate the carboxylic acid, hydroxide the phenol, and the neutral biphenyl comes out of the ether.
If the compound to be removed is an AMINE → hydrochloric acid 1 mol/L, FIRST
Example: pKa(anilinium)=5.10: at pH=1.0 the amine leaves as the ammonium ion
If the compound to be removed is a CARBOXYLIC ACID and a phenol accompanies it → saturated sodium bicarbonate, never sodium hydroxide
Example: logK=+2.15 against −3.85: a factor of 106 between the two
If only a PHENOL is left to remove → sodium hydroxide 1 mol/L, LAST
Example: logK=14.0−10.20=3.80, so K=6.3×103
If the compound is neither acidic nor basic → we do nothing: it stays in the ether and is obtained by evaporation
Example: biphenyl, recovered alone after the three washes
Each aqueous extract is treated by the REVERSE reaction: the acid and the phenol get their proton back with hydrochloric acid, the amine is made neutral again with hydroxide, and the compound reprecipitates.
How the answer is expected to be written
A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.
Writing up a yield calculation that earns every point
When to use it: As soon as the statement gives two reagents and an obtained mass, even if it asks only for the yield.
1Convert each reagent to MOLES: go from volume to mass through the density, then from mass to moles through the molar mass. Write the two amounts one under the other.
2Compare the moles, taking the coefficients of the equation into account, and NAME the limiting reagent: “the alcohol is limiting because 0.1836<0.2096 mol”.
3Calculate the theoretical mass from the limiting reagent alone, never from an average of the two reagents.
4Take the ratio of the mass obtained to the theoretical mass and give it in percent, with the right number of significant figures.
5Check that the result is below 100 percent, and if it is not, say why instead of writing it down as it stands.
Concluding sentence
“The limiting reagent is 3-methylbutan-1-ol, with 0.1836 mol against 0.2096 mol of acetic acid. The theoretical mass of ester is 0.1836×130.18=23.89 g, and the yield is therefore 23.8917.2=72.0 percent.”
The trap: Skipping the moles step because the masses are already given: this is exactly where the marker puts the point, and the order of the masses is the reverse of that of the moles.
Marking: 1 point per amount of substance, 1 point for naming the limiting reagent, 1 point for the theoretical mass, 1 point for the ratio and its unit
Justifying the choice of a technique in a lab report
When to use it: At every purification step described in a report, and in any question that begins with “propose a method”.
1Name the two species to separate, the product on one side and what you want to rid it of on the other.
2Name the physical PROPERTY that differs between them, with the numerical datum that proves it.
3Deduce the technique, and only then give the procedure in two or three steps.
4State the measurable success criterion: melting range, a single retention factor spot on the plate, constant mass.
Concluding sentence
“The product and the coloured impurity differ in their solubility in ethanol as a function of temperature; we therefore purify by recrystallization, and success will be read from a melting range below 1 °C.”
The trap: Starting with the name of the technique. A marker who reads “we recrystallize” without the property exploited cannot know whether the choice is reasoned or recited, and takes off the justification point.
Marking: 1 point for the property named, 1 point for the technique, 1 point for the control criterion
Check before you hand in
Five minutes of checking recover more marks than one more problem started in a hurry.
The drop of water, to never mistake the layers
Before opening the stopcock, place a drop of water at the top of the funnel and watch where it goes. If it passes through the upper layer and joins the one at the bottom, the upper layer is organic.
A drop that merges immediately into the top layer says that layer is aqueous: with dichloromethane, that is exactly what should happen.
Keep all the layers until the final weighing
Do not throw away any layer before you have the dry product on the balance. This is the only check in the chapter that makes a mistake repairable.
A beaker labelled “aqueous, do not throw away” has already saved more than one procedure where the funnel had been read the wrong way round.
The co-spot, the only judge in chromatography
Never compare retention factors measured on TWO different plates. Spot the reagent, the mixture and a blend of the two on the SAME plate.
The same compound gives 0.500 on one plate and 0.398 on the next, a gap of 0.102: without a co-spot you would wrongly conclude two distinct compounds.
The order of magnitude of the yield
A yield is always compared with 100 percent BEFORE being copied down. Above it, the product is wet or impure. Below 20 percent on a standard reaction, it is often the limiting reagent that was wrongly chosen.
108.0 percent signals 1.91 g of solvent or impurity; a yield of 63.0 percent where about 72 is expected signals a theoretical mass calculated on the wrong reagent.
Constant mass, before any conclusion
Weigh, put it back to dry for fifteen minutes, weigh again. As long as the mass drops, what you are weighing contains solvent.
4.95 then 4.82 then 4.81 g: the mass is constant at 4.81 g, and 0.14 g of solvent went off between the first and the last weighing.
The melting range, in five seconds
Read the range before the value: less than one degree, the product is pure; several degrees, recrystallize before going any further.
114.0 to 119.0 °C makes a range of 5.0 °C and a depression of 3.4 °C below the 122.4 °C of the pure product: this solid is not presentable.
The typical problem, taken apart
Isoamyl acetate, from the flask to the weighed mass
20.0 mL of 3-methylbutan-1-ol and 12.0 mL of glacial acetic acid are refluxed with a few drops of sulfuric acid, then the mixture is washed, dried, distilled and weighed.
17.2 g of ester are collected after distillation. A classmate, who skipped the distillation, reports 25.8 g.
The situation of the example: 20.0 mL of alcohol and 12.0 mL of acetic acid heated at reflux with sulfuric acid, before washing, drying, distillation and weighing.
Step 1
m(alcohol)=20.0×0.809=16.18 g, so n=88.1516.18=0.1836 mol. m(acid)=12.0×1.049=12.588 g, so n=60.0512.588=0.2096 mol.
Why
Convert BEFORE comparing. Writing the two amounts one under the other makes the reversal visible: the alcohol weighs more and yet supplies fewer moles, because its molar mass is one and a half times larger.
Step 2
The equation is equimolar, and 0.1836<0.2096: the limiting reagent is 3-methylbutan-1-ol.
Why
This sentence is worth a point by itself, and it closes the door on the trap in the statement, which whispers that the acid is limiting “since less of it is used”.
Step 3
mtheoretical=0.1836×130.18=23.89 g.
Why
The theoretical mass is calculated on the limiting reagent ALONE. The excess acid, 0.0261 mol, produces nothing: it will be removed in the wash, and it is what will release 638 mL of carbon dioxide in the funnel.
Step 4
yield=23.8917.2=0.720, that is 72.0 percent.
Why
The ratio is taken on masses, but only AFTER the limiting reagent has fixed the denominator. The losses that remain are those of the work-up, residual solubility in the washes and forerun fractions in the distillation, not those of the reaction.
Step 5
For the classmate: 23.8925.8=1.080, that is 108.0 percent, hence at least 25.8−23.89=1.91 g of impurities.
Why
Exceeding the theoretical mass is not an achievement, it is a measure of what is not the product. The figure serves as a diagnosis: dry, distil and reweigh to constant mass before writing anything down.
Step 6
Check: 72.0 percent is below 100, the order of magnitude is that of a well-run esterification, and the mass weighed is that of a flask tared empty.
Why
The check takes five seconds and catches the two costliest mistakes of the chapter, the impossible yield and the weighing done on a product that is still wet.
The conclusion, written out
“The limiting reagent is 3-methylbutan-1-ol, with 0.1836 mol against 0.2096 mol of acetic acid. The theoretical mass of ester is 23.89 g and the yield is 72.0 percent. The result of 108.0 percent announced without distillation corresponds to at least 1.91 g of residual solvent or reagent.”
The classic mistake on this problem: Comparing 16.18 g and 12.588 g and concluding that the acid is limiting. The yield then becomes 0.2096×130.1817.2=63.0 percent, nearly nine points off, and all the following questions inherit the error.
Learn by heart
•The property that DIFFERS decides the technique: two phases, extraction; acidity, acid-base extraction; boiling, distillation; solubility as a function of temperature, recrystallization; polarity, chromatography.
•The DENSEST layer is at the bottom. Ether 0.713, water 1.000, dichloromethane 1.326. Keep all the layers until the weighing.
•Acid-base extraction, in order: hydrochloric acid, then bicarbonate, then hydroxide. Bicarbonate takes the carboxylic acid and NOT the phenol.
•Recrystallization: MINIMUM volume of boiling solvent, SLOW cooling, Büchner filtration, wash with COLD solvent.
•On silica, a high Rf signals a WEAKLY polar compound, and the least polar comes out FIRST from the column. Distances are measured from the BASELINE.
•A melting range below 1 °C proves PURITY, not identity. An impurity LOWERS and BROADENS. Identity requires a MIXED melting point.
•The yield is calculated on the LIMITING reagent, identified in MOLES. Above 100 percent, the product is wet, solvated or impure: weigh to constant mass.
Frequently asked questions
How do I know whether the organic layer is on top or at the bottom?
You read the density of the solvent, never a habit. The densest layer is always at the bottom. Ether, at 0.713 grams per millilitre, floats on water; dichloromethane, at 1.326, passes underneath. When in doubt, a drop of water placed at the top of the funnel passes through the upper layer if that layer is organic.
Why are crystals washed with cold solvent?
Because the wash is there to remove the mother liquor left between the crystals, not to redissolve the product. With benzoic acid, ten millilitres of water at 20 degrees carry off 0.029 gram, against 0.680 gram at 95 degrees, which is twenty-three times more. A hot wash undoes what the cooling had just built.
What does a high retention factor mean in chromatography?
On a silica plate, the stationary phase is polar and holds back polar compounds. A high retention factor therefore signals a weakly polar compound, which was little retained and migrated a lot. It is also the one that will come out first from a silica column. Both distances are measured from the baseline.
How do I calculate the yield of an organic synthesis?
You convert each reagent to moles, you identify the limiting reagent by comparing these moles and not the masses, you calculate the theoretical mass from the limiting reagent alone, then you divide the mass obtained by this theoretical mass. The heaviest reagent is not necessarily the limiting one: it all depends on the molar mass.
Does a sharp melting point prove that I have the right product?
No, it proves only that the solid is pure. Thousands of compounds melt at the same place to within a few degrees. For identity, you run a mixed melting point with an authentic sample: if it stays sharp, it is the same compound; if it is depressed and broadened, it is not.
Why use bicarbonate rather than sodium hydroxide?
Because bicarbonate distinguishes and sodium hydroxide does not. It deprotonates the carboxylic acid but not the phenol, with a gap of six orders of magnitude between the two constants. Sodium hydroxide deprotonates both at once, and the extract then contains a mixture that reacidification makes reprecipitate together.
A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.
Something missing from this sheet?A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.
Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space.
Rather not create an account? Write to me directly.
Get in touch for a first session. In front of a mixture, we do not recite a procedure: we name the property that differs, and the technique follows. That is the method set up here.
