Corrected exercises: laboratory techniques in organic chemistry (202-SNU-RE)

The solvent boils in the flask, the vapour rises into the condenser, meets a cold wall, condenses, and the liquid falls back into the flask under gravity. The circuit is closed for MATTER and open for PRESSURE, since the top of the condenser communicates with the air of the fume hood. Nothing leaves, so the mass of solvent stays constant, which is exactly what the word reflux means: a return. The benefit goes further than saving solvent. As long as liquid is boiling, the temperature of the mixture is LOCKED at the boiling point of the solvent, whatever the power of the hot plate. The chemist therefore does not set the temperature with the dial, but by CHOOSING the solvent. Turning the dial changes only the flow of vapour, that is, how vigorous the reflux is. This point deserves to be remembered, because the most common mistake in the lab is to turn the plate up to speed up a slow reaction: you do not gain one degree, you merely send the vapour too high up the condenser, until it is lost.

We need a solvent whose boiling point lies between $100$ and $130$ °C. The table offers only one, toluene, at $111$ °C. Water at $100$ °C sits at the lower limit and would suit a poorly soluble organic compound badly; xylene at $139$ °C exceeds the decomposition temperature of the product. The flask will therefore settle at $111$ °C as long as the toluene boils. Compared with methanol we gain $111-65=46$ degrees. This gain is far from trivial: a rule of thumb says that a rise of ten degrees roughly doubles the rate of a reaction, so $46$ degrees represent a factor of the order of twenty on the reaction time. This is why a procedure always specifies the solvent BEFORE the temperature: the first imposes the second, and changing solvent for solubility reasons also changes, without anyone thinking about it, the working temperature.

The water enters at the bottom because the jacket of the condenser then fills completely, from the bottom up, driving the air ahead of it. Connected at the top, the water runs down the wall and leaves a pocket of air in the lower part of the jacket, precisely where the vapour arrives hottest: the exchange surface is cut short and condensation becomes incomplete. One then observes vapour rising high in the tube, a ring of condensation that climbs instead of staying in the lower third, and eventually a smell of solvent in the fume hood. The second effect is mechanical: a condenser fed from the top empties as soon as the mains pressure drops, whereas fed from the bottom it stays full even if the flow stops. The visual marker to remember is the reflux ring, the sharp boundary between the wet tube and the dry tube: it must stay in the first third of the condenser.

The mass lost is $0.080\times 250\times 0.867=17.34$ g, which corresponds to $20$ mL. This leaves $230$ mL of toluene. Since the amount of reagents has not changed, their concentration is multiplied by $\frac{250}{230}=1.0870$, a rise of $8.7$ percent. Two practical consequences follow. First, the mixture becomes more concentrated, which speeds up the desired reaction but also favours second-order side reactions, dimerizations and polymerizations above all, whose rate depends on the square of the concentration. Second, and this is the more annoying one, if the volume of solvent had been calculated to keep all the product dissolved when hot, there is a risk of crystallization in the flask during the reaction, which removes the product from the medium and spoils the monitoring. The check that costs ten seconds: the liquid level in the flask must be the same at the end as at the start.

A pure liquid heated in a smooth, clean vessel can exceed its boiling point without boiling, for lack of a nucleus around which a bubble can form: this is superheating. When a bubble finally appears, it releases all the accumulated energy at once and throws boiling liquid out of the flask. Boiling chips, being porous, trap air in their cavities and constantly supply these nucleation sites; a magnetic stir bar plays the same role by stirring the liquid and creating surfaces where bubbles can break away. They are never added to a liquid that is already hot, and above all not to a superheated liquid, because the addition instantly triggers all the pent-up boiling and sends the contents of the flask into the fume hood. If you forgot, the rule is to let the flask cool well down before adding the chip. A boiling chip should not be reused either, its pores filling with liquid on first use: a chip that has already been used protects against nothing.

The densest layer sits at the bottom, without exception: a separatory funnel is just a tube in which gravity does the sorting. Diethyl ether, at $0.713$ g/mL, is lighter than water at $1.000$: the organic layer is therefore ON TOP. Dichloromethane, at $1.326$ g/mL, is clearly heavier: the organic layer is AT THE BOTTOM. There is no rule of the type the organic layer floats, and this is the number one source of error of the session: the student who has memorized one case throws the wrong layer into the other. The halogenated solvents, dichloromethane and chloroform, are all denser than water because of the mass of the chlorine atoms; almost all the other common organic solvents are lighter. The correct reflex is to read the density before opening the stopcock, and to KEEP both layers until the final weighing of the product, which makes any mistake repairable.

The density is $\frac{79.6}{60.0}=1.327$ g/mL. The table gives dichloromethane at $1.326$: it is that one. Methanol was not a possible candidate, and for a reason that has nothing to do with its density: it is MISCIBLE with water in all proportions, so it never forms a second layer. This is the trap of the table, which lists extraction solvents side by side with a solvent that extracts nothing. An extraction solvent must satisfy three conditions at once: be immiscible with water, dissolve the product better than water does, and boil low enough to be evaporated afterwards without heating the product. Methanol, ethanol and acetone fail the first; ether, dichloromethane, ethyl acetate and toluene satisfy it. The field test, when in doubt about which layer is which, is to add a drop of water at the top of the funnel and watch where it goes: if it passes through the upper layer, that layer is organic.

With pure water, the difference is $1.000-0.902=0.098$ g/mL. With brine, it is $1.20-0.902=0.298$ g/mL, that is $\frac{0.298}{0.098}=3.04$ times larger. Now the settling rate grows with this difference: three times the difference means a separation that takes one minute instead of ten, and a sharp interface instead of a cloudy veil. This is the main service of brine, and it is decisive with ethyl acetate, whose natural difference of $0.098$ g/mL makes it a poor candidate for fast settling. The second service comes from the salting-out effect: the sodium and chloride ions capture water molecules for their own solvation, which lowers the solubility of the organic product in the aqueous phase and, symmetrically, drives a good part of the dissolved water out of the organic phase. A final brine wash therefore does half the work of the drying agent, which is why the procedure places it just before the magnesium sulfate.

An emulsion is a stable dispersion of fine droplets of one phase in the other, recognizable by its milky appearance and its refusal to separate. Three causes come up again and again. First, shaking that is too violent, which breaks the drops up until their rise becomes slower than diffusion. Second, the presence of a SURFACTANT, soap from a saponification, a carboxylate formed during a basic wash, or simply an amphiphilic impurity: the molecule settles at the interface and stabilizes it. Third, a density difference that is too small, the case of the previous question. The remedies follow from the causes. From now on, mix by slow, repeated INVERSIONS rather than by shaking; add brine, which increases the density difference and breaks up the solvation of the polar heads; and if nothing moves, let it stand, filter through a little Celite to hold back the solid particles that act as a support, or centrifuge. Adding a few extra millilitres of organic solvent also dilutes the surfactant. What you must NOT do is shake harder.

The total organic phase measures $75.0$ mL and weighs $75.0\times 1.326=99.45$ g. The product makes up only $\frac{2.40}{99.45}=2.41$ percent of that mass. The conclusion is that at this stage the balance says nothing: weighing the organic phase amounts to weighing solvent, and the variation of interest, two or three tenths of a gram of product, is lost in the hundred grams of dichloromethane. You therefore NEVER weigh before evaporation, and the evaporation flask is tared empty so that the final weighing covers the product alone. This ratio of forty between solvent and product also explains why one percent of residual solvent is enough to skew a yield: that is the subject of exercise $4$. Note finally that $75$ mL of dichloromethane for $2.40$ g of product is a normal extraction ratio, and that you should not try to save on it at the price of an incomplete extraction.

Start with the ACID, then work up in basicity. Wash $1$ with hydrochloric acid: only $4$-methylaniline is basic, it becomes the $4$-methylanilinium ion, which is charged, so it passes into the water. The other three stay in the ether, benzoic acid and the phenol because an acidic medium does not deprotonate them, biphenyl because nothing touches it. Wash $2$ with bicarbonate: a weak base, it deprotonates only benzoic acid, which leaves as benzoate. Wash $3$ with hydroxide: a strong base, it finally deprotonates the phenol, which leaves as $4$-methylphenolate. Only biphenyl remains in the ether, recovered by simple evaporation. Each aqueous extract is then treated by the reverse reaction: the aniline is made neutral again with hydroxide, the acid and the phenol are made neutral again with hydrochloric acid, and the neutral compound reprecipitates or is re-extracted into ether. The order matters, and the reason is in the next question.

The two reactions are $PhCOOH+HCO_{3}^{-}\rightleftharpoons PhCOO^{-}+H_{2}CO_{3}$ and $ArOH+HCO_{3}^{-}\rightleftharpoons ArO^{-}+H_{2}CO_{3}$. For a proton transfer between two couples, $\log K=pK_{a}(\text{acceptor})-pK_{a}(\text{donor})$. With benzoic acid, $\log K=6.35-4.20=2.15$, so $K=1.4\times 10^{2}$, that is $141$: the reaction is quantitative, more than $99$ percent of the acid goes into the water. With the phenol, $\log K=6.35-10.20=-3.85$, so $K=1.4\times 10^{-4}$: the reaction practically does not take place and the phenol stays in the ether. The ratio of the two constants is $10^{n}$ with $n=2.15-(-3.85)=6$, that is a MILLION. It is this gap of six orders of magnitude that makes bicarbonate the discriminating reagent of the chapter, and it is the only good answer to the question why not use sodium hydroxide right away. The $pK_{a}$ gap between the two compounds, six units, is found exactly in the exponent, which gives an immediate check of the calculation.

The reaction is $ArOH+HO^{-}\rightleftharpoons ArO^{-}+H_{2}O$. The acceptor couple is $H_{2}O/HO^{-}$, whose $pK_{a}$ is $14.0$ in the sense used here, so $\log K=14.0-10.20=3.80$ and $K=6.3\times 10^{3}$. The reaction is quantitative: nearly all the phenol turns into phenolate and migrates into the water. This also explains why sodium hydroxide was not suitable for wash $2$: applied to the mixture, it would also have deprotonated benzoic acid, whose $\log K$ with hydroxide is $14.0-4.20=9.8$, an even more complete reaction. Sodium hydroxide is too strong to DISTINGUISH, and this is a general rule: to separate two species of the same kind, choose the reagent whose strength falls BETWEEN the two, never the most powerful one available. The reasoning applies identically to a mixture of two amines of different basicities, where you would choose a buffer rather than hydrochloric acid.

At $pH=2.0$, the ratio of the forms is $\frac{[A^{-}]}{[AH]}=10^{pH-pK_{a}}=10^{2.0-4.20}=10^{-2.20}=6.31\times 10^{-3}$. The fraction still ionized is therefore $\frac{6.31\times 10^{-3}}{1+6.31\times 10^{-3}}=6.27\times 10^{-3}$, that is $0.63$ percent: $99.37$ percent of the acid has returned to its neutral form, insoluble in water, and precipitates. This represents $2.00\times 0.9937=1.987$ g. But the neutral form itself is not perfectly insoluble: of it, $0.17\times 0.60=0.102$ g remains dissolved in the $60$ mL of aqueous phase. The mass that can be filtered off is therefore $1.987-0.102=1.885$ g, a recovery yield of $94.3$ percent. The two losses do not have the same origin and are not corrected in the same way: the first, tiny, is reduced by acidifying further; the second, ten times heavier, is reduced by cooling in ice and using no more water than necessary. This is already the reasoning of recrystallization.

His first aqueous phase contains the benzoate AND the $4$-methylphenolate, since sodium hydroxide deprotonates both, that is $2.00+2.00=4.00$ g of two compounds mixed together from which he has no simple way left to separate: reacidifying them makes them reprecipitate together. In a single manipulation he has therefore lost half of the mass engaged, in the form of a mixture. The general principle to take from this is that acid-base extraction is carried out from the most SELECTIVE reagent to the strongest, and never the reverse: each wash must remove ONE species, and only one. The sequence hydrochloric acid, then bicarbonate, then hydroxide respects this rule because the three reagents sweep the acidity scale through increasing values of $pK_{a}$, $-6$ for hydrochloric acid, $6.35$ for bicarbonate, $14$ for hydroxide, bracketing the $pK_{a}$ values of the compounds to be separated, $4.20$ and $10.20$.

For an acid-sensitive product we choose SODIUM sulfate. Once hydrated, magnesium sulfate releases a weak but real Lewis acidity at its surface, the magnesium ion being much more polarizing than the sodium ion, and this is enough to isomerize an enol, open a fragile acetal or dehydrate a tertiary alcohol. Sodium sulfate, being neutral, does not have this flaw, at the price of a lower capacity and a slow action, about twenty minutes against two or three. On the other hand it filters far better, its grains being coarse where magnesium sulfate is a powder that clogs the filter. The visual criterion for the end of drying is simple and unambiguous: you add the drying agent in small portions while swirling; as long as there is water left, the grains clump into lumps that stick to the bottom; as soon as the last portion added stays POWDERY and swirls freely when the Erlenmeyer flask is turned, drying is finished. The solution then goes from cloudy to clear.

The molar mass of anhydrous magnesium sulfate is $24.305+32.06+4\times 15.999=120.36$ g/mol. The heptahydrate fixes seven molecules of water, that is $7\times 18.015=126.105$ g of water per mole of salt. So $1.00$ g of anhydrous magnesium sulfate fixes $\frac{126.105}{120.36}=1.048$ g of water: the drying agent absorbs slightly more than its own mass, which makes it the most efficient of the common neutral drying agents. Calcium chloride does even better but forms complexes with alcohols and amines, which rules it out for most organic drying; sodium sulfate, which stops at $Na_{2}SO_{4}\cdot 10H_{2}O$, fixes $\frac{180.15}{142.04}=1.27$ g of water per gram in theory but much less in practice, because its hydration is slow and incomplete at room temperature.

The $100$ mL of ether weigh $100\times 0.713=71.3$ g and contain $0.0150\times 71.3=1.07$ g of water. The minimum mass of drying agent is therefore $\frac{1.07}{1.048}=1.02$ g. Three to five times more is used, that is three to five grams, for a reason of EQUILIBRIUM: hydration is not total, the salt and the residual water coexist, and the final water content depends on the ratio between the drying agent still anhydrous and that already hydrated. With the exact stoichiometric amount, the last grain would be only just hydrated and the organic phase would remain damp. The excess costs little and has only one drawback, the mechanical retention of product by the solid, which is compensated by rinsing the filtered drying agent with a few millilitres of fresh solvent. This loss by retention amounts to a few tenths of a percent, far below the drying gain.

The mass of solvent to be removed is $120\times 1.326=159.1$ g, thirty-three times the mass of the product. If $1.0$ percent of the solvent remains, that makes $0.010\times 159.1=1.59$ g in the flask, which add to the $4.80$ g of product: the relative error is $\frac{1.59}{4.80}=33.2$ percent. In other words, one percent of forgotten solvent becomes an error of thirty-three percent on the yield. This factor of thirty-three is the key of the chapter and by itself explains most of the absurd yields handed in with lab reports, including those that exceed one hundred percent. The remedy fits in a sentence: never weigh a product that has not been under high vacuum, and keep in mind that the smaller the amount of product, the more violent this trap is, the solvent to product ratio increasing when the synthesis is modest.

Lowering the pressure lowers the boiling point, because a liquid boils as soon as its vapour pressure equals the external pressure. Under two hundred millibars, dichloromethane boils around room temperature, so the bath of the evaporator can stay at $40$ °C without the product ever exceeding that value: this is the only way to concentrate a solution without cooking a heat-sensitive compound. The rotation spreads a thin film on the wall of the flask and multiplies the exchange surface, which makes evaporation fast and regular while avoiding superheating. The dryness criterion is WEIGHING TO CONSTANT MASS: you weigh, put it back under vacuum for a quarter of an hour, weigh again, and repeat as long as the mass drops. Here $4.95$, then $4.82$, then $4.81$ g: the difference between the last two weighings, $0.01$ g, is at the level of the balance, so constant mass is reached at $4.81$ g. Between the first and the last, $4.95-4.81=0.14$ g of solvent has gone. Any later characterization presupposes this point has been reached: a wet or solvated product gives a spectrum that is not its own.

For the first pair, $111-56=55$ °C; for the second, $97-82=15$ °C. The workshop rule says that beyond about twenty degrees a simple distillation is enough, and that below that a plate column is indispensable: the first pair is therefore handled with the simple apparatus, the second requires a Vigreux or packed column. The two curves say it without any calculation. The solid line shows a sharp PLATEAU near $58$ °C while the acetone comes over, then an abrupt jump to about $105$ °C when it is exhausted, then a second plateau. Each plateau is a pure fraction and the jump says exactly where to cut. The dashed line, for its part, rises steadily from $82$ to $97$ °C without the slightest plateau: at every instant the distillate is a mixture whose composition changes, and no collected volume is pure. There is nothing to cut, hence nothing to recover: this is the signature of a simple distillation applied to a pair that is too close.

A theoretical plate is one complete cycle of vaporization and condensation. First plate: $y_{1}=\frac{2.50\times 0.500}{1+1.50\times 0.500}=\frac{1.250}{1.750}=0.714$. Second plate, starting again from $x=0.714$: $y_{2}=\frac{2.50\times 0.714}{1+1.50\times 0.714}=\frac{1.786}{2.071}=0.862$. Continuing: $y_{3}=0.940$, $y_{4}=0.975$, $y_{5}=0.990$. The $98.0$ percent threshold is crossed at the FIFTH plate, $y_{4}$ still being $97.5$ percent. A twenty-centimetre Vigreux column is worth about three to five plates, a packed column of the same length about ten: this shows why column height is the parameter you increase when a separation resists. Note the shape of the sequence, which progresses quickly at first and then drags: the last percent of purity cost more plates than the first fifty, and this is true of every purification, recrystallization included.

If $\alpha =1$, the formula gives $y=\frac{x}{1+0}=x$: the vapour has exactly the same composition as the liquid, so no plate enriches anything and no column, however tall, separates the mixture. This is the very definition of the azeotrope, and it is why we say that an azeotrope cannot be crossed by distillation: you have to change method, molecular sieve, third solvent or membrane. In the case of hydrochloric acid, the azeotrope is a MAXIMUM in temperature, $108.6$ °C, that is higher than pure water and than hydrogen chloride: it is therefore the azeotrope that stays at the BOTTOM and the component in excess that comes out. Starting from $200$ g at $10.0$ percent, there are $20.0$ g of hydrochloric acid, and it is WATER that goes to the condenser. When the residue reaches $20.2$ percent, it weighs $\frac{20.0}{0.202}=99.0$ g, so $200-99.0=101.0$ g of water have been removed. The residue will not concentrate any further: the composition of the flask converges toward the azeotrope instead of moving away from it, exactly the opposite of a minimum azeotrope, where it is the DISTILLATE that levels off.

In steam distillation, the two liquids are not miscible, so each exerts its vapour pressure as if it were alone and the mixture boils when the SUM reaches atmospheric pressure: $747+13=760$ torr. The mixture thus boils at $99.5$ °C whereas eugenol alone would boil at $254$ °C, which is exactly the goal, since it would decompose well before. In the vapour, the ratio of amounts of substance is the ratio of partial pressures, so the ratio of masses is $\frac{m_{e}}{m_{water}}=\frac{p_{e}M_{e}}{p_{water}M_{water}}=\frac{13\times 164.20}{747\times 18.02}=\frac{2134.6}{13461}=0.1586$ g of eugenol per gram of water. For $5.00$ g of eugenol, $\frac{5.00}{0.1586}=31.5$ g of water must therefore be distilled, about $32$ mL. The mass yield looks mediocre, one sixth, but it is excellent compared with what ordinary distillation would give, namely a tar. The high molar mass of the organic compound partly compensates for its low volatility, which is why the method works well on essential oils, nearly all of them heavy molecules.

The main fraction represents $\frac{38.0}{50.0}=76.0$ percent of the volume used. The thermometer bulb is placed level with the SIDE ARM, that is just below the opening through which the vapour leaves for the condenser, and not in the liquid. The reason is that we want to read the temperature of the vapour that actually LEAVES, which is in equilibrium with the condensed liquid running down over the bulb; placed too high, the thermometer reads a temperature that is too low because it is no longer bathed in vapour; placed in the liquid, it reads the temperature of the flask, which is that of the mixture and not that of the distillate. Finally, a distillation is never taken to dryness for two cumulative reasons: the empty flask overheats instantly and cracks, and above all the concentrated residue may contain PEROXIDES, accumulated in ethers and alkenes exposed to air, whose detonation is known to be violent. A pool of a few millilitres is therefore always left behind, and old ethers are tested before being heated.

The retention factor is the quotient of the distance travelled by the compound, measured from the BASELINE, by the distance travelled by the solvent front, measured from the same line. The $0.50$ cm must therefore be subtracted in both terms, which is the most frequent mistake of the chapter: measured from the bottom of the plate, spot A would give $\frac{1.86}{7.30}=0.255$ instead of the correct value. The front travelled $7.30-0.50=6.80$ cm. The three compounds travelled $1.36$, $3.40$ and $5.44$ cm, whence $R_{f}(A)=\frac{1.36}{6.80}=0.200$, $R_{f}(B)=\frac{3.40}{6.80}=0.500$ and $R_{f}(C)=\frac{5.44}{6.80}=0.800$. A retention factor is always between $0$ and $1$, has no unit, and is given here to the thousandth, as the answer fields ask, since the distances are read to a tenth of a millimetre.

Silica is a POLAR stationary phase, covered with silanol groups capable of hydrogen bonding. A polar compound adsorbs strongly to it, spends most of its time fixed and advances little; a nonpolar compound, held back by nothing, follows the eluent. The retention factor therefore varies in the OPPOSITE direction to polarity: A, at $0.200$, is the most polar, then B at $0.500$, then C at $0.800$, the least polar. The order of functional groups observed in practice follows this logic: hydrocarbon and halide at the top of the plate, then ether, then ester and ketone, then aldehyde, then alcohol, then amine, then carboxylic acid right at the bottom, often as a streak. This ranking is exactly that of the ability to form hydrogen bonds. Note that the direction is REVERSED on a so-called reversed phase, grafted with $C_{18}$ chains and therefore nonpolar: there, it is the polar compound that migrates fast, and this is the case of most analytical liquid chromatography.

We keep $30$ percent ethyl acetate, which gives $R_{f}=0.38$. The rule for preparing a column is to choose the eluent that gives the target compound a retention factor close to $0.30$: lower, the compound travels through the column too slowly and comes out diluted in a large volume; higher, it travels so fast that it no longer separates from its neighbours, and at $0.62$ everything would come out together in the first fractions. The retention factor from the plate is therefore a PRELIMINARY MEASUREMENT, made in five minutes, which avoids losing a day on a badly chosen column. The more ethyl acetate the eluent contains, the more polar it is, the more it competes with the silica to hold the compound, and the higher the compound climbs: this is the correct reading of the four values given. As for a baseline placed below the level of the eluent, it is fatal: the compound dissolves in the bath of the chamber instead of migrating up the plate, and you get a blank plate with a perfectly sharp front. The liquid level must always stay well below the baseline.

The co-spot is used to compare two spots UNDER THE SAME CONDITIONS. The retention factor is not a physical constant: it depends on the thickness of the layer, on the saturation of the chamber with vapour, on the temperature and on the volume of eluent. Two spots with the same retention factor on two different plates therefore prove nothing. On the co-spot, by contrast, if the two products are identical you see ONE spot, and if they differ you see two, even very close ones. The reaction is finished when the spot of the reagent has disappeared from lane M. The calculation confirms it: on the second plate, $R_{f}=\frac{2.03}{5.10}=0.398$, that is $0.40$, whereas B was $0.500$ on the first. The gap, $0.10$, is enormous on the scale of a plate and would be enough to conclude wrongly that these are two different compounds. The conclusion is that you NEVER compare two plates: you redo the spotting side by side.

Three common visualization methods. First the $254$ nm lamp, on a plate with fluorescent silica: compounds that absorb in the ultraviolet, hence all conjugated and aromatic compounds, quench the fluorescence and show up dark on a green background. Next IODINE vapour, which dissolves in almost any organic matter and gives brown spots, reversible since iodine sublimes again: it is the universal, nondestructive stain. Finally chemical stains, sprayed and then heated, such as potassium PERMANGANATE, which decolourizes wherever it finds an oxidizable group, or ninhydrin, specific to amines and amino acids. A colourless saturated alcohol does not absorb in the ultraviolet and stains poorly with iodine: permanganate is what is needed, the alcohol being oxidizable. Finally, the baseline is drawn in pencil because graphite does not migrate; the ink of a pen, for its part, is a mixture of organic dyes that climbs with the eluent and covers the plate with false coloured spots. And the line is drawn without scratching the silica, since a layer torn away deflects the eluent.

The ratio is $\frac{6.80}{0.17}=40$: water dissolves forty times more benzoic acid at $95$ °C than at $0$ °C, which makes it an excellent recrystallization solvent for this compound. A ratio below five would make the operation pointless, since almost everything would stay dissolved when cold; a solvent in which the compound is insoluble even when hot is no use either. The minimum volume of boiling water is $\frac{5.00}{6.80}\times 100=73.5$ mL. The word MINIMUM is an operating instruction here and not an elegance: you add the hot solvent in small portions until the last crystal disappears, and not one drop more. The two other criteria for choosing a solvent complete the picture: the impurities must either be insoluble when hot, in which case a hot filtration removes them, or on the contrary very soluble when cold, in which case they stay in the mother liquor; and the solvent must boil below the melting point of the solid, otherwise the solid melts instead of dissolving and separates as oily droplets that trap the impurities.

At $0$ °C, the $73.5$ mL still hold $0.17\times 0.735=0.125$ g. The mass of crystals is therefore $5.00-0.125=4.875$ g, a yield of $\frac{4.875}{5.00}=97.5$ percent. These $0.125$ g are an UNAVOIDABLE loss, and this is the point the exercise wants to get across: a recrystallization never gives one hundred percent, even when perfectly carried out, because the cold solution stays saturated. A recrystallization yield between $70$ and $90$ percent is normal in practice, the gap with the theoretical $97.5$ percent coming from excess solvent, from washing, and from product left on the glass. Note finally that the loss is in ABSOLUTE value the same whatever the mass used: recrystallizing $0.30$ g in this volume would leave $0.125$ g in solution, a loss of $42$ percent. Recrystallization is therefore a method that suits very small amounts poorly, where chromatography regains the advantage.

With $150$ mL, $0.17\times 1.50=0.255$ g remains dissolved, so $5.00-0.255=4.745$ g of crystals and a yield of $94.9$ percent. The excess solvent cost him $0.255-0.125=0.130$ g, that is $2.6$ points of yield, for a doubling of the volume. The relationship is direct: the mass lost is proportional to the volume, since the cold solution is saturated whatever its volume. Doubling the solvent doubles the loss. This also shows why the instruction to use the minimum volume is not a whim, and why the hot solvent should be added drop by drop at the end of the dissolution rather than all at once. The remedy, when you have diluted too much, exists: you concentrate by boiling off part of the solvent before cooling, which brings you back to the minimum volume. You can also collect a second crop by concentrating the mother liquor, less pure than the first and not to be mixed with it without recrystallizing it in turn.

The Büchner funnel filters under vacuum, which squeezes the cake much better than gravity filtration and leaves an almost dry solid. The mixture is poured with the pump running, onto a paper that has been wetted beforehand and pressed flat. The wash serves to drive out the mother liquor retained between the crystals, which contains the impurities: without it, the recrystallization is useless. But the wash water redissolves part of the product, and its temperature decides everything. At $20$ °C, the loss is $0.29\times 0.100=0.029$ g; at $95$ °C it is $6.80\times 0.100=0.680$ g, that is $\frac{0.680}{0.029}=23.4$ times more. A hot-solvent wash therefore destroys in ten seconds the benefit of the whole operation. As for cooling, it must be SLOW because the growth rate of the crystals decides their purity: slowly, the molecules have time to line up and the lattice excludes whatever does not resemble it, giving large pure crystals; abruptly, in an ice bath straight out of the heating, you get a fine powder that traps the mother liquor and you recover the impurity along with the product.

The range is $119.0-114.0=5.0$ °C, where a pure compound melts within less than one degree, and the depression of the end of melting is $122.4-119.0=3.4$ °C. The two signals point the same way and say the same thing: the solid is IMPURE. The theory is that of cryoscopy applied to the solid, the impurity behaving like a solute that lowers the melting point of the solvent formed by the product, and that broadens it because the composition of the liquid changes as melting proceeds. The course of action is to recrystallize a second time. On the other hand, and this is the limit you need to know, a sharp melting point proves only PURITY, never identity: thousands of compounds melt between $120$ and $125$ °C. To prove identity you run a MIXED melting point: the sample is ground together with an authentic sample of benzoic acid, and measured. If the mixture still melts sharply at $122.4$ °C, it is the same compound; if it melts depressed and broadened, for example from $108$ to $116$ °C, the two behave as impurities for each other, so they are different.

FALSE. The organic layer is on top if and only if its density is lower than that of the aqueous phase. Diethyl ether, at $0.713$ g/mL, floats on water; dichloromethane, at $1.326$ g/mL, sinks. Chlorinated solvents are systematically denser than water, which makes the sentence an unjustified generalization in half of the extractions of the course. The borderline case deserves attention: a solvent of $1.10$ g/mL ends up at the bottom with pure water, at $1.000$, but on TOP with saturated brine, at $1.20$, so that the same experiment changes geometry in the middle of a procedure. This is exactly why the survival rule is to keep both layers in labelled Erlenmeyer flasks until the final weighing of the product, and to check by adding a drop of water when in doubt.

FALSE, and costly. The wash is done with COLD solvent, ideally chilled in ice, and with the smallest possible volume. The calculation of exercise $7$ puts a figure on it: $10$ mL of water at $20$ °C carry off $0.029$ g of benzoic acid, whereas the same $10$ mL at $95$ °C carry off $0.680$ g, that is $23.4$ times more. A hot wash cancels in a few seconds the whole benefit of the crystallization, and a student who repeats it three times sees the yield collapse without understanding why. The reason is obvious once stated: the recrystallization solvent was chosen to dissolve a lot when hot, so washing with that same solvent hot amounts to redissolving the product. The wash must also be done with the recrystallization solvent and not another one, otherwise you dissolve the product in a solvent where it happens to be very soluble.

FALSE. A sharp melting point, that is a range of less than one degree, proves the PURITY of the solid, nothing more. Thousands of organic compounds melt within the same window of a few degrees, and benzoic acid at $122.4$ °C shares its value with many others. What proves identity is the MIXED melting point: ground together with an authentic sample, the same compound always melts sharply and at the same temperature, whereas two different compounds depress and broaden each other, each one playing the role of impurity for the other. One can also conclude with the spectroscopic techniques, which have their own chapter, but the mixed melting point remains the fastest and cheapest tool, and it requires only five milligrams of product.

FALSE, it is exactly the opposite on a silica plate. Silica is a polar stationary phase: it holds a compound back all the better the more polar it is, so a polar compound advances LITTLE and has a LOW retention factor. A retention factor of $0.80$ signals on the contrary a nonpolar compound, a hydrocarbon, halide or ether, held back by nothing. The order observed in a hexane and ethyl acetate eluent runs from the hydrocarbon down to the carboxylic acid, following the ability to form hydrogen bonds. The sentence is true only on a REVERSED phase, grafted with hydrocarbon chains, where the direction of retention is reversed; but that is not what is used in thin-layer chromatography at the CEGEP level, and the unqualified statement is false.

FALSE, and this is the most frequent mistake in lab reports. The yield is calculated on the LIMITING reagent and in MOLES, never in mass. The counterexample of exercise $9$ is conclusive: $16.18$ g of isoamyl alcohol, that is $0.1836$ mol, react with $12.588$ g of acetic acid, that is $0.2096$ mol. The alcohol is the limiting reagent even though it weighs the most, because its molar mass is much larger, $88.15$ against $60.05$ g/mol. Comparing masses would lead to naming the acid, hence to calculating a wrong theoretical mass and a wrong yield. The correct method takes three steps: convert each mass to moles, divide by the stoichiometric coefficient, keep the smallest quotient. And one last piece of common sense: a yield above one hundred percent does not exist, it signals a wet, solvated or impure product, never an achievement.

The masses are $m_{alcohol}=20.0\times 0.809=16.18$ g and $m_{acid}=12.0\times 1.049=12.588$ g. The amounts of substance are $n_{alcohol}=\frac{16.18}{88.15}=0.1836$ mol and $n_{acid}=\frac{12.588}{60.05}=0.2096$ mol. Since the esterification consumes one mole of each, the limiting reagent is the one with the smaller amount of substance, that is the ALCOHOL. The student is wrong because he compares masses: the alcohol does weigh more, $16.18$ g against $12.588$, but its molar mass is much larger, $88.15$ against $60.05$ g/mol, so the same mass contains clearly fewer molecules. The factor $\frac{88.15}{60.05}=1.468$ is enough to reverse the ranking whenever the ratio of masses stays below this value, which is the case here with $\frac{16.18}{12.588}=1.285$. A balance never weighs molecules, and this is why the first line of a lab report is always a table of amounts of substance.

The reaction gives one mole of ester per mole of alcohol, so $n_{ester}=0.1836$ mol at most and $m_{theoretical}=0.1836\times 130.18=23.89$ g. This would represent $\frac{23.89}{0.876}=27.3$ mL, a useful volume to know before choosing the distillation flask, which should be filled between one third and two thirds. Note that the Fischer esterification is an EQUILIBRIUM, so this theoretical mass will never be reached even with a perfect work-up: it is precisely to shift the equilibrium that the acid is used in excess, which explains the choice of the volumes in the statement. The realistic yield for this pair is of the order of seventy percent. Note finally that the yield is ALWAYS calculated against this theoretical mass, and never against the mass of the crude product before work-up.

The bicarbonate wash removes the excess acetic acid, converted into sodium acetate, which is soluble in water, as well as the sulfuric acid of the catalyst. The excess acid is $0.2096-0.1836=0.0261$ mol, and since each mole of acid releases one mole of carbon dioxide, the gas released occupies $0.0261\times 24.45=0.638$ L, that is $638$ mL. This is considerable in a $250$ mL funnel closed with a stopper: hence the obligation to VENT, that is, to turn the funnel stopcock-up and open it at the very first inversion, then after each shaking. A student who shakes without venting has the stopper ejected and the contents thrown out. As for the layer that contains the ester, it is the UPPER one: isoamyl acetate has a density of $0.876$ g/mL, lower than that of water, and the bicarbonate wash raises the density of the aqueous phase even further.

The yield is $\frac{17.2}{23.89}=72.0$ percent. Two causes of loss belong to the work-up alone. First, the residual SOLUBILITY of the ester in the aqueous wash phases, small but not zero, amounting to a few tenths of a gram over three washes, and reduced by washing with brine rather than water. Second, DISTILLATION losses: the liquid that wets the flask, the column and the condenser, the forerun that is discarded because it still contains alcohol, and the residue that is not distilled to dryness. To these add the losses by transfer and the product retained by the filtered magnesium sulfate, which a rinse with fresh solvent partly recovers. None of these losses is a matter of chemistry: they are reduced by care and by the choice of glassware, and this is why two pairs of students running the same reaction hand in yields twenty points apart.

His apparent yield is $\frac{25.8}{23.89}=108.0$ percent, which is impossible: no reaction creates matter. The minimum mass of impurities is $25.8-23.89=1.91$ g, and it is most probably unreacted isoamyl alcohol, which the distillation would have separated since it boils at $131$ °C against $142$ for the ester, or water left in the product for lack of sufficient drying. What must be done fits in three steps: dry again over magnesium sulfate, distil, collecting only the fraction that comes over at a constant temperature around $142$ °C, then weigh to constant mass. The general lesson of the problem, and the thread of the whole set: a yield above one hundred percent is never an achievement, it is the signal that a purification step was skipped. A marker sees it in a second, and a product whose spectrum were recorded afterwards would give signals that do not belong to the product.

Mixture one: what differs is the BOILING POINT, with a gap of $138-65=73$ °C. This is well beyond the twenty degrees that separate the domain of simple distillation from that of the column, so a simple distillation is enough and will give two sharp plateaus. Mixture two: what differs is the SOLUBILITY AS A FUNCTION OF TEMPERATURE, so this is a recrystallization, the product crystallizing when cold while the impurity, present in trace amount, stays too dilute in the mother liquor to reach its saturation; a little activated charcoal during the hot dissolution also adsorbs the coloured matter. Mixture three: what differs is the ACIDITY, hence the solubility once the species is ionized, and this is an acid-base extraction, a bicarbonate wash being enough to move the acid into the water while leaving the hydrocarbon in the ether. Mixture four: what differs is the POLARITY, the two compounds being neither volatile nor ionizable, so this is column chromatography.

The column is a chromatography on silica, exactly like the plate, and the rule is the same: what climbs high on the plate comes down fast in the column. The order of elution therefore follows DECREASING retention factors, that is $0.62$ first, then $0.41$, then $0.18$. The justification fits in a sentence: the least polar compound is the least retained by the silica, so it spends the least time adsorbed and advances fastest. The compound at $0.18$ will come out late, diluted in a large volume, and often as a streak: this is why the proportion of ethyl acetate is gradually increased during elution, a technique called gradient elution, which pushes the polar compounds without having had to carry them along from the start. Note that the plate serves as a road map for the column: a thin-layer chromatography done in five minutes predicts the whole behaviour of a column that will take two hours.

The recovery rate is $\frac{1.42+0.71}{2.40}=\frac{2.13}{2.40}=88.8$ percent, and $2.40-2.13=0.27$ g is missing. At least two explanations. The first is IRREVERSIBLE ADSORPTION on the silica: very polar compounds, carboxylic acids and amines first, bind strongly enough to the silanols not to come off again, and a fraction of a few percent stays on the column for good. The second is TAILING between fractions: the zones are never perfectly separated, and the intermediate tubes that contain a mixture of the two compounds were set aside because they were pure neither one nor the other, which is a deliberate choice of purity over yield. To these add, more mundanely, losses by evaporation of the product with the solvent when concentrating the fractions, and product left in the glassware. A recovery rate between $85$ and $95$ percent is normal; below $70$, decomposition on the silica, which is acidic, should be suspected.

We need $2.40\times 40=96$ g of silica, a column of decent size for two and a half grams of product: this is the usual order of magnitude, and it always comes as a surprise. The retention factor gap of mixture four is $0.45-0.35=0.10$, which is indeed a difficult separation, the rule being that a gap of $0.10$ or less requires a silica ratio of forty or fifty to one, where a gap of $0.25$ is content with twenty to one. The amount of silica here is the equivalent of the number of plates of a distillation column: the smaller the difference being exploited, the more repetitions of the partition are needed to amplify it. Finally, if the liquid level drops below the top of the silica bed, air gets in, the bed cracks and CHANNELS form: the eluent flows through these preferential paths instead of crossing the whole cross-section, the bands become distorted and the resolution collapses irreversibly. A cracked column cannot be rescued, it is repacked.

The yield is $\frac{1.42}{2.05}=69.3$ percent. The check needed before any spectrum is double, and it is the end point of the set: the product must be DRY and FREE OF SOLVENT. This is proved by weighing to constant mass after vacuum, and not by looking at the vial. The reason is that residual solvent and water do not stay discreet: they add their own signals, a broad band in the region of the hydroxyl groups in infrared, stray peaks in magnetic resonance, and you then spend an hour assigning a functional group the product does not have. The ethyl acetate from the column, in particular, is a classic trap because it carries a carbonyl. Characterization itself has its own chapter; this chapter stops exactly where it begins, that is at the moment when you hold in your hands a few hundred milligrams of solid that is pure, dry and weighed.