Meeting 8 - Condensed Answer Key
1. Online Quiz (no key provided. Consult instructor with questions)
2. a. Fischer esterifications underly an equilibrium with relatively low Keq values (~1-10). Due to the low equilibrium constant for the reaction, the experimenter needs to take advantage of the Le Châtelier Principle.
Unfortunately, the esters and water have significantly higher boiling
points than the alcohols that are used in the lab. This makes it
impossible to remove those compounds from the equilibrium by simple
distillation (as was used in the elimination experiment). Thus, one of the reactants has to be used in excess in this reaction.
Since the carboxylic acids that are used in Chem 30BL are solids, using them in excess would produce a heterogeneous reaction mixture. In the lab, the alcohol is used in a five-fold molar excess because it also acts as a solvent at the same time. A larger excess usually causes serious problems in the phase separation later on
and would have to be distilled off before the workup. The
Fischer
esterification
cannot be used for thermally unstable acids or alcohols that display a
high tendency to eliminate water (i.e., tertiary, allylic, benzylic
alcohols).
b. Water is one of the products in the Fischer esterification. Thus, any water in the system due to wet reagents, wet glassware, etc. will reduce the yield of the ester because the equilibrium will be shifted to the left side.
RCOOH + R'OH <====> RCOOR' + H2O
It is very important to keep the reagent bottles closed because some of them are very hygroscopic (i.e., concentrated sulfuric acid, alcohols).
c. Despite the addition of the catalyst (concentrated sulfuric acid), the reaction still proceeds relatively slowly at room temperature due to its relatively high activation energy. The reflux increases the temperature in the reaction mixture and therefore the rate of conversion. The student should observe that the reaction mixture boils and that the alcohol condenses in the lower third of the air condenser (=reflux ring). In addition, the carboxylic acid should dissolve completely.
d. The addition of water leads to a phase separation because the esters are weakly polar (i.e., methyl benzoate: log Kow=2.12, for more data see also page 93 in course reader) resulting in a poor solubility in a polar solvent/solvent mixture. Many of the esters will settle as a small layer on the bottom or the top depending on its density (i..e., methyl benzoate: 1.08 g/cm3). Please keep in mind that only about 1 mL of ester are formed in the reaction. If a phase separation was not observed, some solid sodium chloride can be added to "salt out" the product.
e. Sodium bicarbonate is used to neutralize the unreacted carboxylic acid and the catalyst (concentrated sulfuric acid) that are dissolved in the organic layer.
H2SO4 + HCO3- ------> HSO4- + CO2
↑ + H2O or
H2SO4 + 2 HCO3- ------> SO42- + 2 CO2
↑ + 2 H2O
The resulting sodium salts (Na2SO4 or NaHSO4, Na+RCOO-) are water-soluble
but very poorly in organic layers. The CO2 formation ceases after all the acids in the organic layer were neutralized. The extraction should be discontinued because the esters dissolve in water more or less as well.
3. a. C4H8(COOH)2 + 2 CH3CH2OH <==> C4H8(COOCH3CH2)2 + 2 H2O
b. In order to determine the yield, you need to identify the limiting reagent first.
| Compound | MW (g/mol) | Density g/cm3 |
| adipic acid | 146.14 | |
| ethanol | 46.02 | 0.789 |
| diethyl adipate | 202.24 | 1.009 |
| sulfuric acid | 98.1 | 1.84 |
Moles of adipic acid= 1.83 g/(146.14 g/mol) = 12.5 mmol
Moles of ethanol = (5.00 mL*0.789 g/mL)/46.02 g/mol = 85.7 mmol
Moles of sulfuric acid = (0.20 mL * 1.84 g/mL)/98.1 g/mol = 3.75 mmol
Sulfuric acid is catalyst in this reaction, which means that the exact molar quantity is not important here. Thus, adipic acid is the limiting reagent and 12.5 mmol of the ester should be obtained theoretically.
Moles of ester isolated = 2.12 g/202.24 g/mol =10.5 mmol
Yield = actual yield / theoretical yield = 10.5 mmol/12.5 mmol *100 % = 83.7 %
c. In order to obtain the boiling point for isopropyl m-toluate at p=60 torr the data given in the reader (yellow reader, page 94) is used. The normal boiling is given in the reader is Tb=239 oC. From the nomograph on page 112 in SKR, one can estimate the boiling point at p=40 mmHg to be Tb=144 oC ± 3oC.
d. The refractive index is a physical constant similar to a melting point and boiling point of a compound. By comparison with the literature, the purity of the sample can be evaluated. A variance of less than Dn= ±0.001 can be considered relatively pure. First, the observed refractive index has to be corrected to the temperature of the literature value (not the other way around!).
nD20= nD18 + (18-20) * 0.00045 = 1.4278- 2*0.00045 = 1.4269 (corr.)
The compound is relatively pure because the corrected refractive index is close to the literature value (nD20= 1.4273, |Δn|<0.001). The lower observed refractive index is most likely due to the presence of some alcohol that decreases the density of the ester.
e. (i) The 1H-NMR spectrum of the ester should exhibit four signals: 4.15 (q, 4 H), 2.60 (t, 4 H), 1.60 (m, 4 H), 1.10 (t, 6 H) ppm.
(ii) The 13C-NMR spectrum should exhibit five signals because there is very little symmetry in the molecule: 173 (carbonyl), 60 (OCH2), 34 (O=CCH2), 25 (CH2) and 14 (CH3) ppm.