IChO 2007 · Theory problem 5

A mysterious mixture (organic hide-and-seek game)

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An equimolar mixture X of three coluorless organic liquids A, B, C was treated by water containing a drop of hydrochloric acid at heating to give (after separation from water) a 1 : 2 (molar ratio) mixture of acetic acid and ethanol without any other components. A catalytic amount (one-two drops) of concentrated sulfuric acid was added o the mixture after hydrolysis and after a long reflux (boiling with reflux condenser) a compound D, a volatile liquid with pleasant smell, was formed in 85% yield. Compound D is not identical with any of A, B, C.

5.1 Draw the structure of compound D?

5.2 Which class of organic compounds does D belong to? Choose the proper variant from those given in the Answer Sheet.

5.3 Even if the reflux is continued twice as long, the yield of D would not exceed 85%. Calculate the expected yield of D if 1 : 1 (molar ratio) mixture of ethanol and acetic acid is taken. Assume that: a) volumes do not change during the reactions; b) all concomitant factors, such as solvent effects, non-additivity of volumes, variation of temperature, etc. are negligible. If you cannot make a quantitative estimate, please indicate whether the yield will be: a) the same (85 %); b) higher than 85 %; c) lower than 85 %.

\(\ce{^1H}\) NMR spectra of compounds A, B, C look very similar and each shows singlet, triplet and quartet with the ratio of integral intensities equal to 1 : 3 : 2.

The same mixture X was subjected to alkaline hydrolysis. A remained unchanged, and was separated. After acidification and short boiling the remaining solution gave 2 : 3 (molar ratio) mixture of acetic acid and ethanol with evolution of gas.

The mixture X (3.92 g) was dissolved in diethyl ether and underwent hydrogenation in the presence of \(\ce{Pd}\) on charcoal catalyst. 0.448 \(\mathrm{dm^3}\) (standard conditions) of hydrogen were absorbed, but after the reaction A and C were isolated unchanged (3.22 g of mixture were recovered) while neither B, nor any other organic compounds, except diethyl ether, could be identified after hydrogenation.

5.4 Determine and draw the structures of A, B, and C.

5.5 Which intermediate compounds are formed during the acidic hydrolysis of C and the basic hydrolysis of B.

The reaction of either B or C with acetone (in the presence of a base) with subsequent acidification by dilute \(\ce{HCl}\) at gentle heating gives the same product, senecioic acid (SA), a compound widely occurring in Nature. Alternatively, senecioic acid can be obtained from acetone by treating it with concentrated \(\ce{HCl}\) and subsequent oxidation of the intermediate product by iodine in alkaline solution. The latter reaction gives, besides sodium salt of senecioic acid, a heavy yellow precipitate E (see the scheme 2).

Figure 1
Figure 2

5.6 Determine the structure of senecioic acid and draw the reaction scheme leading to senecioic acid from acetone.

5.7 Give structure of E

Official English text of the problem.

Answers

  • 5.1: D is ethyl acetate (ethyl ethanoate), \(\ce{CH3C(=O)OCH2CH3}\), SMILES CCOC(=O)C.
  • 5.2: D is a carboxylic ester.
  • 5.3: The expected equilibrium yield is approximately 67% (calculated conversion \(0.6718\)); therefore choose c), lower than 85%.
  • 5.4: A is triethyl orthoacetate, \(\ce{CH3C(OC2H5)3}\), CC(OCC)(OCC)OCC; B is ethoxyethyne (ethyl ethynyl ether), \(\ce{HC#C-OC2H5}\), CCOC#C; C is diethyl malonate, \(\ce{CH2(COOC2H5)2}\), CCOC(=O)CC(=O)OCC.
  • 5.5: C gives malonic acid, O=C(O)CC(=O)O, before decarboxylation. For B, hydration gives ethyl acetate, CCOC(=O)C, which is saponified to acetate ion, CC(=O)[O-], before acidification.
  • 5.6: SA is 3-methylbut-2-enoic acid (senecioic acid), \(\ce{(CH3)2C=CHCOOH}\), CC(C)=CC(=O)O. Acetone undergoes aldol condensation to mesityl oxide, CC(=O)C=C(C)C; haloform cleavage gives sodium senecioate, CC(C)=CC(=O)[O-].[Na+], followed by acidification.
  • 5.7: E is iodoform (triiodomethane), \(\ce{CHI3}\), SMILES IC(I)I.

Solution

5.1 — Identification and structure of D

Acetic acid, CC(=O)O, and ethanol, CCO, undergo acid-catalysed Fischer esterification:

\[ \ce{CH3COOH + C2H5OH <=>[H2SO4] CH3COOC2H5 + H2O}. \]

Thus D is ethyl acetate, CCOC(=O)C, a volatile liquid with a pleasant smell. Water has SMILES O.

For the requested drawing, draw a carbonyl carbon bonded to a methyl group and an oxygen atom; that oxygen is bonded to an ethyl group:

\[ \boxed{\ce{CH3-C(=O)-O-CH2-CH3}}. \]

5.2 — Compound class

The characteristic linkage is \(\ce{-C(=O)-O-}\), with carbon groups on both sides. Therefore D is a carboxylic ester.

5.3 — Equilibrium yield with equimolar reactants

The yield ceases to increase because esterification reaches equilibrium. The catalyst accelerates attainment of equilibrium but does not change its position.

Take initial amounts of \(1.00\ \mathrm{mol}\) acetic acid and \(2.00\ \mathrm{mol}\) ethanol, with no initial ester or water after the stated separation. An 85% yield corresponds to an extent of reaction of \(0.85\ \mathrm{mol}\).

Species Initial amount / mol Equilibrium amount / mol
Acetic acid 1.00 0.15
Ethanol 2.00 1.15
Ethyl acetate 0 0.85
Water 0 0.85

Under the stated assumptions,

\[ K_c= \frac{[\text{ethyl acetate}][\text{water}]} {[\text{acetic acid}][\text{ethanol}]}. \]

The volume factors cancel, giving

\[ K_c=\frac{0.85^2}{(1-0.85)(2-0.85)} =\frac{0.7225}{0.1725} =4.1884. \]

Now start with \(1.00\ \mathrm{mol}\) of each reactant. Let \(y\) be the fraction converted:

\[ K_c=\frac{y^2}{(1-y)^2}. \]

Hence

\[ \frac{y}{1-y}=\sqrt{K_c}, \qquad y=\frac{\sqrt{K_c}}{1+\sqrt{K_c}} =0.6718. \]

Therefore,

\[ \boxed{\text{Yield}\approx67\%}. \]

Removing the initial excess of ethanol lowers the equilibrium conversion. The appropriate qualitative choice is c).

5.4 — Identification of A, B and C

The structures are:

Compound Name and SMILES Description of the required drawing
A Triethyl orthoacetate, CC(OCC)(OCC)OCC A tetrahedral central carbon bonded to one methyl group and three ethoxy groups through oxygen: \(\ce{CH3C(OC2H5)3}\).
B Ethoxyethyne, CCOC#C A terminal carbon–carbon triple bond attached to an ethoxy oxygen: \(\ce{H-C#C-O-CH2CH3}\).
C Diethyl malonate, CCOC(=O)CC(=O)OCC A methylene group between two ester carbonyl groups, each ester bearing an ethyl group.

These assignments satisfy all the independent observations.

Hydrogenation identifies B.

The mass of B in the sample is

\[ m_B=3.92-3.22=0.70\ \mathrm{g}. \]

Using the conventional molar gas volume at standard conditions,

\[ n(\ce{H2}) =\frac{0.448\ \mathrm{dm^3}} {22.4\ \mathrm{dm^3\,mol^{-1}}} =0.0200\ \mathrm{mol}. \]

Diethyl ether, CCOCC, is the only organic product that could be hidden in the solvent. Ethoxyethyne gives exactly this product upon complete hydrogenation:

\[ \ce{HC#C-OC2H5 + 2H2 ->[Pd/C] C2H5OC2H5}. \]

Hydrogen has SMILES [H][H]. Consequently,

\[ n_B=\frac{0.0200}{2}=0.0100\ \mathrm{mol}, \]

and

\[ M_B=\frac{0.70\ \mathrm{g}}{0.0100\ \mathrm{mol}} =70\ \mathrm{g\,mol^{-1}}. \]

This agrees with ethoxyethyne, \(\ce{C4H6O}\):

\[ M(\ce{C4H6O})=4(12)+6(1)+16=70\ \mathrm{g\,mol^{-1}}. \]

The acid/base behaviour identifies A and C.

An orthoester is hydrolysed by aqueous acid but survives alkaline treatment. The orthoester that gives acetic acid and ethanol is triethyl orthoacetate, A:

\[ \ce{CH3C(OC2H5)3 + 2H2O ->[H^+,\ \Delta] CH3COOH + 3C2H5OH}. \]

For C, two equivalent ethoxy groups and a methylene singlet suggest a malonic ester. Hydrolysis followed by decarboxylation gives acetic acid, ethanol and carbon dioxide, O=C=O:

\[ \ce{CH2(COOC2H5)2 + 2H2O ->[H^+,\ \Delta] CH3COOH + 2C2H5OH + CO2 ^}. \]

For B, the overall acidic hydrolysis is

\[ \ce{HC#C-OC2H5 + 2H2O ->[H^+,\ \Delta] CH3COOH + C2H5OH}. \]

Thus one mole of each starting compound gives:

Starting compound Acetic acid / mol Ethanol / mol
A 1 3
B 1 1
C 1 2
Total 3 6

The ratio is therefore \(3:6=1:2\), as required. Carbon dioxide escapes and is not an additional component of the recovered liquid mixture.

In the alkaline experiment, A remains unchanged. After acidification and heating, B and C together give

\[ n(\text{acetic acid}):n(\text{ethanol}) =(1+1):(1+2)=2:3, \]

with carbon dioxide evolved from the malonic-acid intermediate.

The NMR integrations agree.

Compound Singlet Triplet Quartet Integration ratio
A Central-carbon methyl: \(3\mathrm{H}\) Three ethyl methyl groups: \(9\mathrm{H}\) Three ethoxy methylene groups: \(6\mathrm{H}\) \(3:9:6=1:3:2\)
B Terminal acetylenic proton: \(1\mathrm{H}\) Ethyl methyl: \(3\mathrm{H}\) Ethoxy methylene: \(2\mathrm{H}\) \(1:3:2\)
C Methylene between carbonyl groups: \(2\mathrm{H}\) Two ethyl methyl groups: \(6\mathrm{H}\) Two ethoxy methylene groups: \(4\mathrm{H}\) \(2:6:4=1:3:2\)

The ethyl groups give their usual triplet and quartet. The remaining protons have no adjacent proton-bearing carbon that would produce ordinary vicinal splitting.

The remaining masses agree.

Because the mixture is equimolar,

\[ n_A=n_C=n_B=0.0100\ \mathrm{mol}. \]

For A, \(\ce{C8H18O3}\), \(M_A=162\ \mathrm{g\,mol^{-1}}\); for C, \(\ce{C7H12O4}\), \(M_C=160\ \mathrm{g\,mol^{-1}}\). Therefore,

\[ m_A+m_C =0.0100(162+160) =3.22\ \mathrm{g}, \]

and

\[ m_X=0.0100(162+70+160)=3.92\ \mathrm{g}. \]

Both reproduce the measured masses.

5.5 — Hydrolysis intermediates

For C, the important intermediate before decarboxylation is malonic acid (propanedioic acid), O=C(O)CC(=O)O.

Stepwise ester hydrolysis can first give ethyl hydrogen malonate, CCOC(=O)CC(=O)O:

\[ \ce{CH2(COOC2H5)2 + H2O ->[H^+] HOOCCH2COOC2H5 + C2H5OH}, \]

\[ \ce{HOOCCH2COOC2H5 + H2O ->[H^+] CH2(COOH)2 + C2H5OH}. \]

Malonic acid then decarboxylates:

\[ \boxed{\ce{CH2(COOH)2 ->[\Delta] CH3COOH + CO2 ^}}. \]

For B, hydration gives ethyl acetate, D, CCOC(=O)C, followed by saponification to acetate ion, CC(=O)[O-]. The stepwise stoichiometry is

\[ \ce{HC#C-OC2H5 + H2O -> CH3COOC2H5}, \]

\[ \ce{CH3COOC2H5 + OH- -> CH3COO- + C2H5OH}. \]

Here hydroxide ion has SMILES [OH-]. Acidification subsequently gives

\[ \ce{CH3COO- + H+ -> CH3COOH}. \]

For completeness, C initially gives malonate ion, O=C([O-])CC(=O)[O-], under alkaline conditions:

\[ \ce{CH2(COOC2H5)2 + 2OH- -> [CH2(COO)2]^2- + 2C2H5OH}, \]

\[ \ce{[CH2(COO)2]^2- + 2H+ -> CH2(COOH)2}. \]

The resulting malonic acid undergoes the decarboxylation shown above.

5.6 — Senecioic acid and its synthesis from acetone

The yellow precipitate formed with iodine and alkali indicates a haloform reaction. A methyl ketone is cleaved to a carboxylate and iodoform.

Acetone, CC(=O)C, first undergoes acid-catalysed self-aldol addition to diacetone alcohol (4-hydroxy-4-methylpentan-2-one), CC(=O)CC(C)(C)O:

\[ \ce{2CH3COCH3 <=>[HCl] CH3COCH2C(OH)(CH3)2}. \]

Dehydration produces mesityl oxide (4-methylpent-3-en-2-one), CC(=O)C=C(C)C:

\[ \ce{CH3COCH2C(OH)(CH3)2 ->[H^+] CH3COCH=C(CH3)2 + H2O}. \]

Its methyl ketone group undergoes haloform cleavage:

\[ \boxed{ \ce{CH3COCH=C(CH3)2 + 3I2 + 4NaOH -> (CH3)2C=CHCOONa + CHI3 v + 3NaI + 3H2O} }. \]

The carboxylate product is sodium 3-methylbut-2-enoate, CC(C)=CC(=O)[O-].[Na+]. Acidification gives senecioic acid:

\[ \ce{(CH3)2C=CHCOONa + HCl -> (CH3)2C=CHCOOH + NaCl}. \]

Therefore,

\[ \boxed{\text{SA}=\ce{HOOC-CH=C(CH3)2}} \]

is 3-methylbut-2-enoic acid, CC(C)=CC(=O)O.

For the requested drawing, place a carboxyl group on one end of an alkene: its adjacent alkene carbon bears one hydrogen, while the other alkene carbon bears two methyl groups. There is no \(E/Z\) stereochemistry, because one alkene carbon has two identical methyl substituents.

The reactions of B and C independently support this structure.

For B, base-promoted addition to acetone gives 4-ethoxy-2-methylbut-3-yn-2-ol, CCOC#CC(C)(C)O:

\[ \ce{HC#C-OC2H5 + CH3COCH3 ->[base] (CH3)2C(OH)-C#C-OC2H5}. \]

Hydration of the ethoxyalkyne gives ethyl 3-hydroxy-3-methylbutanoate, CCOC(=O)CC(C)(C)O:

\[ \ce{(CH3)2C(OH)-C#C-OC2H5 + H2O ->[H^+] (CH3)2C(OH)CH2COOC2H5}. \]

Dehydration gives ethyl senecioate, CCOC(=O)C=C(C)C, which hydrolyses to SA:

\[ \ce{(CH3)2C(OH)CH2COOC2H5 ->[H^+,\ \Delta] (CH3)2C=CHCOOC2H5 + H2O}, \]

\[ \ce{(CH3)2C=CHCOOC2H5 + H2O ->[H^+] (CH3)2C=CHCOOH + C2H5OH}. \]

For C, Knoevenagel condensation with acetone gives diethyl 2-(propan-2-ylidene)malonate, CCOC(=O)C(=C(C)C)C(=O)OCC:

\[ \ce{CH2(COOC2H5)2 + CH3COCH3 ->[base] (CH3)2C=C(COOC2H5)2 + H2O}. \]

Hydrolysis gives 2-(propan-2-ylidene)malonic acid, O=C(O)C(=C(C)C)C(=O)O:

\[ \ce{(CH3)2C=C(COOC2H5)2 + 2H2O ->[H^+] (CH3)2C=C(COOH)2 + 2C2H5OH}. \]

Decarboxylation yields the same acid:

\[ \ce{(CH3)2C=C(COOH)2 ->[\Delta] (CH3)2C=CHCOOH + CO2 ^}. \]

5.7 — Identification and structure of E

The heavy yellow precipitate is iodoform (triiodomethane):

\[ \boxed{\ce{E = CHI3}}, \qquad \text{SMILES: IC(I)I}. \]

Draw a tetrahedral carbon atom with three single bonds to iodine atoms and one single bond to hydrogen. It is the haloform product of the balanced reaction in 5.6.

The solution was checked against the official solution.