IChO 2000 · Theory problem 4

A Naturally Occurring Compound

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A naturally occurring compound A containing only \(\ce{C}\), \(\ce{H}\) and \(\ce{O}\) has the following elemental composition, percentage mass,

\(\ce{C}\): 63.2 %, \(\ce{H}\): 5.3%, \(\ce{O}\): 31.5%.

4.1 Derive the empirical formula of compound A.

Figure 1

Figure 1

The mass spectrum of compound A is shown in Figure 1.

4.2 What is the molecular formula for compound A?

A solution of A in ether is shaken with an aqueous solution of \(\ce{NaOH}\). After this, no A remains in the ether phase. Another solution of A in ether is shaken with an aqueous solution of \(\ce{NaHCO3}\). A remains in the ether phase.

4.3 Which of the following classes of compounds does A belong to according to these experiments? Mark with an X.

alcohol □ phenol □ aldehyde □ ketone □
acid □ ester □ ether □

Compound A gave rise to formation of a silver mirror with Tollens’ reagent (\(\ce{Ag(NH3)2+}\)).

4-1 Which of the following functional groups does this indicate the presence of in A? Mark with an X.

hydroxy group of an alcohol □ hydroxy group of a phenol □
carbonyl group of an aldehyde □ carbonyl group of a ketone □
carboxylic group □ ester group □
alkoxy group of an ether □
Figure 2

Figure 2a

The \({}^{1}\mathrm{H}\) NMR spectrum of compound A recorded at 300 MHz is shown in Figure 2a (solvent \(\ce{CDCl3}\) (7.27 ppm), reference tetramethylsilane). The signals at 3.9, 6.3 and 9.8 ppm are singlets. Figure 2b is an expansion of the region 6.9 – 7.6 ppm.

Figure 3

Figure 2b

Selected chemical shift and coupling constant values are given in Table 1.

The signal at 6.3 ppm disappears when a drop of \(\ce{D2O}\) is added.

4.5 Which of the following does this indicate? Mark with an X.

  • Exchange of carbon-bonded hydrogen □
  • Exchange of oxygen-bonded hydrogen □
  • Dilution effect □
  • Hydrolysis □

The same signal moves to a lower ppm value upon dilution with \(\ce{CDCl3}\).

4.6 Which of the following does this indicate?

Indicate the true statements (more than one).

  • Increased hydrogen bonding □
  • Decrease in hydrogen bonding □
  • Intermolecular hydrogen bonding □
  • Intramolecular hydrogen bonding □
  • No hydrogen bonding □

4.7 Draw the four possible structural formulas for compound A based on the information given above

4.8 Give structural formulas for the fragments lost corresponding to the peaks at 137 and 123 mass units in the mass spectrum.

4.9 Two of the isomers have a lower \(pK_a\) value than the others. Write the formulas for those.

Table 1. \({}^{1}\mathrm{H}\) Chemical Shift \(\delta\)

Hydrogens attached to carbon

Methyl \(\ce{CH3-C-}\)
\(\ce{CH3-C=O-}\)
\(\ce{CH3-O-R}\)
\(\ce{CH3-OCOR}\)		 0.9 – 1.6 ppm
2.0 – 2.4 ppm
3.3 – 3.8 ppm
3.7 – 4.0 ppm
Methylene \(\ce{CH2-C-}\)
\(\ce{CH2-C=O-}\)
\(\ce{CH2-OR}\)
\(\ce{CH2-OCOR}\)		 1.4 – 2.7 ppm
2.2 – 2.9 ppm
3.4 – 4.1 ppm
4.3 – 4.4 ppm
Methine \(\ce{CH-}\) 1.5 – 5.0 ppm
depending on the substituents. Generally higher than for methyl and methylene
Alkene 4.0 – 7.3 ppm
depending on the substituent
Aldehyde \(\ce{R-CHO}\) 9.0 – 10.0 ppm

Table 1 (continued) \({}^{1}\mathrm{H}\) Chemical Shift \(\delta\)

Hydrogens attached to oxygen

Alcohols \(\ce{ROH}\) 0.5 – 5.0 ppm
Phenols \(\ce{ArOH}\) 4.0 – 7.0 ppm
Carboxylic acids \(\ce{RCOOH}\) 10.0 – 13.0 ppm

Selected spin-spin coupling constants

Alkanes
(free rotation)		 \(\ce{H-C-C-H}\) vicinal 6 – 8 Hz
Alkenes trans
cis
geminal		 11 – 18 Hz
6 – 12 Hz
0 – 3 Hz
Aromates ortho
meta
para		 6 – 10 Hz
1 – 4 Hz
0 – 2 Hz

Official English text of the problem.

Answers

  • 4.1: Empirical formula: \(\boxed{\ce{C8H8O3}}\).
  • 4.2: Molecular formula: \(\boxed{\ce{C8H8O3}}\), from the molecular-ion peak at \(m/z=152\).
  • 4.3: Mark phenol.
  • 4-1 (fourth subquestion): Mark carbonyl group of an aldehyde.
  • 4.5: Mark Exchange of oxygen-bonded hydrogen.
  • 4.6: Mark Decrease in hydrogen bonding and Intermolecular hydrogen bonding.
  • 4.7: The four candidate structures are: 1. 4-Hydroxy-3-methoxybenzaldehyde (vanillin): O=Cc1cc(OC)c(O)cc1 2. 3-Hydroxy-4-methoxybenzaldehyde (isovanillin): O=Cc1cc(O)c(OC)cc1 3. 4-Hydroxy-2-methoxybenzaldehyde: O=Cc1c(OC)cc(O)cc1 4. 5-Hydroxy-2-methoxybenzaldehyde: O=Cc1c(OC)ccc(O)c1
  • 4.8: At \(m/z=137\), loss of a methyl radical, \(\ce{CH3^{.}}\), [CH3]; at \(m/z=123\), loss of a formyl radical, \(\ce{HCO^{.}}\), [CH]=O.
  • 4.9: The two lower-\(pK_a\) isomers are 4-hydroxy-3-methoxybenzaldehyde (vanillin), O=Cc1cc(OC)c(O)cc1, and 4-hydroxy-2-methoxybenzaldehyde, O=Cc1c(OC)cc(O)cc1.

Solution

4.1 — Empirical formula

Take a \(100.0\ \mathrm{g}\) sample. Using atomic molar masses \(12.01\), \(1.008\) and \(16.00\ \mathrm{g\,mol^{-1}}\),

\[ \begin{aligned} n_{\ce C}&=\frac{63.2}{12.01}=5.262\ \mathrm{mol},\\ n_{\ce H}&=\frac{5.3}{1.008}=5.258\ \mathrm{mol},\\ n_{\ce O}&=\frac{31.5}{16.00}=1.969\ \mathrm{mol}. \end{aligned} \]

Dividing by the oxygen amount gives

\[ n_{\ce C}:n_{\ce H}:n_{\ce O} =2.673:2.671:1. \]

Multiplication by three gives approximately \(8:8:3\). Therefore,

\[ \boxed{\text{Empirical formula}=\ce{C8H8O3}}. \]

Its empirical formula mass is

\[ M_{\mathrm{emp}} =8(12.01)+8(1.008)+3(16.00) =152.14\ \mathrm{g\,mol^{-1}}. \]

4.2 — Molecular formula

The molecular-ion peak occurs at \(m/z=152\). For a singly charged molecular ion, this corresponds to a molecular mass of approximately \(152\ \mathrm{u}\), hence a molar mass of approximately \(152\ \mathrm{g\,mol^{-1}}\).

Thus,

\[ \frac{M}{M_{\mathrm{emp}}}\approx\frac{152}{152.14}\approx1, \]

so the molecular and empirical formulas coincide:

\[ \boxed{\ce{C8H8O3}}. \]

The formula alone does not specify connectivity; the possible molecular structures are established in 4.7.

4.3 — Interpretation of the extraction experiments

Hydroxide deprotonates the phenolic group, producing an ionic phenoxide that transfers into the aqueous phase:

\[ \ce{ArOH + OH- -> ArO- + H2O}. \]

Here \(\ce{Ar}\) denotes the substituted aromatic residue. In generic SMILES, with * denoting that residue, the phenol is *O and its phenoxide is *[O-]. Hydroxide is [OH-], and water is O.

Bicarbonate, O=C([O-])O, is too weak a base to produce substantial extraction under the stated conditions. This distinguishes the observed behavior from that expected for an ordinary carboxylic acid, which would react with aqueous bicarbonate.

Therefore, mark phenol. These extraction experiments establish the phenolic group; the other functional groups require the subsequent evidence.

4-1 — Interpretation of Tollens’ test

The silver mirror indicates reduction of silver(I) to metallic silver while the aldehyde is oxidized to a carboxylate:

\[ \ce{ArCHO + 2[Ag(NH3)2]+ + 3OH- -> ArCOO- + 2Ag(s) + 4NH3 + 2H2O}. \]

This equation is balanced in atoms and charge. The aldehyde carbon undergoes a two-electron oxidation, reducing two silver(I) ions.

The generic aldehyde and carboxylate structures are *C=O and *C(=O)[O-], respectively. Diamminesilver(I) has dative-bond SMILES [NH3]->[Ag+]<-[NH3] and approximately linear nitrogen–silver–nitrogen coordination. Ammonia is N.

Thus, mark carbonyl group of an aldehyde. The one-proton singlet at \(9.8\ \mathrm{ppm}\) independently supports this assignment.

4.5 — Disappearance of the 6.3 ppm signal

The phenolic hydrogen exchanges with deuterium:

\[ \ce{ArOH + D2O <=> ArOD + HDO}. \]

Heavy water is [2H]O[2H], the generic oxygen-deuterated phenol is *O[2H], and monodeuterated water is [2H]O.

The resulting oxygen-bound deuterium does not give the original signal in a \({}^{1}\mathrm H\) NMR spectrum. Therefore, mark Exchange of oxygen-bonded hydrogen.

4.6 — Effect of dilution

Hydrogen bonding deshields the phenolic proton, moving its resonance to a higher chemical shift. Dilution reduces association between separate molecules. The resulting decrease in hydrogen bonding increases shielding, moving the signal to a lower ppm value.

Therefore, mark:

  • Decrease in hydrogen bonding
  • Intermolecular hydrogen bonding

A purely intramolecular hydrogen bond would ordinarily be much less sensitive to concentration.

4.7 — Four possible structures

The degree of unsaturation is

\[ \mathrm{DBE}=\frac{2(8)+2-8}{2}=5. \]

A benzene ring accounts for four degrees of unsaturation, and an aldehyde carbonyl accounts for the fifth.

The NMR integrals give the following assignments:

Signal or integration region Integral Assignment
\(9.8\ \mathrm{ppm}\) \(0.94\approx1\) H Aldehyde hydrogen
\(7.56–7.32\ \mathrm{ppm}\) \(1.91\approx2\) H Two aromatic hydrogens
\(7.15–6.92\ \mathrm{ppm}\) \(0.93\approx1\) H One aromatic hydrogen
\(6.3\ \mathrm{ppm}\) \(0.94\approx1\) H Phenolic hydrogen
\(3.9\ \mathrm{ppm}\) \(3.10\approx3\) H Methoxy hydrogens

The phenol and aldehyde account for two oxygen atoms. The remaining oxygen belongs to a methoxy group, \(\ce{-OCH3}\), generic SMILES *OC, which accounts for the three-proton singlet at \(3.9\ \mathrm{ppm}\).

Thus, A is a benzene bearing one hydroxy, one methoxy and one formyl substituent:

\[ \ce{C6H3(OH)(OCH3)(CHO)}. \]

The aromatic splitting constrains their relative positions. For the well-resolved doublet near \(7.04\ \mathrm{ppm}\),

\[ J=(7.059-7.031)\ \mathrm{ppm} \times300\ \mathrm{Hz\,ppm^{-1}} =8.4\ \mathrm{Hz}. \]

This is an ortho coupling according to Table 1. The overlapping resonances near \(7.42\ \mathrm{ppm}\) also show smaller splittings consistent with meta coupling.

The pattern corresponds to three aromatic hydrogens with one ortho relationship, one meta relationship and one para relationship: a 1,2,4-type trisubstituted benzene. In comparison, three consecutive aromatic hydrogens would have two ortho couplings, while the hydrogens of a 1,3,5-trisubstituted ring would have only meta relationships.

The intended additional restriction from the hydrogen-bonding observations is that the hydroxy and formyl groups are not ortho. An ortho arrangement would favor a six-membered intramolecular hydrogen-bonded arrangement between the phenolic hydrogen and aldehyde oxygen.

Applying these restrictions gives four candidates. To draw them, place the formyl group at carbon 1 of a benzene ring and add the other substituents at the positions below. Every remaining ring carbon bears hydrogen.

Candidate Hydroxy position Methoxy position Name and SMILES
I 4 3 4-Hydroxy-3-methoxybenzaldehyde (vanillin) — O=Cc1cc(OC)c(O)cc1
II 3 4 3-Hydroxy-4-methoxybenzaldehyde (isovanillin) — O=Cc1cc(O)c(OC)cc1
III 4 2 4-Hydroxy-2-methoxybenzaldehyde — O=Cc1c(OC)cc(O)cc1
IV 5 2 5-Hydroxy-2-methoxybenzaldehyde — O=Cc1c(OC)ccc(O)c1

These structures require no stereochemical specification.

4.8 — Fragments lost in the mass spectrum

Calculate each loss relative to the molecular ion at \(m/z=152\):

Product peak Nominal mass lost Lost fragment and structure
\(m/z=137\) \(152-137=15\ \mathrm{u}\) Methyl radical, \(\ce{CH3^{.}}\), [CH3]
\(m/z=123\) \(152-123=29\ \mathrm{u}\) Formyl radical, \(\ce{HCO^{.}}\), [CH]=O

The methyl loss corresponds to cleavage of the oxygen–methyl bond of the methoxy group. The formyl loss corresponds to removal of the aldehyde substituent from the aromatic ring; its unpaired electron is on carbon.

These are the assigned net losses from the molecular ion; the mass differences alone do not establish a unique fragmentation sequence.

4.9 — Isomers with lower \(pK_a\)

The lower-\(pK_a\) isomers have the aldehyde group para to the phenolic hydroxy group:

  • I: 4-Hydroxy-3-methoxybenzaldehyde (vanillin) — O=Cc1cc(OC)c(O)cc1
  • III: 4-Hydroxy-2-methoxybenzaldehyde — O=Cc1c(OC)cc(O)cc1

Their conjugate bases are:

  • 4-Formyl-2-methoxyphenolate — O=Cc1cc(OC)c([O-])cc1
  • 4-Formyl-3-methoxyphenolate — O=Cc1c(OC)cc([O-])cc1

In these phenoxide ions, resonance can distribute negative charge through the ring to the para position and onward onto the aldehyde oxygen. This additional stabilization favors deprotonation and lowers \(pK_a\).

For candidates II and IV, the aldehyde is meta to the phenolic group, so this direct resonance stabilization of the conjugate base is unavailable.

The solution was checked against the official solution.