IChO 1987 · Theory problem 5

Organic chemistry questions: chlorination, alcohols, aldohexoses, C4H8 isomers, lecithin, citric acid cycle, NAD+

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5.1 What ratio of primary / secondary / tertiary products can statistically be expected in the high temperature chlorination of methyl butane? Assume that the reaction rate of the substitution is equal for all C-H bonds.

5.2 Which of the following alcohols: pentan-1-ol, pentan-2-ol and 2-methyl-butan-2-ol react with the following reagents listed below?

1) cold, concentrated sulphuric acid; 2) \(\ce{CrO3 / H2SO4}\); 3) \(\ce{ZnCl2 / HCl}\); 4) \(\ce{I2 / NaOH}\)

5.3 Which of the following aldohexose structures are:

1) enantiomer pairs, 2) diastereomer pairs?

Figure 1

5.4 Two hydrocarbon compounds of the same molecular formula, \(\ce{C4H8}\), easily react with hydrogen in the presence of a platinum catalyst. During ozonolysis only ethanal (acetaldehyde) is formed in both cases. The \(\ce{^1H}\)-NMR spectra of the two compounds show two types of protons in a ratio of 3 to 1. Draw the structures of the described compounds.

5.5 Select the components of the group listed below that can be obtained during complete hydrolysis of lecithin (a phospholipid): serine, phosphoric acid, sphingosine, choline, glycerol, myo-inositol, phosphatidic acid and fatty acids).

5.6 Which of the following carboxylic acid can be formed in the tricarboxylic acid cycle (Krebs' citric acid cycle):

maleic acid (cis-butenedioic acid), mandelic acid (a-hydroxy-phenylethanoic acid), malic acid (2-hydroxy-butanedioic acid), tricarb-allylic acid (propane-1,2,3-tricarboxylic acid), oxalacetic acid (2-oxo-butanedioic acid), keto-glutaric acid (2-oxo-pentanedioic acid), fumaric acid (trans-butenedioic acid) and acetoacetic acid (3-oxo-butanoic acid).

5.7 To which structures is the nicotinamide moiety (part) of \(\ce{NAD+}\) converted during the uptake of hydrogen to form \(\ce{NADH}\) (A, B or C)?

Figure 2

Official English text of the problem.

Answers

  • 5.1: Primary : secondary : tertiary monochlorination products \(=\boxed{9:2:1}\), corresponding to \(75.0\%\), \(16.7\%\), and \(8.33\%\).

  • 5.2: The results are:

Alcohol 1) Cold, concentrated \(\ce{H2SO4}\) 2) \(\ce{CrO3/H2SO4}\) 3) \(\ce{ZnCl2/HCl}\), ordinary room-temperature Lucas test 4) \(\ce{I2/NaOH}\)
Pentan-1-ol CCCCCO Reacts Reacts No observable reaction within the usual test time; reacts on heating Negative
Pentan-2-ol CC(O)CCC Reacts Reacts Reacts more slowly Positive
2-Methylbutan-2-ol CC(O)(C)CC Reacts No normal oxidation Reacts rapidly Negative

  • 5.3: Enantiomer pairs: A–C and B–D. Diastereomer pairs: A–B, A–D, B–C and C–D.

  • 5.4: cis-(Z)-But-2-ene C/C=C\C and trans-(E)-but-2-ene C/C=C/C.

  • 5.5: Phosphoric acid OP(=O)(O)O, choline C[N+](C)(C)CCO, glycerol OCC(O)CO, and fatty acids [*]C(=O)O. Here [*] denotes an unspecified fatty-acid hydrocarbon residue.

  • 5.6: L-(S)-Malic acid O=C(O)[C@@H](O)CC(=O)O, oxaloacetic acid O=C(O)C(=O)CC(=O)O, 2-oxoglutaric acid O=C(O)C(=O)CCC(=O)O, and fumaric acid O=C(O)/C=C/C(=O)O.

  • 5.7: C, the 1,4-dihydronicotinamide moiety of \(\ce{NADH}\), [*]N1C=C(C(=O)N)CC=C1. Here [*] represents the unchanged substituent connecting the ring nitrogen to the rest of the cofactor.

Solution

5.1 — Statistical chlorination

Methyl butane is 2-methylbutane, CC(C)CC, with condensed structure \(\ce{(CH3)2CHCH2CH3}\).

With equal substitution rates for every \(\ce{C-H}\) bond, the probability of substitution at a particular kind of carbon is proportional to the number of hydrogens attached to that kind of carbon.

Carbon environment Number of attached hydrogens Product class
Three methyl groups \(3\times3=9\) Primary
One methylene group \(2\) Secondary
One methine group \(1\) Tertiary
Total 12

Therefore, \[ \boxed{n(1^\circ):n(2^\circ):n(3^\circ)=9:2:1}. \]

The corresponding fractions are \[ \frac{9}{12}=75.0\%,\qquad \frac{2}{12}=16.7\%,\qquad \frac{1}{12}=8.33\%. \]

The balanced monochlorination equation is \[ \ce{C5H12 + Cl2 ->[\Delta] C5H11Cl + HCl}. \] Chlorine has SMILES ClCl; hydrogen chloride has SMILES Cl.

There are four constitutional monochlorination products:

Product SMILES Statistical weight
Racemic 1-chloro-2-methylbutane ClC[C@H](C)CC and ClC[C@@H](C)CC 6
1-Chloro-3-methylbutane ClCCC(C)C 3
Racemic 2-chloro-3-methylbutane C[C@H](Cl)C(C)C and C[C@@H](Cl)C(C)C 2
2-Chloro-2-methylbutane CC(Cl)(C)CC 1

The two primary products together account for \(6+3=9\) hydrogens. The enantiomers within each racemic product have equal statistical yields.

5.2 — Reactions of the alcohols

The three alcohols are respectively primary, secondary and tertiary. Their differing oxidation and substitution behaviour follows from the substitution of the carbon bearing the hydroxyl group.

1) Cold, concentrated sulphuric acid

All three alcohols react with sulphuric acid, OS(=O)(=O)O. Protonation and formation of alkyl hydrogen sulfates are possible: \[ \ce{ROH + H2SO4 <=> ROSO3H + H2O}. \] Water has SMILES O.

For the three alcohols, the corresponding sulfate ester structures are:

Alcohol-derived sulfate ester SMILES
Pentyl hydrogen sulfate CCCCCOS(=O)(=O)O
Pentan-2-yl hydrogen sulfate CC(OS(=O)(=O)O)CCC
2-Methylbutan-2-yl hydrogen sulfate CC(C)(OS(=O)(=O)O)CC

These structures do not imply that a single sulfate ester is the exclusive product under every acid concentration and temperature. In particular, the tertiary alcohol also undergoes facile acid-catalysed dehydration: \[ \ce{(CH3)2C(OH)CH2CH3 ->[H2SO4] (CH3)2C=CHCH3 + H2O}, \] \[ \ce{(CH3)2C(OH)CH2CH3 ->[H2SO4] CH2=C(CH3)CH2CH3 + H2O}. \]

The alkenes are 2-methylbut-2-ene, CC(C)=CC, and 2-methylbut-1-ene, C=C(C)CC. Thus the qualitative answer is positive for all three alcohols.

2) Chromium trioxide in sulphuric acid

Chromium trioxide, O=[Cr](=O)=O, oxidizes the primary and secondary alcohols under the usual aqueous acidic conditions.

Pentan-1-ol is oxidized through pentanal, CCCCC=O, to pentanoic acid, CCCCC(=O)O. The overall balanced ionic equation is \[ \ce{3CH3(CH2)4OH + 4CrO3 + 12H+ -> 3CH3(CH2)3COOH + 4Cr^3+ + 9H2O}. \]

Pentan-2-ol is oxidized to pentan-2-one, CC(=O)CCC: \[ \ce{3CH3CH(OH)CH2CH2CH3 + 2CrO3 + 6H+ -> 3CH3COCH2CH2CH3 + 2Cr^3+ + 6H2O}. \]

Here the hydrogen ion is [H+] and the chromium(III) ion is [Cr+3]; sulfate spectator ions have been omitted.

The tertiary alcohol has no hydrogen on its hydroxyl-bearing carbon. It therefore does not undergo the normal alcohol-to-carbonyl oxidation without breaking carbon–carbon bonds.

3) Zinc chloride and hydrochloric acid: the Lucas reagent

Zinc chloride, [Zn+2].[Cl-].[Cl-], promotes replacement of the hydroxyl group by chloride: \[ \ce{ROH + HCl ->[ZnCl2] RCl + H2O}. \]

The corresponding chlorides are:

Starting alcohol Chloride product SMILES
Pentan-1-ol 1-Chloropentane CCCCCCl
Pentan-2-ol 2-Chloropentane CC(Cl)CCC
2-Methylbutan-2-ol 2-Chloro-2-methylbutane CC(Cl)(C)CC

The tertiary alcohol reacts rapidly because formation of a tertiary carbocation is relatively favourable. The secondary alcohol reacts more slowly. The primary alcohol does not give the usual positive room-temperature Lucas test; its substitution is much slower and can be achieved on heating.

The problem specifies no configuration for pentan-2-ol, so the SMILES for its derivatives leave that stereochemistry unspecified.

4) Iodine and sodium hydroxide: the iodoform reaction

Iodine, II, and sodium hydroxide, [Na+].[OH-], give a positive iodoform reaction with a methyl ketone or an alcohol that can be oxidized to one.

Only pentan-2-ol has the required \(\ce{CH3CH(OH)-}\) group. It first forms pentan-2-one; iodination and cleavage then produce iodoform, IC(I)I, and sodium butanoate, CCCC(=O)[O-].[Na+].

The balanced overall equation is \[ \ce{CH3CH(OH)CH2CH2CH3 + 4I2 + 6NaOH -> CH3CH2CH2COONa + CHI3 + 5NaI + 5H2O}. \] Sodium iodide has SMILES [Na+].[I-].

The positive observation is a yellow iodoform precipitate. Pentan-1-ol does not have the required carbon skeleton, and the tertiary alcohol cannot undergo the necessary oxidation to a methyl ketone.

5.3 — Enantiomers and diastereomers

Read each Fischer projection from the aldehyde end, recording the hydroxyl positions at C2, C3, C4 and C5. Horizontal Fischer bonds point towards the viewer; vertical bonds point away.

Structure Hydroxyl positions, C2 to C5 Identity Open-chain stereochemical SMILES
A Right, left, right, right D-Glucose O=C[C@H](O)[C@@H](O)[C@H](O)[C@H](O)CO
B Left, left, right, right D-Mannose O=C[C@@H](O)[C@@H](O)[C@H](O)[C@H](O)CO
C Left, right, left, left L-Glucose O=C[C@@H](O)[C@H](O)[C@@H](O)[C@@H](O)CO
D Right, right, left, left L-Mannose O=C[C@H](O)[C@H](O)[C@@H](O)[C@@H](O)CO

Enantiomers have opposite configurations at every stereogenic centre. Accordingly, \[ \boxed{\text{Enantiomers: A–C and B–D}}. \]

Diastereomers differ at some, but not all, stereogenic centres: \[ \boxed{\text{Diastereomers: A–B,\ A–D,\ B–C,\ C–D}}. \]

More specifically, A–B and C–D differ only at C2 and are C2 epimers. A–D and B–C differ at C3, C4 and C5.

5.4 — Identification of the two hydrocarbons

The molecular formula gives one degree of unsaturation: \[ \mathrm{DBE}=\frac{2(4)+2-8}{2}=1. \]

Hydrogenation and ozonolysis identify this unsaturation as a carbon–carbon double bond. Since ozonolysis gives only ethanal, CC=O, each double-bond carbon must carry one methyl group and one hydrogen. The connectivity is therefore \[ \ce{CH3CH=CHCH3}. \]

This connectivity has two geometrical isomers:

  • cis-(Z)-But-2-ene, C/C=C\C: draw the two methyl groups on the same side of the double bond.
  • trans-(E)-But-2-ene, C/C=C/C: draw the two methyl groups on opposite sides of the double bond.

In each drawing, each alkene carbon also bears one hydrogen.

Both compounds hydrogenate to butane, CCCC: \[ \ce{CH3CH=CHCH3 + H2 ->[Pt] CH3CH2CH2CH3}. \] Hydrogen has SMILES [H][H]; platinum is represented by [Pt].

For example, with a reductive zinc work-up, their ozonolysis can be written: \[ \ce{CH3CH=CHCH3 + O3 + Zn -> 2CH3CHO + ZnO}. \] The additional species are ozone, [O-][O+]=O; zinc, [Zn]; and zinc oxide, [Zn+2].[O-2].

In either isomer, symmetry makes the two methyl groups equivalent and the two alkene hydrogens equivalent. The integrated proton counts are therefore \[ 6:2=\boxed{3:1}, \] as required.

5.5 — Complete hydrolysis of lecithin

Lecithin here means phosphatidylcholine. It contains a glycerol backbone, two fatty-acid ester groups, and a phosphate diester connecting glycerol to choline.

A generic connectivity SMILES is C[N+](C)(C)CCOP(=O)([O-])OCC(OC(=O)[*:2])COC(=O)[*:1].

The mapped wildcards represent the two unspecified hydrocarbon residues. Glycerol stereochemistry is immaterial to identifying the complete-hydrolysis products.

Complete hydrolysis breaks the two carboxylic ester bonds and both phosphate ester bonds. Consequently, one mole of lecithin yields:

Component Amount per mole of lecithin
Glycerol \(1\ \mathrm{mol}\)
Fatty acids \(2\ \mathrm{mol}\)
Phosphoric acid \(1\ \mathrm{mol}\)
Choline \(1\ \mathrm{mol}\)

Writing the neutral zwitterionic phosphatidylcholine as PC, a balanced acid-hydrolysis equation is \[ \ce{PC + 4H2O + H+ -> C3H8O3 + R^1COOH + R^2COOH + H3PO4 + C5H14NO+}. \]

The proton on the left accounts for the charge of the choline cation when the phosphorus-containing product is written as neutral phosphoric acid.

Phosphatidic acid, with generic neutral-acid SMILES O=P(O)(O)OCC(OC(=O)[*:2])COC(=O)[*:1], can arise during partial hydrolysis. It still contains ester bonds and is therefore not an end product of complete hydrolysis.

5.6 — Acids belonging to the citric acid cycle

Four of the listed acids correspond to cycle intermediates:

Selected acid Reason
L-(S)-Malic acid Its conjugate base, L-malate, is formed by stereospecific hydration of fumarate.
Oxaloacetic acid Oxaloacetate is formed by oxidation of malate and regenerated at the end of the cycle.
2-Oxoglutaric acid Also called \(\alpha\)-ketoglutaric acid; it is the cycle’s five-carbon 2-oxo dicarboxylic acid intermediate.
Fumaric acid Its conjugate base, fumarate, is the cycle’s trans-unsaturated dicarboxylate.

For example, using the acid forms to display their carbon skeletons, the hydration is \[ \ce{HOOC-CH=CH-COOH + H2O -> HOOC-CH(OH)-CH2-COOH}, \] where the substrate is trans and the product has the L-(S) configuration.

The subsequent oxidation is \[ \ce{HOOC-CH(OH)-CH2-COOH + NAD+ -> HOOC-C(=O)-CH2-COOH + NADH + H+}. \]

The other choices are excluded as follows:

Excluded acid SMILES Reason
Maleic acid O=C(O)/C=C\C(=O)O It has cis geometry; the cycle contains trans-fumarate.
Mandelic acid O=C(O)C(O)c1ccccc1 This aromatic hydroxy acid is not a cycle intermediate.
Tricarb-allylic acid O=C(O)CC(CC(=O)O)C(=O)O It is not a cycle intermediate; unlike citric acid, it lacks the hydroxyl group.
Acetoacetic acid CC(=O)CC(=O)O It belongs to ketone-body metabolism, rather than being an intermediate of the cycle itself.

For comparison, citric acid has SMILES O=C(O)CC(O)(CC(=O)O)C(=O)O.

5.7 — Reduction of the nicotinamide ring

The correct structure is C.

Number the nicotinamide ring nitrogen as position 1. During reduction, a hydride equivalent—one hydrogen nucleus together with two electrons—is transferred to C4, the carbon opposite the nitrogen.

The balanced reduction half-equation is \[ \ce{NAD+ + H+ + 2e- -> NADH}. \] Equivalently, the hydride-transfer bookkeeping is \[ \ce{NAD+ + H- -> NADH}, \] where hydride has SMILES [H-]; this notation does not imply a free hydride ion in aqueous solution.

The relevant ring structures are:

State Nicotinamide-moiety SMILES
Oxidized pyridinium moiety of \(\ce{NAD+}\) [*][n+]1cc(C(=O)N)ccc1
Reduced 1,4-dihydronicotinamide moiety of \(\ce{NADH}\) [*]N1C=C(C(=O)N)CC=C1

Here [*] represents the unchanged ribosyl linkage to the remainder of the cofactor.

Hydride addition and rearrangement of the ring’s \(\pi\) electrons neutralize the nitrogen. C4 becomes a methylene carbon, and the remaining double bonds are C2=C3 and C5=C6. These are precisely the features shown in C.

In a substrate dehydrogenation, such as malate oxidation above, the second hydrogen is released to the medium as \(\ce{H+}\), explaining the proton drawn alongside C. Structure A instead shows a protonated, positively charged nitrogen; B has the wrong double-bond arrangement.

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.