🧪 TheChemSolverInternational Chemistry Olympiad
Analytical ChemistryIChO

Cinnamon is an important part of many dishes and desserts, including Czech apple strudel, Swedish ci — Analytical Chemistry Chemistry Question

Cinnamon all around

Cinnamon is an important part of many dishes and desserts, including Czech apple strudel, Swedish cinnamon rolls kanelbullar, Indian spicy rice biryani and the popular winter drink mulled wine. There are several compounds in cinnamon which are responsible for its taste and smell, mainly cinnamaldehyde and cinnamic acid and its derivatives. It is noteworthy that (E)-cinnamaldehyde and cinnamic acid are much more abundant in nature than their respective (Z)-isomers. While the former have a honey, cinnamon-like odour, the latter are completely odourless. Let us first explore the syntheses of both stereoisomers of cinnamic acid

24.1.

[VISUAL]

Draw the formulae of isomeric products A and B.

Model Answer

Structure of A: (E)-cinnamic acid (trans-isomer); Structure of B: (Z)-cinnamic acid (cis-isomer).

24.2.

Propose reasonable reaction conditions (X) for the interconversion of cinnamic acid isomers (A → B).

Model Answer

Direct UV irradiation (313 nm, acetonitrile). A conformationally mobile biradical is formed. Under these conditions, B and A are obtained in a 79 : 21 ratio. Alternatively, UV irradiation with sensitizers (e.g. riboflavin), or reagents such as diphenyldiselenide, hydrogen peroxide etc. can be used.

24.3.

Starting from 2-bromoacetic acid, how would you prepare the phosphonate used in the above-mentioned synthesis?

[VISUAL]

Model Answer

Arbuzov reaction with 2-bromoacetic acid and tribenzyl phosphite: Br-CH2-COOH + P(OBn)3 --heating--> dibenzyl phosphonoacetic acid (HOOC-CH2-P(O)(OBn)2).

24.4.

A key intermediate, epoxyacid F, can be prepared from both (E- and (Z)-ethyl cinnamate. (E-Ethyl cinnamate is first reacted with osmium tetroxide in the presence of a chiral ligand. Only one enantiomer of C is formed. The reaction of C with one equivalent of tosyl chloride leads to compound D in which the hydroxyl group at position 2 is tosylated. In a basic environment, tosylate D is converted to compound E. Alternatively, compound E can be prepared in one step from (Z)-ethyl cinnamate by hypochlorite-mediated oxidation. A chiral catalyst ensures the formation of a single enantiomer. Hydrolysis of E then provides acid F.

[VISUAL]

Draw the structures of compounds C, D and E, including stereochemistry. The absolute configuration of all compounds can be deduced from the known structure of acid F.

Model Answer

Structure of C: ethyl (2R,3S-2,3-dihydroxy-3-phenylpropanoate; Structure of D: ethyl (2S,3S-3-hydroxy-3-phenyl-2-(tosyloxy)propanoate; Structure of E: ethyl (2R,3R-3-phenyloxirane-2-carboxylate (cis-epoxide).

24.5.

Epoxyacid F reacts with 10-deacetylbaccatin III (G) in the presence of N,N´-dicyclohexylcarbodiimide (DCC) to provide compound H. A subsequent reaction with NaN3 leads to compound I, which is easily converted to docetaxel (J).

[VISUAL]

Draw the structures of compounds H and I, including stereochemistry.

Model Answer

Structure of H: Ester of 10-deacetylbaccatin III with epoxyacid F at the C13 hydroxyl group; Structure of I: Ester with azide group at C3' and hydroxyl group at C2' of the side chain.

24.6.

What is the role of DCC in the first step? Write the appropriate chemical equation.

Model Answer

The carboxylic acid functional group reacts with DCC to form an O-acylisourea, which serves as the reactive intermediate in reactions with nucleophiles (e.g. alcohols or amines) in acyl nucleophilic substitutions.

[VISUAL]

24.7.

Taxifolin (K) is an inhibitor of ovarian cancer with strong hepatoprotective properties. It belongs to 3-hydroxyflavanone (L) family of natural products.

[VISUAL]

The synthesis of compound L starts with asymmetric dihydroxylation of methyl cinnamate M using osmium tetroxide as catalyst, potassium ferricyanide as oxidant and a chiral ligand. The synthesis continues with the transformation of the ester group in compound N to compound O and subsequent reaction of hydroxyl groups in the presence of an excess of chloromethyl methyl ether (MOM–Cl), yielding compound P. Compound P reacts with a protected aryllithium reagent in a non-stereoselective manner, giving a mixture of two compounds Q and R. The reaction of the mixture of compounds Q and R with PDC yields a single compound S, which upon acidic treatment provides compound T. Finally, the reaction of T with diisopropyl azodicarboxylate (DIAD) and triphenylphosphine proceeds by formal SN2 substitution of one hydroxyl group with the other to furnish target compound L.

[VISUAL]

From the known configuration of product T, decide whether compound M is the ester of (E- or (Z)-cinnamic acid.

Model Answer

The starting compound is (E)-cinnamic acid methyl ester.

24.8.

Draw the structures of compounds N–S and L, with the correct configuration on the benzylic oxygen.

Model Answer

Structures of N, O, P, Q, R, S, and L are drawn with correct stereochemistry.

[VISUAL]

24.9.

Decide whether compounds Q and R are a) constitutional isomers, b) diastereoisomers or c) enantiomers.

Model Answer

The two isomers Q and R are diastereoisomers (diastereomers).

24.10.

Why can we not react compound O with the aryllithium reagent directly?

Model Answer

The acidic hydrogens of the OH groups would decompose the organolithium compound.

24.11.

Draw the structure of the PDC reagent.

Model Answer

Pyridinium dichromate structure.

[VISUAL]

24.12.

After whom is the reaction converting compound T to compound L named?

Model Answer

The reaction is named after Prof. Mitsunobu.

💬
Still have doubts about this question?
Practice more questions like this, completely free.

Practice International Chemistry Olympiad questions like this — free

4,000+ questions across AP Chemistry, USNCO, and IChO — all free, no signup required.