The Spanish jasmine originated from the valleys of Himalaya. This so called Jasminum grandiflorum L. — Physical Chemistry Chemistry Question
No perfume without jasmine
The Spanish jasmine originated from the valleys of Himalaya. This so called Jasminum grandiflorum L., which was grafted on the wild Jaminum officinale L, has the possibility to resist against the cold of the mountain climate [1]. In 1962, E. Demole and E. Lederer successfully isolated and clearly determined the structures of the important flavor component of the essential oil of jasmine [1]. It was actually a mixture of two diastereomers (Z-(3R,7R) methyl jasmonate and (Z-(3R,7S) methyl jasmonate with the structures shown below. [VISUAL] [1]
Due to its important role in perfume industry as well as its limited natural resource, the synthesis of jasmonates has attracted interest of many chemists [2]. The first synthesis of optically pure (Z-(3R,7S)-methyl jasmonate was carried out in 1990 by Helmchen with the following steps [2]:
I. The first step in the synthesis is a Diels-Alder reaction between cyclopentadiene A and ester B of fumaric acid with (S)-ethyl lactate to yield C [2].
A cyclo-addition reaction, a kind of pericyclic reaction, involves bonding between the termini of two π systems to produce a new ring (scheme below) [3]. [VISUAL] The product has two more σ bonds and two less π bonds than the reactants [3]. The Diels-Alder reaction, or so called [4+2], is a common thermally activated cycloaddition whose stereoselectivity is syn addition [3]. In this concerted syn addition, the stereochemical relationships among the substituents are retained in the product(s) [3].
The hydrolysis of C in basic solution followed by an acidification step resulted in the formation of D [4].
20.1 Draw the structures of C and D [4].
Model Answer
In a [4+2] cycloaddition reaction (Diels-Alder reaction), the configuration of the dienophine (B) remained unchanged: the two ester groups –COOLac of compound (C) were placed in different sites in comparison to the six-membered ring [5]. The hydrolysis of these two ester groups in LiOH solution gave the two corresponding trans carboxyl groups [5, 6]. [VISUAL]
II. The reaction between D and I2 in KI solution gave rise to the formation of the five-membered ring lactone of the endo –COOH group [4]. When heated under basic condition, E underwent a decarboxylation-cyclization reaction to produce compound F [4]. The hydrolysis in basic medium followed by oxidation with NaIO4/RuO4 transformed F into γ-keto acid G [4].
20.2 Write down the structures of E to G [7].
Model Answer
When dicarboxylic acid (D) was treated with I2/KI, it was transformed into γ-iodolactone (E) of the endo –COOH [6]. This lactone then underwent a decarboxylation-cyclization in basic solution step to give lactone (F) containing a three-membered ring [6]. The secondary –OH group which resulted from the hydrolysis of (F) was oxidized by NaIO4 to form a carbonyl group [6]. [VISUAL]
III. G was reacted with HI to yield compound H containing only five-membered rings [7]. When H was reduced by Zn/KH2PO4 in tetrahydrofuran, it transformed into carboxylic acid I which was then oxidized in a Bayer-Villiger rearrangement with meta-chloroperoxybenzoic acid (m-CPBA) to give the major product K [7]. K was treated with oxaloyl chloride, followed by a Pd/BaSO4 catalyzed reduction with H2 (Rosenmund reduction) to give L [7]. Compound M was separated from the Wittig reaction mixture between L and the yield Ph3P =CHOCH3 [7].
20.3 Draw the structures of the compounds from G to L [8].
Model Answer
The addition-ring opening step of the cyclopropane ring with HI oriented by the (-C) conjugation effect of the carbonyl resulted in the formation of γ-iodo acid (H) containing only five-membered rings [9]. The reductive elimination of the iodine atom by Zn in acetic acid produced ketoacid (I) which underwent a Bayer-Viiliger oxidation to yield lactone (K) (the main product) with one carboxyl group [9]. The carboxyl group was treated under Rosenmund reduction condition in which lactone (L) with an aldehyde functional group was obtained [9, 10]. Vinyl ether (M) was separated from the Wittig reaction between the aldehyde and the ylide Ph3P=CHOMe [10]. [VISUAL]
IV. Hydrolysis of M in THF/H2O solution of acetic acid produced N which underwent a Wittig reaction with the ylide Ph3P = CHCH2CH3 to form O [8]. The hydrolysis of O in basic solution, followed by a neutralization step and then treatment with diazomethane resulted in the formation of P [8]. In the last step, the target molecule, (Z-(3R,7S)-methyl jasmonate Q, was obtained in the oxidation reaction of P with pyridinium dichromate [5, 8].
20.1 Draw the structures of compounds from N to P [5].
Model Answer
Methyl vinyl ether (M) was hydrolyzed in acidic medium to give lactone (N) containing an aldehyde functional group [10]. From the Wittig reaction between (N) and the ylide Ph3P=CHCH2CH3, lactone (O) with a cis carbon-carbon double bond was separated [10]. The hydrolysis of the lactone (O) followed by treatment with diazomethane produced ester (P) which was oxidized with pyridine dichromate to give the target compound (Z-(3R,7S)-methyl jasmonate [10, 11]. [VISUAL]