Vitamin H or biotin is a highly effective growth promoter which plays an important role in many orga — Organic Chemistry Chemistry Question
Vitamin H
Vitamin H or biotin is a highly effective growth promoter which plays an important role in many organisms, for instance baker’s yeast (Saccharomyces corevisiae) [1]. Human beings have different demand in biotin depending on age [2]. Lack of biotin can lead to diseases such as dermatitis, absence of appetite, fatigue, muscular pain and nerve perturbance [2].
Biotin was first found in 1901 by Wildiers as a growth-promoting vitamin of yeasts [2]. Thereafter, it was found in egg yolk (by Kögl, 1936) and in liver (by Szent-Györgyi, 1936) [2]. The absolute configurations of chiral carbons of biotin were established by Trotter and Hamilton in 1966 by the single-crystal X-Ray diffraction [2]. The molecule of biotin has three chiral carbons [2]. Hence, there are theoretically eight diastereomers [2, 3]. Only the (3aS, 4S, 6aR-(D-(+)-biotin, however, shows the high bioactivity [3].
[VISUAL]
In 1982, researchers from Hoffmann-La Roche published an elegant synthesis procedure for (D-(+)-Biotin from the methyl ester of the amino acid L-Cysteine with the following steps [3].
The thiol group of the methyl ester of L-cysteine was transformed into a disulfide S-S bond (compound A) in an oxidation reaction [3]. A was then treated with hex-5-inoyl chloride to give compound B which was subsequently reduced by Zn / CH3COOH to afford a thiol [3, 4]. This compound was cyclized by the addition of the –SH group to the terminal alkyne under atmospheric condition, resulting in the formation of C with a ten-membered ring containing a (Z) C=C double bond [4].
Write down the structural formulae of A, B and C in the hereunder scheme. [VISUAL]
Model Answer
The chloride acid was reacted with the amino group of the “bis(L-cystein)” (A) to give amide (B) [5]. Zn powder in acetic acid solution reduced the S-S bond of (B) to give an intermediate containing a thiol (-SH) group [5]. Under normal condition, the -SH group added spontaneously to the terminal alkyne group to yield (C) with a ten-membered ring, of which the newly formed C=C double bonds had (Z) configuration [5, 6].
The reduction of C by diisobutylaluminium hydride ((i-Bu)2AlH, DIBAL) resulted in the formation of (D) [4, 7]. The condensation reaction between D and benzylhydroxylamine in dichloromethane produced nitrone E (a nitrone is an organic compound which contains a > C=N+-O- group) [7]. E underwent a 1,3-dipolar intramolecular cyclization reaction to afford polycyclic compound F of which the two heterocycles, isoxazolidine (1,2-oxazolidine) and tetrahydrothiophene, shared a common bond [7]. The cyclization reaction resulted in the (S) configurations of the two carbons at the common bond and the (R) configuration of the carbon connected to the oxygen [7, 8].
Write down the structural formulae of D, E and F.
Model Answer
Diisobutyl aluminium hidride (DIBAL) partially reduced ester (C) into aldehyde (D) which was condensed with benzylhydroxylamine to give nitrone (E) with (E) configuration [6]. In the intramolecular [4+2] cyclization reaction of (E) (note that ‘4’ and ‘2’ are the numbers of π electron of the nitrone and the double bond involved in the cyclization, respectively), the configuration of the double bonds C=C and C=N remained unchanged [6, 9]. The resulting compound (F) had three new chiral centers, two of which were (3aS, 4R) [9]. They were the configurations of the corresponding C3 and C4 in the skeleton of (D-(+)-Biotin [9]. The third chiral carbon which was attached to the oxygen atom had an (R) configuration [9].
When F was reduced with Zn powder in acetic acid, the N-O bond of the heterocycle isoxazolidine was broken to give compound G [8]. The reaction between G and chloroformate in the presence of Na2CO3 in THF resulted in the formation of compound H [8]. The treatment of H in a hot solution of Ba(OH)2 in dioxane, followed by an acidic work-up yielded the bicyclic δ-hydroxy acid I containing all the chiral centers of (D-(+)-Biotin but with an “excessive” -OH group [8, 10].
Draw the structures of G, H and I and explain the formation of I from H.
Model Answer
(F) was reduced by Zn in acetic acid to give (G) containing one –OH group and a second-order amino group [9, 11]. The amino group was reacted with chlorofomate in the presence of Na2CO3 in THF solution produce (H) [11]. Under basic condition, the ten-membered ring of (H) was opened to give δ-hydroxy acid (I) [11]. The configuration (6aR) of (I) resulted from the (R) configuration of the chiral carbon in L-cysteine [11]. Therefore, L-cystein is chosen as the starting material for the synthesis [11].
Compound I was treated with SOCl2 to yield the corresponding chloride acid K, given that the configuration of the carbon attached to the “excessive” -OH is maintained in K [10]. Ester L was formed when K was reacted with methanol [5, 10]. L was reduced with NaBH4 in dimethylformamide at 80oC to give ester M which was hydrolyzed in aqueous acidic solution of HBr to give optically pure (D-(+)-Biotin [5].
Draw the structures of I, K, L and M and the intermediates to explain the influence of the sulfur atom on the stereochemical outcome of K.
Model Answer
The sulfur atom caused an anchimeric effect by which the configuration of the carbon attached to the –OH in compound (I) remained unchanged as this –OH group was replaced by the halogen atom to yield (K) [11, 12]. The halogen atom was then replaced when (K) was reduced with NaBH4 in which the “pentanoic acid” branch of (D-(+)-biotin was formed [12]. The hydrolysis of ester (M) in the aqueous solution of HBr, followed by the removal of the benzyl group resulted in the formation of the target molecule, (D-(+)-Biotin [12].