Routes to Essential Medicines. Peter J. Harrington

      Discussion. The benzimidazole is formed in the final step from the benzene‐1,2‐diamine and N‐methoxycarbonylcyanamide. N‐Methoxycarbonylcyanamide is formed from cyanamide and methyl chloroformate. 4‐(Propylthio)benzene‐1,2‐diamine is formed by reduction of 2‐nitro‐4‐propylthioaniline. A C─S bond is formed by displacement of chloride from 4‐chloro‐2‐nitroacetanilide by sodium propanethiolate. The acetanilide is also hydrolyzed under the chloride displacement reaction conditions. 4‐Chloro‐2‐nitroacetanilide is formed from 4‐chloro‐2‐nitroaniline and acetic anhydride.

      Extended Discussion

      2‐Nitro‐4‐propylthioaniline can also be manufactured from 1‐chloro‐2‐nitrobenzene. Draw the structures of the retrosynthetic analysis of this route. List the pros and cons for both routes. Which route is preferred?

Allopurinol

      Antineoplastics and Immunosuppressives/Cytotoxic and Adjuvant Medicines

      Medicines for Diseases of Joints/Medicines Used to Treat Gout

       Hydrazine is often the source of the two nitrogen atoms in a pyrazole ring.

Discussion. The pyrimidine ring of the pyrazolo[3,4‐d]pyrimidine is formed in the final step by reaction of 3‐aminopyrazole‐4‐carboxamide with formamide. The pyrazole ring is formed from 2‐cyano‐3‐morpholinoacrylamide and hydrazine. The enamine of 2‐cyano‐3‐morpholinoacrylamide is formed from the enol ether by the displacement of ethanol by morpholine. The enol ether of 2‐cyano‐3‐ethoxyacrylamide is formed by the reaction of 2‐cyanoacetamide with triethyl orthoformate.

      Extended Discussion

      The pyrimidine ring is also formed by reaction of ethyl 3‐aminopyrazole‐4‐carboxylate with formamide. Draw the structures of the retrosynthetic analysis of ethyl 3‐aminopyrazole‐4‐carboxylate. List the pros and cons for both routes and select one route as the preferred route.

Amidotriazoate

      Diagnostic Agents/Radiocontrast Media

       For a symmetrical molecule, symmetrical disconnections lead back to symmetrical intermediates and are likely associated with the shortest route.

Discussion. Since some amide hydrolysis is likely under iodination conditions, the diamide is formed in the final step by reaction of the diamine with acetic anhydride. The triiodide is formed by iodination of 3,5‐diaminobenzoic acid.

      Extended Discussion

      List reagents or reagent combinations used for direct iodination of an aromatic ring.

Amikacin

      Anti‐Infective Medicines/Antibacterials/Other Antibacterials

      Anti‐Infective Medicines/Antibacterials/Antituberculosis Medicines

       A single‐enantiomer molecule with multiple chiral carbons is often made by modification of a natural product which has most or all of the chiral carbons already in place.

      Discussion. Amikacin is semisynthetic. Amikacin is formed by acylation of the amino group at C1 of kanamycin A. This selective acylation requires a protection–deprotection strategy since kanamycin A has four amino groups and the amino group at C1 is not the most reactive.

Three of the amino groups of amikacin are released in the final step by benzyl carbamate hydrogenolysis. The amide at C1 is formed by reaction of the amino group with an N‐hydroxysuccinimide ester. Amino groups at C3 and C6′ of kanamycin A are protected as benzyl carbamates (Cbz). Kanamycin A is produced by fermentation.

The N‐hydroxysuccinimide ester is formed from the carboxylic acid. The amino group of the 4‐amino‐2‐hydroxybutanoic acid is protected as the benzyl carbamate. (S)‐4‐Amino‐2‐hydroxybutanoic acid is formed from (S)‐2‐hydroxyglutaramic acid (Hofmann Rearrangement). The amide is formed from the lactone. (S)‐5‐Oxotetrahydrofuran‐2‐carboxylic acid lactone is formed by diazotization of L‐glutamic acid. L‐Glutamic acid is produced by fermentation.

      Extended Discussion

      Draw the structures of three side products which are likely to be formed in the reaction of kanamycin A with two equivalents of benzyl chloroformate. Draw the structure(s) of likely impurities in amikacin as each side product is carried through the amide formation and carbamate hydrogenolysis.

      or

      Draw the structures of the retrosynthetic analysis of the alternative route to (S)‐4‐amino‐2‐hydroxybutanoic acid from L‐asparagine. List the pros and cons for both routes and select one route as the preferred route.

Amiloride

      Diuretic

      A nitrogen substituent on a pyrazine ring carbon is often introduced by displacement of chloride. The substitution is facilitated by the adjacent ring nitrogen and can be further facilitated by an electron‐withdrawing group (NO ₂ , SO ₂ R, COOR, CN) on a para ring carbon.

Discussion. The guanidine group is introduced in the final step by reaction of guanidine with the methyl ester. Chloride at the 5‐position of the 5,6‐dichloropyrazine is displaced by ammonia. The 5,6‐dichloropyrazine is formed by chlorination of methyl 3‐aminopyrazine‐2‐carboxylate. The methyl ester is formed from the carboxylic acid (Fischer Esterification).

      3‐Aminopyrazine‐2‐carboxylic acid is formed by hydrolysis of the pyrimidine ring of lumazine (1H‐pteridine‐2,4‐dione). The pyrazine ring of lumazine is formed by reaction of 5,6‐diaminouracil with glyoxal. The amino group at the 5‐position of 5,6‐diaminouracil is formed by reduction of a nitroso group. The nitroso group is introduced by nitrosation of 6‐aminouracil.