Question

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Answer 1

Olodaterol is a drug which has a a amide group in it. To synthesis this drug, first we need to convert the starting material into a cyclic amide derivative. This reaction has two to three steps. We will write one by one.

The starting material is ''1-(2,5-dihydroxyphenyl)ethan-1-one'' which has two hydroxyl group and one keto group attach to the benzene ring at 1,2,5th position. Here, one of the hydroxyl group present at 5th position of benzene have to protect by reacting with benzyl bromide (BnBr) in the presence of potassium carbonate and methyl isobutyl ketone (MIBK) as a solvent. Hydroxyl group protected compound is treated with nitric acid (HNO₃) in the presence of acetic acid (AcOH) as a proton donor and the nitration reaction at 3rd position occurs to give ''1-[5-(benzyloxy)-2-hydroxy-3-nitrophenyl]ethan-1-one'' as a product.

Further, this nitro group will undergo reduction in the presence of H₂ and PtO₂ (platinum dioxide) as a reducing catalyst to convert nitro group into amine group (-NH₂). This amine group and its adjacent hydroxyl group together reacts with chloroacetyl chloride on heating gives cyclic amide as the product. This amide undergo several chemical reactions to give Olodaterol drug as product.

Here I have mentioned the reaction till the formation of amide:

NO BnO Bno но HNO3, AcOH BnBr, KCO, MIBK он он ње/ "Yo Нас" "O 1-(2,5-dihy droxyphenyl)ethan-1-one 1-[5-(benzyloxy)-2-hydroxy- 3-nitrophenyl]ethan-1-one Reduction H2/PtO2 Cl Cyclisation and amide formation BnO CH CI OH chloroacetvl chloride 2 HCI 1-[3-amino-5-(benzyloxy) 2-hydroxyphenyl]ethan-1-one amide formation 1-[6-(benzyloxy)-3-methylidene-3,4-dihydro- 2H-1,4-benzoxazin-8-yl]ethan-1-one