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Draw the expected product of the reaction of the following sugars with excess methyl iodide and silver oxide.

  1. α-¶Ù-´Ú°ù³Ü³¦³Ù´Ç´Ú³Ü°ù²¹²Ô´Ç²õ±ð
  2. β-¶Ù-²µ²¹±ô²¹³¦³Ù´Ç±è²â°ù²¹²Ô´Ç²õ±ð

Short Answer

Expert verified
  1. methyl 2,3,4,6,-tetra-O-methyl-α-D-fructofuranoside
  2. methyl 2,3,4,6,-tetra-O-methyl-β-D-fructofuranoside


Step by step solution

01

Formation of methyl ether

Hydroxy groups of sugar can be converted to methyl ethers by treating with methyl iodide (CH3I) and silver oxide (Ag2O). CH3-I group can be polarised by silver oxide (Ag2O), which makes the methyl carbon more strongly electrophilic. Attack by the -OH group of carbohydrate is then followed by deprotonation which gives the ether.

02

Pyranose and furanose

Pyranose is a six-membered cyclic hemiacetal while a furanose is a five-membered cyclic hemiacetal.

03

Formation of product

  1. α-¶Ù-´Ú°ù³Ü³¦³Ù´Ç´Ú³Ü°ù²¹²Ô´Ç²õ±ð on treatment with excess methyl iodide and silver oxide gives methyl 2,3,4,6, -tetra-O-methyl-α-D-fructofuranoside as the product.
  2. β-¶Ù-²µ²¹±ô²¹³¦³Ù´Ç±è²â°ù²¹²Ô´Ç²õ±ð on treatment with excess methyl iodide and silver oxide gives methyl 2,3,4,6,-tetra-O-methyl-β-D-galactopyranoside as the product.

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Most popular questions from this chapter

Predict the products formed when the following sugars react with excess acetic anhydride and pyridine.

  1. α-¶Ù-²µ±ô³Ü³¦´Ç±è²â°ù²¹²Ô´Ç²õ±ð
  2. β-¶Ù-°ù¾±²ú´Ç´Ú³Ü°ù²¹²Ô´Ç²õ±ð

Predict the products obtained when D-galactose reacts with each reagent.

(a) Br2 and H2O

(b) NaOH,H2O

(c)CH3OH, H+

(d) Ag(NH3)+2OH+

(e) H2, Ni

(f) excess Ac2O and pyridine

(g) excess CH3I ,Ag2O

(h) NaBH4

(i) Br2 , H2O then H2O2 and Fe2(SO4)3

(j) (1) KCN/HCN; (2) H2 Pd/BaSO4; (3) H3O+

(k)excess HIO4

H. G. Khorana won the Nobel Prize in Medicine in 1968 for developing the synthesis of DNA and RNA and for helping to unravel the genetic code. Part of the chemistry he developed was the use of selective protecting groups for the 5′ OH group of nucleosides.

The trityl ether derivative of just the 5′ OH group is obtained by reaction of the nucleoside with trityl chloride, MMT chloride, or DMT chloride and a base like Et3N. The trityl ether derivative can be removed in dilute aqueous acid. DMT derivatives hydrolyze fastest, followed by MMT derivatives, and trityl derivatives slowest.

(a) Draw the product with the trityl derivative on the 5′ oxygen.

(b) Explain why the trityl derivative is selective for the 5′ OH group. Why doesn’t it react at 2′ or 3′? (c) Why is the DMT group easiest to remove under dilute acid conditions? Why does the solution instantly turn orange when acid is added to a DMT derivative?

Fructose is found in many fruits. From memory, draw fructose in

  1. the Fischer projection of the open chain.
  2. The most stable chair conformation of the most stable pyranose anomer.
  3. The Haworth projection of the most stable pyranose anomer

Question. Retroviruses like HIV, the pathogen responsible for AIDS, incorporate an RNA template that is copied into DNA during infection. The reverse transcriptaseenzyme that copies RNA into DNA is relatively nonselective and error-prone, leading to a high mutation rate. Its lack of selectivity is exploited by the anti-HIV drug AZT (3’-azido-2’,3’-dideoxythymidine), which becomes phosphorylated and is incorporated by reverse transcriptase into DNA, where it acts as a chain terminator. Mammalian DNA polymerases are more selective, having a low affinity for AZT, so its toxicity is relatively low.

  1. Draw the structures of AZT and natural deoxythymidine.
  2. Draw the structure of AZT 5’-triphosphate, the derivative that inhibits reverse transcriptase.
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