Question

Identify the sugar in each description a. An aldopentose that is not D-arabinose forms D-arabinitol when it is reduced with NaBH4. b. A sugar that is not D-altrose forms D-altraric acid when it is oxidized with nitric acid c. A ketose that, when reduced with NaBH4, forms D-altritol and D-allitol

Answer 1

When D-arabinose is treated with NaBH4 followed by workup using dilute acid the aldehyde group is reduced into primary alcohol он но HO i. NaBH он ii. H,O он он H2COH Now, another aldopentose which is not D-arabinose gives the same reduced product. Hence the probable structure of that compound could be that of opposite of arabinose i.e. the terminal CHO and CH2OH groups interchanged он H2C H2COH но HO но i. NaBH i. NaBH4 он ii. H3Ot OH OH H2COH In plane 180 rotation но HOCH2 D-lyxosc So, the other aldopentose is D-lyxose

When D-altrose is heated with HNO3 the aldehyde group and the terminal CH2OH group both are oxidized into carboxylic acid group. Such selectivity arises from the oxidizing capacity of HNO3 which can oxidize only aldehyde and primary alcohol only but not secondary alcohol но HO но он он HNO, он OH он он HO он D-Altraric acid D-Altrose Now, another aldohexose which is not D-altrose gives the same oxidized product. Hence the probable structure of that compound could be that of opposite of altrose i.e. the terminal CHO and CH2OH groups interchanged

он HO но HNO3 HNO3 OH OH но он D-Altrose D-Altraric acid In plane 180° rotation HO HO он D-Tallose So, the other aldohexose is D-Tallose

When a ketose is reduced two products are obtained as a new stereo center is generated at the position of the keto functionality where the reduction takes place. Now, according to the condition the two reduced products obtained are D-altriol and D-allitol. So, the ketose is shown as follows CH2OH он он он CH2OH D-psicose NaBH4 CH2OH CH2OH он HO он он он он он он CH2OH CH2OH So, the ketose under consideration is D-psicose