This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct. Two chiral centres are present in (a) aldotetrose (b) aldohexose (c) aldotriose (d) ketotriose

(a) Aldotetrose - A four-carbon sugar with an aldehyde functional group (b) Aldohexose - A six-carbon sugar with an aldehyde functional group (c) Aldotriose - A three-carbon sugar with an aldehyde functional group (d) Ketotriose - A three-carbon sugar with a ketone functional group

(a) Aldotetrose: Since it has four carbon atoms, its molecular formula is C4H8O4. The carbonyl (C=O) group is present in the aldehyde form at one end of the molecule while the other three carbons have OH groups. This molecule has two chiral centers (C2 and C3). (b) Aldohexose: Its molecular formula is C6H12O6. The carbonyl group is present as an aldehyde at one end of the molecule, while the other five carbons have OH groups. This molecule has 4 chiral centers (C2, C3, C4, and C5). (c) Aldotriose: Since it has three carbon atoms, its molecular formula is C3H6O3. It has an aldehyde group at the end of the molecule, and two OH groups connected to C2 and C3. However, it has only one chiral center (C2). (d) Ketotriose: It has three carbon atoms, and its molecular formula is C3H6O3. It has a ketone group in the middle of the molecule, and the OH group is connected to the C3 carbon. This molecule has no chiral center since the carbonyl group (in the ketone form) is bonded to two similar groups.

Based on the analysis above, the correct option is: (a) Aldotetrose, which has two chiral centers (C2 and C3).