For the aldotetroses I-IV, the combination of true statements, among P-T, is P = I and II are diastereomers and II and III are enantiomers Q = I and IV are mesomers and are optically inactive R = I and III can be interconverted by a base catalysed isomerisation S = only I and IV are HlO4 cleavable T = I and III are D-sugars and II and IV are L-sugars A. Q, R, T B. P, R, T C. Q, R, S, T D. P, Q, S

The correct answer is D. P, Q, S.

P: I and II are diastereomers. This is true because they have the same molecular formula but different configurations at one or more stereogenic centers.

Q: I and IV are mesomers and are optically inactive. This is true because they have two identical chiral centers and the two enantiomers cancel each other out.

S: only I and IV are HlO4 cleavable. This is true because they have an acetal group that can be cleaved by HlO4.

R: I and III can be interconverted by a base catalysed isomerisation. This is false because I and III are mesomers and cannot be interconverted by a base catalysed isomerisation.

T: I and III are D-sugars and II and IV are L-sugars. This is false because I and IV are mesomers and do not have a specific D or L configuration.

Here is a diagram of the aldotetroses I-IV:

The chiral centers are marked with asterisks.

I and II are diastereomers because they have different configurations at the C2 chiral center.

I and IV are mesomers because they have two identical chiral centers and the two enantiomers cancel each other out.

I and IV have an acetal group that can be cleaved by HlO4.

I and III cannot be interconverted by a base catalysed isomerisation because they are mesomers.

I and IV are not D or L sugars because they are mesomers.