Hi Lina,

This question was quite tricky.

The passage essentially is saying to remove the top chiral carbon and look at the bottom three chiral carbons. The top chiral carbon is removed following oxidation, and the top aldehyde is then oxidized again to form a carboxylic acid. Now, you have carboxylic acids on both ends of the molecule. You will notice now that in this case, a lot of the compounds are the same as one another (like 1 with 2 and 5 with 6) and some are actually meso substances (line of symmetry across the middle one of the three chiral carbons) with only three chiral centers. Given that glucose is optically active, you can eliminate any meso-like compounds, and therefore, you can remove 1,2 and 5,6.

Hope that helps!

Cheers,

Molly