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Why reduction of d-glucose is not generally given as compound B of the following figure?enter image description here Is the formation of that compound possible? What should be its name?( since I can't find that compound I don't know it's name). Iam sure that it's not l-sorbitol.

Also I would like to know why almost all sources(on google) shows, on reduction hydroxyl group goes to right hand side (in Fischer projections)?

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  • $\begingroup$ Try to write C-6 in the starting material and C-1/C-6 in the products simply as $\ce{CH2OH}$. Are A and B still different then? $\endgroup$ – Klaus-Dieter Warzecha Nov 9 '16 at 13:42
  • $\begingroup$ @KlausWarzecha No. Thanks, I got it, I was highly confused. Should I delete question? $\endgroup$ – JM97 Nov 9 '16 at 13:47
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Fisher projection shows stereochemical structure of a compound. When we say stereochemical structure, we specify those carbon/atoms that are chiral and the spacial arrangement of the groups attached to it.

Here you have glucose reduced to sorbitol, in which the aldehyde is converted to a primary alcohol, which is an achiral carbon. Thus there will be no meaning for Fisher projection of that carbon. Thus, both the compounds $A$ and $B$ are effectively the same.

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