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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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closed as unclear what you're asking by Jan, Todd Minehardt, Wildcat, ringo, getafix Nov 10 '16 at 12:25

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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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