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An unknown carbohydrate is successively treated with (1) hydrogen cyanide, (2) water, and (3) hydrogen iodide and phosphorus: $$\ce{C6H12O6 ->[HCN] X ->[H2O] Y ->[HI,P]Z}$$ What is the final product $\ce{Z}$?

I could figure out that in $\ce{Y}$ there is a $\ce{CH(OH)(COOH)}$ group instead of a $\ce{CHO}$ group, but I could not find out $\ce{Z}$.

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Your starting material is an unspecified hexose ($\ce{C6H12O6})$. Leaving the stereochemistry aside, it is either an aldose or a ketose.

Reaction scheme

  1. Addition of $\ce{HCN}$ results in the formation of X, a cyanohydrin (α-hydroxy nitrile) with seven carbon atoms ($\ce{C7H13NO6})$.

  2. Hydrolysis of X yields the α-hydroxy carboxylic acid Y ($\ce{C7H14O8})$.

  3. The reaction of Y with $\ce{P_{red}/HI}$ is likely to result in reductive dehydroxylation. Overall, the oxygen atom of the carbonyl group in X has been replaced with $\ce{H}$ and a $\ce{COOH}$ group.
    Z therefore is $\ce{C7H14O7}$.

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