# What is the final product of a carbohydrate treated successively with hydrogen cyanide, water, and hydrogen iodide/ phosphorus?

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 $$\ce{Z}$$.

Your starting material is an unspecified hexose ($$\ce{C6H12O6})$$. Leaving the stereochemistry aside, it is either an aldose or a ketose.
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 $$\ce{C7H14O7}$$.