Pentanone was treated with excess sodium cyanide in HCl (aq) followed by hydrogen gas has over Pd. This produced: A. 2-amino-1-hexanol
B. 1-amino-2-methylpentan-2-ol
C. 1-cyano-1-pentanol
D. 2-aminomethylpentan-1-ol

B. 1-amino-2-methylpentan-2-ol

Explanation:

In this case, the first step, we have the attack of the nucleophile cyanide ( produced by sodium cyanide to the carbon on the carbonyl group (C=O) producing a negative charge in the oxygen.

Then HCl protonates the molecule to produce a cyanohydrin. This cyanohydrin can be reduced by the action of hydrogen gas () in the presence of a catalyst (), producing an amino group. With this in mind, the final molecule is: 1-amino-2-methylpentan-2-ol.

See figure 1 to further explanations

I hope it helps!

im not exactly sure but i think its b!