Question

There are two scenarios and in either case, we start with 5-oxohexanal. A reagent or sequence of reactions "A"
helps convert the substrate into 6-hydroxyhexan-2-one. On the other hand, a reagent or sequence of reactions "B"
converts the starting compound into 5-hydroxyhexanal. A and B are:

• A. A: 1 eq. $NaB{H}_4$ followed by $H_3{O}^{+}$;
  B: 1 eq. $HOC{H}_{2}C{H}_{2}OH$, followed by $NaB{H}_4$ plus $H_3{O}^{+}$
• B. A: 1 eq. $H_2$ Pt in low pressure;
  B: 1 eq. $HOC{H}_{2}C{H}_{2}OH$, followed by $NaB{H}_4$ plus $H_3{O}^{+}$
• C. A: 1 eq. $HOC{H}_{2}C{H}_{2}OH$, followed by $NaB{H}_4$ plus $H_3{O}^{+}$;
  B: 1 eq. $NaB{H}_4$ followed by $H_3{O}^{+}$
• D. None of the above

Answer 1

The correct option is A

A: 1 eq. $NaB{H}_4$ followed by $H_3{O}^{+}$;
B: 1 eq. $HOC{H}_{2}C{H}_{2}OH$, followed by $NaB{H}_4$ plus $H_3{O}^{+}$

Aldehyde is easier to reduce than a general ketone. This could be taken advantage of:
Using 1 equivalent of a hydride donor will selectively reduce the aldehyde.


To the reduce the ketone group selectively, the higher reactiviy of aldehyde could
be put to use. A protective diol could help form an acetal at the aldehyde position.
This could then be reduced as only the ketone part gets reduced. Hydrolysis
does 2 things - work up + hydrolysis of acetal