What are your other substituents? Are they delicate functional groups? If they are relatively robust you might want to try a copper (II) salt in acetonitrile as your oxidising agent.

Chebrolu,

Such transformation is know by

1-H2O2 in formic acid, [Yield ~78%]

---Dodd, Robert H. and Le Hyaric, Mireille, Synthesis,  295-7 (1993)

2-water by a sulfoxylalkyl- substituted N- heterocyclic carbene catalyst [Yield ~92%]

---Yoshida, Masahiro et al., Organic & Biomolecular Chemistry, 7, 4062-4066 (2009)

3-  sodium chlorite [~98%]

---George, Ian R. et al., Tetrahedron, 69, 8209-8215 (2013)

Good luck

Lavanya

The Pinnick Oxidation will work. As Adel has listed above, Tet. 2013, 69, 8209.

One of my group members has a heavily substituted 2-formylpyridine which oxidizes quantitatively under Pinnick conditions. Product is isolated by crystallization.

http://en.wikipedia.org/wiki/Pinnick_oxidation

I recommend the Pinnick Oxidation, too.

http://www.beilstein-journals.org/bjoc/single/articleFullText.htm?publicId=1860-5397-9-229&sn=37

Water solution of KMnO4 - always 100% yield in this reaction, if no another fragile group in molecule. Atter oxydation in alkaline medium salt forms, and MnO2 filtred off, after acidification acid precipitated. If the product is water soluble there are some problems to isolite it, water evaporation is nessesary. 

Formation of the N-oxide may become problematic, but the Pinnick oxidation generally works quite well if you use a scavenger.

Stoichiometric Cu(II) or Ag(I) are also worth considering. Silver mirrored glassware, whilst pretty is a pain to clean.

I would probably avoid the stronger oxidants like KMnO4, Chromium (VI) reagents etc. given that DMF oxone gave poor yields.

Ley oxidation w/ some water, easy work up (via column). very mild oxidation of the hemiacetal

Another vote for the  Pinnick reaction here, although in fairness it should be called the Lindgren reaction.

Agree for all replying Pinnick reaction.