Any one step solution would require oxidation, however, I am not too familiar with this process.  A mild oxidizing reagent would be required...potassium chromate?

In any case, here's another proposal:

Use diatomic chlorine in the presence of uv light, to generate a reaction where the 2-methyl-propane will become a secondary radical (the radical on the most stable, secondary carbon).  After the reaction has stabilized.

Now think of a way to transform the halide to the alcohol; should be easy enough.

Of course there is a secondary carbon in isobutane.

However, I was wrong in my proposing the secondary radical intermediate.  No such reaction takes place, I had it confused with the reaction chlorine radical with an alkene.  

But of course, there is possibility to eliminate HCl from t-butyl chloride, and after addition borane
and oxidation the needed alcohol can be obtained.

The whole question was regarding the details on how such a reaction would bee formed inn the first place.  All I can think of at this moment is to react diatomic chlorine, uv, and the suggested isobutane.  Reaction with diatomic bromine would result in higher yield, however, chlorine, being nonselective would result in the statistical result, that is abstract one of the 9 hydrogen atoms.

mechanism
http://www.mhhe.com/physsci/chemistry/carey/student/olc/ch04radical.html

Elimination of such a secondary halide does produce satisfactory yields, SN2 reactions, among others can take place.  

Substitution of the halide would be the best, two step mechanism (as opposed to three).  However, I'm not arriving at any ideas.

If it were my tedious task, I would throw the hydrocarbon into a plate of bacteria and pray that they can do something magic with it..

Converting 2-methylpropane to 2-methyl-propan-1-ol is essentially converting an alkane into an ethanol.

Refer to http://iglesia.cchem.berkeley.edu/JournalofCatalysis_222_481_2004.pdf

I'm thinking along the line of dehydrogenation as the first step. It allows us to "activate" the alkane, and at the same time, give us more control over the reaction due to the nature of the alkene group. An interesting thing to note about 2-methyl-propane is that removal of hydrogen from any carbon all produces 2-methyl-prop-1-ene

Addition of water across the alkene group will produce the required alcohol.

Free radical reactions are very hard to control and they are rather unpredictable.

OIC. Markonikov's Rule

There is no secondary carbon atom in 2-methyl propane (isobutane).
But of course, there is possibility to eliminate HCl from t-butyl chloride, and after addition borane
and oxidation the needed alcohol can be obtained.

Yes, after making the 2-methylpropene, by using hydroboration-oxidation,

CH3-C(CH3)=CH2 + [1st] B2H6 ---> [2nd] H2O2, NaOH ---> 2-methyl-1-propanol

the desired product is obtained.

Alternative method:

React 2-methylpropene with HBr (hydrogen bromide) with present of peroxide (RO-OR) that gives 1-chloro-2-methylpropane. Then, replace the Cl- by OH-.

(Both methods are anti-Markovnikov!)