Phenol and Deuterated Sulphuric Acid What happens when phenol reacts with $\ce{D2SO4}$ in presence of $\ce{D2O}$?

How many hydrogens in total are replaced by deuterium?

Is the hydrogen of the hydroxyl group and three hydrogens ortho and para to the -OH group replaced by deuterium, and the remaining two unaffected? 

OR are all hydrogens replaced by deuterium? 

I'm not sure what really happens, but as far as I know - even though ortho and para positions are greatly activated, some small percentage of meta attack may also take place. In fact, if it does take place, the final compound will be totally deuterated, as bonds with deuterium are stronger than those with hydrogen, and backward reaction will not be thermodynamically favoured. 

It'd be great if someone could explain what's going on, and if possible please quote sources supporting your answer. Thanks a lot.
The deuteration of phenol and aniline hydrochloride has been investigated by Werstiuk and Kadai by the High Temperature/Dilute Acid (HTDA) method. For phenol, ortho/para substitution is faster than meta (2.7 D/3.0 H vs. 1.3 D/2.0 H). OD/OH exchange was conducted at the end of the reaction. The presence of conc. HCl undoubtedly lowers the efficiency of incorporation of deuterium.
Incorporation of deuterium without the presence of conc. HCl allows for efficient deuteration of aniline hydrochloride. [RND3+ to RNH3+ exchange was conducted.] The authors note that conversion of aniline hydrochloride-d5 to phenol-d5 is efficient via diazotization.


Werstiuk, N. H.; Kadai, T., Can. J. Chem., 1974, 52, 2169.