17a-methyl-17b-hydroxy-1,4-androstadiene-3-one - methylboldenone (Dbol)
17a-methyl-1, 4-Androstadiene-3, 17-diol (M1,4-ADD)
1,4-androstadiene-3-one,17b-diol (Boldenone Undeclynate)
Notice the diol configuration in both M1,4-ADD and Boldenone Undeclynate. Also note the additional hyrdoxy group in methylboldenone. Finally, methylboldenone is configured as a 3-one were are M1,4-ADD and boldenone undeclynate share the same double bond oxygen configuration not found in methylboldenone. These similarities and differences are very important in the enzymatic reduction of each compound, particularly M1,4-ADD.
As far as aromatization, M1,4-ADD, again, configured as a diol, cannot aromatize from the base compound. Obviously, I stated in an earlier post that the liberated parent may aromatize but again it is only marginal in as it will do so in about an equal proportion per mg of converted ph to boldenone as would be expected from boldenone undeclynate b/c they follow the same enzymatic reduction process.
As far as what M1,4-ADD "feels" like I would argue that it "feels" much more like equipoise due to the increase in RBC and massive increase in appetite. Then again, "feelings" only matter to the individual in question. I favor actual chemistry myself.
Pharmaceutical Name: Boldenone (as undecylenate)
Chemical structure: 1,4-androstadiene-3-one,17b-ol
Now lets look at these again
17a-methyl-17b-hydroxy-1,4-androstadiene-3-one - methylboldenone (Dbol)
17a-methyl-1, 4-Androstadiene-3, 17-diol (M1,4-ADD)
1,4-androstadiene-3-one,17b-ol (Boldenone Undeclynate) (not 1,4-androstadiene-3-one,17b-diol)
So obviously Dbol is methyl Boldenone. Boldenone is not a diol, it only has one -OH at the 17th carbon, just like Dbol and M1,4-ADD. Dbol and Boldenone have a double bonded oxygen at the 3rd carbon, M1,4-ADD has a -OH. Dbol and M1,4-ADD are both methylated at the 17th carbon.
M1,4-ADD is a diol because there are -OH at 3 and 17. Unless every other site I saw is wrong you are the only one saying Boldenone is a diol. To do so the way you wrote the formula it would have 2 -OH groups at the 17th carbon, making it much different from the other two. If it did Dbol could not be methyl boldenone as the methyl at 17 would take the place of the 2nd -OH.
All it takes is for the -OH at the 3rd carbon of M1,4-ADD to be converted to a double bonded O and you have Dbol. Seems much more likely to me than cleaving off a methyl that is going to be probably somewhat protected by the -OH and still converting the 3-OH to a 3=O. There is obviously no doubt in your mind that the 3-OH is converted to a 3=O though as boldenone has a =O at the 3rd carbon. Nobody is saying Dbol is converted into boldenone but that would be what would happen if that 17methyl were easily cleaved.
I have no idea of what happens once inside the body but basic organic chem tells me this.