Question

How many isomers do molecules a) and b) have and can they be dissolved in acid enviroment?

a) molecule*HCl

Answer 1

DISCLAIMER: LONG ANSWER!

In `(a)` and `(b)`, the stereocenters will have four groups around carbon that are all different. They turn out to be the `2` carbons connected to `"OH"` and `"CH"_3` in `(a)`, and the `1` carbon connected to `"CH"_3` near the carboxylic acid group in `(b)`.

STEREOISOMERS OF (A) AND (B)

The number of stereoisomers we can have are `bb(2^n)`, where `n` is the number of stereocenters. (We don't have to worry about meso isomers, because the molecules are not symmetrical.)

Let `(a)` have isomers `(i), . . . , (iv)` and `(b)` have isomers `(i)` and `(ii)`.

• `(a, i)` has two stereocenters. The left one has the following priorities:

`"OH" > "CHCH"_3"NH"- > "Ph" > "H"`, `=> bb" R"` stereocenter

where `"Ph"`, phenyl, has a carbon center treated as having three "ghost carbons", as if it were `"C"-("CH"_3)_3` or `"C"-"C"-="C"-`.

Atomic number takes preference in assigning priorities, followed by the highest quantity of substituents of the same atomic number.

And the right one has the following priorities:

`"NHCH"_3 > "CHOHPh" > "CH"_3 > "H"` `=> bb" S"` stereocenter

That makes `color(blue)((a","i))` the `color(blue)(bb((R","S)))` stereoisomer.

By comparison with `(a,i)`, we therefore have that `color(blue)((a","ii))`, `color(blue)((a","iii))`, and `color(blue)((a","iv))` are the `color(blue)bb((R","R))`, `color(blue)bb((S","S))`, and `color(blue)bb((S","R))` stereoisomers, respectively.

• `(b,i)` has one stereocenter, and its configuration from the above logic uses the following priorities:

`"COOH" > "Ph" > "CH"_3 > "H"` `=> bb(R)` stereocenter

Thus, `color(blue)((b","i))` is the `color(blue)(bbR)` stereoisomer, and `color(blue)((b","ii))` is the `color(blue)(bbS)` stereoisomer.

SOLUBILITY IN ACID?

As for whether they can be dissolved in acid, it depends on what the `"pH"` is and in what solvent. In the appropriate solvent, yes, these can be dissolved just fine.

But the most convenient solvent with tabulated `"pK"_a` values is water. In water, `(a)` is probably moderately soluble, and `(b)` is probably poorly soluble, given the relative lack of polar bonds.

A solution made more acidic helps maintain the present form of the species with `"pK"_a > "pH"`.

`"pH" = "pK"_a + log \frac(["A"^(-)])(["HA"])`,

`"pH"` `<` `"pK"_a` implies `["A"^(-)] < ["HA"]`, where `"HA"` is either `(a)` or `(b)` in their present form.

So, you would need to know the `"pKa"` of the hydroxyl proton on `(a)` and the carboxyl proton on `(b)`.

I searched these on Scifinder, and I would estimate the `"pK"_a` of `(a)` to be near `15 - 16`, while the `"pK"_a` of `(b)` is available and is predicted to be about `4.41`.

However, if the solution is too acidic, the solvent itself may become protonated as well, making it a worse solvent for these compounds (which are not charged in their acidic form).