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

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).