Question #32905

First Impressions

`"Tropic acid + HBr" → "C"_9"H"_9"O"_2"Br" + "H"_2"O"`

∴ The formula of tropic acid must be `"C"_9"H"_10"O"_3`.

The index of hydrogen deficiency is `"IHD" ="20-10"/2 = 10/2 =5`.

Anytime I see `"IHD" ≥ 4`, I think of an aromatic ring. Let's try `"C"_6"H"_5`.

There is probably an `"OH"` group (that was replaced by `"Br"`).

From the name tropic acid and a remaining `"IHD"` of 1, I would bet there is a `"COOH"` group.

`"C"_6"H"_5 + "OH" + "COOH" = "C"_7"H"_7"O"_3`.

That leaves `"C"_2"H"_3` (`"CH"_2"CH"` or `"CH"_3"C"`) for the rest of the molecule.

Possible Structures

(a) If there is a monosubstituted phenyl

The possible structures are `"PhCH"_2"CHOHCOOH"` or `"HOCH"_2"CHPhCOOH"` or `"CH"_3"C(OH)PhCOOH"`.

(b) If there is a disubstituted phenyl

If there is a disubstituted phenyl, the possible structures are the ortho, meta, and para isomers of (1-hydroxyethyl)benzoic acid and (2-hydroxyethyl)benzoic acid.

We don’t have enough information to specify the structure of tropic acid.

The reactions

Assume the structure is `"HOCH"_2"CHPhCOOH"`.

(a) Conversion of alcohol to alkyl bromide

(b) Dehydrobromination to atropic acid

(c) Hydrogenation of alkene to form hydratropic acid