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