# Distinguishing between unknown organic compounds

1. Nov 9, 2004

### Karate Chop

hey, if anyone is good with organic chem, could they please list some common and effective tests to distinguish between different organic compounds such as alkanols, acids and alkenes, etc.

2. Nov 9, 2004

### so-crates

Well, there are only about 25 million known organic compounds, so shouldn't be too hard :-)

Old techniques:

Melting point
Boiling point
Refractive index
Solubility
Boiling point elevation/freezing point depression to determine molecular weight

New techniques:

Mass spectroscopy
UV spectroscopy (not as important )
IR spectroscopy
NMR spectroscopy

3. Nov 9, 2004

### chem_tr

In addition to so-crates' post, I can say that there are specialized tests for some functional groups, for example, Lucas' Test for alkanols (tertiary alcohols react readily); NaHCO3 Test for acids (carboxylic acids give positive result); Bäyer's Test (unsaturated groups or reductants give positive result).

See Organic Analysis in your textbooks for more information. If you need the other special analysis methods, call for help.

4. Nov 9, 2004

5. Nov 9, 2004

### Sirus

The polarimeter is also a useful tool in distinguishing organic compounds. Only chiral compounds are optically active, meaning achiral compounds will not give any observed rotation. Also, enantiomers will give observed rotations equal in magnitude but opposite in sign. Google for more info and formulas.

6. Nov 10, 2004

### so-crates

Also keep in mind that racemic mixtures, while they consist of chiral molecules, are not optically aptive. Usually you have to know what compound you are dealing with before you try separating enantiomers since it cannot be done by conventional means like distillation or chromotography.

7. Nov 10, 2004

### chem_tr

Enzymes or reacting with an enantiomer may be a good solution. Pasteur separated tartaric acid enantiomers by a similar way. Alkaloids like brucin or strychnin can be useful.

8. Nov 11, 2004

### Sirus

Correct. Racemic mixtures can be resolved by reacting with a chiral agent, which yields separable diastereomeric products.
$$\underbrace{\left\{\begin{array}{1}R\\S\end{array}\right\}}_{\mbox{enantiomers}}~+~\underbrace{R}_{\mbox{chiral agent}}\longrightarrow\underbrace{\left\{\begin{array}{1}R-R\\S-R\end{array}\right\}}_{\mbox{diastereomers}}$$
Reactions can then be carried out to separate the diastereomers:
$$R-R\longrightarrow R~+~R$$
$$S-R\longrightarrow S~+~R$$

9. Nov 11, 2004

### gravenewworld

LC/MS+NMR=answer to any question you will ever have about structure and unknowns.