[didgerful]

I have four unknown compounds and I need to solve their structures.

Here is some data about each one:

UNKNOWN #1
Iodoform test = positive
Fehling's test = negative
H20 soluble = yes

IR peaks: 1459, 1936, 2963, 3339

UNKNOWN #2
Iodoform test = negative
Fehling's test = positive (red brick precipitate)
H20 soluble = yes (already in aqueus solution)

IR peaks: none given

UNKNOWN #3
Iodoform test = positive
Fehling's test = negative
H20 soluble = yes

IR peaks: 1652, 1767, 1903, 2410, 2659, 2971

UNKNOWN #4
Iodoform test = positive (pale yellow precipitate bottom)
Fehling's test = negative
H20 soluble = yes

IR peaks: 1712, 2123, 2940, 2979, 3416, 3535

Choose unknowns from these structures:

CH2O, C2H6O, C3H6O, C3H8O, C4H8O, C6H12O6

NONE of the unknowns have OH and C=O groups together in the same molecule.

[madscientist]

Read the forum rules. Attempt the question and people here will help 

[didgerful]

Fehling's positive reaction (brick red precipitate) means the unknown is either an aldehyde or alpha hydroxy ketone (a ketone with an OH group on the carbon atom next to it) This is the case for unknown #2

All of the unknowns are water soluble which means they are all polar compounds.

As for the iodoform test, unknown #1 and #3 are positive so they can be methyl ketones or secondary alcohols (must look at the IR peaks to confirm groups). Unknown #4 is also positive, but in a different way. This means unknown #4 could be methyl ketone or secondary alcohol, but might instead be a acetaldehyde. Unknown #2 is the only compound that is negative for the Iodoform test.

As for the IR peaks, unknown #1 is showing alcohol group at 3339. Unknown #4 shows a ketone group at 1712. Other than that, I can't read anything into the other peaks, but maybe some of you can.

Now that I have given you all a little background into my thought process, I would love to see some sharp minds solve this organic chem structures problem.

[Sam (NG)]

Unknown 1 is C2H6O or C3H8O (ethanol or secondary alcohol).  C6H12O6 must have both OH and =O in it so it's not present according to the rules you've stated (unless you have cyclohexane-1,2,3,4,5,6-hexaol).

Unknown 2 is an aldehyde.  Unknown 4 is not acetaldehyde because it's not in the list.

Unknown 3 is a ketone C3H6O or C4H8O

The presence of both 1717 and 3535 in Unknown 4 suggests that there are both OH and C=O groups present, but seeing as you deny this, i would guess that you are looking at keto-enol tautomerism.  This is unusual however because the interconversion usually occurs faster than the timeframe of spectroscopy and the keto form is thermodynamically more stable (suggesting that only the peaks for the ketone would appear on the IR spec).

I'm affraid that's all i can think of at the moment.