[twitchyfides]

http://imgur.com/DGqvPHg

Hello I really need some help at this moment. This is my assignment that is due tomorrow at midnight. I have a computer science final today at 7PM and a Calculus Final at 3 PM on Monday. I need to do well in both those courses to graduate by April.

I need to do this assignment so I can pass Organic Chemistry (assignment is worth 3% and I am at a 47% in the course right now). I know you need to actually work on the problems before asking but I don't have the time right now to study for Organic.

This will take some of you literally a few minutes to do, please help.

[Babcock_Hall]

The forum rules are what they are for a reason.  Please show your attempt, then we can help you.

[twitchyfides]

The forum rules are what they are for a reason.  Please show your attempt, then we can help you.

Well this is what I came up with

 The set of peaks around 7 with an integration of five tell you there is a monosubstituted benzene. The singlet around 10 is an aldehyde proton. The two triplets downfield are two CH2 groups that are next to each other.

[kriggy]

That looks reasonable. So, what is the structure? you have aromatic with 5 hydrogens, 2 CH₂ groups and one aldehyde just put it together

[Babcock_Hall]

The chemical shift of a CH₂ group will show a downfield shift if it is next to an aromatic ring.  That may give some confirmation.

[pgk]

What about the IR band around 3500cm-1?
How can it be explained, when both 1H-NMR and 13C-NMR spectra (and IR, too), absolutely indicate an aldehyde group?
For the moment, if this band confuses you, just ignore it and come back later, after the exams.

[Babcock_Hall]

The peak around 3500 cm^-1 may come from a contaminant.  There is one contaminant that organic chemists often have to work hard to remove.

[pgk]

Given that the compound is absolutely an aldehyde, the IR peak around 3500 cm-1 indicates enol formation, in equilibrium.
Enolic proton cannot be easily observed in 1H-NMR because it usually appears at the aromatic zone (7-8 ppm) at quite lower integration than 1 proton and thus, easily mistaken as a parasite peak.
PS: The given compound has a logp ≈ 2.0, which means that it is not very hygroscopic.

[twitchyfides]

3-phenylpropanal.

THANKS GUYS