Q:

How to identify an unknown compound with spectroscopic data?

I need some help identifying the following unknown compound. Any suggestions or hints regarding the possible compound would be highly appreciated.
So far I am thinking the broad peak on the IR is an alcohol that somehow mixed with possible $\ce{sp^3-CH}$ bonds. And the peak around $\pu{1600 cm-1}$ is I think a $\ce{sp-C-C}$ bond. The compound should have four distinct groups of protons. Beyond this basic analysis, I couldn't really come up with a compound. 

I have updated (07/09/14) the question with a newer NMR (below) and an IR. 

Some other additional info I worked out in the lab:

look: white powder like solid.
melting point: $\pu{116.3-119.0 ^\circ C}$
Solubility:

Water - insoluble
NaOH - insoluble
Sulfuric Acid - soluble

Classification Tests:

Water insoluble phenol test - positive; a dark red precipitate formed. 
Chromic acid test - positive; green precipitate formed. 
Bromine test for unsaturation - positive; bromine discoloration happened until about $\pu{2 ml}$ of bromine was added. White precipitate formed. 

This is my first time doing these tests so please don't consider their results too strictly. My Instructor said the IR is a much safer way to determine functional groups. 

A:

Have a look at typical IR bands (other source) and identify them.

 - 3424 phenolic $\ce{OH}$ stretch
 - 1640 aromatic $\ce{C=C}$ bending or $\ce{C=C}$ alkenyl stretch
 - 1264 aromatic $\ce{C-O}$ stretch
 -  747 aromatic $\ce{C-H}$ bending

Have a look at you spectrum, identify your residual solvent signal.

  7.15 is chloroform $\ce{CDCl3}$

Have a look at the typical $\ce{{}^{1}H-NMR}$ shifts table and identify the protons.

 These are aromatic protons

What is the summary of your first conclusions?

 It's an aromatic compound with an hydroxyl group.

How many protons?

 multiple of 5

Calculate the coupling constants via $$J=\Delta\delta_{\mathrm{ppm}}\cdot\nu_0$$ and identify the splitting pattern. (I can only assume it is a 500MHz spectrometer, hence $\nu_0=500~\mathrm{MHz}$.)
Example: Doublet at 7.55 ppm:
$$J=(7.5615-7.5461)\cdot500~\mathrm{MHz}=7.7~\mathrm{Hz}$$
Solutions:  

 7.55 doublet, $2\ce{H}$, $J=7.7~\mathrm{Hz}$
 7.40 doublet of doublet (fake triplet), $2\ce{H}$, $J=7.6~\mathrm{Hz}$, $J=7.5~\mathrm{Hz}$
 The last one is tricky because there is one peak not picked
 It should be at 7.30 a doublet of doublet, $1\ce{H}$,  ${}^3J=7.7~\mathrm{Hz}$, $J=7.6~\mathrm{Hz}$  

Disclaimer:
Please read long's answer and follow his thoughts. They make a lot of sense and hint that the compound we are looking at is not the one I am proposing. But due to my lack of current knowledge, I cannot make another suggestion.
Combine all you see and you will get a clue which substance it is.

 7.55 ortho
 7.40 meta
 7.30 para
 Phenol.  

A:

The data given is this question has changed sufficiently from the original post that it warrants a new answer to the latest (and much more reliable) data, although should be taken in conjunction with Martin's response, and Long's original response. Let's first look at the IR data:
IR Spectrum:
A much better spectrum, and, compared to the original spectrum, serves as a very good example of how easy it is to introduce unwanted water into a sample. Peaks around 3020, 763 and 692 indicate an aromatic. The 763 and 692 are indicative of a mono-substituted benzene ring. There are also some clear aromatic overtones between 1950-1750. 
NMR Spectrum:
Still only a partial NMR spectrum, and one that we can only assume is correctly calibrated. Linewidths are better, and significantly, we see a small coupling from the peak at 7.15ppm.
The splitting pattern and peak ratio observed is indicative of a monosubstituted benzene ring (see above); 7.30 (1H, dd, H4), 7.41(2H, , H3) and 7.56(2H, d, H2). Long-range (small couplings) are ignored. We also have a peak at 7.15 (1H, d, H5). Typical coupling in these systems is 6.5-8.5Hz for ortho coupling, 1-3 for meta, and <1 for para. Shifted from benzene (7.34ppm), the substituents come at H2 (+0.22), H3 (+0.07), H4 (-0.04). 
The peak at 7.15 could be one of few things. A bis-halide is a possibility, but more likely is an alkene. No real evidence in the IR supports an alkene (although my IR knowledge is limited). However, a symmetrical alkene may be a possibility, as the main -C=C- absorbance would be IR inactive, although Raman active. This gives us two possibilities that come to immediate mind:
cis-stilbene (cis-1,2-Diphenylethylene) m.p. 5-6oC

Additive prediction for the aromatics and alkene proton can be done using many basic NMR texts. I'm using Pretsch, Buhlmann and Badertscher for this case (as it's the only one I have with me at the moment):

H2 : 7.34+0    = 7.33
H3 : 7.34-0.07 = 7.27
H4 : 7.34-0.15 = 7.19
H5 : 5.25+1.38(Argem)-0.07(Artrans) = 6.56

trans-stilbene (trans-1,2-Diphenylethylene) m.p. 123-125oC

H2 : 7.34+0.16 = 7.50
H3 : 7.34+0    = 7.34
H4 : 7.34-0.15 = 7.19
H5 : 5.25+1.38(Argem)+0.36(Arcis) = 6.99

As this is clearly a problem session for you to make a final determination of structure, I will leave it for you to make that judgement. however, as a final point, this whole exercise, I hope, serves to demonstrate the importance of preparing good quality samples in order to collect high quality and reliable data. I hope you can see the difference between your original data and final data. This is a message that I continually preach to all those researchers I work with, both at an undergraduate level and postgraduate. 

A:

This answer aims to build on the general approach that Martin has provided, which overall makes a reasonable summation based on the data provided. IR is not really my specialty, but there is some more information that we can get out of the NMR data which should be helpful, and more reliable (in my opinion) than the IR data. Unfortunately, I am away away from my office for the next week, so cannot provide immediate references to support some statements here, so you'll have to take some things on face value. Let's begin with an overall summary of what data we have:

A partial 1H NMR spectrum, with only some of the peaks integrated. This is an expanded region of what we can assume to be a 500MHz (based on the export path). The linewidths are broad, and there is no clear source to allow confirmation of correct calibration. We therefore need to make two assessments:

The calibration is incorrect, and the peak at 7.15, which has no integration, is in fact the residual CHCl3, and all chemical shifts need to adjust downfield (0.09-0.11 depending on what value for CHCl3 in CDCl3 you use; I use 7.26. For simplicity, let's adjust the chemical shifts downfield by +0.1ppm
The calibration is correct, in which case the peak at 7.15 cannot be discounted, and should therefore have its integral determined. 
A full display NMR spectrum would be very useful here to look for underlying exchange broadened proton signals. This might occur anywhere from about 2-15ppm, and may be very broad such that they appear as a hump in the baseline, but even in CDCl3, we should see them, and 

An IR spectrum which looks to have been run at pretty low concentration. I did not see your original IR spectrum, and wonder why you needed to redo it. What would be nice to know is whether the ratio of intensities for your absorbance peaks are the same for both IR data sets; particularly did the ratio of the broad stretch at 3422 change with respect to absorbances at 3019, 763 and 692?
Your sample is a solid, as you mention in one of your comments. This would be a useful peice of information to have from the start. It is soluble in dichloromethane.

So let's now start with collating information from the data provided.
IR Spectrum.
Clearly, the significant signal is the broad peak at 3422, and this is textbook-indicative of an O-H stretch. It's probably a little too high to consider a N-H group of any sort. So, it could be an alcohol or an acid, but we have no C=O peak, so it leaves us with an -OH group.
We have absorbances at 3019, 763 and 692; all indicative of an aromatic. More specifically, 763 and 692 are indicative of a mono-substituted benzene ring.
As I say though, IR is not really my thing, and that's about all I can get from this spectrum. My biggest concern is the reliability of the OH peak.
NMR Spectrum

The splitting pattern and peak ratio observed is indicative of a monosubstituted benzene ring (see above); 7.30 (1H, dd, H4), 7.39(2H, dd, H3) and 7.55(2H, d, H2). Typical coupling in these systems is 6.5-8.5Hz for ortho coupling, 1-3 for meta, and <1 for para. Looking at the H2 signal at 7.55, we can use our knowledge of coupling constants to determine the frequency of the spectrometer:
7.5615-7.5461 = 0.0154 x Spect.Freq ~ 7.5Hz => 487MHz, so close enough to 500MHz, and confirms our suspicions that it is a 500MHz, as the export path suggests. 
So, we can calculate an accurate ortho coupling for H2-H3 to be:
7.5615-7.5461 = 0.0154 x 500 = 7.7Hz.
Now, mono-substituted benzene rings have been extensively studied and are very well understood; chemical shift data has been widely tabulated, and forms the basis for many chemical shift prediction algorithms. Starting with the benzene chemical shift (7.34ppm) as a basis, it is possible to use the shifts of each group to infer some information about the type of substituent.  The first thing to look for with this type of system is the order of H2 versus H3 (versus naked benzene). An electron-donating group increases shielding, and the ortho proton (H2) is typically found upfield of the meta proton (H3). Electron withdrawing groups decrease shielding, and H2 typically experiences a downfield shift from benzene, and usually resonates downfield from the meta (H3) proton. For the system you have, H2 is downfield of H3, and this is indicative of an electron-withdrawing group.  This is a very strong argument against this system being phenol. Phenol has its H2 protons upfield of H3. This is also what is so confusing about the IR spectrum you have. 
So, let's now consider the possible structure for this unknown compound you have. Let's make the assumption that, as a homework/tutorial problem, this is going to be a fairly simple molecule, with a pretty common substituent. Remember we have two scenarios to consider for our NMR. 
Scenario 1 (corrected for CHCl3 at 7.26ppm): the substituents come at H2 (+0.21), H3 (+0.05), H4 (-0.04). Looking at Pretsch, Buhlmann and Badertscher, this matches incredibly well for the substituent being a phenyl group [H2 (+0.22), H3 (+0.06), H4 (-0.04)]. This would give the structure biphenyl, a white solid, which has a reported H2-H3 coupling of 7.71Hz. Very strong evidence by NMR, but is not supported by -OH stretch in IR data, although all other IR data is in agreement.
Scenario 2 (spectrum already correctly calibrated): If we assume that the spectrum is correctly calibrated, then the CHCl3 residual peak comes under the H4 signal - probably could be the sharp peak which is the second peak from the right in this group. This means that the peak at 7.15 needs to be considered. By eye, its integral is roughly 1. A singlet of chemical shift of 7.15 is typical of a bis-halide, and so we could consider α,α-dichlorotoluene or α,α-dibromotoluene. Both are sufficiently electron withdrawing to give H2 downfield of H3, and  However, the former is definitely a liquid at room temp, and I suspect the latter is also. Therefore, not strong candidates.
I would like to have seen the original IR spectrum, and the full NMR spectrum to have confidence in any prediction. However, if I were just shown the NMR data, I would have confidence in predicting the structure as biphenyl. I hope you can provide the real solution to this eventually.

