A single spot on a TLC plate (even 2D) does not necessarily mean you have isolated only a single compound. If you are only getting doubling up of some of your peaks that is suggestive of the presence of an epimer/diastereoisomer. How well do the 1H-NMR signals integrate? Are your extra peaks present in a one to one ratio? Try obtaining a COSY and a HSQC, this should allow you to determine whether there is in fact more than one compound in your material. Restricted rotation can also cause this, if this is the case, running your NMR at a higher temperature in DMSO should cause the split signals to come together.

If you post the structure of your desired product and the details of the reaction used to obtain it I may be able to provide more specific insight.

May be rotamers also can influence your spectra. To overcome this can you try the NMR at different temperatures. 

Its possible that there are two compounds in the column fraction you collected. You might need chiral column to separate those compounds. 

Double the carbon peaks in spectra mears it may be in the form of isomer i.e.epimer/diastereoisomer. Check purity for TLC. Take Mass spectrum to know the mol. wt. whether it is in dimeric form.

May be rotamers also can influence your spectra. To overcome this can you  try NMR in different non-polar solvents like toluene and different temperatures it will help to overcome this problem

Assuming that your sample is indeed pure, restricted rotation is a possibility. However, if obtaining the spectrum at higher temperatures in DMSO-D6 (carefull not to high!) does not resolve the problem, the non-equivalency is probably due to the presence of an asymmetric center. This can result in two sets of resonances (e.g., two non-equivalent phenyl rings). 2D correlation experiments  should show this and should allow you to assign all the resonances.  

Have you done Nitrogen estimation? If so, does it matches the actual theorotical value? After performing flash chromatography does there any melting point diffrence in two compounds?

@Viraj Ranade, no I haven't performed this estimation. On TLC there is only one visible compound and that too in the case uponn treatment with Iodine. I have submitted the sample for VT-NMR under DMSO. hoping for getting some way.

Remember, whether or not there is restricted rotation, the asymmetric center(s) will cause non-equivalency. The VT NMR experiment may suggest restricted rotation, but the non-equivalency would exist regardless, owing to the chiral center(s); that would be the fundamental reason, not the presence or absence of restricted rotation. 

Perhaps this might help. Consider the hypothetical molecule in which X,Y,Z are different, and R is not H: CXYZ-CHaHb-CHcHd-CHeHf-R

The chirality of CXYZ is explicit, therefore the hydrogens on the adjacent carbon are non-eqivalent (HaHb); this induces chirality to the second carbon, and therefore, the hydrogens on carbon 3 will also be nonequivalent (HcHd); this again induces chirality to the 4th carbon, and, therefore, the two hydrogens on that carbon will again be non-equivalent (HeHf).....

Try to apply similar arguments to your molecule to explain the doubling of resonances. If you would post the structural formula, I could help you in more detail.

One more example: Both 1H and 13C spectra of a  molecule Ph2CXCHYZ show two sets of resonances for the phenyl protons and carbons. Connectivities are readily established by 2D experiments. The CHYZ- is the explicit chiral center. I hope this helps.

Ph

     \

        C X - CHYZ   

     /

Ph

Singla,

You are not showing the exact structure of your compound but you may have two conformers or tautomeric forms for your product present in solution.

Good luck

If the molecule contains a chiral centre then the phenyl carbons and hydrogens will be diastereotopic, and you would expect double the number of signals, see e.g.

http://www.chem.sci.ubu.ac.th/inmr/nmr/chemical_shift/magnetic_eq_and_non_eq_protons.htm

The two phenyls you have drawn are not equivalent.  A plane containing the CH and CX would have a Z and one of the pnenyls on one side and the Y and the other phenyl on the other side.  There is no rotomer in which the phenyls are equivalent.