Question

Please help me with the analysis of an unknown molecule. Please explain your work to get there and not just the final answer:

Formula: CH.CH IR peaks (cm): weak peak at 1570 Mass Spec: base peak (m/z) = 285 UV-Vis = 256 nm (benzene = 255 nm) 13C: 8 164, 162, 160, 158, 157, 150, 146, 144, 142, 139, 84, 74 H: 87.9 (d, 2H), 7.8 (d, 1H), 7.5 (dd, 1H), 7.46 (dd, 1H), 7.44 (d, 1H), 7.42 (d, 2H), 4.2 (d, 1H), 3.7 (d, 1H)

Answer 1

In this question they have given the molecular formula of the unknown compound. That is C₁₄H₁₀Cl₄

Now using the given spectral details, we will collect the information of the unknown compound one by one.

IR spectral details:

There is a weak peak at 1570 cm-1 region. So, this signal corresponding to the C=C stretch of the aromatic ring. Therefore, there must be a aromatic ring or benzene in the unknown compound.

Mass spectral details:

Base peak of the unknown compound m/z = 285. If we calculate the molecular mass of the compound using the given molecular formula, we will get 320 as the molecular mass. So, there are chlorine atoms in the compound. This is a good leaving group. So, leaving of a chlorine atom give rise to a positive charge on the compound. One chlorine atom has 35 as atomic weight. So, 320-35 = 285.

Therefore, the base peak arise because of the removal of a chlorine atom.

UV- Visible spectra : The UV - visible spectral value is 256 nm. It includes benzene group and the chlorine group.

¹³C - NMR :

• There are totally 12 signals given by the compound. In that two signals which are downfield at 84 and 74 ppm are responsible for the alkyl group carbon.
• There are 10 signals appears in the aromatic region. So, there must be minimum two benzene rings present in the compound.

¹H-NMR : This is the important spectra from which we can identify the unknown compound using all the above information.

• As we seen in the 13C-NMR spectra, here also in the downfield region, 4.2 and 3.7 ppm, we can find one hydrogen each in the both signals. Both the signals gives doublet because these two hydrogen atoms present on the neighboring carbon atom of each other (1+1=2).
• In the aromatic region, there are two signals has two hydrogen atoms around 7.9 and 7.42 ppm region. These 4 hydrogen atoms belongs to one benzene group and this benzene is para-substituted.
• Remaining 4 signals present in the aromatic region (ortho-substitured) belongs to another benzene ring. Here each signal consist of single hydrogen atom on it.  
• Both the benzene rings are attached with chlorine atoms. And other two chlorine atoms are attached to the alkyl group.

h н b CH-CI Н CI -нС н CI He 1-chloro-2-[2,2-dichloro-1-(4 chlorophenyl)ethyl]benzene Н H f |Турe of hydrogen hydrogen Number of Splitting Chemical shift (ppm) pattern 1 doublet 3.7 а b 1 doublet 4.2 doublet 7.42 с 7.44 d 1 doublet doublet of doublet 1 7.46 doublet of doublet 1 7.5 1 doublet 7.8 2 doublet 7.9

Therefore, the name of the unknown structure is 1-chloro-2-[2,2-dichloro-1-(4-chlorophenyl)ethyl]benzene. The structure is written above.