Question

what is the structure

C NMR data, are Compound 14 is a low-melting solid (melting point 45-49 "C). The Mass, IR, and H NMR spectra, along with given below. Elemental Analysis C, 85.69, H, 5.53; 0,8.78 Wave Number, om LLLLLL 'H NMR Structure: "C Spectral Data: singlet, 187.0 ppm singlet, 137.8 ppm doublet, 132 2 ppm doublet, 130.1 ppm doublet, 128.2 ppm C ole Company

Answer 1

Using the mass %given in the question, first calculate the empirical formula:

TO CALCULATE EMPIRICAL FORMULA, THE FOLLOWING STEPS ARE FOLLOWED:

• Divide the given masses of elements by their gram atomic mass to get no. of moles of the element. (if mass % is given, we get the number of moles of element/ 100g of compound)
• Express the no. of moles as a ratio. Divide each term by the lowest term in the ratio and round off to whole number/convert each term in the ratio to a whole number by multiplying by a suitable number. This ratio will give the number corresponding to the atom in the empirical formula.

Mass %:

C = 85.69 , H = 5.53, O = 8.78

Gram atomic mass of C = 12g/mole, O = 16g/mole, H =1g/mole

Step 1: dividing by gram atomic mass :

C: 85.69 /12 = 7.13        O: 8.78/16 = 0.55              H: 5.53/1 = 5.53

Step 2: Ratio (C:H:O) = 7.13:5.53:0.55

Dividing by lower number (0.55) and rounding off: 13:10:1

So, empirical formula = C₁₃H₁₀O. Calculating the molar mass, we get : 13x12 + 10x1 + 1x16 = 182

This coincides with the molecular ion peak in mass spectrum. Thus, the molecular formula is C₁₃H₁₀O

Next, calculate the double bond equivalent , using the formula:

DBE = C+1 –(H/2) –(X/2) + (N/2) ,

Where C = no. of C atoms , H = no. of H atoms, X = no. of halogen atoms and N = no. of N atoms

Here, C = 13, H = 10

So, DBE = 9

Next, look at the IR spectrum.

The peak at around 3000 is indicative of aromatic C-H stretching, and the peak at approx. 1700 is indicative of Carbonyl (C=O) stretching.

Looking at 1H NMR:

There is a multiplet above 7ppm, which indicates several aromatic protons. There are no other peaks.

Considering this and the IR data, we can think that two phenyl groups and a carbonyl group is present. No other H containing group is present , as is observed from the 1H NMR spectrum. Considering just two phenyl groups (i.e 2 x C₆H₅) and one carbonyl group (C=O), the molecular formula obtained is C₁₃H₁₀O (this matches with the formula obtained).

Next, look at 13C:

There are 5 peaks, which means that 5 different types of C are present.

But since the compound has 13 C in total, this indicates that the compound is symmetrical somehow, so that the number of unequivalent carbons are reduced. Keeping in mind that we probably have two phenyl groups and one keto group(The carbonyl group should be a ketone, since an aldehyde would give an extra H, and acid/amide carbonyls would lead to a change in molecular formula), we can propose the following structure: singlet doublet H singlet H c This C couples with the H next to it to give doublet С. Η Η H doublet mirror plane Multiplicity of peak is calculated using: 2n1+1, where n = no of atoms of a given type. I = 1/2 for H. For the pink colour C, it is attached to 1 H. So, n=1. Putting in formula, we get multipicity = 2, i.e doublet. The process is similar for blue and green Carbon this structure (benzophenone) explains all the features in the spectrum Its melting point is also within the specified range.