Question

15. The following data is for a compound containing C, H, and N. Answer the questions below:

High Resolution Mass Spectrum: 196.1002 amu IR: 2169 cm "H (400 MHz, CDCl): δ 8.05 (bs, 1H, exchanges with D2O), 7.63 (ddd, J = 0.6, 23, 6.8 Hz, 1H), 7.29 (m 1H), 7.16 (m, 2H), 4.03 (tt, J-1.3, 5.6 Hz, 1H), 2.75 (m, 2H), 2.17 (m, 3H), 1.98 (m, 1H). 13C (100 MHz, CDCl): δ 1357, 135.3, 126.1, 122.1, 121.4, 1202, 1177, 1109, 1035, 27.7, 24.2, 22.6, 20.8.

Calculate the molecular formula of the compound using the rule of 13.

Calculate the units of unsaturation:

Propose structural pieces apparent from the proton spectrum:

Propose structural pieces apparent in the carbon spectrum

Propose a structure that would fit the spectrum

Give two lines of evidence from proton and carbon spectra that would spport your proposed structure

The rule of 13 states that the formula of a compound is a multiple n of 13

Answer 1

15 a. Molecular formula by rule of 13:

According to rule of 13, the formula of an organic compound is a multiple n of 13 (CH) plus a remainder r.

For Mol mass = 196.1002 amu (even number)

On dividing by 13, we get 15 (CH) +r(1) Or C₁₅H₁₅₊₁ = C₁₅H₁₆ (For compund having only C and H)

BUT Since it is given that the compound also contain "N atom" we need to adjust the N atoms in the molecular formula as well, for each N atom, add N (mass 14) and subtract CH2 (mass 14) from themolecular formula.

According to Nitrogen rule, for even mass the molecular formula (like 196 in the question which is even) has an even number of nitrogen atoms (0, 2, 4, etc.).

So possible options for molecular formula can be :

C₁₅H₁₆ (For 0 nitrogen),: Cant be Because the compound has N atoms.

C₁₃H₁₂N₂ (For 2 nitrogens,substract C₂H₄ from C₁₅H₁₆),

C₁₁H₈N₄ (For 4 nitrogens, substract C₄H₈ from C₁₅H₁₆)....and so on.

C₁₃H₁₂N₂ would be the correct molecular formula because in the ¹³C spectrum we are getting signals for 13 Carbons( compund contain 13 Carbons).

15 b. Units of Unsaturation:

Formula for Degrees of Unsaturation = [2 C + 2 + N - H - X] / 2

C =No. of Carbon atoms, N = No. of Nitrogen atoms, H = No. of Hydrogen atoms,, X =No. of halogen atoms,

So DU for C₁₃H₁₂N₂ = (2*13 +2 + 2- 12) / 2 = 18/2 = 9

Degrees of Unsaturation (the sum of rings + multiple bonds )= 9

IR strech at 2169 cm^-1 indicates an alkyne (CC) and/or Nitrile (CN) functional group.

15 c. Findings from Proton spectrum:

Since degree of unsaturation is >4, an aromatic ring is likely to be present in the compound.

Chemical shifts in the range of 6.8-8 ppm on the 1H-NMR is confirming the presence.

In the 1H NMR spectrum,

Signal at 8.05 ppm is acidic and aromatic because it exchanges with D2O. Indicates an -NH proton in an aromatic. ring.

15 d. Findings from 13C spectrum:

13C shows 13 Carbons, out of which 4 are Sp3 hybridized i.e alkane type or non aromatic Carbons and rest 9 are Sp2 hybridized i.e Aromatic or conatins C=C bond or (CC).

If we combine the above facts, the carbon skeleton of the proposed struture comes out as

Degree of Unsaturation = 9

2 Nitrogens

One CN group (2169 IR),

5 Aromatic hydogens (One acidic),

8 aromatic carbons (Sp2 hybridized) and 1 Nitrile (CN) carbon (Sp hybridized).

Contains 7 Hydrogens as Aliphatic region (2 and 3 for Ethyl and 1,1 for Alkene).