When identifying two compounds A and B the following observations were recorded: Both have the molec — Organic Chemistry Chemistry Question
Organic spectroscopy and structure determination
When identifying two compounds A and B the following observations were recorded:
Both have the molecular formula C3H6O. Schematic 1 H-NMR spectra of these compounds at 400 MHz are presented in the following figure.
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The peak positions and the relative intensities of the different lines in the 1 H-NMR spectrum of B are given in the accompanying Table (Note: the values have been altered slightly from the experimental values to facilitate analysis.)
One of these compounds reacts with malonic acid to form a compound known as Meldrum's acid with a molecular formula of C6H8O4 which gives peaks between 0 and 7.0 in its 1 H-NMR spectrum. The IR spectrum shows a peak in the region 1700 – 1800 cm–1. It condenses with an aromatic aldehyde in the presence of a base.
Table 17.1. Peak positions and relative intensities of individual lines in the 1 H NMR spectrum (400 MHz) of B:
Line | (ppm) | Relative intensity
1 | 6.535 | 1
2 | 6.505 | 1
3 | 6.495 | 1
4 | 6.465 | 1
5 | 3.930 | 1
6 | 3.910 | 1
7 | 3.890 | 1
8 | 3.870 | 1
9 | 3.525 | 1
10 | 3.505 | 1
11 | 3.495 | 1
12 | 3.475 | 1
13 | 3.000 | 12
Label the unknown compounds in the bottles with IUPAC names, using the NMR spectra given in the figure.
Model Answer
The given molecular formula is C3H6O. Therefore, the possible structures are:
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The NMR spectrum of compound A shows a single peak which indicates that all the protons in A are equivalent. This holds true only for structure I. The IUPAC name of this compound is 2-propanone.
The NMR spectrum of compound B shows four sets of peaks, which indicate the presence of four non-equivalent protons. This holds true for structures III and IV. However, for structure IV, no singlet peak (see peak at delta = 3) will be observed. So, compound B must have structure III. The IUPAC name is 1-methoxyethene.
In the 1 H-NMR spectrum of B, assign the peak positions to specific protons. Calculate the spin-spin coupling constants for protons of compound B.
Model Answer
C (with Hb and Hc) = C (with Ha and OCH3) representing 1-methoxyethene:
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Three doublets of doublets centred at 6.5 ppm, 3.9 ppm, 3.5 ppm are seen in the spectrum. The assignments in the spectrum are:
- Ha: 6.5 ppm
- Hb: 3.5 ppm
- Hc: 3.9 ppm
Due to the presence of electron donating OCH3, the trans proton Hb has higher electron density and thus more shielded than Hc. Thus, Hb appears upfield as compared to Hc. There is also a singlet line at delta = 3. This corresponds to the H in OCH3.
Coupling constants:
- Ha : 12, 16 Hz; J(Ha, Hb) = 12 Hz, J(Ha, Hc) = 16 Hz
- Hb : 8, 12 Hz; J(Ha, Hb) = 12 Hz, J(Hb, Hc) = 8 Hz
- Hc : 8, 16 Hz; J(Hb, Hc) = 8 Hz, J(Hc, Ha) = 16 Hz
Note: J = (difference in two lines in ppm) * (instrument frequency)
Geminal coupling < cis-vicinal coupling < trans-vicinal coupling
Convert the peak positions of the first four lines into Hz (refer to the Table). What will be the peak positions of these lines in Hz, if the spectrum is recorded on a 600 MHz instrument?
Model Answer
First four lines from Table 17.1 are:
- Line 1: 6.535 ppm
- Line 2: 6.505 ppm
- Line 3: 6.495 ppm
- Line 4: 6.465 ppm
Peak positions in Hz (for 400 MHz instrument):
- Line 1: 6.535 * 400 = 2614 Hz
- Line 2: 6.505 * 400 = 2602 Hz
- Line 3: 6.495 * 400 = 2598 Hz
- Line 4: 6.465 * 400 = 2586 Hz
Peak positions in Hz (for 600 MHz instrument):
- Line 1: 6.535 * 600 = 3921 Hz
- Line 2: 6.505 * 600 = 3903 Hz
- Line 3: 6.495 * 600 = 3897 Hz
- Line 4: 6.465 * 600 = 3879 Hz
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Draw the possible structure of Meldrum's acid.
Model Answer
Compound A (2-propanone) will react with malonic acid in the following manner:
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The structure of Meldrum’s acid is consistent with the 1 H-NMR and IR data. The peak in the IR spectrum at 1700 – 1800 cm-1 is because of the C=O stretching. The presence of peaks only between 0 – 7 in the 1 H-NMR spectrum indicates that the compound doesn’t have any acidic group like COOH or OH.
If compound B reacts, the only possibility is that it will add across the double bond giving a product with molecular formula equal to C6H10O5. This molecular formula does not match with the one stated in the problem.
Meldrum's acid has pKa = 4.83. Explain the acidity of Meldrum’s acid.
Model Answer
The increased acidity is due to active –CH2 group of Meldrum's acid flanked by two –CO groups. The carbanion formed at –CH2 will be stabilised by these –CO groups, which are coplanar.
Give the structure of the condensation product of Meldrum's acid with an aromatic aldehyde.
Model Answer
The condensation product of Meldrum's acid with an aromatic aldehyde has the structure:
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