Boric Ester-Type Molten Salt via Dehydrocoupling Reaction

Figure 1. ¹H-NMR of 5 (solvent: CDCl₃; 10%; r.t.; TMS; 400 MHz).


http://www.mdpi.com/1422-0067/15/11/21080

http://www.mdpi.com/1422-0067/15/11/21080/htm

Int. J. Mol. Sci. 2014, 15(11), 21080-21089; doi:10.3390/ijms151121080

Article

Boric Ester-Type Molten Salt via Dehydrocoupling Reaction

Noriyoshi Matsumi ^1,*, Yoshiyuki Toyota ¹, Prerna Joshi ¹, Puhup Puneet ¹, Raman Vedarajan ¹ and Toshihiro Takekawa ²

¹School of Materials Science, Japan Advanced Institute of Science and Technology (JAIST), 1-1 Asahidai, Nomi, Ishikawa 923-1292, Japan; E-Mails: brothertobrother@gmail.com (Y.T.); joshi.prerna2011@gmail.com (P.J.); rupakagain2008@gmail.com (P.P.); raman@jaist.ac.jp (R.V.)

²Advanced Materials Laboratory, Nissan Motor Co., Ltd., 1 Natsushima-cho, Yokosuka-shi, Kanagawa 237-8523, Japan; E-Mail: t-takekawa@mail.nissan.co.jp

*Author to whom correspondence should be addressed; E-Mail: matsumi@jaist.ac.jp; Tel.: + 81-761-51-1600; Fax: +81-761-51-1665.


 JOSHI PRERNA




Noriyoshi Matsumi

Figure 2. ¹³C-NMR of 5 (solvent: DMSO; r.t.; 100 MHz)



Figure 3. ¹¹B-NMR of 5 (solvent: CDCl₃; r.t.; external standard: B(OCH₃); 128 MHz).



Figure 4. ¹⁹F-NMR of 5 (solvent: CDCl₃; r.t.; external standard: C₆H₅CF₃; 376 MHz).


data
1H-NMR (CDCl3, δ, ppm) 1.26–2.50 (20H, 9-BBN), 3.98 (3H, CH3), 4.33 (2H, –NCH2CH2O), 5.62 (2H, –NCH2CH2OB–), 7.21, 7.35 (2H, –NCHCHN–), 8.87 (1H, –NCHN–). 

13C-NMR (DMSO-d6, δ, ppm) 23.3–33.2 (9-BBN), 36.1, 52.1, 59.8, 119.4 (q, CF3), 123.1–123.8 (N–C=C–N), 137.3 (N–C=N). 

11B-NMR (CDCl3, δδ, ppm) 33.3.






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13C NMR of dimethyl 3,4,5,6-tetraphenylphthalate

he upfield peak at 52.28 is in the expected position for C1, the ester methyl carbon. Down at the other end of the spectrum, the peak at 168.71 is in the expected position for an ester carbonyl carbon (C2).

In the region for aromatic carbons we see 5 low intensity peaks between 143 and 132. These belong to the (unprotonated) quaternary carbons 3, 4, 5, 6, 10. Using diethyl phthalate as a model compound, I would guess that the peak at 132.19 is from C3. The two peaks at 138.73 and 138.60 have the same height and are closely spaced suggesting very similar environments, I might guess them to be C6 and C10, just not sure which is which. Noting that the peak for the carbon corresponding to our C5 is downfield of the peak corresponding to our C4 in the model compound, I would expect the same trend to hold here, just with each of the peaks shifted downfield due to attachment of a phenyl group. So I would assign the peak at 143.27 to C5 and the peak at 139.29 to C4.

This leaves the protonated aromatic carbons for us to consider. The 2 peaks at 125.97 and 126.83 are half the intensity of the other protonated aromatic signals, therefor assign them to C9 and C13 as there are only half as many of these carbons as compared to the other protonated aromatic carbons. It's not clear which peak belongs to C9 and which to C13. This leaves us with the 4 signals due to C7, C8, C11 and C12. , I would expect C7 and C11 to be downfield from C8 and C12. Therefor, I would assign C7 and C11 to the peaks at 129.71 and 130.88, not knowing which is which. C8 and C12 then belong to the signals at 126.88 and 127.41, not knowing which belongs to which.

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Ester infrared spectra

Vinyl acetate

Methyl benzoate

Methyl methacrylate

Methyl salicylate

Ethly butyrate

The C=O of an ester appears near 1750-1735 which can overlap with some ketone C=O stretches. One can usually eliminate ketones by considering the strong and broad C-O peak at 1300-1000. A ketones absorptions will have a weaker and narrower bands. Compare the ethyl butyrate ester above to the ketone below:

Esters can also conjugate on the side of the single-bonded oxygen:

Apparently this form of conjugation interferes with resonance of the carbonyl group, hence increasing the arbsorption frequency of the C=O bond.

α-keto esters: One might expect two C=O peaks due to the two different carbonyls. In practice, it usually manifests as a shoulder on the main C=O peak or a single broadened band.

β-keto esters: These give a strong intensity doublet for the C=O stretches. Since they do not tautomerize to the same extent as β-diketones, one cannot generally observe the OH stretch from the enol form.

The –C–CO 2R stretch characteristic of an ester is visible in the isopentyl acetate IR spectrum in the 1735–1745 cm -1 range. The –C–H stretches are visible just below 3000 cm -1, and the –C–O and –CO 2 stretches appear as several peaks in the 1050–1300 cm -1 range.

The spectra of ethyl acetate and ethyl benzoate are shown above. Note that the C=O stretch of ethyl acetate (1752) is at a higher wavelength than that of the α, β-unsaturated ester ethyl benzoate (1726). Also note the C–O stretches in the region 1300-1000 cm^-1.

1H-NMR and 13C-NMR spectra of 1,2,2-trichloropropane

The molecule has 3 different carbons, so the proton-spin decoupled 13C-NMR spectra will have 3 peaks. The molecule has only two different types of protons that are not coupled so the 1H-NMR will have two unsplit singlet signals.





Off-resonance-decoupled 13C-NMR Spectra of 1,2,2-trichloropropane. The CCl2 group appears as a singlet, the CH2Cl group appears as a triplet, and the CH3 group as a quartet.

The off-resonance-decoupled carbon spectrum will show the spin-spin splitting between the carbon and the hydrogen it has attached to it. The N+1 rule applies so a carbon with 3 hydrogens on it will give rise to a quartet, while a carbon with only 1 hydrogen will have a doublet. Carbons with no hydrogens attached to them will have singlet signals.





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Isopropyl bromide..........1H NMR

Isopropyl bromide.

Absorption by the six methyl protons H(a) appears upfield, split into a doublet by the single adjacent proton H(b). Absorption by the lone proton H(b) appears downfield (the inductive effect of bromine) split into a septet by the six adjacent protons (outside peaks barely visible).

Absorption by the six methyl protons H(a) appears upfield, split into a doublet by the single adjacent proton H(b). Absorption by the lone proton H(b) appears downfield (the inductive effect of bromine) split into a septet by the six adjacent protons (outside peaks barely visible).




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1,2-dibromo-1-phenylethane..1H NMR

The diastereotopic protons H(a) and H(b) give different signals, each split into a doublet by H(c). The downfield peaks of the doublets happen to coincide. [The spectrum shows no splitting due to coupling between H(a) and H(b). With the trial run at high gain, however, the spectrum shows this coupling: each doublet is split into a quartet.]

The four-line pattern of c is due to successive splittings by H(a) and H(b). [ If J(ac) and J(bc) were equal- as they would have to be if, for example, H(a) and H(b) were equivalent - then the middle peaks of c would merge to give the familiar 1:2:1 triplet.]

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1H NMR n-propylbenzene

n-propylbenzene

Moving downfield, we see the expected sequence of signals: a, primary (3H); b, secondary (2H), c, benzylic (2H), d, aromatic (5H). Signals a and c are each split into a triplet by the two secondary protons H(b). The five protons adjacent to the secondary protons - three o none side and tow on the other - are, of course, not equivalent. But the coupling constants J(ab) and J(bc) are nearly identical, and signal b appears as a sextet (5 + 1 peaks). The coupling constants are not exactly the same, however, as shown by the broadening of the six peaks.










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