4-Hydroxy-3-methoxycinnamaldehyde (Coniferaldehyde)

4-Hydroxy-3-methoxycinnamaldehyde (Coniferaldehyde)


C₁₀H₁₀O₃
178.00

ESIMS m/z 178 [M – H]-


λmax 350


IR
IR(cm-¹)
(KBr) 3381 (OH), 
2930 (olefinic -CH=CH-), 
1675 (C=O), 1662, 1459, 
1360-610 (aromatic -CH=CH-)



13 C NMR

(225 MHz, acetone-d6) 

δ: 127.5 (C-1), 
111.6 (C-2), 
151.0 (C-3), 
148.9 (C-4),
116.3 (C-5), 
124.9 (C-6), 
154.2 (C-7),
127.1 (C-8), 
193.5 (C-9), 
56.5 (OCH3)






1H NMR

(900 MHz, acetone-d6) 
δ: 9.64 (1H, d, J = 8.1 Hz, H-9), 
7.57 (1H, d, J = 16.2 Hz, H-7), 
7.38 (1H, s, H-2), 
7.21 (1H, d, J = 8.1 Hz, H-6), 
6.91 (1H, d, J = 8.1 Hz, H-5), 
6.65 (1H, dd, J = 16.2, 8.1 Hz, H-8),
3.93 (3H, s, OCH3)








1H_13C_HMBC




  1H_13C_HSQC



 HH_TOCSY.

APRICOXIB SPECTRAL DATA

Apricoxib, A COX-2 inhibitor.

APRICOXIB

A COX-2 inhibitor.

C19H20N2O3S

Mol wt: 356.439

CAS: 197904-84-0

CS-701; TG01, R-109339, Benzenesulfonamide, 4-[2-(4-ethoxyphenyl)-4-methyl-1H-pyrrol-1-yl]-;

4-[2-(4-ethoxyphenyl)-4-methyl-1H-pyrrol-1-yl]benzenesulfonamide .

Bioorg Med Chem Lett. Oct 15, 2011; 21(20): 6071–6073.

Published online Aug 19, 2011. doi:  10.1016/j.bmcl.2011.08.050

SEE AT

http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3310163/

An efficient synthesis of apricoxib (CS-706), a selective cyclooxygenase inhibitor, was developed using copper catalysed homoallylic ketone formation from methyl 4-ethoxybenzoate followed by ozonolysis to an aldehyde, and condensation with sulphanilamide. This method provided multi-gram access of aprocoxib in good yield. Apricoxib exhibited potency equal to celecoxib at inhibition of prostaglandin E2 synthesis in two inflammatory breast cancer cell lines.

c26. Synthesis of Apricoxib (1): 

Homoallylic ketone (8) (5.0 g, 21.53 mmol) in 180 mL of CH2Cl2/MeOH (1:5) was treated with ozone bubbles at −78°C until a blue coloration persisted. The ---------------------------------DELETED...................................... The crude brown material was purified by silica gel flash chromatography using a gradient of EtOAc in hexane to give apricoxib as white solid (5.5 g, 15.43 mmol, 71%).

m.p. 161–163°C (lit. 135–139°C14).

1H NMR (CDCl3, 300.0 MHz) δ 1.32 (t, J = 6.9 Hz, 3H), 2.1 (s, 3H), 3.92 (q, J = 6.9 Hz, 2H), 4.95 (s, 2H), 6.14 (m, 1H), 6.63 (m, 1H), 6.69 (d, J = 6.6 Hz, 2H), 6.94 (d, J = 6.6 Hz, 2H), 7.13 (d, J = 6.6 Hz, 2H), 7.74 (d, J= 6.6 Hz, 2H).

13C NMR (CDCl3, 75.0 MHz) δ 11.7, 14.8, 63.4, 82.4, 113.2, 114.4, 121.0, 121.1, 124.9, 125.2, 127.4, 129.7, 133.6, 138.7, 144.2, 158.0

M+H Calcd: 357.1273; Found, 357.1252.

01

Click here to view.(2.1M, pdf)   DOWNLOAD TO GET NMR , 13C, COSY

OR

Supplementary Material

1H, 13C, and COSY NMR spectra of compounds 1 and 8.

4-Ethoxybenzaldehyde NMR

 

Example

C₉H₁₀O₂

MW 150

Calculate the degree of unsaturation: the answer is 5. If the degree of unsaturation is 4 or greater, look for an aromatic ring, which has a degree of unsaturation of 4 (3 double bonds plus 1 ring). In addition to an aromatic ring, the molecule can have a carbon-carbon double bond, a carbonyl, or another ring.

IR Spectrum

Look for a carbonyl, since the degree of unsaturation indicates that the compound could have a double bond and we know that the molecule has an oxygen. There is a band at 1697, suggesting an alpha, beta unsaturated aldehyde or ketone(1710-1665). To see if the compound might be an aldehyde, look for bands in the region 2830-2695. In the spectrum below, note the two bands in this region, suggesting that the compound is indeed an aldehyde.

The IR can also help determine whether or not the compound is an aromatic (although the NMR is a better diagnostic method for this). Look for the C–H stretch in aromatics from 3100-3000. There are a couple very small bands in this region.

There is one more oxygen in the molecule, it could be an ether or even an ester (if we are incorrect in assuming an aldehyde is indicated). Ether IR bands are difficult to distinguish from any other C-O stretch band - the C-O stretch of alcohols, carboxylic acids, esters, and ethers all show up in the region 1320-1000 (see Ethers).

Consult the section on Aromatics for more information on IR spectroscopy of aromatics.

NMR Spectrum

From the IR, we know that compound is probably an aromatic and an aldehyde. Aldehydes and aromatics are quite distinctive in the NMR: aldehydes show up from 9-10, usually as a small singlet; aromatic protons show up from 6.5-8.5 ppm. Let's look at the NMR:

The singlet at 9.9 ppm indicates an aldehyde; the 4 protons from 7-8 ppm indicate a di-substituted aromatic ring.

The remaining two peaks represent an ethyl group, -CH₂CH₃: 2 protons split by 3 protons adjacent to 3 protons split by 2 protons. The -CH₂- portion of the ethyl group is shifted downfield further than the benzylic protons this indicates that it is next to an oxygen

Let's draw an 8-carbon 10-hydrogen molecule incorporating these features: an aldehyde, a di-substituted aromatic ring, and an ethyl group:

The molecule is drawn as the para-substituted aromatic; this is indicated by the symmetry of the peaks in the aromatic region.

The following structure with the hydrogens in different colors shows how the protons correlate with the NMR peaks:

NOTE TWO AROM -H ORTHO TO O ATOM APPEAR AT 6.9 PPM

Summary

Example  is 4-ethoxybenzaldehyde:

13 C NMR OF 3-Ethoxy-4-methoxybenzaldehyde(1131-52-8)

Alittle complicated than above example

The interpretation is available in form of numbering

13CNMR

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3-BUTYL PYRIDINE, A simple nmr example

3 Butyl pyridine

Formula: C₉H₁₃N, 135.2062

IR 

IH NMR

SEE

http://www.sigmaaldrich.com/spectra/fnmr/FNMR006274.PDF

 Unsaturation answer

C₉H₁₃N
Rule 3, omit the N and one H, gives C₉H₁₂
9 - 12/2 + 1 = 4 degrees of unsaturation.
Look for an aromatic ring.

IR interpretation

The bands at 3000-2850 indicate C-H alkane stretches. The band at 3028 indicates C-H aromatic stretch; aromatics also show bands in the regions 1600-1585 and 1500-1400 (C-C in-ring stretch), and 900-675 (C-H out-of-plane). The bands in the region 1250-1020 could be due to C-N stretch. The weak, broad banc at about 3500 could be amine N-H stretch or it could be a slight contamination of an impurity (water) in the sample.

 Structure answer

This is the structure. See if you can assign the peaks on your own.

NMR structure interpretation

ETHYL PROPIONATE NMR, carbonyl is dad and oxygen mom, a methyl gets more attention

Dedicated to all moms in the world

C=O group is dad

O atom is mom

Carbonyl is dad and oxygen mom hence c labelled methyl has higher chemical shift  and gets a little more attention

SEE BELOW

NMR IS EASY

A chemical has Formula: C₅H₁₀O₂

C₅H₁₀O₂
Rule 2, omit O, gives C₅H₁₀
5 - 10/2 + 1 = 1 degree of unsaturation.
Look for 1 pi bond or aliphatic ring.

IR

The band at 1740 indicates a carbonyl, probably a saturated aliphatic ester. The bands at 3000-2850 indicate C-H alkane stretches. The bands in the region 1320-1000 could be due to C-O stretch, consistent with an ester.

 Structure answer

This is the structure. See if you can assign the peaks on your own.

NMR answer

C has a higher chemical shift than D because it's closer to a more electron-withdrawing functional group.

Carbonyl is dad and oxygen mom,  hence c has higher chemical shift  and gets a little more attention in proton nmr

13 C NMR

Mass spectrum

RAMAN

WHAT HAPPENS WHEN A CHLORO IS INTRODUCED

THE INTERPRETATION IS BELOW

remember "a" labelled  CH3 appears as a doublet

WHEN THERE IS ONE METHYL

WHEN THERE ONE CH2 SHORT 

WHEN MOM HAS ONE MORE CH2

PROPYL PROPIONATE, try this on your own

1H NMR

see interpretation

 BIGGER ONE THAN OBOVE

13C NMR

APT

COSY

WILL PASTE INTERPRETATION AFTER ONE WEEK...................