DR ANTHONY MELVIN CRASTO,WorldDrugTracker, helping millions, A 90 % paralysed man in action for you, I am suffering from transverse mylitis and bound to a wheel chair, With death on the horizon, nothing will not stop me except God................DR ANTHONY MELVIN CRASTO Ph.D ( ICT, Mumbai) , INDIA 25Yrs Exp. in the feld of Organic Chemistry,Working for GLENMARK GENERICS at Navi Mumbai, INDIA. Serving chemists around the world. Helping them with websites on Chemistry.Million hits on google, world acclamation from industry, academia, drug authorities for websites, blogs and educational contribution

Saturday 5 July 2014

DEPT (Distortionless Enhancement by Polarization Transfer) technique



Proton – coupled 13C-NMR spectra are often difficult to interpret due to large coupling constants and overlaping of signals. For this reason, 13C-NMR spectra are taken with proton–decoupled mode in which C/H ratios is lost.
To provide this information while retaining signal strength, DEPT (Distortionless Enhancement by Polarization Transfer) is developed.
In DEPT experiments, methyl, methylene, and methine protons can be distinguishable. There are several variations on the experiment.


sub-spectrum
technique
CH
DEPT 90o
CH2
DEPT 45o - DEPT 135o
CH3
DEPT 45o + DEPT 135o - 0.707DEPT 90o
C
comparing the DEPT with the BB decoupled spectrum




The types of carbon observed with various of DEPTs.
1.DEPT 45o signals of all protonated carbons
2.DEPT 90o signal of CH groups
3.DEPT 135o negative signal of CH2, positive signal of CH and CH3, and no signal of C with no attached H


Nomally, only two DEPT experiments are sufficient, DEPT 90o and DEPT 135o.
 we can distinguish C, CH, CHand CH3 because;
         - there is no signal of C with no attached H         - 
CH2 shows negative signal whereas CH and CH3 show positive signal
         - CH carbons absorb at lower field and lower signal intensity than CH3carbons




Example 1: DEPT spectrum of isobutyl acetate

Interpretation :
d (ppm)
type of signal in DEPT
represent
22 (b)
positive
2 CH3
24 (c)
positive
CH3
24 (a)
positive
CH
37 (d)
negative
CH2
62 (e)
negative
CH2
170 (f)
not present
C of  >C=O




Example 2: DEPT spectrum of caryophyllene oxide

Interpreting a NMR Spectrum


Interpreting a NMR Spectrum
The following 1Hnmr spectrum of a C10H12O2 compound was obtained on a 90 MHz spectrometer.
The 1Hnmr spectrum of the unknown C10H12O2 compound is shown above. Seven structural formulas of some possible candidates for this compound are presented below. After reviewing the chemical shifts, coupling patterns and integration values shown in this spectrum, select the structure most likely to give this spectrum,  ( A through G ) 
THE FOLL POINTS WILL HELP YOU
 
1.  Ignoring the TMS signal at δ = 0, how many discrete groups of proton signals are present in this spectrum?

2.  What is the multiplicity ( sdtq ) of the highest field signal from this sample?

3.  The sample has a singlet at δ = 3.8 ppm. In units of Hertz (Hz) how far is this signal from the TMS signal?
4.  What structural feature is suggested by the singlet at δ = 3.8 ppm?
          A CH3–C=O     B –CH2–     C –O–H     D –O–CH3     E C–CH3     F C=C–H 

5.  From multiplet line separations (Js), which of the other signals is coupled to the quartet at δ = 2.9 ppm?
          A δ = 1.2 ppm     B δ = 3.8 ppm     C δ = 6.9 ppm     D δ = 7.9 ppm


6.  Using the integrator trace and the formula of the sample, assign a whole number ratio to the sample signals as follows:
          7.9 ppm signal ;     6.9 ppm signal ;     3.8 ppm signal ;     2.9 ppm signal ;     1.2 ppm signal 
 
 
 
 
Structure Analysis

The compound's formula, C10H12O2, has less hydrogen than decane (C10H22). The difference of 10 hydrogens suggests the presence of five double bonds and/or rings in the structure. Since a benzene ring has three double bonds and one ring, and the nmr signals at δ = 7.9 and 6.9 ppm are consistent with aryl hydrogen resonances, we conclude that this compound has a benzene ring.

Integration indicates the presence of two methyl groups, a O–CH3 at δ = 3.8 ppm, and a C–CH3 at δ = 1.2 ppm. The former is of course a singlet, but the latter is a triplet (J = 8 Hz) due to coupling with a –CH2– moiety, which in turn is split into a quartet (signal at δ = 2.9 ppm). From the chemical shifts, we conclude there is an ethyl group bonded to an sp2 hybrid carbon atom.

If we examine the structural formulas of the seven candidate compounds, we can eliminate A, C, F & G because they do not have a –O–CH3 group and cannot account for a singlet at δ = 3.8 ppm. Structure D does not have an ethyl group, so we are left with B & E as the most likely source of this spectrum.

It is not a simple job to distinguish B & E by 1Hnmr alone. The simplicity of the aromatic proton region indicates para-disubstitution of the benzene ring. If the two substituents are similar in nature, the ring proton signals would have similar chemical shifts. If the substituents are very different in nature, these signals would also have significantly different chemical shifts. Since the aromatic doublets in this spectrum are separated by 90 Hz (1.0 ppm), we may tentatively select E (oxygen and carbon substituents) over B (two different carbon substituents).

A better means of distinguishing these compounds is by infrared spectroscopy. Ester carbonyl stretching frequencies are usually higher than equivalent ketone frequencies by roughly 25 cm-1. Also the 13Cnmr carbonyl signal from esters (δ = ca. 170 ppm) is usually observed at higher field than ketone carbonyl signals (δ = ca. 200 ppm).
The spectrum is in fact that of E.


1H NMR
13 C NMR


IR


















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Friday 4 July 2014

p-Anisaldehyde, or 4-methoxybenzaldehyde. NMR




p-Anisaldehyde, or 4-Methoxybenzaldehyde.



RN:
123-11-5
MF:
C8H8O2
MW:
136.15032
bp (°C):
248 - 249
mp (°C):
-1
density:
1.121
nd:
1.571 - 1.574
cLogP: 1.565
cLogS: -1.958
Polar surface: 26.3
NEW: 3D model: 
Show


This 13C spectrum exhibits resonances at the following chemical shifts, and with the multiplicity indicated:
Shift (ppm)
190.8130.2
154.6114.5
132.055.4

55.4........... -0CH3
114.5.........TWO AROM -CH ORTHO TO O ATOM
130.2........AROM C OF -CHO
132..........TWO AROM -CH ORTHO TO -CHO
154.6..........AROM  C OF -OGP
190.8....... CH=O GP









1H NMR



FIRST SIGNAL IS 3H OF CH3

SECOND IS DOUBLET OF TWO H OF AROM RING ORTHO TO O ATOM

THIRD IS DOUBLET OF TWO H OF AROM RING ORTHO TO -CHO GP

LAST IS LONE H OF CHO

IR





ANTHONY MELVIN CRASTO
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Thursday 3 July 2014

2-Amino-5-bromobenzaldehyde








2-Amino-5-bromobenzaldehyde (3) has the following physical and spectroscopic properties: mp 7476 °C; 

1H NMR  (CDCl3, 400 MHz) δ:6.14 (br s, 2 H), 6.56 (d, J = 8.8, 1 H), 7.37 (dd, J = 8.8, 2.3, 1 H), 7.58 (d, J = 2.4, 1 H), 9.79 (-CHO , s, 1 H).

 13C NMR  (CDCl3, 100 MHz) δ: 107.4, 118.1, 120.1, 137.5, 138.0, 148.8. 192.9( C=0). 

HRMS (EI) m/z: Calcd. for C7H6BrNO [M]+: 198.9633, found: 198.9628. 

IR (ATR, cm-1): 3424, 3322, 1649, 1614, 1545, 1468, 1390, 1311, 1185. 

Anal. calcd for C7H6BrNO: C, 42.03; H, 3.02; N, 7.00. Found: C, 42.30; H, 3.02; N, 7.03.



 2-Amino-5-bromobenzyl alcohol (2) has the following physical and spectroscopic properties: mp 112 113 °C;

 1H NMR  (acetone-d6, 400 MHz) δ: 4.19 (t, J = 5.5, 1 H), 4.55 (d, J = 5.5, 2 H), 4.82 (br s, 2 H), 6.65 (d, J = 8.5, 1 H), 7.12 (dd, J = 8.5, 2.4, 1 H)7.22 (d, J = 2.4, 1 H). 

13C NMR  (acetone-d6, 100 MHz) δ: 62.6, 108.3, 117.6, 128.7, 131.1, 131.3, 146.8. HRMS (ESI-TOF) m/z: Calcd. for C7H8BrNNaO [M+Na]: 223.9681, found: 223.9687. 

IR (ATR, cm-1): 3381, 3201 (br), 1473, 1408, 1340, 1268, 1192, 1079. 

Anal. calcd for C7H8BrNO: C, 41.61; H, 3.99 N, 6.93. Found: C, 41.78; H, 3.94; N, 6.91.







ANTHONY MELVIN CRASTO
THANKS AND REGARD'S
DR ANTHONY MELVIN CRASTO Ph.D
MOBILE-+91 9323115463
GLENMARK SCIENTIST ,  INDIA
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