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
Showing posts with label MALUKU ISLANDS. Show all posts
Showing posts with label MALUKU ISLANDS. Show all posts

Sunday 5 July 2015

ISOAMYL ALCOHOL




Compound  has a composition a 68.2% C, 13.6% H and 18.2% O by mass.




IR







  • A notable feature of the 1H spectrum is a multiplet of nine at d1.7. 
splitting pattern of nine peaks must be produced by eight adjacent protons (8+1)

  • The 13C spectrum shows the reference peak and four peaks 


13C Spectrum



Peak  is due to the carbon atom in position d as it has a chemical shift of d60.7 (corresponds to a -C-O- functional group) and is connected to an even number of protons (2).
Peak  is due to the carbon atom in position c as it has a chemical shift of d41.5 (corresponds to a secondary carbon group) and is connected to an even number of protons (2).
Peak  is due to the carbon atom in position b as it has a chemical shift of d24.6 (corresponds to a tertiary carbon group) and is connected to an odd number of protons (1).
Peak  is due to the two carbon atoms in position a as it has a chemical shift of d22.5 (corresponds to a primary carbon group). It can be seen that the carbon atoms are also connected to an odd number of protons (3).
............
1H NMR

3-Methyl-1-butanol

FormulaC5H12O
Formula Weight88.2 gmol-1

Peak Table

PeakIntensityMultipletProton/s
1Reference Peak
25.8 (2)Tripletd
33.2 (1)Multiplet of 9b
46.1 (2)Quartetc
53.1 (1)Singlete
618.9 (6)Doubleta

1H spectrum

This spectrum, like all other simple 1H spectra, can be assigned by using the following criteria;
  1. Convert the given intensities of the peaks to integrals.
    • Add all the peak intensity values together and divide by the number of hydrogen atoms in the molecule. This gives the average intensity per hydrogen atom.
    • Divide each peak intensity by this average term to give a whole number (integer) value.
  2. Compare the chemical shift of the peak with a correlation chart to get an idea which functional groups could be present.
  3. Determine the multiplicity by examining the fine structure of each peak. If there are (n) adjacent hydrogens then the peak will have a (n+1) splitting pattern.

1H spectrum (expansion)


Notes

  • Peak  is due to protons in position d as it has a chemical shift of d3.6 (corresponds to a -OH containing functional group) and is a triplet (split by two adjacent hydrogens - Hc).
  • Peak  is due the proton in position b as it has a chemical shift of d1.7 (corresponds to a tertiary -CH- group) and is a nontet (split by eight adjacent hydrogens - Ha and Hc).
  • Peak  is due to protons in position c as it has a chemical shift of d1.4 (corresponds to a -CH2 group) and is a quartet (split by three adjacent hydrogens - Hb and Hd).
  • Peak  is due to protons in position e as it has a chemical shift of d1.3 (corresponds to hydroxyl proton) and is a singlet (no splitting).
  • Peak  is due the proton in position a as it has a chemical shift of d0.8 (corresponds to a -CH3 group) and is a doublet (split by one adjacent hydrogens - Hb).



Maluku islands,   Indonesia