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.
A 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).
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1H NMR
3-Methyl-1-butanol
Formula | C5H12O | |
Formula Weight | 88.2 gmol-1 |
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Peak Table
Peak | Intensity | Multiplet | Proton/s |
1 | Reference Peak |
2 | 5.8 (2) | Triplet | d |
3 | 3.2 (1) | Multiplet of 9 | b |
4 | 6.1 (2) | Quartet | c |
5 | 3.1 (1) | Singlet | e |
6 | 18.9 (6) | Doublet | a |
1H spectrum
This spectrum, like all other simple 1H spectra, can be assigned by using the following criteria;
- 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.
- Compare the chemical shift of the peak with a correlation chart to get an idea which functional groups could be present.
- 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)
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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).
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