Chemistry
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Question “Chemistry”
Deduce the structure of an unknown compound using the data.
Molecular Formula: C6H12O
IR: 1705 cm−1
H1
NMR: no absorptions greater than ? 3 ppm
C13
NMR: ? 24.4,
? 26.4,
? 44.2,
and ? 212.6 ppm.
Resonances at ? 44.2
and 212.6 have very low intensity.
Answer
C6H12O
Degree of Unsaturation = (nC x 2-nH) (2) = ( 2×6 + 2-12) /2 = (1 (where O is considered to be 0H).
IR peak at 1705cm-1 indicates C=O
Now HNMr peaks are all below 3ppm idnicates. We have simple alkanes, and CH groups adjacent C=O
There is no aldehyde peak that is near 9 and . Our C=O is ketone
C–NMR showed 4 peak which indicated 4 different types of carbons. Low intensity refers to Carbon with no Hydrogens attached.
CNMR peak at 212.6ppm indicates that C=O
Based on this
Conclusion
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