Identify the compound A (C5H10O) with the proton NMR spectrum shown. Compound A has IR absorptions…
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Question “Identify the compound A (C5H10O) with the proton NMR spectrum shown. Compound A has IR absorptions…”
Identify the compound A (C5H10O) with the proton NMR spectrum
shown. Compound A has IR absorptions at 3200�3600 cm�1 (strong,
broad), 1676 cm�1 (weak), and 965 cm�1, and also has 13C NMR
absorptions (attached protons in parentheses) at ? 17.5 (3), ? 23.3
(3), ? 68.8 (1), ? 125.5 (1), and ? 135.5 (1). Compound A may be
resolved into enantiomers; draw one molecule of A, omitting
wedge/dash bonds.
Answer
The problem is based on the concept of NMR spectroscopy of proton.
The structures of molecules can be determined by using proton NMR spectroscopy. Different molecular environment of protons is responsible for different values in NMR.
Double bond equivalents of a molecule give the number of double bonds in it. The formula for double bond equivalent (DBE) for a molecule is as follow.
{\rm{DBE}} = 1 + \frac{1}{2}\sum {n\left( {v - 2} \right)}Here, n
For the molecule {{\rm{C}}_5}{{\rm{H}}_{{\rm{10}}}}{\rm{O}} {\rm{DBE}} = 1 + \frac{1}{2}\left[ {{n_{\rm{C}}}({v_{\rm{C}}} - 2) + {n_{\rm{H}}}({v_{\rm{H}}} - 2)} \right]
Substitute 6 for {n_{\rm{C}}} \begin{array}{c}\\{\rm{DBE}} = 1 + \frac{1}{2}\left[ {6(4 - 2) + 12(1 - 2)} \right]\\\\ = 1 + \frac{1}{2}\left[ {6(2) + 12( - 1)} \right]\\\\ = 1 + \frac{1}{2}\left[ {12 - 12} \right]\\\\ = 1 + \frac{1}{2}\left[ 0 \right]\\\end{array}
DBE becomes
\begin{array}{c}\\{\rm{DBE}} = 1 + 0\\\\ = 1\\\end{array}There is one double bond. So, the molecule is unsaturated.
In IR region, strong and broad absorption at 3200{\rm{ c}}{{\rm{m}}^{ - 1}}
By Using Carbon-13 NMR values and absorption data, the structure of the molecule is found to be as follow.
Ans:
The structure of the molecule is as follows:
Conclusion
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