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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.

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Concepts and reason

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.

Fundamentals

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.

DBE=1+12n(v2){\rm{DBE}} = 1 + \frac{1}{2}\sum {n\left( {v - 2} \right)}

Here, nn is the number of an atom and ν\nu its valency.

For the molecule C5H10O{{\rm{C}}_5}{{\rm{H}}_{{\rm{10}}}}{\rm{O}}

DBE=1+12[nC(vC2)+nH(vH2)]{\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 nC{n_{\rm{C}}} , 4 for vC{v_{\rm{C}}} , 12 for nH{n_{\rm{H}}} and 1 for vH{v_{\rm{H}}} .

DBE=1+12[6(42)+12(12)]=1+12[6(2)+12(1)]=1+12[1212]=1+12[0]\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

DBE=1+0=1\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 3200cm13200{\rm{ c}}{{\rm{m}}^{ - 1}} to 3600cm13600{\rm{ c}}{{\rm{m}}^{ - 1}} indicates the hydrogen bonded OH stretching and weak absorption at 1676cm1{\rm{1676 c}}{{\rm{m}}^{ - 1}} indicates the alkene stretching. Also, absorption at 965cm1{\rm{965 c}}{{\rm{m}}^{ - 1}} indicates trans disubstituted alkene.

By Using Carbon-13 NMR values and absorption data, the structure of the molecule is found to be as follow.

23.3(t)
CH3
135.5(s)
НС— ОН
Н.
125.5(s)
68.8(s)
H
НаС
17.5(t)
(E)-pent-3-en-2-ol

Ans:

The structure of the molecule is as follows:

CHз
НС— ОН
На
н
Н.С

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