Question

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 δ 17.5 (3), δ 23.3 (3), δ 68.8 (1), δ 125.5 (1), and δ 135.5 (1). Compound A may be resolved into enantiomers; draw one molecule of A, omitting wedge/dash bonds.

Answer 1

1) MF: C5H10O. Corresponding hydrocarbon formula is C5H10 (no compensation for O). With 5 C expected saturated hydrocarbon formula is C5H12 (general formula CnH2n+2). This indicate deficiency of 2 H and hence unsaturation site = 1.

2)IR spectral data analysis:

i) Strong and sharp absorption in the range 3200 – 3600 cm^-1 is for O-H stretch i.e. i.e. –O-H group present.

ii) A weak absorption at 1676 cm^-1 is for sp2 C=C stretch i.e. olefinic C=C present.

iii) 965 cm^-1 absorption indicate presence of trans 1,2-disubstituted double bond.

3) 13C NMR data and analysis,

13C chemical shift and attached protons

i) 17.5, 3H indicate –CH3 piece

ii) 23.3, 3H indicate –CH3 piece

iii) 68.8 , 1H indicate –CH piece bonded to electronegative atom like O hence probably –CH(OH) piece present.

iii) 125.1 1H and 135.5 1H indicate olefinic C=C group.

On the basis of all this information structure proposed is,

(E)-pent-3-en-2-ol

135.5 on the side of –OH group

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