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Find the structure of a compound with formula C5H10O and shows the following 1HNMR.

 

Hint, it is not benzene ring as there are not enough carbons. Ten hydrogen atoms so look at the integration. There are four groups of signalss; the integrations from left to right are ratio 2: 3: 2: 3. i.e.  CH2: CH3: CH2: CH3.

The triplet centred at 2.3 ppm must have 2 neighbours, the singlet at 2.0 ppm has no neighbours, the sextet at 1.50 ppm must have 5 neighbours and the triplet at 0.9 ppm has 2 neighbours. What is missing from the formula is C=O. Putting the puzzle together gives CH3CH2CH2C=O(CH3)

Confirm against the tables:

 

Table A.1 CH3 next to -C =O R expected at 2.1 ppm (found 2.1 ppm), CH2 also next to -C=O R at 2.3 ppm (found 2.5 ppm). Table A.2 CH2 once removed from the ketone –CH2C-C(=O)R at 1.55 ppm (found 1.5 ppm) and the end methyl expect to be at 0.8 ppm (Table A.1). Chemical shifts check out, integration check out, coupling pattern checks out. Structure is proven.

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