The 1H-NMR of a compound with molecular formula C6H12O2 consists of four signals: 1.1 (triplet, integrating to 3 Hydrogens), 1.2 (doublet, integrating to 6 Hydrogens), 2.3 (quartet, integrating to 2 Hydrogens) and 5.0 (septet, integrating to 1 Hydrogen). Propose a structural formula for this compound consistent with this information.

The formula is C₆H₁₂O₂. A six-carbon alkane would have the formula C₆H₁₄. The deficiency of two H atoms indicates the presence of either a ring or a double bond.

2. Information from the spectrum

(a) Triplet-quartet

A 3H triplet and a 2H quartet is the classic pattern for a CH₃CH₂- (ethyl) group

(b) Septet-doublet

A 1H septet and a 6H doublet is the classic pattern for a -CH(CH₃)₂ (isopropyl) group

(c) The rest of the molecule

The ethyl and isopropyl groups together add up to C₇H₁₂.

The rest of the molecule must have the formula CO₂ and one unit of unsaturation. That must be a C=O group.

The compound is either

CH₃CH₂-COO-CH(CH₃)₂ or (CH₃)₂CH-COO-CH₂CH₃.

(d) Well, which is it?

The O atom of the ester function should have the greatest effect on the H atom on the adjacent carbon atom.

The CH of an isopropyl is normally at 1.7. The adjacent O atom should pull it down perhaps 3.2 units to 4.9.

The CH₂ of an ethyl group is normally at 1.2. The adjacent O atom should pull it down to about 4.4.

We see a signal at 5.0 but none near 4.4. The compound is isopropyl propionate.

3. Summary

My peak assignments are shown in the diagram below.