Step 1. Start with the molecular formula
The formula is #"C"_7"H"_14"O"_2#. The formula of an alkane with seven carbon atoms is #"C"_7"H"_16#.
The two missing hydrogens tell us the compound must contain a double bond or a ring.
Step 2. Analyze the IR spectrum
The strong peak at #"1730 cm"^"-1"# confirms the presence of a carbonyl #"C=O"# group.
There is no evidence of #"OH"# or stretching above #"3000 cm"^"-1"#, so there is no #"OH"# group.
The other #"O"# atom may be part of an ester group. This is consistent with the high wavenumber for the carbonyl group and the strong band at #"1200 cm"^"-1"# (#"C-O"# stretch?).
Step 3. Analyze the #""^1"H"# NMR spectrum
The peaks and tentative assignments are:
#ulbb(δ"/ppm"color(white)(m)"H atoms"color(white)(m)"Multiplicity"color(white)(mmll)"Assignment"color(white)(mmmm))#
#0.9color(white)(mmmmm)3color(white)(mmmm)"triplet"color(white)(mmmmmm)"CH"_3color(white)(l)"adjacent to CH"_2#
#1.3color(white)(mmmmm)6color(white)(mmmm)"doublet"color(white)(mmmmmll)("CH"_3)_2color(white)(l)"adjacent to CH"#
#1.7color(white)(mmmmm)2color(white)(mmmm)"multiplet"color(white)(mmmmm)"CH"_2#
#2.2color(white)(mmmmm)2color(white)(mmmm)"triplet"color(white)(mmmmmml)"CH"_2color(white)(l)"adjacent to CH"_2#
#5.0color(white)(mmmmm)1color(white)(mmmm)"septet"color(white)(mmmmmml)"CH adjacent to "("CH"_3)_2"#
Typical patterns are
#"1H:6H doublet-septet = -CH"("CH"_3)_2#
#"3H doublet = CH"_3"CH"_2"-"#
#"2H doublet = -CH"_2"CH"_2#
My tentative guess is isopropyl butyrate.
#color(white)(ll)"CH"_3"CH"_2"CH"_2"C(=O)OCH"("CH"_3)_2"#
#δcolor(white)(l)0.9color(white)(m)1.7color(white)(ll)2.2color(white)(mmmmll)5.0color(white)(ll)1.3#
Step 4. Analyze the #""^13"C"# NMR spectrum
The spectrum accounts for all seven carbon atoms.
#ulbb(δ"/ppm"color(white)(m)"No."color(white)(m)"Assignment")#
#color(white)(ll)13color(white)(mmmm)1color(white)(mm)"CH"_3#
#color(white)(ll)18 color(white)(mmmm)1color(white)(mm)"CH"_2#
#color(white)(ll)21 color(white)(mmmm)2color(white)(mm)("CH"_3)_2#
#color(white)(ll)36 color(white)(mmmm)1color(white)(mm)"CH"_2"C=O"#
#color(white)(ll)68 color(white)(mmmm)1color(white)(mm)"CHOR"#
#172color(white)(mmmm)1color(white)(mml)"RCOOR"#
This confirms that the compound is isopropyl butyrate.
#color(white)(ll)"CH"_3"CH"_2"CH"_2"C(=O)CH"("CH"_3)_2"#
#δcolor(white)(l)13color(white)(m)18color(white)(m)36color(white)(ml)172color(white)(mll)68color(white)(m)21#
