An organic compound whose percentage composition is 83.63% carbon and 16.37% hydrogen with a molecular mass of 86.11, presents the following IR spectrum.

Determine the structure of the compound.

**1. Determination of the empirical formula**

\(C:\frac{83.63}{12.01}=6.96;\;\;\; H:\frac{16.37}{1.008}= 16.24\). Dividing by the smallest value \(C:\frac{6.96}{6.96}=1;\;\;\;H:\frac{16.24}{6.96}=2.33\) . Multiplying by 3 we get integer values C:3 H:7. Therefore, the empirical formula is \((C_3H_7)_n\).

**2. Determination of the molecular formula**

\(n=\frac{86.11}{3\times 12.01+7\times 1.008}=2\). The molecular formula is: C

_{6}H_{14}**3. Determination of the unsaturation index.**

Comparing with the formula \(C_nH_{2n+2}\) it is observed that the alkane of 6 carbons has 14 hydrogens. Therefore, our molecular formula lacks unsaturations (alkane).

**4. Write the possible isomers.**

Since it is a saturated compound, it will contain neither cycles nor double bonds.

**5. Study of the spectrum to determine which isomer it belongs to.**

The indicated band, due to the symmetric bending vibration of the methyl CH bonds, is split for the isopropyl and tert-butyl groups. In the case of isopropyl, two bands of equal intensity appear at 1380 and 1370 cm

^{-1}. In the case of tert-butyl the 1380 band has half the intensity. As can be seen in our spectrum, the two bands are of different intensity, a fact that indicates the presence of a tert-butyl group in the molecule. Of the 5 isomers proposed, only one presents this group, 2,2-Dimethylbutane. Spectra for hexane and 2-methylpentane are included below.