Question

I started with 2-methylcyclohexanol and reacted with h3PO4 to get alkene isomers. based on these GC results, what is the percent of each isomer produced?

Qualitative Analysis Report User Chromatograms Fragmentor Voltage Collision Energy 0 Ionization Mode EI x107+El BPC Scan Dehydrogenation_00x_1_2.D 4.68 1.4 37029905 11 1.2 2 0.8 0.6 0.4 0.2 3.99 9296884 4.41 2267816 6.73 382818 6 6.2 6.4 6.6 6.8 3.6 3.8 4 4.2 4.4 4.6 4.8 5 5.2 5.4 5.6 5.8 Counts vs. Acquisition Time (min) Integration Peak List Peak Start RT End Height Area Area % 1 3.95 3.99 4.05 3312724 9296884 25.11 2 4 .05 4.08 4.12 334155 742651 2.01 3 4.38) 4.41 4.45 1267842 2267816 6.12 4 4.63 4.68 4.77| 16255627 37029905 100 5 6.71 6.73 6.76 400725 382818 1.03

Answer 1

The reaction was:

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As you can see, there are only two possible products, so two peaks should appear in the chromatogram. Since there are more than 2 peaks, we can assume that there is some sort of secondary product or contamination. We will consider these to be the least abundant, so we will work with the two biggest peaks in terms if area, which are peak 1 and peak 4. Amongst these, the addition of their relativa abundances is 125.11 (100 + 25.11). So we can calculate the fraction of each as:

$1:\: \frac{25.11}{125.11}=0.20$

20% of the least abundant isomer (which is the one most to the right, since it is the least substituted alkene.

$2:\: \frac{100}{125.11}=0.80$

80% of the most abundant isomer, which is 1-methylcyclohexene, the most substituted isomer, as predicted by Saytzeff's rule.