Question

The GC spectrum below is for the products of the dehydration of 2-butanol, Complete the table with the name of each product. (Hint: see the boiling point info in the manual handout). Calculate the % of each product formed in this reaction. 5) % yield Area Product Name Peak I30 Peak 2 180 Peak 3 100 The dehydration of 2-butanol in the experiment produced more than one alkene product. When multiple alkenes were formed in this reaction, they formed in unequal amounts. Did the experimental results match what theory would predict for the % yield of each product? Explain the reason for the different yields obtained. 6) 7) On a separate sheet of paper (to use the space you need), draw the stepwise dehydration mechanism for each the following compounds from this week's lab. (Correct curved arrows bonds, lone pairs & charges needed for full credit). a. 2-butanol b. 2-methyl-2 propanol Butene Isomer Boiling point (°C) -6.3 0.88 3.7 -6.9 1-Butene (E)-2-Butene (Z)-2-Butene 2-Methylpropene Table 10.1. Boiling points of butene isomers.

Answer 1

A mixture of gaseous alkene products are obtained on dehydration of 2-butanol, which are then separated and identified by Gas chromatography (GC). In GC, the component with a higher boiling point has a lower vapour presure and hence, the component spends more time in the gas phase resulting in higher Retention time (R). In short, component with a higher boiling point has a higher R, and vice versa As the table illustrates that 1-Butene has a lower boiling point followed by (E)-2-Butene, whereas (Z)-2- Butene has a higher boiling point, which may result in it's higher retention time As Peak 1 appears first, it has a lower R+and thus coresponds to the lower boiling point component which is 1-Butene. Next the Peak 2 corresponding to (E)-2-Butene pops up. Finally, the component with a higher Rt appears, which is (Z)-2-Butene. This peak is observed at last due to ts higher boiling point Area of each peak can be calculated by multiplying width and height of the corresponding peak. As the area of all the peaks are given. addition of the peak area is necessary to calculate the % yield Peak 1 - 30 mm2 Peak 2 180 mm2 Peak 3 100 mm2 Total area = Peak 1 Peak 2 + Peak 3 = (30 + 180 + 100) mm2-310 mm2 % of Peak 1-30/310 X 100 : 9.68% % of Peak 2 180310 X 100-58.06% % of Peak 3-100/310 X 100 :-32.26% TOTAL-100% The following table correlates the Rt, boiling point and the yield of the components obtained by GC analysis of dehydration of 2-butanol Boiling | R(as per | point (C) literature) Yield (%) Peak 1 Peak 2 Peak 3 0.75 0.20 0.20 6.3 0.88 3.7 9.68 58.06 32.26

Ans 6 As observed in the GC chromatogram of the 2-butanol, three peaks are obtained in different ratio. The smallest peak corresponds to 1-Butene, followed by (Z)-2-Butene. The largest peak corresponds to (E) 2-Butene. The observed result is correlated to the stability of the individual component 1-Butene has the smallest peak in GC chromatogram due to its less stability in the gas sample with an yield of 9.68% The trans isomer, (E)-2-Butene is the most stable and abundant component in the mixture and hence appears as the largest peak in the chromatogram with an yield of 58.06%. The cis isomer. (Z)-2-Butene is slightly less stable than the trans isomer and thus appears as a smaller peak than the (E)-2-Butene isomer (yield-32.26%) Ans. 7. (a) Dehydration of 2-Butanol It undergoes acid catalyzed E1 elimination reaction via formation of carbocation to yield various 1somers 1-Butene -H (E)-2-Butene (Z)-2-Butene (b) Dehydration of 2-methyl-2-propanol он -H 2-methylprop-1-ene