1. Student A conducted the dehydration of ethanol at 180 °C and obtained a gaseous
product. Student B conducted the same experiment at 140 °C and obtained a low-
boiling liquid. What were the products that each of the students made? Draw
mechanisms for the formation of these products. What could account for the
formation of two different products?
2. What would be the effect of decreasing the carrier gas flow rate, raising the column
temperature, have on the retention time?
3. What would be the effect of raising the column temperature and decreasing the
carrier gas flow rate on the ability to resolve two closely spaced peaks?
4. From your knowledge of the dehydration of tertiary alcohols, the alkene 2-
methylpropene should predominate in the product of the dehydration of what
alcohol?
1. Heating to 180 the product is ethylene gas. Heating to the lower T produces diethyl ether. We can't draw mechanisms on the boards.
2. Unclear as to what you mean. Obviously its a GC question. What do you think would happen?
3. This should follows from 2.
4. Check out the dehydration of t-butyl alcohol.
So for number 1 why does it form two products? Is it just because of the change in heat?
Alkene formation is a 1,2-elimination reaction. Ether formation is a nucleophilic substitution with the protonated alcohol as substrate and a second molecule of alcohol as nucleophile.
Can you explain why it forms diethyl ether?
1. To determine the products formed by each student and explain the difference, we need to understand the reaction and the conditions used.
a) Dehydration of ethanol at 180 °C: At this higher temperature, ethanol undergoes dehydration to form ethene gas (C2H4) as the product. The mechanism involves the removal of a water molecule from ethanol, resulting in the formation of a double bond. The reaction can be represented as follows:
CH3CH2OH (ethanol) --> CH2=CH2 (ethene) + H2O
b) Dehydration of ethanol at 140 °C: At a lower temperature, ethanol can undergo dehydration to form a low-boiling liquid called diethyl ether (C2H5OC2H5). The mechanism involves the removal of a water molecule from ethanol, forming an ether linkage between two ethanol molecules. The reaction can be represented as follows:
2CH3CH2OH (ethanol) --> C2H5OC2H5 (diethyl ether) + H2O
The difference in the products obtained by the two students can be explained by the difference in temperature used. At higher temperatures, the reaction favors the formation of ethene gas, while at lower temperatures, the reaction favors the formation of diethyl ether.
2. The effect of decreasing the carrier gas flow rate and raising the column temperature on the retention time in gas chromatography will vary. Generally, decreasing the carrier gas flow rate will increase the retention time, while raising the column temperature will decrease the retention time.
The carrier gas flow rate affects the rate at which the sample components move through the column. A lower flow rate slows down the movement of the components, resulting in a longer retention time. On the other hand, raising the column temperature increases the thermal energy of the components, facilitating their movement through the column, thus reducing the retention time.
3. Raising the column temperature and decreasing the carrier gas flow rate can have a positive effect on the ability to resolve two closely spaced peaks in a chromatographic separation.
Increasing the column temperature enhances the separation by providing more energy to the sample components, increasing their volatility, and improving their mobility through the column. This greater mobility enables better separation of closely spaced peaks.
Similarly, decreasing the carrier gas flow rate allows for a longer interaction time between the sample components and the stationary phase in the column. This prolonged interaction time can result in improved separation and resolution of closely spaced peaks.
4. The dehydration of tertiary alcohols typically leads to the preferential formation of the corresponding tertiary alkene. In this case, the dehydration of 2-methyl-2-propanol (t-butyl alcohol) would predominantly yield 2-methylpropene.
The mechanism for the dehydration of t-butyl alcohol involves the elimination of a water molecule from the tertiary alcohol, resulting in the formation of the alkene. The reaction can be represented as follows:
(CH3)3COH (t-butyl alcohol) --> (CH3)2C=CH2 (2-methylpropene) + H2O
The stability of the intermediate carbocation formed during the dehydration determines the selectivity for the formation of the major product. In the case of tertiary alcohols, the resonance stabilization of the resulting carbocation makes the reaction more favorable, leading to the preferential formation of the corresponding alkene.