Student A conducted the acid-catalyzed dehydration of ethanol at 180°C and obtained a gaseous product. Student B conducted the same experiment at 140 °C and obtained a lowboiling 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?(Hint: Consider what other reactions might compete with an elimination under

these conditions.)

ethylene is the gas.

diethyl ether is the other.
Here is a site where you can read about them.
http://en.wikipedia.org/wiki/Ethanol#Reactions
The boards don't support drawing mechanisms or diagrams. I think I may be able to find a mechanism on the web. If so I'll post a link.

http://www.chemguide.co.uk/mechanisms/elim/dhethanol.html

To determine the products obtained by each student in the acid-catalyzed dehydration of ethanol, we need to consider the reaction conditions and possible competing reactions.

Student A conducted the experiment at 180°C and obtained a gaseous product. At this temperature, the acid-catalyzed dehydration of ethanol primarily leads to the formation of ethene (C2H4) as the major product. The reaction mechanism involves the protonation of ethanol followed by the loss of a water molecule, resulting in the formation of an ethyl carbocation. This carbocation then undergoes deprotonation to yield ethene. The reaction can be represented as follows:

H2CCH2OH + H+ → H2C=CH2 + H2O

Further details on the reaction mechanism can be found in organic chemistry textbooks or online resources.

On the other hand, Student B conducted the experiment at 140°C and obtained a low-boiling liquid product. At this lower temperature, the primary competing reaction to the acid-catalyzed dehydration of ethanol is the formation of diethyl ether (C2H5OC2H5). The reaction mechanism involves the protonation of ethanol followed by the formation of an ethyl carbocation. Instead of undergoing deprotonation to produce ethene, the ethyl carbocation can react with another ethanol molecule to form diethyl ether. The reaction can be represented as follows:

H2CCH2OH + H+ → H2C+CH2OH
H2C+CH2OH + H2CCH2OH → H3CCH2OCH2CH3

This reaction leads to the formation of a low-boiling liquid product, which is diethyl ether.

The difference in products obtained by Student A and Student B can be attributed to the different reaction temperatures. At higher temperatures (180°C), ethene is the major product of the acid-catalyzed dehydration of ethanol. However, at lower temperatures (140°C), diethyl ether becomes the major product due to the competing reaction between the carbocation and another ethanol molecule.

It is important to note that various factors such as reaction conditions, reaction time, concentration of reactants, and the choice of catalyst can influence the product distribution in a chemical reaction. Experimental control of these factors can be crucial in obtaining desired products selectively.