what will form when ethile alchole is denydrared
Answers
During the dehydration of ethyl alcohol in presence of sulphuric acid, the alcohol group from one carbon and hydrogen group from other carbon is eliminated and the double bond is formed between the carbons forming an alkene.
Answer:
Ethanol can be dehydrated to give ethene by heating it with an excess of concentrated sulphuric acid at about 170°C. ... The acids aren't written into the equation because they serve as catalysts.
Explanation:
he dehydration of ethanol
The facts
Ethanol can be dehydrated to give ethene by heating it with an excess of concentrated sulphuric acid at about 170°C.
Concentrated phosphoric(V) acid, H3PO4, can be used instead.
The acids aren't written into the equation because they serve as catalysts. If you like, you could write, for example, "conc H2SO4" over the top of the arrow.
The mechanism
A problem!
You will find two versions of the mechanism for the dehydration of primary alcohols on the web and in various textbooks.
One of these is exactly the same as the mechanism for the reaction involving propan-2-ol and other secondary or tertiary alcohols (known technically as an E1 mechanism), but the other is different (known as an E2 mechanism).
The more reliable sources give the E2 mechanism for the dehydration of primary alcohols including ethanol. I am going to treat this as the "correct" version, and have added a note as to why I think is better following the mechanism.
The correct version in full
If you have read the page on the dehydration of propan-2-ol, you will know that it involves the formation of a carbocation (a carbonium ion).
If ethanol used the same mechanism, you would get a primary carbocation formed, CH3CH2+, but this is much less stable than a secondary or tertiary carbocation.
That would lead to a very high activation energy for the reaction.
The alternative mechanism avoids the formation of the carbocation, and so avoids the high activation energy.
We are going to discuss the mechanism using sulphuric acid.
In the first stage, one of the lone pairs of electrons on the oxygen picks up a hydrogen ion from the sulphuric acid. The alcohol is said to be protonated. That is exactly the same as happens with propan-2-ol and the other secondary and tertiary alcohols.
In the mechanism we have already looked at with propan-2-ol, the next thing to happen was loss of water to form a carbocation, followed by removal of a hydrogen ion from the carbocation and the formation of a double bond.
In this case, instead of happening in two separate steps, this all happens at the same time in one smooth operation. By doing that, you avoid the formation of an unstable primary carbocation for primary alcohols.
No simplified version of this!
I am not giving a simplified version of this mechanism just in terms of hydrogen ions. If you don't show something removing the hydrogen ion from the protonated alcohol, you are really missing an important feature of the reaction.
Why do I think this is most probably the correct version?
There is experimental evidence to support it. While I was researching this, I came across two academic papers (here and here). These suggest that there is experimental evidence for both propan-1-ol and ethanol dehydrating using a mechanism which doesn't involve a carbocation as the intermediate. I haven't been able to find any experimental evidence for a dehydration mechanism involving carbocations in the case of primary alcohols.
Reliable sources quote the mechanism above. For example, the first page of a Google search for dehydration ethanol mechanism includes two university sources from LibreTexts from the University of California, and from the University of Calgary.
Interestingly, if you read sources giving this mechanism, they will usually explain why it is different for primary alcohols rather than secondary or tertiary ones - in terms of the stability of the carbocations. Sources quoting the other mechanism tend to quote it without comment.