Which of the following products of the acid-catalyzed elimination of 2-pentanol is
the Hoffmann elimination product?
A. trans-2-butene
B. cis-2-butene
C. 1-butene
what is the reaction between 2-methyl-2-butene and hydrogen iodide?
To determine the Hoffmann elimination product from the acid-catalyzed elimination reaction of 2-pentanol, we need to understand the mechanism involved.
Step 1: Protonation of the alcohol
In the presence of an acid catalyst, such as sulfuric acid (H2SO4), the alcohol group of 2-pentanol is protonated, forming an oxonium ion.
Step 2: Loss of water
The oxonium ion loses a water molecule to form a carbocation. This is the rate-determining step of the reaction.
Step 3: Deprotonation
In the presence of a strong base, such as hydroxide ion (OH-), the carbocation deprotonates to form an alkene.
Now, let's analyze the given options:
A. trans-2-butene: In this case, the double bond appears between the second and third carbon atoms. It is not the Hoffmann elimination product because trans-2-butene has a more substituted (higher degree) alkene.
B. cis-2-butene: In this case, the double bond appears between the second and third carbon atoms. Like trans-2-butene, it is not the Hoffmann elimination product because cis-2-butene is more substituted (higher degree) alkene.
C. 1-butene: This option is the Hoffmann elimination product because it is the least substituted alkene among the given options. 1-butene has a lower degree of substitution compared to trans-2-butene and cis-2-butene.
Therefore, the correct answer is C. 1-butene.