In the reaction of ethene with H-Br, what role does ethene play based on acid-base definitions?

The reaction shows ethene ($\text{H}_2\text{C}=\text{CH}_2$) reacting with $\text{H}-\text{Br}$. The curved arrow indicates that the pi electrons from the double bond of ethene attack the hydrogen atom of $\text{H}-\text{Br}$. Simultaneously, the bond between hydrogen and bromine breaks, with the electrons going to bromine.

Let's analyze the role of ethene based on acid-base definitions:

• Brønsted-Lowry definition:

  • An acid is a proton ($\text{H}^+$) donor.
  • A base is a proton ($\text{H}^+$) acceptor. In this reaction, ethene accepts a proton from $\text{H}-\text{Br}$. Therefore, ethene acts as a Brønsted-Lowry base.

• Lewis definition:

  • An acid is an electron pair acceptor.
  • A base is an electron pair donor. Ethene uses its pi electrons (an electron pair) to form a new bond with the hydrogen atom. Therefore, ethene acts as a Lewis base.

Given the options:
A. Brønsted-Lowry acid (Incorrect, ethene accepts a proton, it doesn't donate one)
B. Brønsted-Lowry base (Correct, ethene accepts a proton)
C. Lewis acid (Incorrect, ethene donates an electron pair, it doesn't accept one)
D. none of the above

Since ethene accepts a proton, it is correctly described as a Brønsted-Lowry base.

The final answer is $\boxed{\text{B}}$