1. How many mono-chlorinated products of 2-methyl butane will be formed upon chlorination in the presence of sunlight?
  2. Find the percentage of major product formed.

Six products were formed: two pairs of enantiomers and two achiral chloroalkanes.

Product List

Until here my answer was correct, confirmed by my teacher.

For the second part that is the percentage yield I had an issue with the enantiomers. I calculated the relative yield ratios by multiplying probability factor and rate factor.

For cases A and B, where enantiomers are being formed, I further multiplied $p_\mathrm{f} \times r_\mathrm{f}$ by half since there's an equal likelihood of each of the enantiomers to form. My teacher said this is wrong, as we do not multiply it by half.

So, my question is why?


1 Answer 1


I'll say that this is just a matter of convention, and not really worth worrying about. Your underlying method and concept is 100% correct. Your final question boils down to:

Are enantiomers separate chemical compounds?

To elaborate, if the question was:

Find the percentage of major product formed, counting the enantiomers separately.

Your approach would have without doubt been correct.

But since the question only says:

Find the percentage of major product formed.

It is difficult to ascertain what the question wants. Enantiomers themselves have different chemical reactions - so they may be interpreted as different chemical products. But, your teacher has interpreted them as a single compound, presuming that optical isomers are just one compound.

Again, I guess this is just convention, because I haven't been yet able to find references to favor either definition.

Finally, to satisfy both the parties involved, and write a foolproof answer that guarantees a complete score, you can frame your results like so:

  • Percentage of (R)-2-chloro-3-methylbutane = $x$
  • Percentage of (S)-2-chloro-3-methylbutane = $x$
  • Total percentage of 2-chloro-3-methylbutane = $2x$

where $x$ is your calculated value for a single enantiomer.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.