Chlorohydrin formation of E-2-pentene (3)

In the previous pages we have derived two products from the reaction of chlorine water with 2-pentene, the chlorohydrin formation.
These were two different products, both of them chiral (A and B below).
As you may expect from a reaction between achiral compounds, the mirror images of these products are formed as well. They result from the attack by chlorine at the opposite side of the pentene molecule, followed by addition of water to either C2 or C3.
Alltogether we may expect two (Cl top/bottom) times two (water to C2/C3) is four products.

All four are displayed below. Rotate the molecules in such a way that the carbon skeletons all have the same orientation, and check that:

A B C D
Questions:
5. Which pairs are enantiomeric pairs?
  1. A and B ; C and D
  2. A and C ; B and D
  3. only A and B
  4. only A and C

6. Are they formed in equal amounts?
  1. A and B are formed in equal amounts, as are C and D
  2. A and C are formed in equal amounts, as are B and D
  3. only A and B are formed in equal amounts
  4. only A and C are formed in equal amounts
  5. none are formed in equal amounts

7. Are all four compounds formed in equal amounts?
  1. Yes, all in equal amounts
  2. No, A/C and B/D are different compounds, not necessarily formed in equal amounts
  3. No, A/B and C/D are different compounds, not necessarily formed in equal amounts
  4. No, only A and C are formed in equal amounts

Next we see how the products simplify if:

  • the substrate is symmetrical (E-2-butene) and
  • a second chlorine attacks, instead of water.

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