Consider the above chemical reaction. The total number of stereoisomers possible for Product 'P' is _____________.

Solution

Chemistry ma’am · 4 steps

Bromine adds anti, so cis-2-butene gives only the racemic pair, never the meso form.

1. Identify P. Br₂ in CCl₄ adds across the double bond of cis-2-butene. P is 2,3-dibromobutane, CH₃-CHBr-CHBr-CH₃, with two stereocentres at C2 and C3.
2. Anti addition. The first Br forms a cyclic bromonium ion on one face of the alkene. Br⁻ then attacks from the opposite face, so the two Br atoms end up anti to each other.
3. Cis plus anti. Anti addition to a cis alkene gives the threo pair, (2R,3R) and (2S,3S). Br⁻ opens the bromonium ion equally well at either carbon, so both enantiomers form in equal amounts as a racemic mixture. The meso (2R,3S) form would need anti addition to the trans alkene.
4. So. P exists as two stereoisomers, the (2R,3R) and (2S,3S) enantiomers, so the answer is 2.

The common slip: 3. Counting every stereoisomer of 2,3-dibromobutane, including meso, gives 3. Anti addition to the cis alkene never forms the meso compound.

The idea to keep. For Br₂ addition, remember the pattern: cis alkene with anti addition gives a racemic pair, and trans alkene with anti addition gives meso. Count only what the reaction actually forms.

Answer from the NTA final answer key. Solution by PrepWiser.