Which of the following reactions will not produce a racemic product?

To decide whether a reaction gives a racemic mixture, check two points: (i) does the reaction create at least one chiral centre, and (ii) if it does, are the two possible enantiomers formed in equal amounts? A 50 : 50 pair of enantiomers is a racemic mixture.

Below each option is examined on the above lines.

Option A: $$CH_3CH(OH)CH_3 \xrightarrow{SOCl_2} CH_3CH(Cl)CH_3$$

The conversion of a secondary alcohol to the chloride with $$SOCl_2$$ (no pyridine) follows the $$S_{N}i$$ mechanism.
 • A chlorosulfite intermediate is formed, which is achiral and can rotate freely about the C-O bond.
 • When $$Cl^-$$ makes the internal (front-side) attack, an equal amount of each enantiomer of 2-chlorobutane is produced.
Therefore the product is a racemic mixture.

Option B: $$cis\!-\!CH_3CH\!=\!CHCH_3 \xrightarrow{Br_2/CCl_4} BrCH_2CHBrCH_2CH_3$$

Addition of $$Br_2$$ to an alkene proceeds by anti addition through a cyclic bromonium ion.
For the cis alkene the two new stereocentres (C-2 and C-3) end up with opposite configurations in the two products: $$RS$$ and $$SR$$. These are mirror images formed in 1 : 1 ratio, so the product is racemic.

Option C: $$CH_3CH_2CH=CH_2 \xrightarrow{HBr} CH_3CH_2CH(Br)CH_3$$

The electrophilic addition is Markovnikov: H⁺ adds to the terminal carbon giving a planar secondary carbocation $$CH_3CH_2\!-\!\overset{+}{C}HCH_3$$. The nucleophile $$Br^-$$ attacks this planar centre equally from either face, producing ($$R$$)- and ($$S$$)-2-bromobutane in equal amounts. Hence a racemic mixture is obtained.

Option D: $$trans\!-\!CH_3CH\!=\!CHCH_3 \xrightarrow{Br_2/CCl_4} CH_3CH(Br)CH(Br)CH_3$$

Again the reaction goes through anti addition of bromine. For the trans alkene, anti addition puts the two bromine atoms on opposite sides of the former double bond in such a way that the product, 2,3-dibromobutane, possesses an internal mirror plane:

$$\displaystyle CH_3-CH(Br)-CH(Br)-CH_3$$

This single compound has the absolute configuration $$2R,3S$$ (or equivalently $$2S,3R$$); because the molecule contains a plane of symmetry, it is a meso form and is achiral. Only one achiral product is formed, so no racemic mixture appears.

Thus, the only reaction that does not produce a racemic product is:

Option D which is: addition of $$Br_2$$ to trans-2-butene giving meso-2,3-dibromobutane.