Given below are 4 reactions. Two of these reactions will give product which is an equimolar mixture of the d and 1 forms. Identify these 2 reactions. [A] 2- Methylpropene + $\mathrm{HI} \rightarrow$ --------- [B] But-1-ene + $\mathrm{HBr} \rightarrow$ ----------- [C] 3-Methylbut-1-ene + $\mathrm{HI} \rightarrow$ ----------- [D] 3- Phenylpropene + $\mathrm{HBr}$ (Peroxide) $\rightarrow$ -----------

For a racemic (equimolar d and l) mixture, the product must have a chiral center formed from an achiral reactant via a planar intermediate (carbocation).

[A] 2-Methylpropene + HI $\to$ Markovnikov addition gives tertiary carbocation $\to$ 2-iodo-2-methylpropane $CH_3 - C(I)(CH_3)_2$. Central carbon has three identical methyl groups $\to$ achiral. No racemic mixture.

[B] But-1-ene + HBr $\to$ Markovnikov addition gives secondary carbocation $\to$ 2-bromobutane $CH_3 - CHBr - CH_2CH_3$. C-2 is bonded to H, Br, $CH_3$, and $C_2H_5$ $\to$ chiral center. Attack from either face equally probable $\to$ racemic mixture.

[C] 3-Methylbut-1-ene + HI $\to$ Markovnikov addition gives secondary carbocation $CH_3 - CH^+ - CH(CH_3)_2$ $\to$ iodide attacks to give 2-iodo-3-methylbutane with chiral center at C-2 $\to$ racemic mixture.

[D] 3-Phenylpropene + HBr (peroxide) $\to$ anti-Markovnikov (radical) addition $\to$ Br adds to terminal carbon $\to$ 1-bromo-3-phenylpropane $C_6H_5 - CH_2 - CH_2 - CH_2Br$ $\to$ no chiral center $\to$ achiral.

Reactions giving racemic mixture: B and C. Answer: (1)