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 +HI⟶−‌‌‌‌ [B] But-1-ene +HBr⟶‌‌‌‌ [C] 3-Methylbut-1-ene +HI⟶‌‌‌‌ [D] 3- Phenylpropene +HBr (Peroxide) ⟶‌‌‌‌
In order to determine which of the reactions will produce an equimolar mixture of the d and I forms, we need to analyze each reaction and understand the concept of racemic mixtures (equal mixture of enantiomers). When a reaction forms a chiral center in a compound, and if there is no preference in the spatial arrangement of the substituents around that chiral center, it will result in a racemic mixture. Let's examine each reaction: 1. Reaction [A]: 2-Methylpropene +HI⟶ This reaction will form 2-iodo-2-methylpropane. Since the product does not have a chiral center, it will not give an equimolar mixture of enantiomers. 2. Reaction [B]: But-1-ene +HBr⟶ This reaction will form 2-bromobutane. Since the formation involves a planar carbocation intermediate and substitution can occur from either side with equal probability, it will produce a racemic mixture of 2bromobutane (having a chiral center at C2). 3. Reaction [C]: 3-Methylbut-1-ene +HI⟶ This reaction will form 2-iodo-3-methylbutane. Again, the formation of a planar carbocation intermediate and subsequent substitution will lead to a racemic mixture because the resultant product has a chiral center. 4. Reaction [D]: 3-Phenylpropene +HBr (Peroxide) ⟶ Reaction in the presence of peroxide will follow an anti-Markovnikov addition, resulting in the formation of 1 bromo-3-phenylpropane. This compound does not have a chiral center, hence it does not produce a racemic mixture. Based on the analysis, the reactions that result in an equimolar mixture of d and I forms (racemic mixture) are: Option C: B&C