JEE MainChemistryHydrocarbons
Propyne is subjected to the following sequence of reactions: (i) NaNH ₂ (ii) CH ₃ I (iii) Na in liquid NH ₃ (iv) Br ₂ in CCl ₄ The final major product obtained is:
Options
- AA meso compound
- BA racemic mixture
- CAn optically active pure enantiomer
- D1,2-dibromobutane
Correct answer
A. A meso compound
Step-by-step solution
Steps (i) & (ii): Propyne reacts with NaNH ₂ to form a propynide ion, which then acts as a nucleophile to attack CH ₃ I via an S_N2 mechanism, yielding 2-butyne. Step (iii): Birch reduction of 2-butyne using Na in liquid NH ₃ stereoselectively produces trans-2-butene. Step (iv): The addition of Br ₂ in CCl ₄ to an alkene proceeds via an anti-addition mechanism. The anti-addition of halogens to a symmetrical trans-alkene yields a meso compound (meso-2,3-dibromobutane). Answer: A meso compound