Question

Would you please give me step by step instruction on the following.

What is the major product of the reaction of 1 mol of propyne with the Br₂ CH₂Cl_2.Does the information "1 mol" make the problem different than if we just said Br2. The text book shows the answer as the Br ending up on the 1st C. but that's the least substituted so I 'm unclear as to why it's there versus C # 2. You have already answered that the 1st C is the most substituted. it looks to me like the least substituted. Also, the internet is showing both 1 Br & 2 Br's attached please tell which is right.

Thank you

Answer 1

Reaction of 1 mol of propyne with Br2 in CH2Cl2 gives, 1,1,2,2-tetrabromoprapane as the major product. The reaction consumes 2 mols of Br2 per mole of propyne.

If instead only 1 mole of Br2 was employed, we would get, 1,2-dibromopropene as the major product.

Addition of Br2 does not show Markovnikov's addition rule. 1 Br atom gets added to each side of alkyne bond. So 2Br atom gets added on each side by the two step addition of 2 mol Br2 to propyne. The first step is mechanism is formation of bromonium ion intermediate of propyne, which gets opened up by Br- as nucelophile. The 1,2-dibromopropene formed again reacts with another mol of Br2 to form another bromonium ion which is opened by Br- anion to the tetrabromopropane product.