3. (15 pts) Give the structure of the products that you would expect from the reaction of 2 butyne with:
a. One molar equivalent of $\mathrm{Br} 2$
b. One molar equivalent of $\mathrm{HBr}$
c. Two molar equivalents of $\mathrm{HBr}$

Question

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Identify the reaction for part (a)**: The reaction of 2-butyne with one molar equivalent of $\mathrm{Br_2}$ is a halogenation reaction, specifically an addition reaction where the bromine adds across the triple bond. ‖ ‖ 2 ⋮ **Predict the product for part (a)**: Since only one equivalent of $\mathrm{Br_2}$ is used, the reaction will result in a dibromoalkene. The product will be 2,3-dibromo-2-butene due to anti-addition of bromine across the triple bond. ‖ ∻a Answer∻ ⚹ The product is 2,3-dibromo-2-butene. ⚹ ∻Key Concept∻ ⚹ Halogenation of alkynes ⚹ ∻Explanation∻ ⚹ Alkynes react with halogens to form dihaloalkenes through anti-addition. ⚹ ‖ 1 ⋮ **Identify the reaction for part (b)**: The reaction of 2-butyne with one molar equivalent of $\mathrm{HBr}$ is a hydrohalogenation reaction, where the hydrogen and bromine add across the triple bond. ‖ ‖ 2 ⋮ **Predict the product for part (b)**: The addition of $\mathrm{HBr}$ to an alkyne follows Markovnikov's rule, where the hydrogen adds to the carbon with more hydrogens, and the bromine adds to the carbon with fewer hydrogens. The product will be 2-bromo-2-butene. ‖ ∻b Answer∻ ⚹ The product is 2-bromo-2-butene. ⚹ ∻Key Concept∻ ⚹ Hydrohalogenation of alkynes ⚹ ∻Explanation∻ ⚹ Hydrohalogenation adds hydrogen and halogen across the triple bond following Markovnikov's rule. ⚹ ‖ 1 ⋮ **Identify the reaction for part (c)**: The reaction of 2-butyne with two molar equivalents of $\mathrm{HBr}$ is a sequential hydrohalogenation reaction, where two molecules of $\mathrm{HBr}$ add across the triple bond. ‖ ‖ 2 ⋮ **Predict the product for part (c)**: The first equivalent of $\mathrm{HBr}$ adds following Markovnikov's rule to give 2-bromo-2-butene. The second equivalent of $\mathrm{HBr}$ adds to the double bond, again following Markovnikov's rule, to give 2,2-dibromo-butane. ‖ ∻c Answer∻ ⚹ The product is 2,2-dibromo-butane. ⚹ ∻Key Concept∻ ⚹ Sequential hydrohalogenation of alkynes ⚹ ∻Explanation∻ ⚹ Two equivalents of $\mathrm{HBr}$ add to an alkyne to form a tetra-substituted alkane following Markovnikov's rule twice. ⚹◊What is the chemical equation for the reaction of 2-butyne with one molar equivalent of Br$_2$?⍭ Generate me a similar question◊

Sia