Question

Show how hex-1-yne might be converted tod. 1,1,2,2-tetrabromohexane.

Accepted Answer

Identify the starting material and the target product. The starting material is hex-1-yne, which is an alkyne, and the target product is 1,1,2,2-tetrabromohexane, which is a fully saturated alkane with four bromine atoms attached to the first and second carbons. Recognize that the reaction involves the addition of bromine (Br₂) to the alkyne. This process occurs in two steps: (1) addition of bromine to the triple bond to form a dibromoalkene, and (2) further addition of bromine to the double bond to form the tetrabromoalkane. In the first step, treat hex-1-yne with one equivalent of bromine (Br₂) in an inert solvent like CCl₄. This will result in the formation of 1,2-dibromohex-1-ene. The bromine atoms will add across the triple bond in an anti-addition manner, forming a trans-dibromoalkene. In the second step, treat the 1,2-dibromohex-1-ene with another equivalent of bromine (Br₂) in the same solvent. This will result in the addition of two more bromine atoms across the double bond, forming 1,1,2,2-tetrabromohexane. The addition will again proceed via anti-addition. Finally, confirm the structure of the product. The resulting compound, 1,1,2,2-tetrabromohexane, will have two bromine atoms attached to the first carbon and two bromine atoms attached to the second carbon, with the rest of the molecule being a fully saturated alkane chain.

Show how hex-1-yne might be converted to
d. 1,1,2,2-tetrabromohexane.

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Key Concepts

Alkyne Reactivity

Halogenation

Stereochemistry