Write a halogenation reaction of the following alkyne with Br 2 and name the product formed.

Write a halogenation reaction of the following alkyne with b r 2, and name the product form. So since our alkyne has 2 pi bonds, we're gonna react it with 2 moles of Br2. We're gonna say here that the first mole would add, adding 2 BR's, 1 to each triple bonded carbon, and we'd sacrifice the PI bond. Then the second mole would come in, attach again to get rid of the last PI bond. So we'd be left with only single bonds within our structure. Now it's important to see that there's a carbon here and a carbon here. Now if we drew this out, this part stays the same, and then we're connected to 1, which is this carbon here, 2, which is this carbon here, and that carbon is still connected to the rest of this structure here. So here it'd be connected to that carbon, which is still connected to this methyl group here, then it's connected to that. Okay. So it's important that you draw this as best as you can. Now our goal is to find the longest carbon chain here, and by looking at this we can see that the longest carbon chain is this portion right here. And we start numbering it from the end closest to a substituent, which would be this end. 1, 2, 3. Now, here we have to add our Bromines. Each triple bonded carbon is no longer triple bond, and each one has 2 Bromines. 4, 5, 6, 7, 8, 9. Nine carbons means that the end of the name is no name. And what do we have here? We have bromos because of the bromines, and we have 1, 2, 3 methyl groups. Alphabetical order for substituents, bromo b comes before methyl m. So we'd say here, nonin is the end of the name, We'd say here that bromo comes first, so they're on carbons 45, so this would be 4, 4, 5, 5, 4 of them which is tetra, bromo, remember we look at the b for alphabetical order, not the numerical prefix of tetra. And we have methyl groups on 2, 3, and 6, so 2, 3, 6 Tri Methyl no name. This would be our final product and its name.

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22. Organic Chemistry

Halogenation Reactions

General Chemistry

22. Organic Chemistry

Halogenation Reactions

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Multiple Choice

Write a halogenation reaction of the following alkyne with Br₂ and name the product formed.

Product structure of the halogenation reaction of an alkyne with Br2.

Illustration of the halogenation reaction mechanism for alkynes.

Diagram showing the addition of Br2 to an alkyne, forming a dibrominated product.

Visual representation of the final product from the halogenation of an alkyne with Br2.

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Verified step by step guidance

Identify the alkyne structure in the image. The alkyne is a hydrocarbon with a carbon-carbon triple bond.

Understand that halogenation involves the addition of halogen atoms (in this case, bromine, Br2) across the triple bond of the alkyne.

Recognize that the reaction with Br2 will first convert the alkyne into a dibromoalkene, where two bromine atoms add across the triple bond, reducing it to a double bond.

If excess Br2 is present, the dibromoalkene can further react to form a tetrabromoalkane, where two additional bromine atoms add across the double bond, resulting in a single bond.

Name the final product based on the number of carbon atoms and the position of the bromine atoms. The name will reflect the saturated nature of the final product and the presence of bromine substituents.