[Neoskatedog]

Hey,

Sorry about the title, I didn't know what to name it as. So, here is a question from my textbook that I'm having trouble with.

Select the compound in each of the following pairs that will be converted to the corresponding alkyl bromide more rapidly on being treated with hydrogen bromide. Explain the reason for your choice.

1-methylcyclopentanol or trans-2-methylcyclopentanol

The answer is 1-methylcyclopentanol. And I know why. It's because the first molecule has the OH group attached to a tertiary carbon, which bromide needs to be attached to. HBR splits, and the H attaches to the OH group creating an H2O. Then the H2O departs from the ring and then BR replaces it. I think this is correct?

My question is, how do I know from the IUPAC name of the molecule that the OH is attached to a tertiary carbon (the same carbon that methane is attached to). And why is the OH group not attached to the same carbon as the methane group in trans-2-methylcyclopentanol.

Sorry if my post is confusing; I am not the greatest writer.

[Archer]

The IUPAC name numbers the position the methyl group relative to the -OH in cyclohexanol.

So 1-methyl means that the OH and CH₃ are on the same carbon 2-methyl means that they are on adjacent carbons.

[spirochete]

The answer is 1-methylcyclopentanol. And I know why. It's because the first molecule has the OH group attached to a tertiary carbon, which bromide needs to be attached to. HBR splits, and the H attaches to the OH group creating an H2O. Then the H2O departs from the ring and then BR replaces it. I think this is correct?

You are essentially correct in your description of the mechanism but you haven't quite explained why this mechanism favors conversion of the tertiary alcohol over conversion of a secondary alcohol.

[Neoskatedog]

The IUPAC name numbers the position the methyl group relative to the -OH in cyclohexanol.

So 1-methyl means that the OH and CH₃ are on the same carbon 2-methyl means that they are on adjacent carbons.

Oh, wow that is so simple. Thank you for the clear explanation.

You are essentially correct in your description of the mechanism but you haven't quite explained why this mechanism favors conversion of the tertiary alcohol over conversion of a secondary alcohol.

Hmm.. Well I think tertiary alcohols are favored in this mechanism because the intermediate R₃C radical is more stable than R₂CH radical and so on. However, I am unsure as to why Br almost exclusively bonds with tertiary carbons while Cl will bond with primary's and secondary's. My book writes about how the activation energy for a Br to substitute with an alcohol on a primary carbon is much higher than a Cl doing the same, but I don't know why there is such a large difference in their activation energies.

Edit: and to explain why R₃C radical is more stable than R₂CH radical, it is because the former has more of a spread out distribution for the radical because the R₃ core atoms pull on it. And also because the H's in the R₃ atoms share the charge. This is called hyper conjugation I believe.

Double Edit: I just realized that radicals are only an intermediate with Br₂ and Cl₂ reactions, not HBr and HCl. But I assume that the same idea still applies; The more R groups a carbocation has attached to it, the more stable it is. Therefore, tertiary carbons react more often than secondary and primary carbons. However, I still do not know why Br exclusively reacts with tertiary carbocations while Cl can react with secondaries.

[spirochete]

The IUPAC name numbers the position the methyl group relative to the -OH in cyclohexanol.

So 1-methyl means that the OH and CH₃ are on the same carbon 2-methyl means that they are on adjacent carbons.

Oh, wow that is so simple. Thank you for the clear explanation.

You are essentially correct in your description of the mechanism but you haven't quite explained why this mechanism favors conversion of the tertiary alcohol over conversion of a secondary alcohol.

Edit: and to explain why R₃C radical is more stable than R₂CH radical, it is because the former has more of a spread out distribution for the radical because the R₃ core atoms pull on it. And also because the H's in the R₃ atoms share the charge. This is called hyper conjugation I believe.

Double Edit: I just realized that radicals are only an intermediate with Br₂ and Cl₂ reactions, not HBr and HCl. But I assume that the same idea still applies; The more R groups a carbocation has attached to it, the more stable it is. Therefore, tertiary carbons react more often than secondary and primary carbons. However, I still do not know why Br exclusively reacts with tertiary carbocations while Cl can react with secondaries.

Correct. Mixing a primary alcohol with HBr would require that the mechanism go by Sn2, which would be rough because Br- is not the greatest nucleophile.

Are you sure HBr reacts exclusively with Tertiary? Tertiary should definitely go faster than secondary because the mechanism is Sn1 for tertiary/secondary carbons. I'd think that HBr might even be more reactive (thus working with secondary) because it is a stronger acid than HCl and the nucleophile is not even included in the rate limiting step.

[Neoskatedog]

My understanding for how Br works exclusively with tertiary carbons might only be true for a Br₂ reaction. In the book it explains how those reactions are endothermic with secondary carbons, yet exothermic with chlorides. I suppose that this does not imply it goes the same with an HBr reaction. I am sorry if I have confused you, my knowledge of this material is very minute.

Going back to my original question "Why does Br₂ react exclusively with tertiary carbons (for Br₂)" I suppose I should just accept the heat of enthalpy/endothermic explanation and be done with it. I have a long ways to go to prepare for my upcoming exam and I should not be bothered learning the 1%.

Thank you both for helping me. I greatly appreciate it.

[spirochete]

My understanding for how Br works exclusively with tertiary carbons might only be true for a Br₂ reaction. In the book it explains how those reactions are endothermic with secondary carbons, yet exothermic with chlorides. I suppose that this does not imply it goes the same with an HBr reaction. I am sorry if I have confused you, my knowledge of this material is very minute.

Going back to my original question "Why does Br₂ react exclusively with tertiary carbons (for Br₂)" I suppose I should just accept the heat of enthalpy/endothermic explanation and be done with it. I have a long ways to go to prepare for my upcoming exam and I should not be bothered learning the 1%.

Thank you both for helping me. I greatly appreciate it.

Yes I think you were mixing up a couple of different issues when you spoke before. Br2 is less reactive than Cl2 in radical additions because the C-Br and H-Br bonds formed are less strong the C-Cl and H-Cl bonds. This lack of reactivity essentially makes Br2 more selective, causing it to mainly substitute at the more substituted hydrogen in an alkane. If one of the H's is tertiary that will usually be the one that gets substituted. Cl2 is less selective and will replace all kinds of H's with less discrimination. This is explained in detail with an analysis of the Hammond postulate.

Edit: I should add that a great many of your questions may be answered by thoroughly studying the relevant mechanisms and thinking about them logically.