Dat-Destroyer Orgo pg. 25 #70

[Mrhyde]

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Dat-Destroyer Orgo pg. 25 #70 (Current Edition)

How can E be the wrong answer if we are looking at a primary carbocation if OH where to become H2O and leave,

The explanations thinking is IF the OH group where to become H2O and leave then a rearrangement would happen BUT this is wrong because rearrangements only happen on secondary carbocations because it cant happen on a primary since you would form a primary carbocation which cant happen thus on a primary like in this case we would get an SN2 to avoid rearrangements so choice E has to be correct ?

Basically : For primary Alcohols the SN2 pathway Dominates

 

[rdub]

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[DAT Destroyer]

Dat-Destroyer Orgo pg. 25 #70 (Current Edition)

How can E be the wrong answer if we are looking at a primary carbocation if OH where to become H2O and leave,

The explanations thinking is IF the OH group where to become H2O and leave then a rearrangement would happen BUT this is wrong because rearrangements only happen on secondary carbocations because it cant happen on a primary since you would form a primary carbocation which cant happen thus on a primary like in this case we would get an SN2 to avoid rearrangements so choice E has to be correct ?

Basically : For primary Alcohols the SN2 pathway Dominates

Not a chance. Although a primary carbocation is very unstable, it is seen occasionally. For example, it can be generated in magic acid, FSO3H·SbF5. This is the solvent that was used by Nobel Prize scientist George Olah. to study carbocations. However, this point is moot. The question asked for a unlikely rearrangement. Choice C would not occur since a higher energy specie would be generated. Do not overthink this question.

BTW......Your statement that rearrangements only happen on secondary carbocations is also untrue.Tertiary systems can rearrange if greater stability can result. For example, if ring strain can be relieved. This is seen often in advanced organic chemistry. If you would like to explore this enchanting topic in greater depth, the Physical Organic text by Felix Carroll does a wonderful job.

Hope this help.

Dr. Romano

 

[Mrhyde]

Not a chance. Although a primary carbocation is very unstable, it is seen occasionally. For example, it can be generated in magic acid, FSO3H·SbF5. This is the solvent that was used by Nobel Prize scientist George Olah. to study carbocations. However, this point is moot. The question asked for a unlikely rearrangement. Choice C would not occur since a higher energy specie would be generated. Do not overthink this question.

BTW......Your statement that rearrangements only happen on secondary carbocations is also untrue.Tertiary systems can rearrange if greater stability can result. For example, if ring strain can be relieved. This is seen often in advanced organic chemistry. If you would like to explore this enchanting topic in greater depth, the Physical Organic text by Felix Carroll does a wonderful job.

Hope this help.

Dr. Romano

Thank you 🙂 the ring strain makes sense for a tertiary, I never thought of that.

When you say:

"The question asked for a unlikely rearrangement. " ---> What do you mean by that? the question just asked which reagents would be best?

"Choice C would not occur since a higher energy specie would be generated. Do not overthink this question." ---- > Your right, I am over thinking it, but I still don't understand why for example in these cases the H-X does SN2 when the scenario looks similar to your question- Specifically number 2 in this picture:

From:
http://www.masterorganicchemistry.com/2015/02/27/making-alkyl-halides-from-alcohols/


-Thank you for always helping me !

 

[tommyngu]

I don't have access to this question, but from what I understand if the question is asking for the best product then the Cl in the tertiary position would be the best answer. Yes, primary substrates favor Sn2, but that doesn't mean only Sn2 products are created.
Sn1 would also occur with that unstable primary carbocation being taken care of via a hydride shift... which actually would result in a mixture of products with the Cl in the primary and tertiary position, but obviously the product with the Cl in the tertiary would be more stable.
Furthermore, this reaction contains water, so the solvent would be polar protic which favors Sn1. The nucleophile is also weak and so favors Sn1, too.

 

[Mrhyde]

I don't have access to this question, but from what I understand if the question is asking for the best product then the Cl in the tertiary position would be the best answer. Yes, primary substrates favor Sn2, but that doesn't mean only Sn2 products are created.
Sn1 would also occur with that unstable primary carbocation being taken care of via a hydride shift... which actually would result in a mixture of products with the Cl in the primary and tertiary position, but obviously the product with the Cl in the tertiary would be more stable.
Furthermore, this reaction contains water, so the solvent would be polar protic which favors Sn1. The nucleophile is also weak and so favors Sn1, too.

Hi, thanks for replying .... then how can you explain the explanation from the picture above (the 2nd example) . Its basically saying that is the major product

 

[tommyngu]

Oh, sorry about that.
From what I've gathered so far in general primary substrates will go through the Sn2 mechanism.
Secondary and tertiary substrates favor Sn1 and also allow for rearrangements via hydride or alkyl shifts.
Since the substrate is isobutanol in this question, you would think it would go through Sn2 and produce isobutyl chloride since it's a primary alcohol. But I'm thinking that because of the tertiary C2 on the substrate which is directly adjacent to the C the nucleophile would hit (in other words the beta carbon), it rules out Sn2. So you would get either 2-chlorobutane from a concerted methyl shift or t-butyl chloride if hydride shifts can be concerted (I'm not sure of this) via Sn1.
Or... maybe in this case where the beta carbon is hindered, the primary carbocation does form but is quickly replaced by a hydrogen via a hydride shift giving you the t-butyl chloride.
similar to this:

What is the answer anyway?

As for the picture showing Sn2 being the dominant mechanism for butanol + HBr, the alkyl portion is unbranched. So you have an unhindered primary substrate which would favor Sn2 (assuming concerted hydride shifts are not possible).
And I think the question only asks for the best product which could mean the most stable and not necessarily the dominant product. I can't interpret the question as I don't have the question.

 

[FutureDent20]

Hi, thanks for replying .... then how can you explain the explanation from the picture above (the 2nd example) . Its basically saying that is the major product

I brought this up before and Dr. Romano explained it well. It's an exception that you need to commit to memory. I see that you're doing O-chem odyssey. If you recall it mentions if you see branching be suspicious of a possible shift. So in this case even though primary alcohol proceed via SN2 when I see branching with primary alcohol I think about it as if its doing SN1 where OH gets protonated and leaves forming a primary (least stable) carbocation follow by hydride shift to form a more substituted (more stable) carbocation (in this case tertiary). finally Cl attacks and you get both enantiomers (R and S) 2-chloro-2-methylbutane.