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Here's my attempt:

the electrophile is a tertiary carbocation, so that rules out sn2. I believe the nucleophile is a rather weak base since carboxylic acids are stable. However, it is polar, plus the temperature is low. Thus it seems likely that it proceeds by SN1? So would "I" be the correct answer?

  • $\begingroup$ After the formation of the carbocation, IV also seems to be a likely candidate to me.So I would go with I and IV (major product's'). $\endgroup$ Nov 10, 2015 at 4:59
  • $\begingroup$ I think more (IV) via SN1 at room temperature, with some (I) via E1 elimination. Heat to get more (I) than (IV). $\endgroup$
    – MaxW
    Nov 10, 2015 at 7:40
  • 3
    $\begingroup$ SN1 does not lead to product 1, which is an elimination product. Just because it proceeds via a carbocation does not mean it is SN1. $\endgroup$ Nov 10, 2015 at 8:45

1 Answer 1


The commenters are correct; this reaction, based on the steric hindrance of the tertiary alkyl halide and the propionic acid, I would expect a mixture of SN1 and E1, and minimal if any second-order reactions to the mechanism.

Specifically, the propionic acid would slowly approach the bromide substituent, and the acidic carboxyl proton (pKa ~ 5) would begin to associate with the bromide, which is a good leaving group. This slow formation of a tertiary carbocation (stabilized by hyperconjugation of the central empty p orbital through the C-H sigma bonds) then leads to the propionate binding to the carbocation with no rearrangements occurring.

Even though this reaction is at room temperature, E1 should not be ruled out, as there is no reason why a carboxylic acid cannot grab a proton from the 9 available ones on the alkyl halide. Due to the necessity for the proper orientation for beta-elimination though, it is plausible that E1 is less likely.

Regardless, I would project both SN1 and E1 to be the most likely reactions to occur. However, if you were to pick only one, I would pick SN1 based on the reaction temperature (and call E1 a "secondary" major product if that makes sense).

  • 1
    $\begingroup$ E1 does not require the antiperiplanar conformation. That is for E2. $\endgroup$ Nov 10, 2015 at 9:26
  • $\begingroup$ Alright; adjusted that part. $\endgroup$
    – timaeus222
    Nov 10, 2015 at 9:28
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