What is the major product formed by heating 2-ethoxypentane with $\ce{HI}$?


The options are:

Option 1: (2S)-2-methylbutan-1-ol

Option 2: 1-iodobutane

According to me the correct answer should be option-1 because 2-methylbutyl carbocation is more stable than ethyl carbocation as there are more +I groups attached in 2-methylbutyl carbocation. But according to my book the second option is correct. Is my answer correct? If not please explain


Strictly speaking, none of those answers are correct, provided your stem is right. The reaction will start with protonation of the ether to give you the following species:

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And the iodide nucleophile can attack either side via $S_{N}2$ mechanism. Since the left side is more accessible, it'll probably attack there first giving you:

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Then, since we're still in the presence of HI and it is a strong acid, it protonates the alcohol and converts it into a corresponding alkyl iodide via either $S_N2$ like in the example above or $S_N1$:

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A general rule of thumb here for the future reference is that in a hydrolysis of an ether with a strong hydrohalogen acid (HI, HBr), you get a mix of two corresponding alkyl halides with halogen where the oxygen used to be.

  • $\begingroup$ can you please explain how the left side is more accessible ? Is it because of less steric hindrance? $\endgroup$ Jul 4 '16 at 15:01
  • $\begingroup$ precisely so. Remember, that things constantly move, groups rotate around single bonds, so on the left you have a relatively small ethyl, while on the right you have a huge propeller of a group that will knock off anything that comes (or attempts to do so) close to attack that secondary carbon with the oxygen. $\endgroup$ Jul 4 '16 at 20:37

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