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

2-ethoxypentane

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

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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:

enter image description here

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 $S_N1$:

enter image description here

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.

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  • $\begingroup$ can you please explain how the left side is more accessible ? Is it because of less steric hindrance? $\endgroup$ – TheNewGenGamer 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. $\endgroup$ – ChemistryHelpCenter Jul 4 '16 at 20:37

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