1
$\begingroup$

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

$\endgroup$

1 Answer 1

4
$\begingroup$

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:

enter image description here

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.

$\endgroup$
2
  • $\begingroup$ can you please explain how the left side is more accessible ? Is it because of less steric hindrance? $\endgroup$ Jul 4, 2016 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, 2016 at 20:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.