Identify the final product (B) in the following reaction:


I'm not sure how $\ce{NaNH2}$ and $\ce{CH3I}$ will react with the different types of alcohol present here. According to my solution,ether was formed in place of $\ce{CH2OH}$ in the end, whereas the answer says that $\ce{CH2OH}$ group will remain unaffected. What am I doing wrong? Can anyone explain to me the full mechanism?

  • $\begingroup$ Tertiary alcohols are usually more reactive than primary alcools. $\endgroup$
    – Maurice
    Mar 29, 2020 at 10:18
  • 3
    $\begingroup$ But the pKa of primary alcohols is lower see evans.rc.fas.harvard.edu/pdf/evans_pKa_table.pdf. In real life I would not expect much selectivity in this reaction. $\endgroup$
    – Waylander
    Mar 29, 2020 at 11:27
  • $\begingroup$ t-BuOH is more acidic than primary alcohols in the gas phase but not in solution. $\endgroup$
    – user55119
    Mar 29, 2020 at 15:44
  • 1
    $\begingroup$ Per my answer, a more likely explanation relates to an issue with reagents and/or procedure. $\endgroup$
    – AJKOER
    Mar 29, 2020 at 15:58
  • $\begingroup$ I had since edited my answer, and in particular, suggest checking the quality of the NaNH2 given its sensitivity to moisture/water/air. $\endgroup$
    – AJKOER
    Mar 29, 2020 at 16:45

1 Answer 1


I suspect, possibly just to a minor extent, that you have also managed to produce an expected photolysis product here (see below).

Interestingly, the photo sensitive methyl iodide, in the presence of a suitable catalyst ($\ce{CO2, Cu,}$..) with lab light (rich in UV) may have commenced in a radical reaction chain in the presence of ethanol:

$$\ce{CH3I ->[$h \nu$] ^.CH3 + I^.}$$

$$\ce{C2H5OH + I^. -> HI + C2H5O^.}$$

$$\ce{C2H5O^. + C2H5O^. -> (C2H5)2O2}$$

$$\ce{(C2H5)2O2 + ^.CH3 -> (C2H5)2O + CH3O^.}$$

....(the reaction chain continues)

where, I have stopped at the formation of Ether, albeit possibly created in very small amounts but detectable by its smell, nevertheless.

Repeat the experiment controlling for light/catalyst and see if you can still smell the Ether.

My advice, any time working with a photoreactive iodides, be mindful of light exposure.

[EDIT] The implication of my analysis, check your product yields, reagents and procedure, as a minor photo-induced side product may not (but still could) be connected to your results. In particular, I would start with the $\ce{NaNH2}$ as per Wikipedia:

$\ce{NaNH2 + H2O → NH3 + NaOH}$

$\ce{4 NaNH2 + 7 O2 → 2 Na2O + 4 NO2 + 4 H2O}$

which suggests that either exposure to moisture/water/air can degrade the compound.

  • 5
    $\begingroup$ I can't see how it answers OP question? $\endgroup$
    – Zenix
    Mar 29, 2020 at 15:16
  • $\begingroup$ I have edited my comments. My answer clearly presents an understandable analysis and further notes it may NOT be even related to the poor results. So, look elsewhere and do remove the light factor. $\endgroup$
    – AJKOER
    Mar 29, 2020 at 15:51
  • 1
    $\begingroup$ No step is photochemically-induced free radical reaction. First there will be acid-base reaction, followed by SN reaction and finally oxidation.. These type of questions aren't to know possible issues with reactants or procedure, rather what final product wil be and why? $\endgroup$
    – Zenix
    Mar 29, 2020 at 16:33
  • $\begingroup$ Zenix: My edited answer points to NaNH2 (which readily degrades). Your comment is definitely valid if it is even apparent that the reaction is at all proceeding. But with only a minor photolysis side product being created, this suggests to me that the reaction may, indeed, not be proceeding! $\endgroup$
    – AJKOER
    Mar 29, 2020 at 16:54
  • 2
    $\begingroup$ The issue is that the reaction hasn't actually been done in reality and is just a hypothetical paper-pen reaction.. So, it is ideal conditions outside the reaction.. $\endgroup$ Aug 29, 2020 at 14:41

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.