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

question

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?

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  • $\begingroup$ Tertiary alcohols are usually more reactive than primary alcools. $\endgroup$ – Maurice Mar 29 '20 at 10:18
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    $\begingroup$ 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 '20 at 11:27
  • $\begingroup$ t-BuOH is more acidic than primary alcohols in the gas phase but not in solution. $\endgroup$ – user55119 Mar 29 '20 at 15:44
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    $\begingroup$ Per my answer, a more likely explanation relates to an issue with reagents and/or procedure. $\endgroup$ – AJKOER Mar 29 '20 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 '20 at 16:45
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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.

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    $\begingroup$ I can't see how it answers OP question? $\endgroup$ – Zenix Mar 29 '20 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 '20 at 15:51
  • $\begingroup$ 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 '20 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 '20 at 16:54
  • $\begingroup$ On further thought, the ethanol employed (the source of the Ether per a radical mechanism) may not be absolute, the source of water. $\endgroup$ – AJKOER Mar 29 '20 at 16:59

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