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Reactivity of different alcohols with NaNH2 and CH3I

Chemistry Asked on October 5, 2021

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?

One Answer

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.

Answered by AJKOER on October 5, 2021

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