What is the mechanism for iodoform test given by 2-aminoalkanes?

Question

I found 2-aminoalkanes in the list of compounds that give positive iodoform test. I was surprised as they have no keto methyl or methyl carbinol group.

Can someone help with the mechanism?

Answer 1

There is a comment elsewhere in this post saying:

"(The mechanism for iodoform test given by 2-Aminoalkanes might be) similar to the iodoform reaction of secondary alcohols. It is oxidized to imine, maybe then similar procedure."

I won't necessarily agree with that argument. To have a positive iodoform test, it is necessary to acitivate the methyl group (making its three α-hydrogens acidic). That means $\ce{-CH(NH2)\!-}$ of $\ce{RCH(NH2)CH3}$ should be oxidized to $\ce{-C(=NH)\!-}$ group by the available oxidizer, $\ce{I2/NaOH}$. Even if this oxidation happens, without transforming it to the corresponding keto group, those $\alpha$-hydrogens are not active enough to remove by $\ce{NaOH}$.

The reality is it is not easy to oxidize amines compared to alcohol in this situation. In their 1969 publication (Ref.1), E. J. Corey and coworkers have said it the best:

In contrast to the generation of carbonyl compounds by oxidation of primary and secondary alcohols, which is used with great frequency in synthesis and for which many selective reagents are known, the efficient oxidation of primary amines to carbonyl compounds has only been realized by the fore mentioned process.

One of the oxidation method used in the past has been $\ce{NBS/base}$ followed by adjusting the resultant solution to $\mathrm{pH} \ 2$ at $\pu{0 ^\circ C}$ to convert a primary amine to corresponding ketone (Ref.2):

At least, this example shows that the conditions needed to make a ketone from the corresponding imine is to make the solution acidify. Even if the conditions used in the iodoform test $(\ce{I2/NaOH})$ could have been able to oxidize the 2-aminoalkane to the corresponding imine (I doubt it though; vide supra), you still needed to acidify the imine to convert it to the corresponding ketone. Therefore, this point alone, it is safe to conclude that this iodoform reaction is not happening with any of 2-aminoalkanes.

Nonetheless, I am not able to find any reference showing iodine (a milder oxidizer) or $\ce{NaIO}$ $(\ce{I2/NaOH})$ oxidizing amine in literature. Even, $\ce{ICl}$ gives an adduct with amines such as 1-aminotetraline or 1-amino-1-phenylethane, which is used to prepare lactones from corresponding alkene carboxylic acid (Ref.3). Thus, I can say that my reasoning is good enough to convince that this reaction is not possible.

References:

1. Elias J. Corey and Kazuo Achiwa, "Oxidation of primary amines to ketones," J. Am. Chem. Soc. 1969, 91(6), 1429–1432 (DOI: https://doi.org/10.1021/ja01034a027).
2. Elias J. Corey, Niels H. Andersen, Robert M. Carlson, Joachim Paust, Edwin Vedejs, Isidoros Vlattas, and Rudolph E. K. Winter, "Total synthesis of prostaglandins. Synthesis of the pure $dl$-$\mathrm{E_1},$ -$\mathrm{F_{1\alpha}},$ -$\mathrm{F_{1\beta}},$ -$\mathrm{A_1},$ and -$\mathrm{B_1}$ hormones," J. Am. Chem. Soc. 1968, 90(12), 3245–3247 (DOI: https://doi.org/10.1021/ja01014a053).
3. Jürgen Haas, Stewart Bissmire, and Thomas Wirth, "Iodine Monochloride–Amine Complexes: An Experimental and Computational Approach to New Chiral Electrophiles," Chemistry, A European Journal 2005, 11(19), 5777-5785 (DOI: https://doi.org/10.1002/chem.200500507).

Answer 2

Addendum to Mathew Mahindaratne's answer (though not appropriate to edit his post for this purpose). As Rishi pointed out, let's try to find a new path for the reaction to occur.

$\ce{RNH2 + NaOH <=> ROH + NaNH2}$

That is a secondary amine forming $2^\circ$ alcohol through $\mathrm{S_N}$2 path. But $\ce{NH2^-}$ is a very poor leaving group as compared to $\ce{OH-}$, so the equilibrium lies largely to the left and hence the above reaction practically does not occur.

Source

Another path is the reaction below, but even that cannot occur because amines such as isopropyl amine are very weak acids, with $\mathrm{p}K_\mathrm{a}$ around 35, so you will need a much stronger base than $\ce{OH-}$ to deprotonate them.

$\ce{RNH2 + OH- -> RNH- + H2O }$

Source