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In an isonitrile (isocyanide) group, the carbon atom has a negative charge (I know that this is due to the lone pair of electrons) instead of the nitrogen atom, but in my point of view, this seems a little unstable. So, why doesn't the isocyanide group convert into a cyanide group in which carbon doesn't have a negative charge (carbon valency fully satisfied)?

$$\ce{R-N+\bond{3}C- -> R-C\bond{3}N}$$

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    $\begingroup$ isocyanide has two resonance structures of which one has double bond between C and N and also doesn't involve charge separation. $\endgroup$
    – JM97
    Commented Sep 19, 2016 at 5:44
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    $\begingroup$ @JM97 That resonance structure is formally a carbene, I would prefer to think of an isonitrile as a stabilised carbene rather than think of a carbene as a stabilised isonitrile. $\endgroup$ Commented Sep 19, 2016 at 6:15
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    $\begingroup$ No dupe, but related: chemistry.stackexchange.com/questions/45255/… $\endgroup$ Commented Sep 19, 2016 at 9:15
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    $\begingroup$ Not a dupe, but it does answer the question, I suppose. $\endgroup$ Commented Sep 19, 2016 at 10:22

1 Answer 1

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Isonitriles will, given the correct conditions, spontaneously transform into nitriles. The reaction mechanism is very interesting; it is basically a cationic [1,2]-migration of the $\ce{R}$ group under retention with a three-membered cyclic transition state. To reach this transition state, the $\ce{C-N}$ bond must be broken and the alkyl residue must form a cation-π interaction with the $\ce{C#N}$ triple bond. This is a non-classical carbonium cation, as shown in scheme 1.

Rearrangement mechanism
Scheme 1: Mechanism of the [1,2]-migration and sp³-π interaction in the intermediate carbonium ion.

The reaction rates were extensively studied by Meier et al.[1] They note that the typical migration preference of $\text{tertiary} > \text{secondary}>\text{primary} > \text{methyl}$, indicative of cationic carbenium intermediates and observed e.g. in Criegee, Beckmann and Lossen rearrangements, does not apply here. Instead, a migratory aptitude similar to the Curtius or pinacol rearrangements is observed. Another interesting observation is ‘the almost complete lack of a bridgehead effect on the [reaction] rates’, further indicating that a planar carbenium transition state cannot be part of the mechanism and that bond angles and lengths will probably remain unchanged en route to the transition state.[1] When phenylic isonitriles react, next to no substituent effect is observed, indicating that the cation is not delocalised into the phenyl ring.

All of this evidence indicates towards the reaction being ‘properly prototypical for a sigmatropic 1,2-rearrangement via a nonpolarized, tight, hypervalent three-centered transition state.’[1] The authors go on to state that this is in perfect agreement with ab initio calculations, but do not provide an appropriate source.

Isonitriles were known in the 19th century; I am not able to find a definite first description but A. Gautier noted their reactions with acids in 1867.[2] In 1868, he goes on to be the apparant first to describe an isonitrile–nitrile rearrangement.[3] Weith, however, was probably the first to actually observe the reaction as such; he notes:[4]

Zu dem Ende wurde nach der  H o f m a n n ’schen Methode dargestelltes und durch Destillation möglichst von Anilin befreites Cyanphenyl 2—3 Stunden lang im geschlossenen Rohr auf 200–220° erhitzt. Nach dem Erkalten fand sich eine beträchtliche Menge dunkel gefärbter nadlicher Krystalle vor, der Röhreninhalt besass nicht mehr den Geruch des Cyanphenyls, dagegen den des Benzonitrils. [sic! to all apparant errors]

Thus, phenylisonitrile prepared by the Hofmann method and purified from aniline by distillation was heated in a sealed tube to 200–220 °C for 2–3 h. After cooling, a significant amount of dark-coloured, needle-like crystals were found; the tube’s content no longer smelt like phenylisonitril but like benzonitrile.

Therefore, for the rearrangement to take place temperatures above $150~\mathrm{^\circ C}$ appear to be required — which also explains why isonitriles are kinetically inert enough to be isolated under standard conditions and even with 19th century methods. (Gautier used $\ce{AgCN}$ to synthesise them from iodoalkanes.)[3] An activation energy as high as this makes sense, since, as noted above, the $\ce{C-N}$ bond must be broked and a significant rearrangement must take place for the carbonium ion to form.

Some isonitriles, namely those which upon dissociation form very stabilised cations such as benzylisonitrile, can also react according to a different path that is basically an $\mathrm{S_N1}$ rearrangement or can be considered in a chain-like structure. This, however, is a rather special case as Meier et al. note.[1]


References:

[1]: M. Meier, B. Müller, C. Rüchardt, J. Org. Chem. 1987, 52, 648. DOI: 10.1021/jo00380a028.

[2]: A. Gautier, Liebigs Ann. Chem. 1867, 142, 289. DOI: 10.1002/jlac.18671420304.

[3]: A. Gautier, Liebigs Ann. Chem. 1868, 146, 124. DOI: 10.1002/jlac.18681460108.

[4]: W. Weith Ber. Dtsch. Chem. Ges. 1873, 6, 210. DOI: 10.1002/cber.18730060180.

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