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$\ce{N}$ is more electronegative than $\ce{Cl}$ (albeit by a very small difference) but it should ideally dissociate as $\ce{Cl+}$ and $\ce{-NO}$ (similar to the dissociation of $\ce{Cl}$ in $\ce{HOCl}$). Moreover, due to the stronger bonding between $\ce{N}$ and $\ce{O, N}$ should also be pulling electrons even more tightly to itself due to the very electronegative $\ce{ O}$ by its side.

I was doing a problem on the electrophilic addition of $\ce{NOCl}$ on alkene and ended by getting the wrong product (according to the book) because I took $\ce{Cl+}$ and $\ce{-NO}$ in the mechanism at the very beginning. Could someone explain?

I've attached pictures of what I have done and what the book has done too. Please correct me if I am wrong.

What I did:
enter image description here
What the book did:
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    $\begingroup$ Note: Electronegativity of Cl >$\approx$ Electronegativity of N. $\endgroup$
    – MollyCooL
    Apr 5, 2018 at 18:26
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    $\begingroup$ @MollyCooL Could you further explain why? Doesn't N form hydrogen bonds but not Cl? Shouldn't that be considered as a push for N to be more electronegative. $\endgroup$
    – Neha Malcom
    Apr 5, 2018 at 18:30
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    $\begingroup$ The regiochemistry of Markovnikov addition of ClN=O to propene gives the answer. BTW the chloronitroso cmpd. your book gives as a answer likely tautomerizes to the oxime of 2-chloropropanal. $\endgroup$
    – user55119
    Apr 5, 2018 at 18:39
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    $\begingroup$ Of course N supports H-bonding but Cl doesn’t. That in no way means N is more electronegative. @NehaMalcom $\endgroup$
    – MollyCooL
    Apr 6, 2018 at 0:03
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    $\begingroup$ Regarding your drawing: a positive charge "(+)" never attacks a double bond or lone pair. Always the other way around. $\endgroup$
    – pH13 - Yet another Philipp
    Apr 6, 2018 at 8:22

2 Answers 2

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Firstly, the correct comparison to make to determine which is more likely of $\ce{NO+}+\ce{Cl-}$ and $\ce{NO-}+\ce{Cl+}$ is the ionization potential vs. the electron affinity of the the two species. The electronegativities are correlated with these other metrics, but it is important to emphasize that those are really the important things.

Relatedly, nitrosonium $\ce{NO+}$ is stabilized by resonance and so this is a considerably stabilized cation (i.e. IP not too high). On the other hand, the IP of $\ce{Cl}$ is huge while the EA is small; it would much rather get an electron than lose one. The hyponitrite anion $\ce{NO-}$ actually usually dimerizes for reasons outside the bounds of the question. So $\ce{NO+}+\ce{Cl-}$ is way lower in energy.

In general, a red-flag should go up when you see halogens with positive charges. They do happen, but very rarely, especially at the level of chemistry in an intro organic course.

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    $\begingroup$ Actually there can be halogens with positive charges. But, only if these are pi-bonded to something else so as to keep their full octet, like a contributing structure to a $\ce{CR2Cl+}$ ion. $\endgroup$
    – Oscar Lanzi
    Apr 6, 2018 at 9:44
  • $\begingroup$ You don't even need to get that exotic. There are $\sigma$-bonded examples. ICl (iodine monochloride) is a bench stable source of $\ce{I+}$ and is a superb reagent for turning Aryl-TMS to Aryl-I. I stand by that halogens with positive charges should set off red-flags, especially at intro organic (which is the level of most Chem SE users). You will be served best by treating such invocations with caution. $\endgroup$
    – levineds
    Apr 6, 2018 at 18:46
  • $\begingroup$ Not a positive formal charge, though, which is what I thought you meant. Nucleophilic attack on ICl could be Sn2. In any event "short of an octet" is a bigger problem for halogens, even fluorine, than a positive charge per se. $\endgroup$
    – Oscar Lanzi
    Apr 6, 2018 at 18:49
  • $\begingroup$ The original question involved invoking fully dissociated $\ce{NO-}$ and $\ce{Cl+}$ so that it what I'm talking about with respect to charges. ICl is principally an electrophilic reagent (although I'm told it can add to double bonds, I've only ever used it for electrophilic aromatic substitution). $\endgroup$
    – levineds
    Apr 6, 2018 at 18:59
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NOCl cannot be considered as NaOCl for the calculation of oxidation staes owing to the following : (i)Na is highly electropostive and it exhibits +1 state always (ii) O is neither peroxide or superoxide in NOCl ; it is normal oxide and it is in -2 state (iii)O exhibits positive oxidation state only with F (iv)N is next to O in electronegativity and it exhibits -3 to +5 states (v)NOCl adds to unsymmterical alkenes obeying Markovnikov's rule i.e., NO+ attacks double bonded carbon containing more hydrogens to create a secondary carbocation which is attacked by chloride ion to give the major product (vi)So the calculation is x - 2 = +1 and x =+3 In nitrosyl chloride, the oxidation state of N is +3

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