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Chemical compounds that contain halogens (group 17 in the Periodic Table) as part of their chemical structure. This tag should only be applied where the halide group is one of the main focuses of the question.

2
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I think the term you want is "interhalogen compounds", not halogen halides. Personally, I have never heard of them being prepared via the reaction of the hydrogen halides. Most common synthetic …
answered Jan 12 '16 by orthocresol
10
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Is methanol even a strong enough reagent for such a reaction? Methanol is a very poor nucleophile, so the rate of the $\mathrm{S_N2}$ pathway will be extremely slow. This is much likelier to be a …
answered Nov 13 '16 by orthocresol
4
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dipole in the molecule $I$ is the first ionisation energy of the molecule If you plug all the numbers in for the hydrogen halides $\ce{HX}$, you will find that the magnitude of the dispersion forces … quantitatively. In fact, this trend is hardly unique to the hydrogen halides. The hydrides of the Group 15 and Group 16 elements behave exactly the same way: $$\begin{array}{|cc|cc|} \hline \text{Species …
answered Jan 6 '16 by orthocresol
19
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The answer simply has to do with the accessibility of the high +6 oxidation state. In Cr, the 3d electrons drop in energy extremely rapidly as you remove electrons. So, it is much harder to remove mu …
answered Apr 4 '16 by orthocresol
4
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As a disclaimer, it is worth noting that the reaction of an alcohol with thionyl chloride can proceed via multiple different mechanistic pathways (Wikipedia has a pretty good overview), depending on t …
answered Aug 25 '17 by orthocresol
31
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I'll present a LCAO-MO argument. But first let's debunk a myth. $\ce{SF6}$ has "hypervalent" sulfur and the 3d orbitals on sulfur participate in bonding No. This is not true. I would close one e …
answered Apr 20 '16 by orthocresol
41
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TL;DR Xenon hexafluoride has a fluxional structure in the gas phase, with multiple rapidly interconverting conformers. The three most important conformers have $C_\mathrm{3v}$, $O_\mathrm{h}$, and $C_ …
answered Sep 8 '17 by orthocresol
1
vote
Without looking up data, I would firstly assume that the $\ce{-CHBrCH3}$ group acts like any other alkyl group, and that path I would be preferred (by Markovnikov's rule). Bromine is electron-withdraw …
answered Sep 14 '16 by orthocresol
5
votes
It does appear. The graphs you're looking at online probably just dumbed it down. Here's some proof if you like, a compound synthesised by yours truly, with molecular formula $\ce{C7H7BrN2O2}$. I cli …
answered Nov 25 '16 by orthocresol
3
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This is tricky and one way to think about it that I can offer is to treat $\ce{ClO3-}$ as a combination of $\ce{Cl^5+}$ and $\ce{O^2-}$. Since the problem is that the starting material is undergoing d …
answered Apr 29 '16 by orthocresol
9
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I am reluctant to post this as an answer because it is not really backed up by much research, but since nobody has said anything yet, here's my theory: It is probably to do with the fact that the …
answered Apr 12 '16 by orthocresol
3
votes
With primary alkyl iodides the (simplified) mechanism is likely similar to a $\mathrm{S_N2}$ process, but with attack of $\ce{t-Bu}$ at iodine instead of carbon. Note that this is not necessarily t …
answered Aug 2 '18 by orthocresol
11
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Rule P-45.2.2 in the 2013 IUPAC Recommendations reads: The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefix …
answered Feb 16 '17 by orthocresol
5
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In general, iodides stabilise lower oxidation states and fluorides stabilise higher oxidation states, e.g. $\ce{CuF2}$ versus $\ce{CuI}$. This can be explained with some thermodynamics. Consider $$\c …
answered Sep 30 '18 by orthocresol
4
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In comparing the reactivity of fluorine or iodine, there are a couple of factors to consider. Part of it is the strength of the $\ce{X–X}$ bond, which is a barrier that tends to discourage them from r …
answered Apr 21 '18 by orthocresol

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