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I already know that the answer is III. I'm just looking for further explanation on how the others compare to each other, mainly IV vs V.
I know that iodine is less electronegative than chlorine, making I the least acidic.
Only having one chlorine atom, while the rest have two, puts II next in line.
III is the most acidic, due to the two chlorine atoms being on the same carbon, and closest to the sulfur group.

My question comes in for the comparison of IV and V. While V has two chlorine atoms on the same carbon, they are both 3 carbons away from the sulfur group. Meanwhile, IV has one chlorine atom closer to the sulfur group, but the other isn't on the same carbon atom.
Which one of these is more effected by induction? Proximity to the sulfur group (IV) or being grouped together on a single carbon (V)?

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  • $\begingroup$ Why do you think structure I is least acidic? Consider the following facts: In water, HI is more acidic than HCl, and 2-Iodoethanol is more acidic than 2-Chloroethanol. How does (or does it?) Chlorine being more electronegative than Iodine make the chloro-substituted compounds more acidic? $\endgroup$
    – Zachr
    Mar 1, 2017 at 23:15
  • $\begingroup$ I'm still pretty new to this so I may be wrong. But I was under the impression that polarizability is only the stronger influence on pKa in binary acids. So HCl is more acidic than HF, because Cl is the larger and more polarizable atom (the larger the atom, the easier it is to stablize the negative charge). However, in more complex structures where the halogens aren't directly attached to a hydrogen, inductive effects are the stronger influence. So, for example, when comparing fluoracetic acid to chloroacetic acid, the former has a pKa of ~2.59, while the latter of ~2.86. $\endgroup$
    – Rapid99
    Mar 2, 2017 at 3:07
  • $\begingroup$ Fluoroacetic acid is more acidic than Chloroacetic acid in water, but not in DMSO, or the gas phase where there're no solvent effects. Now, in terms of chemical bonding, a more electronegative substituent will usually make the bond homolytically much stronger. A calculation of C-H bond enthalpies in Fluoroaceticacid vs Iodoacetic acid would reveal, C-H bond enthalpies in the former are larger than the latter. Yet, the acidity values are different. It is clearly more complex than "polarizability" and "electronegativity", it could be important to think about how these affect acidity. $\endgroup$
    – Zachr
    Mar 2, 2017 at 7:10

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Inductive effects decrease strongly with increasing distance. Thus, the further away an inductively acting group is, the lower its influence will be. For practical intents and purposes, the otherwise very comprehensive Brodwell $\mathrm{p}K_\mathrm{a}$ database does not even list any β-halo compounds; only the difference between α-hydrogen, carbon and halogen is typically given.

Thus, we can expect compound IV to have practically the same acidity as compound II, and that compound V’s acidity is comparable to the unhalogenated pentan-4-thiol. Considering that all chlorines are very far away from the acidic proton, the effects would be smaller still.

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