I know that $\mathrm{H^b}$ is most acidic due to conjugation (resonance).
But I am confused as to how to compare $\mathrm{H^a}$ and $\mathrm{H^c}$. I think $\mathrm{H^a}$ should be more acidic than $\mathrm{H^c}$ due to hyperconjugation.
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4$\begingroup$ Have you been taught the relative acidity of alkanes, alkenes and alkynes? chemistry.stackexchange.com/questions/32341/… $\endgroup$– orthocresolCommented Jan 9, 2016 at 12:38
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$\begingroup$ @orthocresol so the correct decreasing order should be $\mathrm{H^b}$ , $\mathrm{H^c}$ and $\mathrm{H^a}$ $\endgroup$– user101522Commented Jan 9, 2016 at 15:53
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1$\begingroup$ Yes, according to this link the allylic proton (b) has a pKa ~36, the vinylic proton (c) ~43, and the sp3 proton (a) ~50. $\endgroup$– orthocresolCommented Jan 9, 2016 at 16:25
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5 Answers
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When comparing acidities of protons, we need to find out what the corresponding conjugate bases are.
The conjugate base upon removing $\ce{H_\mathrm{b}}$ is an allyl anion. There is (and should be) no question that it is the most stable carbanion of the three we are going to consider.
The conjugate base upon removing $\ce{H_\mathrm{a}}$ is an $\mathrm{sp^3}$ configured carbanion while the conjugate base upon removing $\ce{H_\mathrm{c}}$ is an $\mathrm{sp^2}$ configured carbanion. This is nice, because it allows us to directly compare the two using their hybridisation.
In general, lone pairs (and that explicitly includes those coming from anions) want to be in the lowest-lying atomic or molecular orbital possible. Since an s orbital is lower in energy than a p orbital from the same shell, lone pairs prefer s orbitals. If there is no pure s orbital available — as would be the case in practically all carbon-centred lone pairs — an orbital with the highest possible s character is preferred instead. If we look at the configuration $\mathrm{sp^3}$ we can say that it has $25~\%$ s character while $\mathrm{sp^2}$ has $33~\%$ s character. Since $33~\% > 25~\%$, the $\mathrm{sp^2}$ configured carbanion is more stable — which is also supported by corresponding $\mathrm{p}K_\mathrm{a}$ values.
The method outlined above is a very crude approximation building on a high amount of simplifications but gives us the correct result in practically all carbon-centred anion cases.
Thus, the acidity order is: $$\ce{H_\mathrm{b}} > \ce{H_\mathrm{c}} > \ce{H_\mathrm{a}}$$
Note: Ron noted in a comment above that the acidity of an $\mathrm{sp^2}$ hydrogen almost parallels that of an allylic hydrogen. So while most courses may teach this simple ordering, you should keep that in mind as additional information.
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To address the question of hyperconjugation, ${H^b}$ presents the classic example of hyperconjugation between the ${\sigma}$ bond and ${\pi^*}$. No hyperconjugation is possible between ${\pi^*}$ and the ${sp^2}$ orbital attached to ${H^c}$. However, some very weak interaction between ${\sigma}$ bond between $\ce{CH^a}$ and ${\pi^*}$ could lead to some weakening of the sigma bond and enhanced acidity.
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If we were to remove $\ce{H^c}$, we would get a structure similar to a vinyl cation. The vinyl cation is highly unstable and hence it would be more difficult to remove $\ce{H^c}$,i.e. $\ce{H^c}$ is less acidic than $\ce{H^a}$. Also hyperconjugation is greater for $\ce{H^a}$ than it is for $\ce{H^c}$.
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$\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax.For more information in general have a look at the help center. $\endgroup$– Martin - マーチン ♦Commented Feb 5, 2017 at 9:38
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1$\begingroup$ An $\mathrm{sp^2}$ $\ce{C-H}$ bond is more acidic than an $\mathrm{sp^3}$ $\ce{C-H}$ bond, hence Hc is more acidic than Ha. Allylic protons are a bit more acidic than olefinic protons but they are close, so the final order of increasing acidity would be Ha < Hc <= Hb $\endgroup$– ronCommented Feb 5, 2017 at 15:18
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Hb is the most acidic hydrogen.
When it is extracted a negative charge appears on the carbon attached to it, this negative charge can then undergo resonance with the adjacent double bond. This resonance leads to stabilization of the system. Therefore the compound would prefer losing Hb.
Hence Hb is the most acidic
When comparing between Ha and Hc, Ha would be the more acidic as when we remove Hc there is a vinylic carbanion that is being formed. And Vinylic Carbanions are very unstable. Thus extricating Hb is more feasible thus making it more acidic.
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1$\begingroup$ There is no confusion about Hb being the most acidic. The question is actually asking about which of the other two hydrogens (a and c) is more acidic. $\endgroup$– f''Commented Jul 24, 2016 at 16:32
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To check acidity see conjugate base by removing H+. So when you remove H+ from double bond you get a sp2 carbanion which is more unstable than sp3 carbanion. Now as weaker the conjugate base stronger is acid. hence hc>ha. Hope its clear.
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8$\begingroup$ You have it the wrong way round, the sp2 carbanion is more stable than the sp3 carbanion. $\endgroup$ Commented Jan 10, 2016 at 7:09
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$\begingroup$ Yes orthocresol is correct , you are wrong . $\endgroup$ Commented Jan 10, 2016 at 7:28
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1$\begingroup$ but if we go by conjugate acid base theory the weaker conjugate base means strong acid, isnt it? Can you please explain.. $\endgroup$ Commented Jan 10, 2016 at 11:18