# Determining Acidity in organic compounds [closed]

Among the following compounds,the strongest acid is:

(a) $\ce{C2H2}$
(b) $\ce{C6H6}$
(c) $\ce{C2H6}$
(d) $\ce{CH3OH}$

My doubt is how do we see acidity in such cases.....arhenius or lewis definition or we just see acidic hydrogen in such cases

• Arrhenius definition. – Weijun Zhou Jan 23 '18 at 7:16
• Arrhenius definition is about water, so it is no good here. Use Brønsted instead. – Ivan Neretin Jan 23 '18 at 7:20
• @IvanNeretin but who will they be donting hydrogen to? – Hydrous Caperilla Jan 23 '18 at 7:36
• Who cares? Their relative acidic strength does not depend on the nature of the base you use. – Ivan Neretin Jan 23 '18 at 7:52
• Yeah, that's about right, except that you don't have ethanol here. – Ivan Neretin Jan 23 '18 at 8:13

To judge the acidity in any case, just check out the stability of the conjugate base of the acids given. More the stability of the conjugate base, more is the acidic character in the acid.

So according to what I said, the $$\ce{CH3OH}$$ is most acidic compound among all of them. This can be easily seen if we compare the stabilities of their conjugate bases.

For (a): If ethyne loses a proton, its conjugate base will be $$\ce{HC#C-}$$. The negative charge resides on a $$\mathrm{sp}$$ hybridized carbon.

For (b): If benzene loses a proton, in its conjugate base, the negative charge resides on a $$\mathrm{sp^2}$$ hybridized carbon.

For (c): If ethane loses a proton, its conjugate base will be $$\ce{H3C-CH2-}$$. The negative charge resides on a $$\mathrm{sp^3}$$ hybridized carbon.

For (d): If methanol loses a proton, its conjugate base will be $$\ce{H3C-O-}$$. The negative charge resides on an oxygen atom(no need to take care of the hybridization).

We can see that the point of difference among all the conj bases above is the atom on which the negative charge resides. As we know greater the electronegativity of the element holding the negative charge, more is its stability. Also more the $$\mathrm{s}$$-character of the atom, more is its electronegativity.

So the order of electronegativities of the atoms holding the -ve charge is: $$\ce{O}>\mathrm{sp}>\mathrm{sp^2}>\mathrm{sp^3}$$.

Now the order of the stability of conjugate base is: $$\mathrm{(d)>(c)>(b)>(a)}$$

And hence the order of their acidic character is: $$\mathrm{(d)>(c)>(b)>(a)}$$

The acidity of a substance is quantified by the energy it takes to remove a proton from it, $$RT \ln(K_\mathrm{a}) = RT \mathrm{p}K_\mathrm{a}$$. Thus to find which of these substances is the most acidic you would look up the individual $$\mathrm{p}K_\mathrm{a}$$ values and list the compounds in ascending order of $$\mathrm{p}K_\mathrm{a}$$. The compound with the smallest $$\mathrm{p}K_\mathrm{a}$$ is the 'strongest' acid and I put that in quotes because these are all very weak acids indeed - all weaker than water. But compare the structure of water, $$\ce{HOH}$$, to $$\ce{CH3OH}$$. Pretty similar, no? If this suggests to you that the acidity of $$\ce{CH3OH}$$ may be close to that of water then your surmise would be correct. The other compounds listed are much, much weaker acids.