# Effect of isotopes on H-Bonding

The following question was asked in one of the assignments my teacher has given.

Acetone ($$\ce{Me2CO^16}$$) on treatment with $$\ce{H2O^18}$$ gives a mixture of $$\ce{Me2CO^16}$$ and $$\ce{Me2CO^18}$$, the latter being in slight excess. This may be explained by

(A) hydration of acetone is an equilibrium process

(B) $$\ce{C – O^18}$$ bond is slightly stronger than $$\ce{C – O^16}$$

(C) $$\ce{Me2CO^18}$$ forms stronger hydrogen bonds than $$\ce{Me2CO^16}$$ with water

(D)hydration of acetone is irreversible

I have doubt about option (C). According to me, (C) should be correct, but it is given incorrect. Why the $$\ce{^18O...H}$$ is weaker than $$\ce{^16O...H}$$ bonding?

• Without thinking further about it, the question actually doesn't state what you have deduced. Just because the option is marked false, it doesn't mean that the reversion of the statement would be correct. – Martin - マーチン Sep 12 '20 at 9:11
• Is the reason given as B? Notice they ask for which option is the explanation and not the validity of the statements. – Safdar Sep 12 '20 at 9:21
• Can any conclusions really be drawn since the ratio of acetone to water isn't specified? – MaxW Sep 12 '20 at 11:08
• @MaxW I think OP's point is to just compare the strength of $\ce{^18O...H}$ bond vs $\ce{^16O...H}$ bond, which is a good question. – Aniruddha Deb Sep 12 '20 at 16:11
• @AniruddhaDeb But $\ce{^18O...H}$ is not weaker than the $\ce{^16O...H}$ bond or atleast you can't conclude that from the fact that that is not the answer. Seems like more of a lack of understanding what the question asked for (which is ambiguous enough). – Safdar Sep 12 '20 at 16:27

$$\ce{(CH3)2C^16O + H2^18O <=> (CH3)2C^18O + H2^16O }$$
The equilibrium constant of this reaction will depend on how the bond strengths of the $$\ce{O-H}$$ bond in water and the $$\ce{O=C}$$ bond in acetone is affected by isotope exchange. The isotope-dependent strength of hydrogen bonds might be a secondary effect, as well.