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question

The answer given is (D) but I think (B) should be the answer as the concerned benzylic carbon in option (D) is an unstable tetrahedral complex and it should break. Besides, the medium is acidic so it should catalyse this process.

In generalising, my question is: are lactols stable?

Edit: I found online that there is an equilibrium between lactols and corresponding hydroxyaldehyde. So, why lactol is favoured more in this case?

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  • $\begingroup$ If there's something wrong with the question then please suggest edits. $\endgroup$
    – Satya
    Commented May 5, 2021 at 5:17
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    $\begingroup$ What are tetrahedral complexes and why must they be unstable? I heard of no such law. Then again, you might point out that hemiacetals should be prone to hydrolysis, which is generally true, but not when the alcohol and aldehyde are both contained in one molecule. $\endgroup$
    – Ivan Neretin
    Commented May 5, 2021 at 12:04
  • $\begingroup$ That is a term one of my teachers uses. I checked the meaning online and it does not match with the meaning implied. By unstable tetrahedral complex I meant that a carbon on which there's a group which has leaving tendency and -OH which can kick it out. I do not know another word to edit it to. $\endgroup$
    – Satya
    Commented May 5, 2021 at 12:23
  • $\begingroup$ Sir, the other part that you talked about "...but not when the alcohol and aldehyde are both contained in one molecule." I did not understand that. Why does that happen. $\endgroup$
    – Satya
    Commented May 5, 2021 at 12:29
  • $\begingroup$ This happens just because the molecule (B) happens to contain an alcohol group and an aldehyde group, as simple as that. $\endgroup$
    – Ivan Neretin
    Commented May 5, 2021 at 12:33

1 Answer 1

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You may notice that B is an intermediate to eventually reach D. Indeed, it is quite possible, performing the reaction in the lab, B would spontaneously yield D with only little chance to isolate this intermediate as such.

$\ce{NaBH4}$ reduces the aldehyde to an alcohol. Then, mediated by the acid, the actetal is converted into an aldehyde to yield B.

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The aldehyde in B already is more reactive than a ketone would be, but the presence of acid (still present from converting the acetal) may increase the reactivity of the carbonyl group even further. The subsequent cyclization is favored

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because the chance that alcohol and aldehyde are close enough to form a new bond is much higher for an intramolecular reaction, than for an intermolecular one. This is a general rule worth to retain. In addition, the formation of five- and six-membered rings equally is energetically favorable.

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    $\begingroup$ It is also worth noting that cyclic hemiacetals (lactols) are often stable. See the cyclic forms of most simple sugars for examples. $\endgroup$
    – Ben Norris
    Commented May 5, 2021 at 12:43
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    $\begingroup$ @BenNorris Based on your answer given here, supporting a postulate by computations, I notice a little bit of envy with me for those capable to perform such a reaction in the lab .and. to use such an in silico yardstick to estimate quantitative likelihoods in advance. $\endgroup$
    – Buttonwood
    Commented May 5, 2021 at 12:56

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