Question about the cyclic hemiacetal form of compounds

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D would not be a valid answer if you think about what forms or what happens that leads to formation of cyclic hemiacetal. In the course of formation of cyclic hemiacetal, the -OH oxygen has 2 lone pair and the C=O carbon has partial positive charge on it and so, in a nutshell the -OH oxygen attacks the carbon with partial positive charge and this leads to formation of hemiacetal. Now,the -OH is attached to 4th C atom and so during the course of ring formation, the 5th carbon will always lie outside the ring and this the reason why D cannot be a valid answer.

A scheme of this critical mechanism step is shown below. The carbon atoms are numbered for clarity. Additionally, the methyl group on carbon 5 is circled. Since the mechanism of the ring closure does not involve carbons 2-5, the structure cannot change at those positions.

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Samantha
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Samantha

Updated on January 21, 2023

Comments

  • Samantha
    Samantha 10 months

    enter image description here

    Why is D not a valid answer for this question?

    • user55119
      user55119 almost 5 years
      D is formed from 5-hydroxypentanal, a primary alcohol. Your compound is 4-hydroxypentanal. B is the answer.
    • Mathew Mahindaratne
      Mathew Mahindaratne almost 5 years
      Short and simple, if you count atoms from $\ce{O}$ (of $\ce{OH}$ group) to carbonyl $\ce{C}$, you'd find the number is 5. Thus, it makes 5-membered cycle. So, it can't be D.
  • Abner Alfred Thompson
    Abner Alfred Thompson almost 5 years
    Thanks for additional editing to the answer to make it more easier to understand. Actually I'm quiet new to this website so I'm figuring out how to do things here.So thank you.
  • chemicalromance
    chemicalromance almost 5 years
    Hi and welcome to the site! Don't doubt to take the tour for a quick intro and use the help center for any doubts you might have.