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Can anyone also explain me the mechanism of this reaction?

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I think it might be formation of acetal or hemiacteal reaction but I am not sure about it.

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  • $\begingroup$ Why do you consider a the formation of a hemiacetal under anhydrous conditions? $\endgroup$
    – Klaus-Dieter Warzecha
    Commented May 16, 2016 at 7:44
  • $\begingroup$ @KlausWarzecha sir actually I don't know. I am just having a guess because I think ester or ether will not formed from this reaction. Can you please explain me the mechanism and the product? $\endgroup$
    – Murtuza Vadharia
    Commented May 16, 2016 at 8:00

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Please have a closer look at your starting materials and the reaction conditions.

  1. The reaction is performed under protic, ahydrous conditions.
  2. One of the starting materials is a primary alcohol.
  3. The other starting material apparently looks like a cyclohexene where one carbon adjacent to the $\ce{C=C}$ double bond has been replaced by an oxygen atom. This is a dihydropyrane.

Now, let's try a mind game:

  1. Supposed that your second starting material is cyclohexene, what would happen under anhydrous, protic conditions?

    The proton would probably add to the $\ce{C=C}$ double bond to form a cyclohexyl cation. Since the starting material is symmetric, there is only one secondary cation formed. Note that the situation would be different if you would start with methylcyclohexene! Then, a secondary or a tertiary cation could be formed. Remember inductive effects to judge which one is favoured.

  2. How would the primary alcohol come into play here?

    It would probably add to the cyclohexyl cation and yield an ether.

  3. In reality, your second starting material is a dihydropyrane, think cyclic vinyl ether. How will the lone pairs on the oxygen atom effect the protonation?

    If in doubt, draw! Your $\ce{C=C}$ double bond now has two markedly different carbon atoms. Protonation can result in two different cations. I suggest to draw some resonance structures to figure out which one is favoured. Remember the lone pairs ;-)

  4. What happens in the presence of a primary alcohol now?

    Again, it will add to the cation.

  5. How does the final structure look like?

    I'm certain that you have figured this out by drawing and once you have, you will easily identify the structure class :)

Note that the reaction in question is quite useful to reversibly protect alcohols in sequences of chemical transformations.

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