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When nonane-2,8-dione is treated with base, the product 1-(2-methylcyclopent-1-en-1-yl)ethan-1-one forms according to the following scheme:actual reaction The arrow denotes the hydrogen that is first attacked by the base. But why does the reaction not take place like this? —this does not occur Why is the arrow-marked hydrogen less acidic than the previous one?

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    $\begingroup$ Its a reversible process and the 6 ring is thermodynamically favoured over the 8 $\endgroup$ – Waylander May 25 '18 at 14:58
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    $\begingroup$ Somewhat relevant:chemistry.stackexchange.com/questions/89790/… $\endgroup$ – Αντώνιος Κελεσίδης May 25 '18 at 15:15
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    $\begingroup$ There is an error in the reation scheme drawn - Nona-2,8- dione cannot give a [5] ring, it will be [6]. $\endgroup$ – Waylander May 28 '18 at 11:58
  • $\begingroup$ @Waylander, Yes, I have made a very big error. I will repost the images within a short time. $\endgroup$ – Shoubhik Raj Maiti May 28 '18 at 14:58
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The two protons are of similar pKa and you can get deprotonation at both sites with equal ease, however it is important to understand that most intramolecular aldol reactions are reversible and thus give rise to the thermodynamic product i.e. the most stable product. [6]-ring formation is always preferred over [8]-ring formation 1, 2. If you prepare the [8] product by other means, and subject it to the aldol conditions you will get the [6] product. The only exception to this is when you set up the reaction to be under kinetic control e.g. by using LDA, then you can get a mixture.

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  • $\begingroup$ It is not a [6] ring, it is a [5]- ring $\endgroup$ – Shoubhik Raj Maiti May 27 '18 at 12:34
  • $\begingroup$ You can't get a 5-membered ring from the starting material you have drawn $\endgroup$ – Waylander May 27 '18 at 15:42
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Maybe the hydrogen in the second case is the allylic carbon containing hydrogen, so it is more stable. More stable means less reactive, which indicates less acidic. Allylic carbon means sp2 carbon beside another carbon.

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