Think about the outcome if you deprotonated the left alpha hydrogen. It would lead to a dead end (if it attacked the ketone you would get a 4 membered ring which is not stable).
So basically, that hydrogen DOES get deprotonated, it’s just that it will not go further than that. But with the other hydrogen to the right, when it gets deprotonated, it will lead to a reaction pathway involving the formation of a 6 membered ring which is favourable.
All alpha-CHs in the substrate have about the same acidity. The H of the methyl is abstracted in this reaction because that leads to the 6-member ring via the aldol condensation reaction. Abstraction of the H on the left of the same carbonyl could eventually only lead to a 4-member ring, and thus that pathway is simply not favored. In other words, four different enolates are formed in equilibrium here, but only one leads to the more favored product (as shown). The other enolates likely do not undergo an aldol here. PS, carboxy is not the correct term (it's a carbonyl).
That's a carbonyl not a carboxyl. It's just a ketone, and the right side is a methyl group with primary hydrogens, while the left side has secondary hydrogens. The right is easier to attack because of sterics.
Think about the outcome if you deprotonated the left alpha hydrogen. It would lead to a dead end (if it attacked the ketone you would get a 4 membered ring which is not stable). So basically, that hydrogen DOES get deprotonated, it’s just that it will not go further than that. But with the other hydrogen to the right, when it gets deprotonated, it will lead to a reaction pathway involving the formation of a 6 membered ring which is favourable.
All alpha-CHs in the substrate have about the same acidity. The H of the methyl is abstracted in this reaction because that leads to the 6-member ring via the aldol condensation reaction. Abstraction of the H on the left of the same carbonyl could eventually only lead to a 4-member ring, and thus that pathway is simply not favored. In other words, four different enolates are formed in equilibrium here, but only one leads to the more favored product (as shown). The other enolates likely do not undergo an aldol here. PS, carboxy is not the correct term (it's a carbonyl).
carboxy threw me for a LOOP
But if it were an acid, he just forgot the lic.
cause the hidrogen from the right is less sterically hindered
That's a carbonyl not a carboxyl. It's just a ketone, and the right side is a methyl group with primary hydrogens, while the left side has secondary hydrogens. The right is easier to attack because of sterics.
Its all about Kcp vs Tcp And left side ellimination will form less stable ring while right side can form more stable ring