Question

H3C–COOEt + H3C–COOEt 1. NaOEt, EtOH 2. H3O+ H3C–CO–CH2–COOEt + EtOH Esters, like aldehydes and ketones, are weakly acidic. Treatment of an ester with 1 equivalent of a strong base, such as sodium ethoxide, causes a reversible carbonyl condensation to occur that forms a ?-ketoester. This reaction is known as the Claisen condensation. The tetrahedral intermediate thus formed expels an alkoxide leaving group to give an acyl substitution product, a ?-ketoester. The steps leading up to this point are all reversible. The best yields for the Claisen condensation are obtained when the starting ester has more than one acidic ?-hydrogen. In these cases, a second deprotonation of the ?-carbon drives the reaction equilibrium toward the product, thus requiring a full equivalent of base instead of a catalytic amount. Subsequent protonation with dilute aqueous acid results in a high yield of the final product. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions: (OEt attached to tetrahedral intermediate shown with lone pairs) H3C | C–(OEt) | C(=O)OEt ? H3C–C(=O)–CH2–C(=O)OEt + EtO– (Show lone pairs on OEt and the two H atoms on the ?-carbon as indicated)

          H3C–COOEt   +   H3C–COOEt

1. NaOEt, EtOH
2. H3O+

H3C–CO–CH2–COOEt   +   EtOH

Esters, like aldehydes and ketones, are weakly acidic. Treatment of an ester with 1 equivalent of a strong base, such as sodium ethoxide, causes a reversible carbonyl condensation to occur that forms a ?-ketoester. This reaction is known as the Claisen condensation.

The tetrahedral intermediate thus formed expels an alkoxide leaving group to give an acyl substitution product, a ?-ketoester. The steps leading up to this point are all reversible. The best yields for the Claisen condensation are obtained when the starting ester has more than one acidic ?-hydrogen. In these cases, a second deprotonation of the ?-carbon drives the reaction equilibrium toward the product, thus requiring a full equivalent of base instead of a catalytic amount. Subsequent protonation with dilute aqueous acid results in a high yield of the final product.

Draw curved arrows to show the movement of electrons in this step of the mechanism.

Arrow-pushing Instructions:

(OEt attached to tetrahedral intermediate shown with lone pairs)

H3C
  |
C–(OEt)
  |
C(=O)OEt

?

H3C–C(=O)–CH2–C(=O)OEt   +   EtO–

(Show lone pairs on OEt and the two H atoms on the ?-carbon as indicated)

H3C–COOEt + H3C–COOEt

1. NaOEt, EtOH
2. H3O+

H3C–CO–CH2–COOEt + EtOH

Esters, like aldehydes and ketones, are weakly acidic. Treatment of an ester with 1 equivalent of a strong base, such as sodium ethoxide, causes a reversible carbonyl condensation to occur that forms a ?-ketoester. This reaction is known as the Claisen condensation.

The tetrahedral intermediate thus formed expels an alkoxide leaving group to give an acyl substitution product, a ?-ketoester. The steps leading up to this point are all reversible. The best yields for the Claisen condensation are obtained when the starting ester has more than one acidic ?-hydrogen. In these cases, a second deprotonation of the ?-carbon drives the reaction equilibrium toward the product, thus requiring a full equivalent of base instead of a catalytic amount. Subsequent protonation with dilute aqueous acid results in a high yield of the final product.

Draw curved arrows to show the movement of electrons in this step of the mechanism.

Arrow-pushing Instructions:

(OEt attached to tetrahedral intermediate shown with lone pairs)

H3C
|
C–(OEt)
|
C(=O)OEt

?

H3C–C(=O)–CH2–C(=O)OEt + EtO–

(Show lone pairs on OEt and the two H atoms on the ?-carbon as indicated)

Added by Lisa K.

Question

Please give Ace some feedback