What is the product of an acyl substitution reaction

Key Concepts

Acyl Substitution Reaction

This is a fundamental mechanism in organic chemistry where a nucleophile attacks an acyl compound, forming a tetrahedral intermediate that subsequently reverts to a carbonyl compound after expelling a leaving group. The overall transformation allows the substitution of one group on the acyl carbon with another, and is central to the synthesis and interconversion of carboxylic acid derivatives.

Tetrahedral Intermediate Formation

During nucleophilic acyl substitution, the nucleophile’s attack on the carbonyl carbon generates a tetrahedral intermediate. This intermediate is a transient species where the planar sp2 hybridized carbonyl carbon temporarily adopts a tetrahedral geometry, setting the stage for the elimination of a leaving group and the reformation of a double bond.

Leaving Group Ability

The ability of a group to depart from the tetrahedral intermediate as a leaving group plays a crucial role in determining the reaction pathway and product. A good leaving group is typically a weak base, facilitating the forward reaction. Conversely, if the incoming group forming the tetrahedral intermediate is a stronger base than the resident leaving group, it can revert back, potentially causing an equilibrium or no reaction at all.

Role of Relative Basicity in Reaction Outcome

The overall outcome of an acyl substitution reaction depends on the relative basicity of the groups involved. If the new group introduced is a stronger base compared to the original substituent, it tends to remain bound, often reversing the reaction. If it is a weaker base, it is more likely to be expelled as a leaving group, leading to the formation of a distinct carboxylic acid derivative. When their basicities are similar, the reaction might reach an equilibrium, potentially yielding a mixture of products.

Transcript

00:01 Okay, this problem is analyzing the difference in product formation.

00:04 If i have in a tetrahedral intermediate of a carboxylic acid derivative, the stronger base come in as a substituent.

00:11 So let's go ahead and analyze that.

00:13 If i have a tetrahedral intermediate of a carboxylic acid derivative, the basic form is going to be this right here, in which i have two opposing substituents.

00:21 I'm going to have, for example, a chlorine versus, for example, an nh2.

00:25 So which one is the stronger base in this case? is it going to be my nh2, or is it going to be my chloro? so we're analyzing chloride versus nh2.

00:36 Okay, so just analyzing the pqas of the conjugate acids of these, we'll soon find out that nh2 minus is the stronger base.

00:43 So if we have that difference in basicity between nh2 and chlorine, that means we're going to form a product, a very specific product, and that is going to be a case in which my electrons from the oxygen come down to form a carbonyl.

00:55 And as i form a carbonyl here, this carbon has exceeded its octet, so we need to get rid of electrons.

00:59 And it needs to move those electrons onto the leaving group.

01:02 So are we going to make my chlorine into a leaving group, or are we going to make my nh2 into a leaving group? we always want to form the weakest base possible.

01:09 So if we have a situation in which we can form either a strong base or a weak base, we are always going to want to form that weak base because it's the most favorable in terms of equilibrium.

01:18 So in this case, because we're going to get rid of a chlorine, that would form a weak base instead of the opposite in which i would to form my nh2 as a leading group, which would form a strong base.

01:29 So we always want to form the weakest base possible as that leaving group.

01:33 So as far as the product goes, we're going to get this.

01:35 And this is just specific to this example in terms of this product, but the same situation would occur across all difference in basicities.

01:43 Okay, so we're going to form the one in which the strongest base is still attached to that carbonyl.

01:48 Okay, moving on to this one, in which we have a weaker base come in.

01:52 So if we have a tetrahedral intermediate of that carboxylac acid derivative, again, we're going to get these two, and then we're going to get the difference in the substituents in basicity.

02:01 So we're going to get, let's just say we have oh this time, and then we'll keep nh2.

02:07 Okay, so which one is the weaker base? okay, so we're analyzing oh minus compared to nh2 minus.

02:14 Okay, so just analyzing the pca is the conjugate acids of these.

02:17 I know that the conjugate acid of oh minus is water, which has a pkk of 15 .7, and then the conjugate acid of my nh2 minus is nh3, which has a pkk of approximately 36...

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