Predict the products of the following nucleophilic acyl substitution reactions: a. b. c. d.

Name: Problem 5, Chapter 21: Organic Chemistry
Uploaded: 2020-05-19T20:58:45-08:00
Duration: 6 min 59 s
Channel: Stefan Kaliszuk
Description: Predict the products of the following nucleophilic acyl substitution reactions: a. b. c. d.

Predict the products of the following nucleophilic acyl...

Key Concepts

Nucleophilic Acyl Substitution Mechanism

This is a fundamental reaction process in organic chemistry where an acyl derivative reacts with a nucleophile. The reaction typically proceeds through an addition-elimination sequence, wherein the nucleophile attacks the electrophilic carbonyl carbon to form a tetrahedral intermediate, which then collapses to expel the leaving group, ultimately producing the substitution product.

Acyl Derivatives

Acyl derivatives are compounds containing a carbonyl group attached to a leaving group, such as in acid chlorides, esters, or anhydrides. Their reactivity in nucleophilic acyl substitution reactions is due to the polar nature of the carbonyl group and the ability of the leaving group to depart, which makes them prone to nucleophilic attack.

Tetrahedral Intermediate

A key stage in the nucleophilic acyl substitution mechanism is the formation of a tetrahedral intermediate. This intermediate is formed when the nucleophile adds to the carbonyl carbon and temporarily disrupts the planar structure of the carbonyl group, setting the stage for the subsequent elimination of the leaving group.

Leaving Group Ability

The success and rate of nucleophilic acyl substitution reactions are strongly influenced by the ability of the leaving group to depart. A good leaving group stabilizes the negative charge after bond cleavage, which facilitates the collapse of the tetrahedral intermediate and the formation of the final substitution product.

Transcript

00:01 So we have to predict the result of these carboxylic acid or carbonylacic acid derivatives, nucleophilic substitutions.

00:12 So we're going to start same way every time.

00:17 We're going to take our nucleophile and we're going to attack the central carbon, kicking these electrons up and going into our tetrahedral intermediate.

00:29 So let's see.

00:32 H3 and then oh.

00:36 Now that we're in the tetrahedral intermediate, we have to compare the stability of our leaving groups as they go into solution.

00:44 So the alkyl chain and the negatively charged oxygen are not going to be very stable in solution.

00:52 Because a doubly negative charged double negatively charged oxygen is very unstable.

00:57 Same thing as a negatively charged carbon.

01:00 So that leaves our only choices with the hydroxide and the methoxide.

01:07 So, and if we compare them side by side, three minus an oh, h, the hydroxide is only stabilized by the hydrogen, which can donate only so much electron density.

01:25 Meanwhile, the methoxide is bonded to a carbon, which is bonded to three hydrogens, which can more effectively stabilize, the negative charge on the oxygen.

01:38 So this is going to be our preferred leaving group.

01:42 So we recreate the double bond and we kick out our methoxide.

01:53 And that creates a normal carboxylic acid and a methoxide.

02:05 And since the methoxide pca, well the alcohol pca, is higher than the carbacilic acid pkha.

02:17 It's going to abstract the hydrogen and we're only left with the carboxylate.

02:24 Okay, on to b...

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