Draw a stepwise mechanism for each reaction .

Name: Problem 64, Chapter 9: Organic Chemistry
Uploaded: 2019-05-01T19:46:59-08:00
Duration: 4 min 58 s
Channel: Nima Gharibi
Description: Draw a stepwise mechanism for each reaction .

Key Concepts

Stepwise Reaction Mechanism

A stepwise reaction mechanism breaks down a chemical process into several discrete elementary steps rather than a single concerted event. Each of these steps usually involves the formation and consumption of intermediates, and they are typically represented by a sequential reaction diagram that shows the progression from reactants to products.

Reaction Intermediates

Reaction intermediates are transient species formed during one or more of the intermediate steps in a stepwise mechanism. They are not present in the overall balanced equation but are crucial for understanding the pathway of the reaction, as their formation and further reaction often determine the overall reaction course and kinetics.

Arrow-Pushing Formalism

Arrow-pushing is a graphical method used to illustrate the movement of electrons during each step of a reaction mechanism. It involves drawing curved arrows to indicate electron flow, helping chemists visualize how bonds are broken and formed, and it is essential in constructing and understanding stepwise reaction mechanisms.

Transition States

Transition states represent the highest energy points along the pathway of an elementary step in a reaction mechanism. They are fleeting configurations that occur during bond making or breaking, and although they cannot be isolated, understanding them is critical as they determine the activation energy and influence the rate of the reaction.

Rate-Determining Step

The rate-determining step in a stepwise mechanism is the slowest step that limits the overall rate of the reaction. It acts as a bottleneck for the reaction process, and its kinetic properties often dictate the reaction's speed and response to changes in conditions such as temperature and concentration.

Transcript

00:01 Hello, today we are doing problem 9 .64 and this problem gives us two parts and it asks us to determine the stepwise mechanism for each of the following reaction.

00:13 So the first reaction is on page one and we'll begin with this.

00:17 So as you see here we start with some sort of cyclic ether and we go into some sort of straight chain a hydrocarbon and alkaliate, dialkal halide.

00:26 So obviously the first thing that we see here is that we're starting with two equivalents of a.

00:31 So we know right away h will dissociate h plus and i minus.

00:38 When we look at our starting material, we see that we have oxygen which can act as a good nucleophile.

00:42 So right away we know that oxygen will pick up one of those protons, the first equivalent proton, to protonate itself.

00:54 Now we have oxygen with three bonds, giving it a formal positive charge.

00:58 And remember, oxygen does not like positive charge.

01:00 It likes to being neutral or negative.

01:02 So one of these sigma bonds adjacent to auction will break.

01:05 Forming a carbocatin intermediate and neutralizing this alcohol.

01:10 However, which one is going to break? well, the one that forms the most stable carbocataon.

01:14 So if we see, if we break the right sigma bond, we form a tertiary carbotidine.

01:20 If we were to break the left sigma bond, we form a primary carbocation.

01:25 And remember, primary carbocatines do not form.

01:27 They are not stable.

01:28 So we right away know that the right one will break.

01:33 So now we have our tertiary carbocation with our.

01:39 Alcohol.

01:41 So now this first equivalent of h .i.

01:44 We use the proton of it, but now we have the i minus...

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