It has been found that 3-substituted methyl 3-hydroxy-2-methylene alkanoates give rise to a pref

It has been found that 3-substituted methyl...

Key Concepts

Substituent Effects

Substituent effects involve the influence of different groups (such as alkyl versus aryl) on the reaction mechanism. The size and electronic properties of substituents can alter the degree of steric hindrance in the transition state, thereby affecting the energy profile and ultimately leading to a preference for either the E or Z configuration depending on the substituent's characteristics.

Stereoselectivity

Stereoselectivity is the preference for the formation of one stereoisomer over another in a chemical reaction. This outcome is often dictated by the transition state's geometry, where steric hindrance and substituent orientation lead to a more stable pathway for the formation of the preferred isomer.

Claisen Rearrangement

A Claisen rearrangement is a [3,3]-sigmatropic pericyclic reaction in which an allyl vinyl ether (or related structure) undergoes a concerted rearrangement via a cyclic six?membered transition state. This concept is important because the transition state’s cyclic nature heavily influences the final stereochemistry of the product.

Transition State Structure

The analysis of a transition state involves understanding the geometry and energy of the molecular configuration at the point of highest energy along the reaction coordinate. Here, it is used to predict how steric interactions among substituents dictate which stereoisomer (E or Z) is preferentially formed in the rearrangement.

Steric Interactions

Steric interactions refer to the repulsive forces arising when bulky groups approach each other too closely. In a transition state, these forces can increase the activation energy by destabilizing certain conformations. By minimizing these unfavorable interactions, the reaction pathway is biased toward a structure that leads to one stereoisomer over another.

Transcript

00:01 Hello, today we're doing problem 9 .54, and this gives us this reaction scheme saying that we started from this starting material, which they call v, and when you're reacted with pakl and puridine, we get the alkene elimination product, w.

00:19 But why is it that when we reacted with h2s -o -4, we get an isomeric mixture of three products, x, y, and z.

00:30 And it asks us to draw the stepwise mechanism for each reaction, explain why the difference occurs.

00:36 So if we remember the reaction mechanism of poccalimperidine, elimination will occur through an e2 mechanism.

00:43 So your nucleophile will attack when you're same, at the same time as your alkyne forming, at the same time as your leaving group leaves.

00:51 So that means that it's three arrows all in one step.

00:54 And because of that, you can't really do any carbocatin rearrangements.

00:59 You don't even have a carbonyl intermediate.

01:02 So that explains why we only get this one product.

01:05 Conversely, when we use a strong acid such as h2s .s .o4 sulfuric acid, we go through an elimination through an e1 mechanism.

01:13 So a carbocatine intermediate is formed, and because that carbocation intermediate is formed, we can do hydride shifts or r shifts to generate the most stable carbocatine intermediate.

01:24 C rcadam possible.

01:25 So if we have a secondary carbocatin, we can do a proton, a hydride shift, or in a r shift, to make a tertiary carbonatine, which would be more stabilized due to its inductive stabilization.

01:39 So let's begin with the mechanism of pockel, and we'll understand why we only get this one product.

01:44 So if i redraw pockel, we see that we have an electrophilic phosphorus atom here.

02:01 And obviously in our alcohol we have a very good nucleophile being this oxygen so we can show the nucleophilic attack to pop off one of these chlorine atoms giving us our intermediate and remember periodine is used as our proton mop so this proton here will be deprotonated by pyridine just neutralize everything and help the reaction go to completion so now we made a very good leaving group from this bad leaving group that we started from.

03:13 So because of that, we can do our elimination step and from our elimination step, we need to identify where the beta carbons are.

03:21 So obviously the carbon that is bound to our leaving group is called alpha carbon.

03:26 So the carbons right adjacent to that are called beta carbons.

03:29 So we see we have two beta carbons.

03:31 In order to do our elimination reaction, we need to find the beta protons.

03:36 So on the right, we already have four functional groups.

03:38 So there is no point.

03:39 Protons on this beta carbon.

03:41 However, here we have two equivalent protons that can be used for this elimination.

03:47 So another molecular perioding will be used at this point.

03:50 The nucleophilic nitrogen will attack and picking up this proton, making a bond.

03:56 When you make a bond, you break a bond.

03:58 So that sigma bond will break to form your alkyne.

04:02 Now we made a new bond.

04:03 So therefore, when you make a bond, you need to break a bond.

04:06 So our good leaving group will leave...

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