The chart below shows the regio- and stereoselectivity observed for oxymercuration reduction of

step 1 Hello everyone. Let us see the following question. An outline of racemic synthesis of alpha mult

The chart below shows the regio- and stereoselectivity...

Key Concepts

Steric and Conformational Effects

Steric hindrance and the conformational properties of cyclic systems, such as cyclohexenes, play key roles in influencing both the regiochemical and stereochemical outcomes. Bulky substituents on the ring can affect the formation and stabilization of the mercurinium ion intermediate, as well as dictate the approach of the nucleophile, thereby controlling the product ratios in terms of both regiochemistry and stereochemistry.

Stereoselectivity

Stereoselectivity arises during the opening of the cyclic mercurinium ion when the nucleophile attacks from the side opposite to the mercury moiety, resulting in anti addition. This controlled stereochemical outcome explains why only specific spatial orientations of substituents are observed in the final products, directly linking the mechanism to the observed stereoselective product ratios.

Oxymercuration Reduction Mechanism

This concept describes a two?step electrophilic addition process to alkenes where an electrophilic mercury species initially reacts with the double bond to form a cyclic mercurinium ion intermediate, which is subsequently opened by a nucleophile. The mechanism avoids carbocation rearrangements that are common in related hydrohalogenation reactions, ensuring predictable addition products.

Mercurinium Ion Intermediate

The mercurinium ion is a three?membered, cyclic intermediate that forms upon electrophilic attack on the alkene. Its cyclic, bridged structure leads to the reaction proceeding via an anti?addition pathway when the nucleophile attacks from the opposite side of the leaving mercury portion, directly impacting the stereochemistry of the final products.

Regioselectivity

Regioselectivity in oxymercuration reduction is determined by the relative stability of the intermediate in the mercurinium ion and the subsequent nucleophilic attack. The nucleophile preferentially attacks the more substituted carbon of the intermediate if it offers greater stability, leading to a Markovnikov orientation in the final product. This concept explains the differences in product distribution based on substituent effects and electronic stabilization.

Transcript

0:00 Hello everyone.

00:01 Let us see the following question.

00:03 An outline of racemic synthesis of alpha multistratin.

00:08 So we are talking about preparation of alpha multistrating.

00:17 It is a pheromone of elmbach beetle is shown below.

00:22 So let us see the synthesis given in the question.

00:25 So here is the preparation of multistrating.

00:29 So based on the structure 1, 2, 3 and the structure of alpha multistratin, we have a questions.

00:36 So, the question a is related to, so suggest a reagent for the one step synthesis of one from two butin.

00:46 So, one is this.

00:48 This is 1, 2, 3, 4, but 2 in 1 .4 diol.

00:55 So the following structure, how it can be prepared from butene.

01:00 So, so, this is 1, 2, 3, 4.

01:03 So, there was double bond here.

01:06 So hence there is oh group and also there is oh group.

01:10 How it is prepared from buteim that is there has to be triple bond and then there is a oh group here and then there is a oh group.

01:22 So the triple bond, all kinds are converted to alkymes by hydrogenation.

01:28 So, hydrogenation happens from the same side.

01:32 So, if it is added from the same side, we use poisoned platinum with the baso4.

01:38 This is called lindelars, lindelars, catalyst, which helps in partial reduction.

01:48 So, this is the process used as partial hydrogenation or partial reduction.

01:55 Why, we have to poison the palladium, otherwise this unsaturated.

02:00 Completely lead to saturated.

02:02 That means alkanes will be formed.

02:04 So, since we want double bond here, we have to use partial reduction method that is using lindelars -katlish.

02:11 So, the right answer for the question a is poisoned palladium, yeah, that is lin -lars -katlist.

02:17 So, let us see the question b here.

02:19 So the question b is give the i -u -p -s -e name of compound 1.

02:24 So this is the compound 1 according to the question.

02:27 The i -u -p -s -e name of this compound is 1, 2, three, four, there is double bond here, there is oh group here, and there is also oh group here.

02:38 So, we can see the oh groups are on the same side.

02:41 So, i will maintain the same.

02:43 So the oh groups are, sorry, the oh groups are on the same side.

02:48 If the oh groups are maintained on the same side, that we is z is is so much.

02:53 So it is z, okay? so then it is also diol, right? so, position 1, 2, 3, 4.

03:02 So, it is position 1 .4 diol, then it is but to 2 in.

03:07 So it is but 2 in.

03:12 Position 1 .4 it is diol.

03:15 So we have to write it as z.

03:19 1 .4 diol but 2 in...