Predict the product and give the stereochemistry resulting...

Predict the product and give the stereochemistry resulting from reaction of each of the following nucleophiles with $(R)-2$ -bromooctane: $\begin{array}{lll}\text { (a) }-\mathrm{CN} & \text { (b) } \mathrm{CH}_{3} \mathrm{CO}_{2}^{-} & \text {(c) } \mathrm{CH}_{3} \mathrm{~S}^{-}\end{array}$

Predict the product and give the stereochemistry resulting from reaction of each of the following nucleophiles with $(R)-2$ -bromooctane:
$\begin{array}{lll}\text { (a) }-\mathrm{CN} & \text { (b) } \mathrm{CH}_{3} \mathrm{CO}_{2}^{-} & \text {(c) } \mathrm{CH}_{3} \mathrm{~S}^{-}\end{array}$

Key Concepts

Stereochemical Inversion (Backside Attack)

Due to the backside attack characteristic of SN2 reactions, the configuration at the chiral center is inverted during the reaction. This means that if the starting material has an (R) configuration, the product will have an (S) configuration, provided no other stereochemical influences are present.

Leaving Group Ability

A key aspect of SN2 reactions is the quality of the leaving group. A good leaving group, like bromide in this case, is one that can stabilize the negative charge after dissociating from the substrate, which facilitates the progress of the nucleophilic substitution by lowering the activation energy of the reaction.

SN2 Reaction Mechanism

The SN2 (bimolecular nucleophilic substitution) mechanism involves a one-step process where the nucleophile attacks the electrophilic carbon atom from the side opposite to the leaving group. This results in a simultaneous bond formation with the nucleophile and bond breaking between the carbon and the leaving group, making the reaction rate dependent on both the substrate and the nucleophile concentrations.

Transcript

00:01 So we have the following organic compound here.

00:02 We have r2 -bromo -octane, and we would like to know what product we're going to get if we react this organic compound with different nucleophiles.

00:12 So first step here is to draw what that is going to look like, our r2 -bromal octane.

00:17 So we can go ahead and draw our 8 -carbon chain.

00:23 It looks like that.

00:24 And we can draw our bromal group in the 2 position with the r configuration.

00:27 This is going to be denoted by a wedge.

00:32 So we can quickly assign priorities to make sure that we are in the right configuration here.

00:38 So broma group is prior to number one.

00:40 Carbon connected to the long chain is going to be number two.

00:43 Methyl group is number three, and the hydrogen that is facing away from us.

00:47 But it's not shown here is going to be in group priority number four.

00:51 So we do see that we are in the r configuration here.

00:57 Now what would happen if we reacted this with the nucleophile cyanide? so let's go ahead and draw the r2 -bromal octane in the our configuration here.

01:13 All right.

01:14 So we do know that cyanide is a very good nucleophile, a very strong nucleophile, and we have our alkylo halide in the secondary position.

01:24 So it looks like we are going to proceed.

01:26 This reaction will proceed via sn2.

01:32 And so our resulting product is going to look like this.

01:38 And now this is going to proceed with inversion of configuration.

01:41 And so our cyanide group that comes in is not going to be going away from us.

01:49 So it's going to be denoted by dash -dash line.

01:55 And we see that our cyanide group remains the same priority as our leaving alkylo halide, which in this case is going to be veralemein.

02:04 So we will have an inversion of absolute configuration as well.

02:08 And so whereas we had an r configuration here, we will have an s configuration as well...

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