a. Give the structure of the products obtained from the reaction of each enantiomer of cis -1-ch

A. Give the structure of the products obtained from the...

Key Concepts

Substitution versus Elimination Competition

In reactions involving strong bases and nucleophiles like sodium methoxide, both substitution (SN2) and elimination (E2) pathways are viable. The outcome depends on factors such as the nature and concentration of the base/nucleophile, the substrate's structure, and the reaction conditions. Differences in these factors can lead to different product distributions and reaction rates for substitution and elimination.

Stereochemistry and Optical Activity

Understanding stereochemistry is crucial when dealing with enantiomers, diastereomers, and geometric isomers (cis and trans). The spatial arrangement of the substituents on a cycloalkane affects not only the type of reaction pathway (substitution or elimination) but also the optical activity of the products. A product may be optically active if it is chiral, but reactions can also generate achiral or meso compounds depending on symmetry and stereochemical outcomes.

Steric Effects in Cyclic Systems

Steric hindrance plays a vital role in cyclic substrates, where the relative positions of substituents (cis versus trans) can affect the accessibility of reaction sites. For instance, cis substituents may impose greater steric crowding, potentially slowing down substitution reactions, while also influencing the specific hydrogen atoms accessible for eliminations. This effect explains differences in reactivity and the nature of the products formed in cycloalkane reactions.

SN2 Reaction Mechanism

This is a bimolecular nucleophilic substitution mechanism where the nucleophile attacks the electrophilic carbon from the side opposite the leaving group, leading to an inversion of configuration at that carbon. It is highly sensitive to steric hindrance, and the reaction rate depends on both the substrate and the nucleophile concentration. In the context of cyclic systems, the spatial arrangement of substituents (e.g., cis or trans) can significantly affect the feasibility and outcome of the reaction.

E2 Reaction Mechanism

The E2 mechanism is a bimolecular elimination process that occurs in a single concerted step. In this mechanism, a base abstracts a ?-hydrogen while the leaving group departs simultaneously, forming a double bond. The reaction is often influenced by the stereochemical alignment of the leaving group and the ?-hydrogen (antiperiplanar geometry) and competes with substitution pathways under strong basic conditions.

Transcript

00:01 Hi, so this video is going over problem 39 in organic chemistry 4th edition, chapter 11, and this is talking about the competition between substitution and elimination reactions.

00:13 And we're given an example here of the cis isomer of 1 -clero -2 -isopropo cyclopentane.

00:22 And this is reacted in sodium methoxide and methanol, high concentration of sodium.

00:31 Metoxide.

00:33 So we're asked to find the problems and then we have to ask some or answer some different questions about the products.

00:40 So first of all, when we're talking about substitution and elimination reactions, we have to deduce which ones are possible, right? so we have a secondary alcoholide here on both of these.

00:57 So we can't eliminate any of the reactions because sn1, sn2, e1, e2 can occur on secondary.

01:06 So now we have to look at the nucleophile.

01:08 So a strong nucleophile or a strong base will tend towards sn2 and e2, whereas a weak nucleophile, weak base will tend towards sn1 and e1.

01:21 So here, our sodium, we can cross off and put a negative charge there.

01:25 So this methoxide, this negatively charged oxygen attached to a methyl group, is a strong nucleophile and a strong base.

01:37 So we know that we can do e2 and sn2 on both of these.

01:44 And on a secondary nucleophile or a secondary alcoholide, that means that e2 will be the major product most likely.

01:52 But we have to kind of go through the process.

01:57 So for e2, if we're finding our e2 product, we have our zeytsev and our hoffman product.

02:06 Our zaytsev is the most stable and that's the most, the double bond goes on the most substituted side, whereas our hoffman product goes on the less substituted side.

02:16 And there are a couple ways we can get our hoffman product.

02:23 So if you have a week, a week base, no, a week leaving group, sorry.

02:35 So something like fluorine, which we don't have.

02:39 We have a chlorine here.

02:41 Another way is if we have a bulky base, so something like this turputoxide.

02:51 And we don't.

02:53 We just have methoxide that's not bulky, so we can do our z8 sub product.

02:58 And then another thing is if we don't have a trans hydrogen on that one side.

03:07 So we can kind of draw out a 3d representation of this to try to figure out if they're trans.

03:22 So we're told there's cis, right? so we have our cl and our isopropyl are sticking up in this one.

03:36 And our cl and our isopropyl sticking down in the second one.

03:42 So our hydrogen that we're going to eliminate has to be trans to the halide.

03:48 So we do have a trans -hydrogen to that halide.

03:59 And remember, we're looking at the trans -hydrogen on the most substituted side.

04:05 Because we do have a trans -hydrogen on the least substituted side as well.

04:12 I guess right here.

04:18 But for our zaytsev product, which is the double bond on the most substituted side, we want to see if we have a trans -hydrogen on that side that we can take off, because that is going to be our most stable product.

04:30 So if we started with the transcyclopentane, that would mean that we don't have a trans -hydrogen on the most substituted side, and we can't make our z8 -subproduct.

04:40 We have to go with our hoffman product for e2.

04:46 So we know that we can eliminate that halide and make it on, or make our double bond on that most substituted side for our e2 product.

05:00 So we can go ahead and do that.

05:03 That's our e2 zsav product.

05:08 And we probably will make our hoffman product a little bit as well.

05:12 But our major product would be our zaytsev.

05:16 So we're going to go ahead and put our zetsv product.

05:22 And then we can also do some s &2.

05:25 Remember that our e2 is going to be major when we have a secondary alkaliid, and we can make that zathev product.

05:37 Our e2 is going to be major.

05:41 But our sn2, our sn2, remember, we invert our stereochemistry.

05:58 It would not be co.