[hell0joe]

My professor is very picky about stereochemistry regarding mechanisms so I was hoping I could get some clarification here:

In an SN1 reaction, I have seen sources drawing 2 enantiomer products to form a racemic mixture, while other sources like my textbook do not bother most of the time dash/wedge products and only use straight lines (generating 1 product).

This has made me confused about SN1 reactions, specifically do SN1 reactions always produce two enantiomeric products? If so, are there cases where they don't? And is my textbook just being lazy in not showing the stereochemistry?

Thanks

[Dan]

In an SN1 reaction, I have seen sources drawing 2 enantiomer products to form a racemic mixture, while other sources like my textbook do not bother most of the time dash/wedge products and only use straight lines (generating 1 product).

Not defining the stereochemistry in the product implies it is mixed or irrelevant in the context of the discussion. I think it can be a perfectly acceptable thing to do, but do what your professor wants you to do.

This has made me confused about SN1 reactions, specifically do SN1 reactions always produce two enantiomeric products? If so, are there cases where they don't?

Consider S_N1 substitution of the bromide in this molecule:



What are the products and are they enantiomers?

[hell0joe]

In an SN1 reaction, I have seen sources drawing 2 enantiomer products to form a racemic mixture, while other sources like my textbook do not bother most of the time dash/wedge products and only use straight lines (generating 1 product).

Not defining the stereochemistry in the product implies it is mixed or irrelevant in the context of the discussion. I think it can be a perfectly acceptable thing to do, but do what your professor wants you to do.

This has made me confused about SN1 reactions, specifically do SN1 reactions always produce two enantiomeric products? If so, are there cases where they don't?

Consider S_N1 substitution of the bromide in this molecule:



What are the products and are they enantiomers?

Looking at the structure, Br would be the leaving group and would be substituted with the nucleophile. Since the substrate has a chiral carbon, I think that it would lead to 2 different configurations once the nucleophile attacks? Therefore enantiomers?

[sjb]

Yes the bromine is the leaving group. What is the intermediate? Consequently, what are the two products? Are these enantiomers?

[hell0joe]

The intermediate would be the tertiary carbocation. I am not quite sure what you mean by what the two products are as we don't have a specific nucleophile here, but I think one would be R and the other S, so enantiomers.

[Dan]

The intermediate would be the tertiary carbocation. I am not quite sure what you mean by what the two products are as we don't have a specific nucleophile here, but I think one would be R and the other S, so enantiomers.

What is the definition of enantiomer?

The starting material has two stereogenic centres - it's configuration is (3S,4S). Let's say you substitute the Br in (3S,4S)-3-bromo-3,4-dimethylhexane with water via S_N1 - what do you get?

[hell0joe]

An enantiomer is the mirror image of a structure that is non-superimposable on the original, I think. I'm not sure how water would play a part here. Would it dissociate into H+ and OH-, leading to formation of HBr? And then OH- would be the nucleophile?

[Dan]

An enantiomer is the mirror image of a structure that is non-superimposable on the original, I think. I'm not sure how water would play a part here.

Draw the products, are they non-superimposable mirror images?

[carotis]

in SN1 reactions you have to focus on cis/trans products, so a mixture of retention and inversion. On the other hand SN2 reactions always go under inversion (Walden inversion). The very reason is the intermediate (in SN2 reaction there is only the "attack from the back-end". So you always have to consider both products. Hence SN2 reactions induce inversions, and inversion (cis/trans products also might appear in SN1 reactions, you might think about what factors would favour an inversion of its product, for instance concentration of the nucleophile or its strenght. 

[Babcock_Hall]

@OP, Besides enantiomers is there another class of stereoisomers?

[hell0joe]

@Babcock_Hall Diastereomers

[Babcock_Hall]

How many centers of chirality (stereogenic carbon atoms) are there in the product?

[hell0joe]

2, if I'm not mistaken.

[sjb]

How many configurations were affected by the reaction?