[pfnm]

A problem from McMurry, Chapter 11:

SN2 reactions take place with inversion of configuration, and SN1 reactions take place with racemization. The following substitution reaction, however, occurs with complete retention of configuration. Propose a mechanism.

I thought:

The alpha-hydrogen is very electrophilic (because of the electronegativity of the Br). It should be a very acidic hydrogen.

In very dilute OH-, I thought:

1. OH- abstracts the acidic hydrogen from the alpha-carbon
2. The electrons from C-H bond break, and move to the alpha carbon
3. A resonance structure could be drawn with a new pi bond formed between the carbonyl carbon and the alpha carbon, and a pi bond broken between the Carbonyl carbon and oxygen
4. The nucleophilic structure abstracts a proton from H2O
5. A pair of unshared electrons moves back down to reform the carbonyl-oxygen pi bond

But this just reforms the original product.

Could someone explain where I am going wrong?

Thanks alot

(My first attachment is the question from McMurry, the second is my reaction mechanism)

[discodermolide]

A problem from McMurry, Chapter 11:

SN2 reactions take place with inversion of configuration, and SN1 reactions take place with racemization. The following substitution reaction, however, occurs with complete retention of configuration. Propose a mechanism.

I thought:

The alpha-hydrogen is very electrophilic (because of the electronegativity of the Br). It should be a very acidic hydrogen.

In very dilute OH-, I thought:

1. OH- abstracts the acidic hydrogen from the alpha-carbon
2. The electrons from C-H bond break, and move to the alpha carbon
3. A resonance structure could be drawn with a new pi bond formed between the carbonyl carbon and the alpha carbon, and a pi bond broken between the Carbonyl carbon and oxygen
4. The nucleophilic structure abstracts a proton from H2O
5. A pair of unshared electrons moves back down to reform the carbonyl-oxygen pi bond

But this just reforms the original product.

Could someone explain where I am going wrong?

Thanks alot

(My first attachment is the question from McMurry, the second is my reaction mechanism)

Think about a double inversion via an alpha lactone!

[stevet]

Also, consider which proton is the most acidic in the system, and will be removed first.

[pfnm]

Hi,

Thanks for the advice guys,

I've written a new reaction scheme.

The steps are:

1. -OH nucleophile abstracts H from the carboxylic acid (the most acidic hydrogen in the system), O-H bond breaks, pair of electrons move to the oxygen
2. The anionic O is a strong nucleophile and attacks the weakly electrophilic alpha carbon, in an SN2 reaction - the Br is the leaving group. Inversion of configuration occurs, and an epoxide type strained-ring structure is formed

Some confusion at this point:
Either the acid protonates the oxygen in the epoxide ring, then a new OH- nucleophile attacks the alpha-carbon to form a hydroxyl group, opening the ring..and lastly the O- is protonated.

Or

The way I drew the mechanism

3. OH- nucleophile attacks the alpha-carbon, by SN2. The ring opens and inversion has occurred (so the configuration at the alpha carbon has been retained)
4. Acid protonates the O- to form the final product.

Would the acid protonate the epoxide-type ring first? Or is the way I've drawn this acceptable?

Thanks again

[discodermolide]

Hi,

Thanks for the advice guys,

I've written a new reaction scheme.

The steps are:

1. -OH nucleophile abstracts H from the carboxylic acid (the most acidic hydrogen in the system), O-H bond breaks, pair of electrons move to the oxygen
2. The anionic O is a strong nucleophile and attacks the weakly electrophilic alpha carbon, in an SN2 reaction - the Br is the leaving group. Inversion of configuration occurs, and an epoxide type strained-ring structure is formed

Some confusion at this point:
Either the acid protonates the oxygen in the epoxide ring, then a new OH- nucleophile attacks the alpha-carbon to form a hydroxyl group, opening the ring..and lastly the O- is protonated.

Or

The way I drew the mechanism

3. OH- nucleophile attacks the alpha-carbon, by SN2. The ring opens and inversion has occurred (so the configuration at the alpha carbon has been retained)
4. Acid protonates the O- to form the final product.

Would the acid protonate the epoxide-type ring first? Or is the way I've drawn this acceptable?

Thanks again

This mechanism is OK. Only thing is the intermediate is an alpha-lactone, not an epoxide or epoxide like.

[nox]

umm the poster did draw an alpha-lactone

[discodermolide]

umm the poster did draw an alpha-lactone

Yes I know that but he referred to it as epoxide or epoxide like which it is not!

[pfnm]

No worries, I wasn't sure what it was called - I haven't encountered alpha-lactones until now, so thanks for letting me know.