Ochem reaction question [Kaplan Section tests 2]

[FeinMS]

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Consider the two reversible reactions:
I. HOCH2CH2COOH + H+(H2O) <===> substance X
II. HOCH2CH2CH2CH2COOH + H+(H2O) <===> Substance Y
Which forward reaction would be expected to have the larger equilibrium constant?

A. I, because X is more stable than Y
B. II, because Y is more stable than X
C. I, because X is less stable than Y
D. II, because Y is less stable than X.

A&E and my opinion in white below
I thought the H+ H2O will do E1 reaction. As E1 products, X has conjugation while Y doesn't, so I chose A.
The correct answer is B because the reaction doesn't follow E1 but it follows lactone formation. (protonation of carbonly oxygen, nucleophilic attack by alcohol) Because Substance Y has less ring strain and more stable, the answer is B.
Am I totally wrong to think of E1 in this case? How do we know if OH can reach the carboxyl carbon?

 

[Idrinkwater]

I though along the same reasoning as you- that it is an elimination rxn. Lactone formation- it looks like what is being said is that the longer the hydrocarbon chain, the less ring strain in any possible cyclic ester (lactone). I will check on this, and come back to this thread.

I Forgot what a lactone was wow.

 

[SSerenity]

Consider the two reversible reactions:
I. HOCH2CH2COOH + H+(H2O) <===> substance X
II. HOCH2CH2CH2CH2COOH + H+(H2O) <===> Substance Y
Which forward reaction would be expected to have the larger equilibrium constant?

A. I, because X is more stable than Y
B. II, because Y is more stable than X
C. I, because X is less stable than Y
D. II, because Y is less stable than X.

A&E and my opinion in white below
I thought the H+ H2O will do E1 reaction. As E1 products, X has conjugation while Y doesn't, so I chose A.
The correct answer is B because the reaction doesn't follow E1 but it follows lactone formation. (protonation of carbonly oxygen, nucleophilic attack by alcohol) Because Substance Y has less ring strain and more stable, the answer is B.
Am I totally wrong to think of E1 in this case? How do we know if OH can reach the carboxyl carbon?

The answer is B.
Draw out the compound. A 4 membered ring would be insanely unstable. This is a nucleophillic attack, not an E1 reaction. Remember this order of stability: 5-Member Rings > 6-Membered Rings > 4 > 3

5 Membered ring formation is the most stable. 6 Is also pretty good, definitely more stable than 4.

 

[FeinMS]

The answer is B.
Draw out the compound. A 4 membered ring would be insanely unstable. This is a nucleophillic attack, not an E1 reaction. Remember this order of stability: 5-Member Rings > 6-Membered Rings > 4 > 3

5 Membered ring formation is the most stable. 6 Is also pretty good, definitely more stable than 4.

My question is why the reaction follows nucleophilic attack instead of E1. Thanks tho

 

[SSerenity]

My question is why the reaction follows nucleophilic attack instead of E1. Thanks tho

Sorry, didnt see the part in white. I see, so you thought this was an E1 with alcohol as our leaving group?

I think we would just need a stronger base to pull that proton off. H2O in the presence of acid would not be strong enough to do that.

 

[Schenker]

Sorry, didnt see the part in white. I see, so you thought this was an E1 with alcohol as our leaving group?

I think we would just need a stronger base to pull that proton off. H2O in the presence of acid would not be strong enough to do that.

Base strength is not of much importance for E1 reactions, as the rate limiting step is carbocation formation.

My question is why the reaction follows nucleophilic attack instead of E1. Thanks tho

An E1 reaction is unlikely to occur, because the carbocation intermediate would be unstable. Primary carbocations are unstable:

I. +CH2CH2COOH (unstable)
II. +CH2CH2CH2CH2COOH (unstable)

Thus, the more likely reaction is the one that proceeds by nucleophilic attack, i.e. the one that form the lactone.

Simply put, because this is an acid-catalyzed reaction, there will be a positively charged intermediate somewhere along the reaction pathway. The most stable positively charged derivative of each parent compound is the one where the carbonyl oxygen is protonated.

 

[FeinMS]

Base strength is not of much importance for E1 reactions, as the rate limiting step is carbocation formation.



An E1 reaction is unlikely to occur, because the carbocation intermediate would be unstable. Primary carbocations are unstable:

I. +CH2CH2COOH (unstable)
II. +CH2CH2CH2CH2COOH (unstable)

Thus, the more likely reaction is the one that proceeds by nucleophilic attack, i.e. the one that form the lactone.

Simply put, because this is an acid-catalyzed reaction, there will be a positively charged intermediate somewhere along the reaction pathway. The most stable positively charged derivative of each parent compound is the one where the carbonyl oxygen is protonated.

Wow nice point! I can't believe I missed that...