Quick Orgo Question

[xjoohn]

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This is from the acs study guide:

The answer is B.

I was deciding between B or D. If I'm understanding correctly, the answer is B because -OCH3 takes up more space than the ethyl substituent. But then I read somewhere that a methyl takes up more space than an hydrooxyl (am I wrong here?). So why is is that -OCH3 is bigger than ethyl?

Thanks

 

[Labrat07]

I think you're overthinking the expoxide reaction. For what I know, the nucleophile attack the more substituted carbon (acidic) and less sub (basic). There's no where it mention about anything about conformational changes.... I'd just pick the safest best with ethyl at its orginal axial.

You're confused about epoxide and stability of equatiorial/axial. they're not relate in this situation

 

[dechristine]

This is an SN1 reaction because CH3OH is a weak nucleophile. Because CH3OH is a weak nucleophile, it will attack the more substituted side of the epoxy ring. In this case, the more substituted side is the side with the ethyl group. When CH3OH attacks that side of the epoxy ring, the epoxy ring breaks.

The reason the -OCH3 ends up in the down position is because the CH3OH attacks from the bottom of the ring. It couldn't possibly attack from the top because the epoxy ring takes up too much space. So, although the ethyl group starts in the down position, it gets "pushed" to the up position after the CH3OH attacks.

 

[Labrat07]

Thank you Dech. Understand it clearly now

 

[Cawolf]

The only error in the drawn mechanism is that the epoxide oxygen is protonated first. This formal charge on the oxygen lowers the LUMO of the tertiary ring carbon allowing for the SN1-style attack. It is also incorrect to call it a SN1 reaction as the mechanism of attack is concerted. It would be more correct to call it a "SN1-style" attack as the regioselectivity is clearly SN1 but the concerted mechanism is representative of SN2.

 

[xjoohn]

Thanks guys,

I have another question:

 

[Cawolf]

What is your question?

It is a simple Grignard reaction. You can use the forum as an answer key, but you will be more successful if you at least post what you don't understand.

 

[xjoohn]

Oops forgot the mention the question.

I used the grignard's reaction to get the answer B, but the answer is actually D. Can anyone explain to me why?

 

[Cawolf]

Oops!

Not a simple Grignard reaction as I clearly overlooked.

The key point is that Grignard reagents are strong bases and will react with acidic protons competitively or even preferably over nucleophilic attack at a carbonyl. This is why alcohols, acids, water, etc are not stable conditions for a Grignard reagent.

In this case the Grignard will immediately deprotonate the alcohol to form methane and then the aqueous workup simply converts the alkoxide back into the alcohol.

Your solution would be correct if the molecule did not also have an alcohol.

 

[xjoohn]

Thanks,

I have another question about a different problem:

Here, the answer is D. I understand all the reactions except 3; can someone explain to me what steps are involved in this?

Thanks.

 

[Labrat07]

3 is a dehydration problem. You have alcohol + H = carbocation to alkene.

You take out the oh . you have carbocation at 1st carbon. Since there's a hydrogen on carbon 2. It'll do a hydride shift and carbocation move to carbon 2 = more stable. Then we have tert.

My question is number 2. I assume the final product for two is secondary only

 

[Cawolf]

Right, 2 will only form secondary carbocations.

 

[Cawolf]

Also, OP, I doubt EAS problems are on the MCAT. Are you using really old materials?

 

[ciruji]

I want to add that this is a Sn1 rxn because we are in an acidic solution. Since the O will get protonated, it is going to become unstable with the positive charge and is going to break inn a way that leaves the more stable "carbocation"(no carbocation is formed) .

 

[Cawolf]

@ciruji

It is in no way an Sn1 reaction. What do you mean no carbocation is formed? The oxygen protonates (as you say) and leave as water while undergoing a concurrent hydride shift to form a free carbocation.

The carbocation formation may be identical, but the substitution is in an aromatic ring - making it EAS.

 

[ciruji]

@ciruji

It is in no way an Sn1 reaction. What do you mean no carbocation is formed? The oxygen protonates (as you say) and leave as water while undergoing a concurrent hydride shift to form a free carbocation.

The carbocation formation may be identical, but the substitution is in an aromatic ring - making it EAS.

I'm talking about the epoxide substitution (OP original problem).

 

[Labrat07]

Right, 2 will only form secondary carbocations.

Thanks Cawolf. As always, you're the best.

 

[Teleologist]

The only error in the drawn mechanism is that the epoxide oxygen is protonated first. This formal charge on the oxygen lowers the LUMO of the tertiary ring carbon allowing for the SN1-style attack. It is also incorrect to call it a SN1 reaction as the mechanism of attack is concerted. It would be more correct to call it a "SN1-style" attack as the regioselectivity is clearly SN1 but the concerted mechanism is representative of SN2.

How does protonation lower the LUMO on (not on, but involving) carbon?

 

[Cawolf]

@Teleologist

No expert, but here is my take on it.

Nucleophilic attack at carbon occurs at the antibonding sigma orbital (LUMO) that forms the bond with the oxygen. This is most likely as the electronegative oxygen pulls electron density towards it's end of the sigma bond and leaves a relatively electrophilic sigma* orbital. Protonation of the oxygen develops further cationic character on the carbon as the formally charged oxygen pulls more electron density towards it's end of the sigma bond. This shift in electron density shifts the energy level of the sigma* orbital closer to the HOMO (lowers it) of the nucleophilic lone pair on methanol.