Slew of organic chemistry concept/practical questions

[thebillsfan]

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1. What is an enediol and the enediol rearrangement?
2.Why are internal alkenes more stable than terminal alkenes?
3.How can an Sn1 reaction NOT depend on concentration of the acid? Since protonation does need to occur, so I higher concentration of acid would cause more protonation and higher rate of carbocation formation.
4. Does the branching of alkanes lower or increase MP? I always thought that it lowered it b/c it reduced the surface area for VDW forces. However I've also heard the reasoning that lowering the surface area, ie making it more spherical, would increase packing and INCREASE VDW forces. So i'm not really sure.

Anyway, I'd appreciate any help to these question. Thanks and good luck!

 

[raab32]

Enediol rearrangements allow the conversion of glucose to fructose. Glucose (http://upload.wikimedia.org/wikipedia/commons/2/27/D-glucose-chain-2D-Fischer.png) is an aldose because it has a terminal carbonyl group, making it an aldehyde sugar. Fructose, on the other hand, (http://chem.pdx.edu/~wamserc/CH331F97/Egifs/E2_3a.gif) is a ketose because carbon 2 is a ketone. We want to essentially move the carbonyl group one carbon over and the enediol rearrangement allows this to happen.

Here's a basic sketch of the reaction: http://www.grossmont.net/tomolmstead/Chem232_2007/carbohydrates_study_guide_files/image030.gif. An enediol is an alkene with two hydroxyl groups attached to it, and it is an intermediate in the glucose --> fructose conversion. The reaction proceeds by converting glucose to an enolate anion, which then accepts a proton to form the enediol. However, we can create a more stable compound than the enediol, so we form a different enolate, where the soon-to-be carbonyl carbon is now C2 instead of C1. With some electron rearrangement to form the more stable carbonyl, we now have a carbonyl at C2 instead of C1, which gives us fructose from glucose.

This is essentially how high fructose corn syrup is made... a catalyzed reaction of corn syrup (which is made from corn starch) undergoing these rearrangements.

 

[G1SG2]

2.Why are internal alkenes more stable than terminal alkenes?
They have a higher degree of substitution, which is good because alkyl groups are electron donating.

3.How can an Sn1 reaction NOT depend on concentration of the acid?
Since protonation does need to occur, so I higher concentration of acid would cause more protonation and higher rate of carbocation formation.
Protonation does not always need to occur. Consider t-butyl chloride. The Cl- leaves spontaneously. Also, if there is protonation (such as in the case of an alcohol), you can have as much as acid as you want, but if you don't have enough substrate to protonate, who cares about how much acid you have? SN1 depends only on the concentration of the substrate.

4. Does the branching of alkanes lower or increase MP? I always thought that it lowered it b/c it reduced the surface area for VDW forces. However I've also heard the reasoning that lowering the surface area, ie making it more spherical, would increase packing and INCREASE VDW forces. So i'm not really sure.

It increases the melting point because branching allows it to pack better. The lowering of the surface area and reducing van der waals forces is correct for lowering the boiling point. Increase packing resulting from branching is the reason for the higher MP.

 

[thebillsfan]

2.Why are internal alkenes more stable than terminal alkenes?
They have a higher degree of substitution, which is good because alkyl groups are electron donating.

With this logic, are ketones more stable than aldehydes?
Also, do internal alkenes have a higher bp then terminals? If that's the case, then don't necessarily understand how that would lead to greater intermolecular bonding between the molecules.

 

[thebillsfan]

Enediol rearrangements allow the conversion of glucose to fructose. Glucose (http://upload.wikimedia.org/wikipedia/commons/2/27/D-glucose-chain-2D-Fischer.png) is an aldose because it has a terminal carbonyl group, making it an aldehyde sugar. Fructose, on the other hand, (http://chem.pdx.edu/~wamserc/CH331F97/Egifs/E2_3a.gif) is a ketose because carbon 2 is a ketone. We want to essentially move the carbonyl group one carbon over and the enediol rearrangement allows this to happen.

Here's a basic sketch of the reaction: http://www.grossmont.net/tomolmstead/Chem232_2007/carbohydrates_study_guide_files/image030.gif. An enediol is an alkene with two hydroxyl groups attached to it, and it is an intermediate in the glucose --> fructose conversion. The reaction proceeds by converting glucose to an enolate anion, which then accepts a proton to form the enediol. However, we can create a more stable compound than the enediol, so we form a different enolate, where the soon-to-be carbonyl carbon is now C2 instead of C1. With some electron rearrangement to form the more stable carbonyl, we now have a carbonyl at C2 instead of C1, which gives us fructose from glucose.

This is essentially how high fructose corn syrup is made... a catalyzed reaction of corn syrup (which is made from corn starch) undergoing these rearrangements.

So does this happen spontaneously in a basic solution of glucose, considering fructose being an ketone is probably more stable than glucose?

 

[G1SG2]

With this logic, are ketones more stable than aldehydes?
Also, do internal alkenes have a higher bp then terminals? If that's the case, then don't necessarily understand how that would lead to greater intermolecular bonding between the molecules.

Yeah, aldehydes are more susceptible to nucleophilic attack than ketones because aldehydes are relatively electron poor (1 electron donating alkyl group instead of 2 for the ketone). In that sense, I guess we can say ketones are "more stable".

I'm not too sure about the relative boiling points. I would think that internal alkenes have a lower boiling point since the double bond makes kinks within the chain, reducing van der waals forces. I'm not sure how that would correspond to the stability of internal alkenes. Maybe someone else can chime in.

 

[austinap]

Yeah, aldehydes are more susceptible to nucleophilic attack than ketones because aldehydes are relatively electron poor (1 electron donating alkyl group instead of 2 for the ketone). In that sense, I guess we can say ketones are "more stable".

I'm not too sure about the relative boiling points. I would think that internal alkenes have a lower boiling point since the double bond makes kinks within the chain, reducing van der waals forces. I'm not sure how that would correspond to the stability of internal alkenes. Maybe someone else can chime in.

I think you need to be careful when talking about stability, as there are really two types of stability: thermodynamic stability and kinetic stability. Aldehydes being more reactive than ketones is an example of kinetic stability. Aldehydes react more quickly, and thus are kinetically less stable than ketones. However, this in itself implies nothing about the relative thermodynamic stability of the two functional groups.

Think of it this way: to compare kinetic stability of the two groups, you'd need to consider two different reactions:

1. CH3CH2CHO ----[CN-]----> CH3CH2C(OH)CN
2. CH3COCH3 ----[CN-]-----> CH3C(OH)(CN)CH3

The relative kinetic stability can be obtained by taking ratios of the two reaction rates. To determine thermodynamic stability, you'd need to look at the equilibrium constant of the isomerization of the two reactants:

CH3CH2CHO <----------> CH3COCH3


As far as the BP of internal vs. terminal alkenes, internal double bonds typically lead to lower BP because the kinking usually makes it harder for the molecules to pack tightly. Numbers for how much this matter are outside the realm of the MCAT.

As far as the branching vs. melting point question, it depends on the specific alkane. Small alkanes typically show increased melting points with increased branching since they can form extremely tight-packed structures (think of this almost light a crystal lattice type structre). However, this trend isn't general, and larger alkanes show a decrease in melting point with increased branching, since this makes it harder for the molecules to pack tightly together.

For these trends, the thing to think about is surface contact. Can you pack the molecules together in such a way that their surfaces contact significantly? That will lead to an increased melting temperature. The opposite also holds.

 

[amberisma]

3.How can an Sn1 reaction NOT depend on concentration of the acid? Since protonation does need to occur, so I higher concentration of acid would cause more protonation and higher rate of carbocation formation.

The reaction depends on the rating determining step. The RDS involves forming a carbocation, so it only depends on the substrates ability to form a carbocation. Only the substrate is involved in the rate determining step, hence Sn1. So no matter how much or little acid there is, the reaction will only proceed as fast as the leaving group is able to separate from the substrate.