Which is more acidic

[chiddler]

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A CH with 3 aromatic rings attached, or a CH with 3 butanes attached?

I would specifically like to know about the benzenes. As constituents, how much do they increase acidity in an alkane? And why?

If I remember correctly (probably not 😛), -Ar groups are slightly electronegative so they would stablize and make the alkane more acidic by inductive effect.
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A second question: Pi bonds are made of overlapping P orbitals. How does hybridization contribute here? Where are the sp2 electrons in an alkene?

<3

 

[MedPR]

A CH with 3 aromatic rings attached, or a CH with 3 butanes attached?

I would specifically like to know about the benzenes. As constituents, how much do they increase acidity in an alkane? And why?

If I remember correctly (probably not 😛), -Ar groups are slightly electronegative so they would stablize and make the alkane more acidic by inductive effect.
--

A second question: Pi bonds are made of overlapping P orbitals. How does hybridization contribute here? Where are the sp2 electrons in an alkene?

<3

1. I would assume that a substituent benzene would increase the acidity due to resonance stabilization. A benzene carbocation is more stable than a butane carbocation. Thus, comparatively the benzene is "more likely" to donate an acidic hydrogen (or accept an electron pair) than is butane. The same should be true for the Hydrogen bound to the central carbon. No clue by how much though.

2. Double bonds are 1 sp2 bond (the sigma bond) and 1 pi bond (the electrons above and below the sigma bond). In addition to the sigma bond in C=C, the substituents of each carbon (4 in total) are bonded with an sp2 hybridized bond. So, Ethene (C2H4) has 1 pi bond, and five sp2 sigma bonds.

Are your questions from a book -- are there definitive answers?

 

[study pray love]

this kind of stuff is gray territory to me.
if we don't have to know about aromatics for the mcat, would we have to know about aromatic acidic stabilization?

(for the record i knew the answer =p)

 

[MedPR]

this kind of stuff is gray territory to me.
if we don't have to know about aromatics for the mcat, would we have to know about aromatic acidic stabilization?

(for the record i knew the answer =p)

Assuming we are right about benzene substituents being more acidic than butane substituents, the concept is resonance, which is on the MCAT.

 

[chiddler]

1. I would assume that a substituent benzene would increase the acidity due to resonance stabilization. A benzene carbocation is more stable than a butane carbocation. Thus, comparatively the benzene is "more likely" to donate an acidic hydrogen (or accept an electron pair) than is butane. The same should be true for the Hydrogen bound to the central carbon. No clue by how much though.

2. Double bonds are 1 sp2 bond (the sigma bond) and 1 pi bond (the electrons above and below the sigma bond). In addition to the sigma bond in C=C, the substituents of each carbon (4 in total) are bonded with an sp2 hybridized bond. So, Ethene (C2H4) has 1 pi bond, and five sp2 sigma bonds.

Are your questions from a book -- are there definitive answers?

1. A benzene carbocation is not stable at all because it breaks aromaticity! Nor is it likely to accept an electron pair for the same reason, right?

2. Specifically regarding the double bond: sp2 is made of a hybrid of s and p orbitals. Pi bonds are made of p orbitals. Where do the p orbitals come from if all the valence p's are hybridized?

The first question is not directly book inspired, the second question was prompted by an EK1001 question.

thanks!

 

[Ssina]

not all of the p orbitals are hybridized, only two out of three are hybridized, so we have 3 sp2 hybridized (one for the sigma bond and two for C-H bond) and a unhybridized p orbital for the double bond

 

[Ssina]

1. A benzene carbocation is not stable at all because it breaks aromaticity! Nor is it likely to accept an electron pair for the same reason, right?

I also think it will be more acidic because of the resonance, even though, it'll break the aromaticity. After all, the choices are resonance (via phenyl groups) or inductive (from the alkyl groups) and resonance always wins.

e.g. triphenylmethane (wiki) with pka 31 vs. tri-t-butylmethane (couldn't find a pka value for this one, but this is just an example)

 

[MedPR]

1. A benzene carbocation is not stable at all because it breaks aromaticity! Nor is it likely to accept an electron pair for the same reason, right?

2. Specifically regarding the double bond: sp2 is made of a hybrid of s and p orbitals. Pi bonds are made of p orbitals. Where do the p orbitals come from if all the valence p's are hybridized?

The first question is not directly book inspired, the second question was prompted by an EK1001 question.

thanks!

1. I guess it does break aromaticity according to Huckel's rule, but wouldn't it be able to take the electrons from the CH bond, so the positive charge ends up on the central carbon = tertiary carbocation stabilized by 3 benzenes? Obviously whether or not the benzene will donate an electron is not really determined by how it can rearrange, but that was my thought process.

2. sp2 implies 3 orbitals hybridizing. 1 s, and 2 p, leaving one "empty" p orbital which is actually filled by electrons forming the pi bond.

 

[chiddler]

I also think it will be more acidic because of the resonance, even though, it'll break the aromaticity. After all, the choices are resonance (via phenyl groups) or inductive (from the alkyl groups) and resonance always wins.

e.g. triphenylmethane (wiki) with pka 31 vs. tri-t-butylmethane (couldn't find a pka value for this one, but this is just an example)

good point. the image was also very helpful. Also, it wouldn't be tri-butylmethane! Draw it out it'll be 5,5-dibutylnonane. Not to be a semantic prick, but correcting for learning.

 

[Ssina]

👍 yeah definitely for your example.. I actually meant a carbone bounded to 1 hydrogen and three tert-butyl groups since I used triphenylmethane

 

[MT Headed]

Also, to contribute to the pedantry, if an H+ leaves a carbon, you will have a carbanion (C-), and not a carbocation (C+).

 

[MedPR]

Also, to contribute to the pedantry, if an H+ leaves a carbon, you will have a carbanion (C-), and not a carbocation (C+).

Lol.. I'm in worse shape for organic than I thought. Completely missed that.