CrO3

RogerRabbit, they stated "aqueous CrO3" ie. excess CrO3 I assumed.

 

Although CrO3 is considered a weak oxidizer, many organic profs agreed that it would oxidize a primary alcohol to a ald and again to Carb. Acid. Most of the time if you want just aldehyde ox. You also add PCC, which limits amount of O2 available in reaction mix.

 

I believe the solvent was CH2Cl2 and not aqueuous. does that change the answer?

 

Yay, spent the last ~4 mins on this ques, changed from aldehyde to COOH. But that doesn't change my state of bliss 'cause I thought I repeatedly indicated that it was "aqueous"? Have I not been clear? Heheh k/d...

 

roger, wrong!

it's COOH! I have consulted my orgo book and teachers on this.

 

<a href="http://www.rdc.ab.ca/rdc/organic_chemistry/chemistry/251_seminars/alcohol_reactions.pdf" target="_blank">http://www.rdc.ab.ca/rdc/organic_chemistry/chemistry/251_seminars/alcohol_reactions.pdf</a>

pwned!

 

ahemm...Blitz...I believe the solvent was CH2Cl2.
I think it's kinda funny how we all are going insane over this and AAMC is prolly thinkin "hehehe suckazz" with the 8 million they made in one day. <img border="0" title="" alt="[Eek!]" src="eek.gif" /> <img border="0" title="" alt="[Eek!]" src="eek.gif" />
keep it real!!
Tweetie

 

ahem, tweet, aldehyde is gotten with ch2cl2 ONLY with PCC....dude, CrO3= mighty oxidizer..read that link again..

 

Hey guys,

You didn't need outside info to answer this one... the answer was in the passage. It said that CrO3 oxidized "aldehydes and alcohols" in the flow chart, which means that both must be taken to COOH. There's no reason an alcohol would stop at aldehyde if aldehydes themselves are oxidized. Maybe the passage was wrong, because it sounds like 1o alcohols don't always go to COOH. BUT, in the passage, it said they do.

Does anyone else remember the flow chart of reactivities from the passage, to corroborate this?

 

Yes, I do - that is exactly why I chose carboxylic acid. How can you identify an aldehyde with Cr03 if the aldehyde is not oxidized to a carboxylic acid?

 

according to the chart, in AgNO3, alkane would have formed a ppt, cos it's not soluble. However, this isnt' the case. Alkyl halide forms the ppt, even though it's soluble in AgNO3..I put alkane cos of the chart alone,and am kicking myself. boo

 

H2Cr207 and Cr02 both will oxidize a primary alcohol all the way to a carboxylic acid. The aldehyde is formed as an intermediate but is further oxidized into the COOH.

If you want to create an aldehyde from a primary alcohol you have to use PCC or PDC (I think its PDC...).

By the way.... HF is far more reactive than HBr, HCl, and HI. If you spill concentrated HCl on you hand you can wash it off but if you spill concentrated HF on your hand it will only stop when it reaches the bone.

Flourine is so electronegative that it doesn't care if there is a H+ there or not so it is very very very acidic.

 

USC Premed-- I agree with you.

Blitz-- The only way an alkane will form a precipitate in solution is if it reacts, since alkane solids don't exist at STP. The precipitate formed was not because of a failure of an alkane to dissolve in AgNO3, but because of the alkyl halide reacting with AgNO3. If you think about it, the alkane layer and the aqueous AgNO3 layer wouldn't mix, they'd just form two phases.

esoteric-- HI is much more acidic than HF. Check your Ochem text. This is because I- is much more stable than F-. Both anions form in solution pretty readily, but much more I- is formed (pH of 1M HI &gt; pH of 1M HF) HOWEVER, since the F- is less stable, it is more reactive, which explains the violent reactivity you described.

 

Ok the only thing I'm wondering about now is...
Do I ever have to look at another frickin ochem reaction for as long as I live? I hope they don't cover this shizzat during medical school.

 

I know what you mean. The thing I'm pissed about is that after studying for the MCAT, I finally get a lot of stuff that I didn't get in undergrad Ochem. Oh well Med school shouldn't have too much Ochem, thank God.

 

</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by Alli Cat:
<strong>USC Premed-- I agree with you.

Blitz-- The only way an alkane will form a precipitate in solution is if it reacts, since alkane solids don't exist at STP. The precipitate formed was not because of a failure of an alkane to dissolve in AgNO3, but because of the alkyl halide reacting with AgNO3. If you think about it, the alkane layer and the aqueous AgNO3 layer wouldn't mix, they'd just form two phases.

esoteric-- HI is much more acidic than HF. Check your Ochem text. This is because I- is much more stable than F-. Both anions form in solution pretty readily, but much more I- is formed (pH of 1M HI &gt; pH of 1M HF) HOWEVER, since the F- is less stable, it is more reactive, which explains the violent reactivity you described.</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">ya, I know. That's why I'm so stupid. it was an easy question <img border="0" title="" alt="[Frown]" src="frown.gif" />

 

So are we 100% sure it was HF and not HI that was the most reactive? I'm still not sure about this one!!

 

If you're on good terms with your Ochem professor, you could ask him/her <img border="0" title="" alt="[Wink]" src="wink.gif" /> The way I understand it, I- is a much more stable anion than F-, so it's less likely to react. However, I'm pretty sure the Q wasn't as straightforward as "Which halo acid is most reactive?" I put down HI, and while I don't remember the exact Q, I remember going through this reasoning and getting HI as the answer.

Good luck Mish!

 

In the argument over HF or HI being the most acidic, I'll throw in my $0.02.

The strongest acid is the one with the most stable conjugate base, correct?!?

That would make the Iodine ion is much MUCH more stable than the Flouride ion, thereby making HI more acidic than HF...I could be wrong on this one...but that's just my opinion.

 

Acid Ka
HF 7.2 X 10^-4
HCl 1 X 10^6
HBr 1 X 10^9
HI 3 X 10^9

So, it all depends on how you interpret this question. If you equate most reactive with the largest Ka or dissociation constant, then you would choose HI as the answer. I- is a more stable base than F-, thus making it much easier for HI to donate a proton. If you look at reactivity from an Ochem standpoint (addition of HX to an alkene), than HF is clearly the right answer. Personally, I don't even remember what the question asked anymore, but I put HF. It was in the Bio section right? I guess I was in orgo mode, so I wasn't thinking about it from a Gen Chem standpoint.

 

Ok I'm back. I checked out some websites and I figured out where I was in error. HF is more reactive.....but less acidic. The bond dissoc. energy for H-F is huge compared to H-I.
I have no idea what the mcat problem said but I'm sure you guys knew what you were doing. Good Job! It's too bad the problem didn't include what type of reaction we were dealing with...

See ya.

-esoteric