July 29, 2010
11:01 PM sleepy:
what is going on with the 2 NH2 groups?
11:00 PM sleepy:
i have a question about flashcard rxn #6 on pg 4
10:59 PM sleepy:
hi dr. nerz
10:45 PM nerz:
ethylene glycol can be used to protect a carbonyl in a grignard
10:44 PM nerz:
alpha and beta both form anomers are in equilibrium
09:59 PM robyn:
dr nerz, what is the point of using an ethylene glycol? why would we use it in a synthesis?
09:53 PM Room 336:
Does it matter if we make an alpha vs. a beta in that situation?
09:52 PM Room 336:
new question: flashcard sheet 7, reaction 8
09:48 PM Room 336:
sorry, nevermind re: ketal formation.
09:44 PM Room 336:
Different question: for ketal formation...
09:39 PM Room 336:
Thanks Dr Nerz
09:38 PM Room 336:
I was thinking the two ketones would be like dienes in terms of resonance
09:38 PM Room 336:
Sigh... I feel dumb. That is very true
09:28 PM nerz:
you can't delocalize from the benzene to teh center carbonyl
09:28 PM nerz:
well you can't draw any resonance onto the the center carbonyl from the aromatic
09:22 PM Room 336:
Is the middle ketone, in fact, NOT in resonance with the rest of the molecule?
09:21 PM Room 336:
ooooh.. wait. Are carbonyls in general termini for conjugation the way aldehydes are termini for chains?
09:18 PM nerz:
friedel crafts do not work on deactivated rings in general.
09:18 PM Room 336:
I would have thought there was conjugation to both of the other carbonyls, which are in turn conjugated with the ring.
09:18 PM nerz:
robyn you are right about the friedel crafts. Other deactivated rings do react
09:17 PM Room 336:
...But why isn't the middle carbonyl also conjugated with the molecule?
09:17 PM nerz:
not likely to form carboxylic acid with base
09:16 PM Room 336:
yes, that one
09:16 PM nerz:
the other two carbonyls are conjugated with the aromatic ring. and reacting to form the hydrate would break up their conjugation
09:16 PM nerz:
in that case, the center carbon is the most delta plus.
09:15 PM nerz:
are you talking about the hydrate formation.
09:07 PM Room 336:
Would adding a little NaOH turn the reacted carbonyl into a carboxylic acid?
09:06 PM Room 336:
Hi Dr Nerz, could you please explain reaction 5 on Flashcard #7? Why is only one carbonyl attacked? Why not the other two ketones? Would adding a
08:27 PM Robyn Rubenstein:
also, will a benzene with an electron withdrawing group undergo sulfonation, nitration, and halogenation?
08:10 PM Robyn Rubenstein:
with a Br on it undergo friedel crafts? because Br is a deactivating group
08:09 PM Robyn Rubenstein:
Hi dr nerz. I am confused about friedel crafts because i know that a deactivated ring will not undergo a friedel crafts reaction, but will a benzene w
08:28 AM the prophet:
Any latest on hydrogen
07:09 AM nerz:
this is a chat room for bryn mawr organic chemistry students, we can help peoplw with organic chemistry questions, but it is only open an hour a day
July 28, 2010
10:40 PM scott:
anyone home?
10:39 PM scott:
anyone home?
10:27 PM scott:
given the balanced equation H3 PO4(aq) + 3NaOH(aq)->Na3 PO4(aq) + 3H2O what volume of .100M H3PO4 is required to exactly react with 15 mL of .154 M Na
10:23 PM scott:
pretest questions need a formula or how to do it fof my notes please anyone can help me?
09:56 PM nerz:
this is what we did when we worked with sugars before
09:56 PM nerz:
this is what we did when we worked with sugars before
09:54 PM nerz:
no it is just the cyclization of the sugar. This happens in water. water is the solvent.
09:51 PM nerz:
i better look at it, just a minute
09:50 PM atk:
okay so i'd just attack the carbonyl with h20?
09:48 PM nerz:
I believe that problem is just the cyclization of a straight chain sugar. Some chemistry involves the straight chain
09:43 PM atk:
like on the flashcards p7 #8
09:43 PM atk:
you have to convert it to a chair form in order to see how it reacts, right?
09:43 PM atk:
hi dr. nerz - when a sugar is in the haworth form,
09:34 PM nerz2:
at high enough temperrature
09:34 PM nerz2:
at high enough temperrature
09:34 PM Lusty Cup:
(and thanks for the answer) :-)
09:34 PM nerz2:
the carbonyls would definitely form as an intermediate
09:30 PM Lusty Cup:
Indeed I do! :-)
09:25 PM ner:
lusty cup do you work in the lusty cup?
09:25 PM ner:
eventually with enough heat it would probably break the chain into two carboxylic acids
09:24 PM ner:
when the addition reaction to the vicinol diols is done it is done normally at room temperature.
09:24 PM ner:
they could be, but it is dependent on conditions.
09:20 PM Lusty Cup:
due to being 2ndary alcohols in the presence of a strong oxidizing agent
09:20 PM Lusty Cup:
I was wondering why those alcohols aren't immediately turned into ketones by the KMnO4
09:19 PM Lusty Cup:
I know alkene oxidation is done with KMnO4 and H2O to make cis-vicinal diols
09:18 PM Lusty Cup:
Hi Dr Nerz, quick question:
09:12 PM ner:
you are welcome
09:11 PM Mooney:
Thank you!
09:10 PM ner:
we have been sticking because we have been attacking aldehydes and ketones
09:09 PM ner:
yes to both question, there will be no flipping on this test
09:09 PM Mooney:
carbon bearing the carbonyl? 2) during "flipping" is the double bond to oxygen reformed as the leaving group leaves?
09:08 PM Mooney:
Hi Dr. Nerz. I have a question about the "sticking" vs. "flipping" when reducing a carbonyl with a nucleophile. 1) Does the nucleophile attach to the
01:01 PM mari juana:
NaAlCO3(OH)3(s) + HCL (aq) --> ?
July 27, 2010
10:17 PM Anne:
will it be different for the LAH because the solvent is diethyl ether
10:12 PM Anne:
okay. but what about the diethyl ether?
10:11 PM nerz-storems:
in water it will be AL(OH)4
10:09 PM Anne:
one more question. for the reduction with LAH, after it has gone 4 times, does it end up as Al(OH)4? or can it be Al(OR)4 as in any group?
10:07 PM Lusty Cup2:
great thanks
10:06 PM nerz-storems:
yes regarding the sticking and flipping
10:06 PM Lusty Cup2:
(when attacked by a nucleophile)
10:06 PM Lusty Cup2:
Is it ALWAYS true that with an aldehyde or ketone, it sticks, and with an acyl halide or ester, it flips?
10:05 PM Lusty Cup2:
Dr Nerz just want to confirm something regarding stick and flip
10:05 PM nerz-storems:
yes, the ether is the solvent for lah and the acid is added at teh end to protonate
10:01 PM Anne:
oh whoops i think you are right
10:00 PM Lusty Cup2:
Anne I think it was diethyl ether
10:00 PM Anne:
Okay. but there is h3o+ present at the end to protonate the O- right?
09:59 PM Lusty Cup2:
That does make sense. I'll review that list again. Thanks :-)
09:59 PM nerz-storems:
you can do Na BH4 in an alcohol because it is less reactive.
09:59 PM nerz-storems:
you can do Na BH$ in an alcohol because it is less reactive.
09:58 PM nerz-storems:
with lah, you can't be in a protic solvent like an alcohol, it would react with it
09:58 PM nerz-storems:
cations are more stable that secondary and so forth
09:58 PM Anne:
Great thank you. another question about reduction with LiAlH4...the solvent has to be ethanol right? but we also put in h30+ to protonate?
09:58 PM nerz-storems:
it should make sense that alkyl groups are stabilizing to carbocations as tertiary
09:57 PM nerz-storems:
lusty cup it is not so bad, the only one that is op that does not have electron pair is alkyl
09:56 PM nerz-storems:
anne no you don't need to know that mechanism
09:54 PM Anne:
I know we vaguely looked at how a chromate ester gets hooked on
09:51 PM Lusty Cup2:
yeah, definitely EDG. Thanks for that :-)
09:50 PM Lusty Cup2:
I fear it may be time to get that chart out again... *sigh*
09:50 PM Anne:
Do we need to know the mechanism for oxidation of diphenols into quinones
09:49 PM Lusty Cup2:
I thought all EDGs had an electron pair to donate except halogens
09:49 PM nerz-storems:
you need an electron withdrawing group ortho and or para to the leaving group
09:47 PM nerz-storems:
Lusty cup, about last nights question that I missed the isopropyl group is an EDG
09:47 PM Lusty Cup2:
I had a question; let me go find it. :-)
09:47 PM Lusty Cup2:
Hi Dr Nerz
09:43 PM nerz-storems:
sorry I got distracted and lost track of tiem
July 26, 2010
10:16 PM Lusty Cup:
Why did we determine it to be an Elimination/Addition reaction?
10:16 PM Lusty Cup:
I wrote in my notes that NAS was unacceptable because it requires ortho/para, but isn't the LG actually para to the isopropyl group?
10:14 PM Lusty Cup:
then Br2 with FeBr3 to make 4-bromocumene
10:14 PM Lusty Cup:
first we reacted benzene with 2-chloropropane to make cumene
10:14 PM Lusty Cup:
first we reacted benzene with 2-chloropropane
10:13 PM Lusty Cup:
oops, just kidding. we were to synthesize 4-isopropylaniline
10:12 PM Lusty Cup:
We were to synthesize 4-bromoanaline from benzene
10:10 PM Lusty Cup:
I'm having a hard time distinguishing NAS from Elim/Addn. In my notes I wrote that NAS requires ortho/para
10:08 PM Lusty Cup:
Hi Dr Nerz, are you still there?
10:01 PM nerz-storems:
have a good night
10:00 PM atkhan:
great. thanks dr. nerz
09:59 PM nerz-storems:
if you need any more help with it let me know. I will do one tomorrow in the problem session
09:45 PM nerz-storems:
the diels alder traps the benzyne
09:43 PM nerz-storems:
by having it react with a diene
09:43 PM nerz-storems:
the trapping intermediates means that the triple bond or benzyne like intermediate is demonstrated
09:42 PM nerz-storems:
then the nh2 adds and it can add to any of the three carbons of the two triple bonds
09:41 PM nerz-storems:
this means the two directions for alkyne formaiton, involving three carbons
09:41 PM nerz-storems:
normally, the triple bond can form on either side of the halide making two triple bonds
09:41 PM atkhan:
ooo okay thanks
09:40 PM nerz-storems:
the base NH2 pulls off the beta hydrogen, the electrons flows in kicking out the leaving group froming a triple bond
09:40 PM nerz-storems:
the base NH2 pulls off the beta hydrogen, the electrons flows in kicking out the leaving group froming a triple bond
09:39 PM atkhan:
where is the benzyne coming from in the elim/addn rxn?
09:38 PM atkhan:
but i don't understand the 'trapping the intermediate' part we did
09:37 PM atkhan:
and i understand why there are 3 possible positions to attack based on the benzyne rxn
09:37 PM atkhan:
the cyclohexadiene
09:37 PM atkhan:
i understand what happens once a benzyne reacts with the cyclohexene
09:36 PM atkhan:
oh also - can you explain when benzyne forms in an elim/addn rxn?
09:31 PM nerz-storems:
they donate density making hte ring richer in electrons
09:30 PM nerz-storems:
they also block out the possibility of beta hydrogens
09:30 PM nerz-storems:
they are electron donating groups which supports eas not nas
09:29 PM atkhan:
i mean in terms of electron denstiy
09:29 PM nerz-storems:
i actually had it on the board to day but it was erased, but at ten I will start posting things
09:29 PM atkhan:
what affect do the methyls have on the benzene ring?
09:29 PM nerz-storems:
well I am going to list it on an email tonight and I will try to list it on the board on wed
09:28 PM atkhan:
ohh but there aren't any B Hs
09:28 PM atkhan:
ohh but there aren't any B Hs
09:28 PM nerz-storems:
to do the elimination one must have beta hydrogens adn that example had none
09:28 PM nerz-storems:
is not possible. Therefore the only option is elimination edition
09:27 PM nerz-storems:
without the electron withdrawing groups ortho and or para, the normal nas reaction
09:27 PM nerz-storems:
the only option is elimination edition. NH2 and NH3 are nucleophilic, but with
09:26 PM nerz-storems:
if the reaction is nucleophilic and there is a leaving group and there are no groups ortho and or para
09:25 PM atkhan:
could you explain the steps again to figure that out?
09:25 PM nerz-storems:
ok
09:25 PM atkhan:
i'm trying to walk through that again and can't reason out why its no rxn
09:25 PM Lusty Cup:
sum up all the material we've covered so far. You mentioned doing it in class today, but we ran out of time.
09:24 PM atkhan:
with br and then two methyls ortho to it and an aldehyde meta to it
09:24 PM Lusty Cup:
Hi Dr Nerz, I was hoping at the beginning of Lecture on Wednesday you could perhaps
09:24 PM atkhan:
it was the one that was no rxn
09:23 PM atkhan:
hi dr. nerz - you know the first problem we did in class this morning?
09:17 PM nerz-storems:
I am here, sorry I was late had to pick sons up from airport
July 25, 2010
10:29 PM Robin Shatz:
I appreciate your help. Good luck with your studies.
10:29 PM Robin Shatz:
thanks anyway for trying
10:28 PM sydney hyder:
sorry, i have to go. i may have confused you more than anything. i'm just a student too...
10:28 PM Robin Shatz:
so if there is a chiral center in a reaction mechanism, then it will produce a racemic mixture. Is that it?
10:28 PM sydney hyder:
http://en.wikipedia.org/wiki/Racemic_mixture
10:28 PM sydney hyder:
which your book hinted that there is because it showed both the dashed and wedged
10:27 PM Robin Shatz:
ok
10:27 PM sydney hyder:
it is only a racemic mixture if there is 50/50 of the chirality
10:27 PM Robin Shatz:
do you know why a mechanism, like the addition of bromine, would produce a racemic mixture?
10:27 PM sydney hyder:
wait!
10:27 PM sydney hyder:
because the carbon is a chiral center it can be a racemic mixture
10:27 PM Robin Shatz:
oh
10:26 PM sydney hyder:
the questionmark is as in im not 100% sure?
10:26 PM sydney hyder:
umm yes?
10:24 PM Robin Shatz:
is it just because carbon is a chiral center?
10:24 PM Robin Shatz:
sorry to keep bothering you, but do you know why the addition of a bromine to an alkene produces a racemic mixture?
10:22 PM sydney hyder:
yes
10:20 PM Robin Shatz:
so does chiral mean that 4 different groups are attached to the carbon?
10:20 PM sydney hyder:
oops sorry
10:20 PM sydney hyder:
chiral = racemic
10:20 PM sydney hyder:
chiral = racemic
10:20 PM sydney hyder:
chiral = racemic
10:20 PM sydney hyder:
chiral = racemic
10:20 PM sydney hyder:
the book is trying to show how the carbon attached to the bromine is asymmetric and is chiral
10:19 PM sydney hyder:
so a racemic mixture has to do with chirality. if the molecule is chiral than it can be racemic
10:18 PM Robin Shatz:
so the fact that the bromine can attack the carbon from the front or the back means that it is a racemic mixture?
10:17 PM Robin Shatz:
it's an addition of 1-methylcyclobutene
10:16 PM Robin Shatz:
there is no h20 involved in the reaction
10:16 PM Robin Shatz:
i am doing an addition of bromine to an alkene
10:15 PM sydney hyder:
it could attack on the front or the back of the molecule
10:14 PM sydney hyder:
the only reason why it has different dash/wedge is because the br can attack from either side
10:14 PM Robin Shatz:
thank god because I was not following your explanation
10:14 PM sydney hyder:
so its a br2 and h20 reacting with butene
10:13 PM Robin Shatz:
and there's a methyl group coming off of the bottom carbon
10:13 PM sydney hyder:
ive been doing the whole wrong thing
10:13 PM sydney hyder:
omg sorry, we are doing aromatic rings in class and i thought you were doing that reaction
10:12 PM Robin Shatz:
it's on a butene
10:12 PM sydney hyder:
is this on a benzene ring or a cyclohexane?
10:11 PM Robin Shatz:
and the methyl group is on a dash
10:11 PM sydney hyder:
therefore you will only have a bromine coming off a carbon that has a single bond to another carbon and a double bond to a carbon
10:11 PM Robin Shatz:
So, in the answer from the book, it has a bromine on a dash opposite the methyl group, a bromine on a wedge next to the methyl group and the methyl gr
10:10 PM sydney hyder:
aromatic ring
10:10 PM sydney hyder:
a water will come along, grab the hydrogen and the hydrogen will donate its electrons back into the ring recreating the conjugated bonds of the aromat
10:09 PM sydney hyder:
so on the ring you will have a carbon with a hydrogen and a bromine coming off
10:09 PM sydney hyder:
the double bonds from the ring will grab the br+ and loose for a second its sp2 shape and become sp3 therefore loosing its conjugated bonds
10:08 PM sydney hyder:
the br2 is activated usually with a catalyst (AlCl3) to become Br+
10:07 PM sydney hyder:
the methyl group on the benzene ring is an electron donating group which means the br2 is going to add to the ring most likely on the complete opposit
10:05 PM Robin Shatz:
I have not studied that concept yet.
10:05 PM Robin Shatz:
what do you mean by rearomatize?
10:05 PM Robin Shatz:
sorry, I am still in my 1st quarter of o-chem
10:04 PM sydney hyder:
*once
10:04 PM sydney hyder:
one you rearomatize you regain sp2 hybridization
10:03 PM sydney hyder:
since it is a benzene ring it should be sp2/trigonal planar all the way around.
10:03 PM Robin Shatz:
what do you mean by that?
10:02 PM Sydney Hyder:
i dont think there is any stereochemistry?
10:02 PM Sydney Hyder:
i dont think there is any stereochemistry?
10:01 PM Robin Shatz:
hi Sydney, would you know the answer to the question that I asked at 9:43pm? Thanks
09:56 PM Sydney Hyder:
for an NAS reaction, do the things ortho and para need to be actual functional groups?
09:56 PM Sydney Hyder:
sorry, didn't see that it posted on the top
09:56 PM Sydney Hyder:
dr nerz?
09:55 PM Sydney Hyder:
dr nerz?
09:55 PM Sydney Hyder:
dr nerz?
09:45 PM Robin Shatz:
what would be the answer for that problem and why?
09:45 PM Robin Shatz:
I don't understand how you know which bromine is on a wedge, which bromine is on a dash, and if the methyl group is on a wedge or a dash?
09:43 PM Robin Shatz:
here's the problem that I am stuck on: the addition of Br2 to 1-methylcyclobutene
09:39 PM Robin Shatz:
I don't understand what that means. Can you explain what you just wrote in simpler terms? I am studying for a final for this Friday. I will send you
09:38 PM nerz-storems:
because there it is an anti additon to a planar functional group and there are two equivalaent anti additions that are enantiomeric
09:30 PM Robin Shatz:
specifically, why does the addition of a bromine to an alkene produce a racemic mixture?
09:26 PM nerz-storems:
radicaks or when the chemistry takes place on a double bond.
09:26 PM nerz-storems:
radicaks or when the chemistry takes place on a double bond.
09:26 PM nerz-storems:
not sure what your question means, but a reaction results in a racemate in many situations when there are carbocatoins involved
09:24 PM Robin Shatz:
Dr. Nerz are you there?
09:17 PM Robin Shatz:
how do you know if a mechanism is racemic?
09:14 PM Robin Shatz:
how do you know if a mechanism is racemic?