<p>@iloveochem: thanks for that extra information. </p>
<p>can you answer an additional question about NH2NH2 and OH-?</p>
<p>what if this was treated with NH2NH2, OH-, and heat?</p>
<p><a href=“ImageShack - Best place for all of your image hosting and image sharing needs”>ImageShack - Best place for all of your image hosting and image sharing needs;
<p>what would be reduced? only the C=O bond closest to the benzene ring? would nothing happen to the NO2 group?</p>
<p>@calbear: i wish i could answer that for you but i haven’t gotten there yet… sigh</p>
<p>I believe… both C=O’s will get reduced, since Wolff-Kishner isn’t selective. To reduce only the C=O closest to the ring, you would use Pd/C, NH4+, HCO2-.</p>
<p>@calbear: regarding your HCl question, I don’t think it really matters lol. it’s just a matter of having HCl molecules in there in the first place. </p>
<p>i think if you’re asking questions like that, you’ll be fine lol</p>
<p>@iloveochem: in my notes, i have it as only reducing the C=O closest to the benzene. however, i also circled the other C=O and put an asterik next to it, LOL. i think it means i made a mistake and told myself to correct it later. unfortunately, i don’t think i ever did. 
</p>
<p>Zn(Hg) & HCl would reduce everything though right? because i have in my notes that Zn(Hg) & HCl and NH2NH2, OH-, heat do the same thing as each other.</p>
<p>thanks!!!</p>
<p>edit: actually, would Zn(Hg) & HCl reduce the NO2 as well?</p>
<p>@batman: i didn’t read the book. i did last year for frechet’s class, but someone i know told me it was unnecessary (for pedersen at least)
so i haven’t been doing it…</p>
<p>I can’t even answer some of these questions because I’m only on problem set 3. luld.
Question, why is a 1,3-butadiene aromatic if you take 2 electrons away? or did I write something wrong?</p>
<p>Mech I think you are confusing (NH2NH2, OH, heat) with (NH4^+, HCO2^-, pd/c). NH4^+, HCO2^-, pd/c is a reagent that is selective for allylic carbonyl groups. In lecture his example was a carbonyl next to a benzene ring (hence the allylic part).</p>
<p>cuz then it fulfills the 4N+2 rule
N becomes 0</p>
<p>Oh duh. N can be 0. Good looking out.</p>
<p>@calbear: okay i see that in my notes, NH4+ with HCO2- and Pd/C reduces only aryl carbonyls</p>
<p>so what does NH2NH2, OH, and heat reduce? only the aryl carbonyls or other C=Os as well?</p>
<p>It reduces everything with a C=O…like EVERYTHING. The thing is hella not selective.</p>
<p>lolol thanks</p>
<p>i wish i could help you with some of your questions calbear, i feel bad for not returning the favor to you :(</p>
<p>edit: do you have any questions from problem sets 4 and earlier? lol</p>
<p>lmao hella not selective</p>
<p>also for the original equation, i found a similar problem in the notes…:O</p>
<p>Uhh whats the rule for heteroatoms in calculating aromaticity? Like S, N, and O. Is it when it has a double bound it has no contribution, but if not db than lone pairs count as 2? See you all at the BioE review session at 101 Morgan Hall @ 9am tmrw!..if i wake up in time haha</p>
<p>i’m not sure what specifically you’re referring to, but it all depends on hybridization. for example, if you see -N- (and there are two lone pairs on it), then one of the lone pairs is going to be sp2. the other goes into the p orbital and is then counted as pi electrons.</p>
<p>Yeah that sounds right. I’m hoping he just uses O though. Because I get confused with Ns and Ss in a molecule</p>
<p>Shooooot are you guys doing the Vollhardt stuff too? :O</p>
<p>wait what’s the difference between
NH2NH2, OH-, and heat
and NH2NH2, OH-, no heat</p>
