<p>don’t say stuff like that. that just makes me sad.</p>
<p>allyson and kathy, hubba hubba</p>
<p>i hate terrence too, dude needs to get his head out of his ass</p>
<p>just say that to him on the last day of office hours. just run in with a mask to his OH and say GET OFF YOUR ****** HIGH HORSE and run out.
Allyson is indeed hubba hubba. But a paper bag might be needed for the face. i mean it’s not bad, but it’s certainly scary sometimes like when she gets flustered in OH. I get scared and start thinking she might just punch the next person that asks her the same question.</p>
<p>^HAHAHA yeah she kinda gets frustrated in lab since everybody always asks the same thing, but I love how it’s easy to get the answer out of her.</p>
<p>I actually hope Terence reads this thread haha. It’s like the 4th result when you google “terence choy berkeley”</p>
<p>anyone know if he’s gonna put grades on bspace? or will we have to wait for bearfacts?</p>
<p>how do you know relevant pkas for pI?</p>
<p>jesus man, 2 days…</p>
<p>take the average of the pka-------> (of H going to zwitterion) + pka (of zwitterion)</p>
<p>i mean like how do you know which pkas to use before calculating pI. like 201B…I’m confused on the charges. And Tyr can only be deprotonated right? Lol. After hardcore studying, I feel as though I’ve forgotten everything.</p>
<p>You have to go through the sequence one by one and look at the pKas of all the side chains, as well as the amino and carboxylic pKas for the ends. Use the given pH and pKas along with the henderson-hasselbalch equation to determine which form is favored at that particular pH, which will give you the net charge after summation. </p>
<p>Does anyone know why you need twice the amount of equivalents you’d think you need when doing a transesterification with an amine? For example on page 55 it says 4 equivalents of the amine when you’d only need two, and on 94 ad 197(b) it shows two equivalents when you only need one.</p>
<p>does anyone understand 197a? how do all the MeO’s just disappear…?</p>
<p>equiether: i think it’s to deprotonate the H+'s from the amides</p>
<p>flutter: that’s an acetal so when you add H3O+ it turns into an aldehyde, the alcohol generates from step 1 then attacks</p>
<p>my incorrect guess would be that H3O kicks them off, but I would never have guessed that both MeOs would be gone and replaced by alcohol.</p>
<p>oh well there we go</p>
<p>okay that makes sense now, thanks!</p>
<p>one more that i don’t get at all - 234c? so the intramolecular step has to happen first right? then what… does NaBH4 reduce esters?</p>
<p>thought sodium borohydride reacted with the aldehyde on the bottom. made it an alcohol which then proceeds to attack ester…</p>
<p>Oh wow for some reason i thought it would reduce it to a methyl group ugh stressing out :(</p>
<p>NaBH4 doesn’t do crap to either esters (no reaction at all) or carboxylic acids (simply deprotonates)</p>
<p>LAH does though, to carboxylic acids it will generate the dianion and then the O2- will leave forming an aldehyde, which then gets further attacked by H-; then for esters, it will attack and then kick out the OR leaving group, which generates another aldehyde and gets attacked again.</p>
<p>Hey for futterfly’s question what does H2SO4 and heat do? Protonate the MeO?</p>
<p>i’m confused by my notes: so what happens after getting attacked as an aldehyde (LAH/H3O with ester)? I wrote down that it forms an anion, so does it become an alcohol? And then LAH withOUT H3O completely kicks off O2-?</p>