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<p>Why doesn't the benzene grab that proton from the H2SO4? My answer key has the molecule on the right grabbing the proton instead.</p>
<p>My logic: if the benzene grabbed the proton, a secondary allylic carbocation would be formed with a lot of conjugation in the molecule. I would expect this to be more stable than the tertiary carbocation that is formed if the molecule on the right gets protonated.</p>
<p>So why does the benzene not get protonated?</p>
<p>Thanks!</p>
<p>benzene is aromatic and more stable than the 2- methylcyclohexene. it takes the H+ to form the most stable carbocation (Markovnikov rule): the tertiary carbon gets the positive charge and reacts w/ the benzene by EAS rxn. The negatively charged OSO3H takes the Hydrogen that the cyclohexane is on and aromaticity is restored.</p>
<p>monday night…</p>
<p>thanks for the response anon.</p>
<p>but i’m wondering why the benzene can’t take the H+? due to conjugation (allylics), wouldn’t the resulting carbocation be more stable than the tertiary carbocation that is formed from the protonation of 2-methylcyclohexene?</p>
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<p>from everything that pedersen has said, benzene doesn’t have an allylic carbon on it. Allylic carbons are ones that are sp3 and can rehybrize to sp2 (w/ a p-orbital) all the carbons in the benzene ring are sp2 hybridized and have only one hydrogen on it. </p>
<p>think about it like this, could you react NBS w/ benzene? </p>
<p>besides that, he said benzene is generally unreactive because breaking aromaticity would require a lot of energy. Breaking a double bond that has no conjugation is easier than breaking a double bond that is conjugated like benzene. </p>
<p>Edit: and he said that a positive change on a benzene ring isn’t stabalized</p>
<p>Edit edit: and think about it, it seems a bit counterproductive that the benzene would take a hydrogen, break aromaticity, get the new R group on the carbocation, and give up the same hydrogen to make sulfuric acid. From everything we’ve learned about EAS rxns, positive charges react with the benzene ring.</p>
<p>yes but the original benzene was way way more stable so there is no way that reaction would take place</p>
<p>alright thanks a lot anon and flutter, that makes a lot of sense now :)</p>
<p>pg 217 A…what’s up with the beginning of the reagents? =</p>
<p>that’s so you convert the molecule into a diene, and only then can you do a Diels-Alder reaction</p>
<p>i dont get this… omg. few more hours to study.</p>
<p>**** is that from 3a? I don’t know **** from 3a. How does that make it into a diene?</p>
<p>Uhh no that’s from 3B. NBS makes Br2 and with hv (light) it makes a radical halogenation rxn so the Br radical is gonna attack an allylic* hydrogen and make HBr and the other bromine attacks the remaining radical on the methyl group to form a bond with a bromine. The butoxide than makes a concerted attack on the other allylic hydrogen on methyl group below making the molecule into a diene (takes off hydrogen, and than arrows go crazy making 2 double bonds and pushing Br off molecule). What PS are you guys on btw?</p>
<p>i don’t remember butoxides at all in lecture…? but thanks for the lesson
Edit: lulz. I had to go back to frechet’s lectures notes. gah</p>
<p>**** me, i coulda had an extra day and a half to study for this if i didn’t have a midterm on friday gahhhhhhhh</p>
<p>Yeah Bio took my time too and it’s not helping me that physics 8b is on thursday…ugh, I just want to sleep. I don’t think he ever used butoxide in lecture, but he uses it in the workbook and also in SLC. NBS,hv and butoxide is pretty handy if you want a double bond. Also why does he only have one HCl when doing diazonium rxns in the workbook. He made it pretty clear in lecture to use 2 to make the oxygen into a LG (water). Is he just lazy or what?</p>
<p>can you provide an example of that calbear? what page/problem #?</p>
<p>also, does anyone know if NH2NH2, OH-, and heat (the combination of these 3) only reduces aryl carbonyls? i have in my notes that it reduces everything but i think i’m wrong…</p>
<p>NH2NH2, OH-, heat reduces any ketone/aldehyde. it’s called the wolff-kishner reduction, and you guys will probably learn about it soon.</p>
<p>have any of you been reading the book?</p>
<p>and i think mech was referring to you HCL question calbear</p>
<p>Oh. For HCl he only uses one in 40A and 116B. I read the book for the Diels-Alder section…it was ok.</p>
<p>And MechRocket, for your original question ‘does benzene grab a proton?’, the answer is yes, it does. But elimination will follow, regenerating a benzene, and this reaction will simply not be observable–in contrast, if he used DCl or D2SO4, you’d get a benzene with its H’s replaced by D’s. (Check lecture notes.) Thus, the only meaningful reaction for our purposes is EAS with the other molecule. Hope this made sense.</p>
<p>So in problem set #5 he has stuff on wavelength (I thought he said it was for our enjoyment) and stuff on hemiacetals (I thought it wasn’t gonna me on MT1). Did I mishear him?</p>
