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not quite - the most reactive functional group almost always sets the anchor point in the molecule in IUPAC, except for carboxylic acids, which is more a reflection on this historical naming of these compounds. When we developed our own material for this course, we found that you can get into false equivalencies when looking at naming as many textbooks happen to pick examples where the priority is usually closest to the end of the chain making the numbering also lower, so it looks like the higher rule is the numbering and not the reactivity. this becomes apparent for more complicated molecules, but with more complicated molecules it also becomes apparent that while IUPAC is well meaning, the reality is there are a myriad of flavors for naming and they look a lot like IUPAC but aren't. a great example of this is "diphenylmethanol" - this looks a lot like IUPAC but the longest chain is actually the benzene ring so the name is really (1-phenyl-1-hydroxymethyl)-benzene. the theme that shows up with these alternative naming schemes is that simpler is better. as such we use IUPAC until it becomes unnamable, then we just jump to whatever works easiest.

@@StubbornAlchemist oh alright thanks alot

glad you liked it, thanks!

glad to hear it!

thanks I appreciate it

Thanks for the feedback

I'm glad you like it! And thank you for the kind words!

@bananadobanana6035 thanks for the positive feedback, I really appreciate it

Any HX (so X = Br, Cl, I) can do anti-markovnikov if you use peroxides or light to encourage a free radical mechanism. in fact most markovnikov reactions have some anti-markovnikov equivalent, so water addition (HX where X = OH) can be done anti-markovnikov using hydroboration oxidation. thanks for asking

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Law of Mass Action dates back to the 1850s and is the progenitor to a lot of this, but in a lot of ways it's an empirical approach to this concept. Looking at reversible reactions they could see that adding reactants drives the forward reaction, but there was no good explanation for this. So it would be misleading to say that this it the law of mass action - law of mass action gives this outcome but doesn't provide a framework beyond empirical experiments. The more rigorous definition of the reaction quotient, Q, is the better explanation, with the requirement that Q = Kc (or Keq, which is the more modern take on K) at equilibrium

@@StubbornAlchemist I am sorry you lost me there; are you trying to approve or disapprove my question? Because I thought this is the stepping stone towards the Law of Mass Action because we know the rates are equal and the ratio of of the rate constants is turn out to be Kc.

details details

@@StubbornAlchemist No, I don't give details of this dropping of a patient in a scanner ;-)

I believe you just add all the orders up for both the forward and reverse terms, though it's not been my experience that the overall order is meaningful for reversible reactions and honestly it's not a calculation I've ever done.

@@StubbornAlchemist so let say if the forward reaction has order 1 and reverse reaction has order 2, the overall order shld be 1 - 2 = -1?

@@stevelow7900 I think that is correct, but to be honest I don't think there is a physical meaning for overall order for reversible reactions. for irreversible the order sort of makes sense as it can directly translate into pressure dependence, which is directly observable. but for reversible reaction the observed order will depend on the ratio of products and reactants. sorry to give you such a non-committal answer but I just haven't seen much on this topic.

@@StubbornAlchemist Oooooooo I was quite shocked too haha bcoz I saw this qn in my uni assignment but it wasn't covered in lecture and I can't find anything on internet that talks about the overall order for reversible reaction. Anyways, thank you soooo much for the detailed explanation!!

@@stevelow7900 glad I could help

tertiarys are more stable, I just misspoke.

I don't need the pic, but sounds like you have good taste in Spidermen

but which version of spiderman? Tom Holland? The Japanese direct to video?

@@StubbornAlchemist No sir is not Japanese version Is marval movie spider man Tom Holland Don't you know that I am just shearing my feelings which is for him My name is mnu please reply

Hello sir can I talk to you May you please reply to me You are brilliant teacher and I love chemistry subject and I love Tom Holland Please 🙏 reply

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hey buddy, the staff at Guitar Center warned you

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Aware of since I took a graduate course in NMR, but for this level I try to keep it simple enough that not too many important details get glossed over. Relying on boltzmann distributions to describe spin populations seems a bit optimistic for a sophomore level organic course. but thanks for your feedback

it's better to think about it that way - they can both technically be rotating, but really it's just one turning relative to the other one

A quarter of quarters would be four peaks each split into 4 peaks - so 16 total peaks - whereas a septet would be just 7. When it splits that much you'll generally see people write "multiplet" because, while it's truly a quartet of quartets, in practice each peak would only be 4-8% of a single proton signal (so less than 1% of the signal of two methyl groups creating it), so unless your signal-to-noise (S/N) is incredible, it'll just be a bunch of bumps against the baseline.

As for the definition you've provided, this is the generally used one in chemistry, though I know you can come across a few others in computation chemistry and mathematics. In the lowest energy configuration I would agree that the angle would be gauche (60 degrees) as it would minimize steric interference between atoms on the adjacent carbons. I have a couple of videos on this; th-cam.com/video/7QACswGsSO8/w-d-xo.html and th-cam.com/video/aFbtBNMKYP4/w-d-xo.html

so helpful

thanks!

@@Jennilovesyouu14 thanks I really appreciate the feedback

For BF3 there are no dihedral angles because there is only one central atom but the bond angles are 120-degrees. For C2H2 there are also no dihedral angles because the entire molecule is linear.

For elimination to occur by an E2 mechanism the leaving group and the associated H on the adjacent carbon have to anti-periplanar - meaning on opposite sides of the carbon-carbon bond (which is being converted to a double bond). So technically yes, but for linear alkanes this can be overcome as the molecule can rotate around single bonds, this is will just have implications on your final stereochemistry. For rings, this means a hydrogen must be trans to the leaving groups for E2. If you have E1, such as acid catalyzed dehydration of alcohols, then it doesn't matter as the hydrogen isn't removed at the same time as the leaving group.

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I really appreciate it! Thanks!

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Glad it helped!

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pretty much any group that have more than 3 carbons in it, such as tert-butyl and isopropyl groups, or any long chains that come off the ring.

@@StubbornAlchemist thanks alot

@@vibessz21 you're welcome

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what do you need explained?

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@@vibessz21 oh well, thank you then I appreciate it

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here you go th-cam.com/video/XXS23VYCjFM/w-d-xo.html

correct - you'll end up with both chiralities, so both wedged or both dashed

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they are a weaker oxidizing agent, which is why they can't convert a primary alcohol into a carboxylic acid and instead only into aldehydes. They are effectively Jones reagent (chromic acid) that has been poisoned with chlorine to make it a weaker AO, much like the Lindlar catalyst is a poisoned Pd catalyst that can't hydrogenate as well.

thanks!

Clausius-Clapeyron assumes equilibrium has been established, which is reasonable if let the pressures equilibrate in contact with a heat source (e.g., measuring vapor pressure by measuring the pressure of a sealed vessel in a hot water bath). For transient systems, it's a let a more complicated and you'd have to know the dH (heat of vaporization) to even solve those kinds of problems.

@@StubbornAlchemist I don't have sufficient knowledge to understand that, but actually I figured it out. n/V (and, hence P/RT) is constant here and deltaH is changed from enthalpy to enthalpy per mole somewhere in the derivation. Hope I am right.

Thanks

you're welcome. glad this helped