Kinetic vs Thermodynamic Product - 1,2 vs 1,4 Addition of HBr to 1,3- Butadiene

[@TheOrganicChemistryTutor]

Access The Full 40 Minute Video: www.patreon.com/MathScienceTutor Direct Link to The Full Video: bit.ly/3mwEi8R

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[@cowgang7395]

Absolutely this.

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[@800-high9]

4:26 The reaction to make the kinetic product IS reversible. When bromide leaves the 1,2 product it makes the same intermediate as when the bromide leaves the 1,4 product.

[@obedapiri1581]

Found this of Great Help

[@mahla4461]

I love you. Thank you for explaining this so clearly!

[@HydrogenTwoO]

Thank you, this is extremely helpful :)

[@sakshishukla3693]

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[@fauzansyahru1362]

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[@Esther-ub9wf]

I don't undestand in the first part of the video why the kinetic products would be formed in low temperature. In my logic as an increase of temperature is an kinetic factor, the reaction would be faster and the product formed should be the one formed quicker. Right? If anyon ecould help I would appreciate it very much

[@Esther-ub9wf]

never mind, I just got it. For anyone wondering: the increase of temperature often leads to an increase of the reaction rate constant which enables the thermodynamic equilibrum to be reached faster. which means that the longer one waits, the more we form the thermodynamic product. At the end this product will be in majority.

[@brieanaschaff9843]

When you have a diene that isn't symmetrical how do you know which double bond is going to undergo the reaction?

[@nikhilpundir1433]

One which is more substituted would react

[@mustafakeskes9378]

Think this problem like this: which is more stable for example a tertiary or a secondary carbonation

[@hazeld8016]

@@nikhilpundir1433 wouldn’t the less substituted double bond react because less substituted double bonds are less stable?

[@nikhilpundir1433]

@@hazeld8016 less stable doesn't mean that they would be more reactive. Go through the reaction mechanism and you would see that in the first step alkene will give it's electron to Hydrogen of hbr to form a carbocation and the most stable carbocation would be formed from more substituted alkene isn't it

[@syd5604]

@@nikhilpundir1433 pleaseeee help im so confused about which resonance structure to use. the positive charge and double bond can just be chasing each other around the structure. im currently working with cyclic compounds and i really don’t understand how my teacher is picking which structure to use, im probably just dumb and missing something but i can’t find any good answers online

[@bareerabuhary]

All of your videos were extra ordinary.there are inch by inch explanations for our doubts and these explanation are much helpful and support us. Really heartfelt thank you for sharing these clips for the children like us🫠

[@800-high9]

9:47 The resonance with the secondary cation does NOT form faster. Both resonance structures refer to the same intermediate - there is only one intermediate, the allyl cation. The secondary resonance structure has a larger contribution to the ground state. It is NOT of lower energy, it is not more stable. "Lower energy" and "more stable" refer to two different structures where you can measure or estimate the energy of each structure. Again, there is only one intermediate - the allyl cation.

[@therosefamily5281]

am I correct? so at 4:02. so because its low -40c its always going to be 1,2 addition and b/c its high 60c its always going to 1,4 addition? they are not going to change position ?

[@lilliannader5536]

Ooooh i was studying that part suddenly you shock me with that video!

[@sathish6637]

Thank you brother

[@isatahawakamara2072]

Thank you so much Sir

[@fionaruss9573]

I dont understand how the kinetic product is mono substituted versus the thermodynamic product being disubstituted. I can't visual the two versus one R groups

[@pralay5156]

Very helpful video 👍

[@worshipthecasket]

This makes me so happy I’m not in chemistry anymore.

[@cowgang7395]

This shit is cool tho, not like maths🤢

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How do you feel about him not uploading full tutorial vids now, these are all only half videos. I dont see many mentioning this

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[@calinmarie5974]

how do we know whre to put the alkene

[@GreenBean-kr6fk]

Resonance structures

[@shiken69420]

Thank you

[@songohan393]

Why is the primary halogenalkane the more stable product thermodinamically?

[@shiken69420]

More no. of alpha hydrogens that causes it to form more hyperconj structures

[@worshipthecasket]

But your vids were helpful !

[@kouitystanislaszingbe4070]

What about isolated diene reaction ??

[@chungsfamily3244]

Smart human

[@fauzansyahru1362]

Often to upload sirrr :)))

[@lukelupiano6264]

this video doesnt make sense at the end, primary cation simply do not exist and cannot be synthesised so the mechanism has to be different

[@sailek.741]

But why is the second product the thermodynamic stable product?? The video did just end very abrupt

[@marleahknights6772]

I may be wrong, but because the thermodynamic reaction is the slower product formation, it has time to select for the more stable product (the di-substituted alkene). In the kinetic reaction, it happens so quickly that the Br- reacts with whatever reactant is closest and more immediately stable.

[@iconicguy5717]

Because it is a more substituted alkene. Hence it is more stable

[@AshmitThakur1704]

is it normal to study this in high school? here in india we do study it :/

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[@a.od-reszki4674]

I'm still diene

[@bayleekennedy505]

SAME

[@a.od-reszki4674]

@@bayleekennedy505 i'm actually sitting over my Murry and triene to learn about diene

[@sakshishukla3693]

First comment

[@jackmeyers1175]

You're absolutely great, but can you please stop saying "produc-" and throw that last 't' in there. Drives me crazy. You're very helpful though!

[@ryanouzgane3831]

It’s product, not produck

[@ananyapal14_11]

Do you know a term called "accent"?

[@ryanouzgane3831]

@@ananyapal14_11 yes, and completely omitting a letter is not the same thing as an accent

[@erikverhelst5694]

@@ananyapal14_11 not an accent I think he’s got a speech impediment

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@@erikverhelst5694 Just like " A bou o wa " ?