You do realize you're a lifesaver for us Indians right? We have very high level engineering college entrance exams and we study what a college student in america would study in their whole course in our high school! Its too overwhelming for us and your website that organizes all the topics, along with the notes, the quizzes, its too fantastic! I've been trying to understand sn1,sn2,e1,e2 for the past whole month and you taught me in just two days :)
Thanks so much for this video! Really helped with understanding the kinetics of SN2. Just a quick question about E2, why do higher temperatures stabilize the pi bond between carbon attached to the leaving group and an adjacent carbon?
Miss tell me please how to remember these chemical reaction of alkanes,alkenes,alkynes alkyles,benzene,aldehydes,ketones etc its reation and preparations.tell me if you know any trick or method.
So why do both the CN and the Br have a negative charge in the transition state? I understand the Br because it is leaving and therefore becoming a negative charge but why does the CN have a negative charge? Wouldn't it be a neutral charge because it is making a bond with the carbon
When you draw out CN in a lewis structure, both atoms have a lone pair and are triple bonded to each other. Looking at this, when you find the formal charge of both atoms, C = -1 while N = 0, thus making the molecule negative The formal charge formula can be expressed with; Formal charge = VE of atom - (Non binding electrons + number of bonds) Using carbon in this molecule as an example, Carbon 4 valence electrons, has 1 lone pair (2 electrons) and a triple bond (3 bonds) connecting to nitrogen. So, Formal Charge = 4 - (2 + 3) = -1 You can do this for nitrogen with the same steps N= 5 valence electrons FC = 5 - (2+3)= 0 =no charge In the transition state, they haven't so much bonded, but are in the middle of the reaction. You can still view them as separate entities floating around. A simplified way of thinking about these reactions are this; if you know you are working with either and SN1 or SN2 reaction, then you know that the reactant must have a halogen, the what it is reacting with must be a base. In SN2 reactions, the base must have a negative charge. When doing a SN2 reaction, the halogen will break off in the transition state, leaving the original reactant as an ion and leaving you with a negative Br and the still negative CN (because it hasn't connected yet). After the transition state, the base will attach itself to the reactant, at that point it will lose its negative charge and become neutral. TL:DR CN is still negative in the transition state because it hasn't connected yet. You can tell because in the transition state, the halogen is breaking off and leaving you with an anion. If the CN was connected in the transition state, then the transition state would be the end of the energy diagram because that is the product, but instead, the transition state is just the highest point of energy because it takes much more energy to separate atoms from their bonds than to attach them.
The cyanide ion has a negative charge on its carbon to begin with. Approaching the transition state of an SN2 reaction between methyl bromide and cyanide would create a partial bond between the two carbons, as well as a partial bond between the carbon and bromine. At this transition state, no atom is wholly negative. The carbon of CN would be partially negative as it begins to form its new bond, and the Br would be partially negative as it prepares to break its bond.
You do realize you're a lifesaver for us Indians right? We have very high level engineering college entrance exams and we study what a college student in america would study in their whole course in our high school! Its too overwhelming for us and your website that organizes all the topics, along with the notes, the quizzes, its too fantastic! I've been trying to understand sn1,sn2,e1,e2 for the past whole month and you taught me in just two days :)
Wow, that's crazy! I'm so glad that my videos are a help to you!
Hello mam.. I m from India... I really impressed with your teaching technic .. keep it up ... thank dear...
Thanks and you're welcome!
Great explanation dear💯💯
Thank you!!
Americans might read this at college but we Cambridge A level students are reading it at Year 11 of high school. Thanks a lot
You're very welcome!
Thanks for the effort
My pleasure
Great video, as always!
Glad you liked it!
Thanks so much for this video! Really helped with understanding the kinetics of SN2. Just a quick question about E2, why do higher temperatures stabilize the pi bond between carbon attached to the leaving group and an adjacent carbon?
You're very welcome! I cover the E2 stability in the E2 videos on my channel
Good explanation thanks
You are welcome
Miss tell me please how to remember these chemical reaction of alkanes,alkenes,alkynes alkyles,benzene,aldehydes,ketones etc its reation and preparations.tell me if you know any trick or method.
Try active writing to help with memorizing: leah4sci.com/memorize
Thanks ma'am for this video...l am from India and you..:🇮🇳🇮🇳🇮🇳🙂
My pleasure 😊
this is great
Thanks!
leah do you sell any course of complete organic chemistry?
yes I do: leah4sci.com/join
great video, thanks a lot!
You are welcome!
What do I have to do to become a scientist?
Learn, study, and discover all the branches of science and mix them together, and/or expertise at one.
Yeah pretty much
So why do both the CN and the Br have a negative charge in the transition state? I understand the Br because it is leaving and therefore becoming a negative charge but why does the CN have a negative charge? Wouldn't it be a neutral charge because it is making a bond with the carbon
When you draw out CN in a lewis structure, both atoms have a lone pair and are triple bonded to each other. Looking at this, when you find the formal charge of both atoms, C = -1 while N = 0, thus making the molecule negative
The formal charge formula can be expressed with;
Formal charge = VE of atom - (Non binding electrons + number of bonds)
Using carbon in this molecule as an example, Carbon 4 valence electrons, has 1 lone pair (2 electrons) and a triple bond (3 bonds) connecting to nitrogen.
So, Formal Charge = 4 - (2 + 3) = -1
You can do this for nitrogen with the same steps
N= 5 valence electrons
FC = 5 - (2+3)= 0 =no charge
In the transition state, they haven't so much bonded, but are in the middle of the reaction. You can still view them as separate entities floating around. A simplified way of thinking about these reactions are this; if you know you are working with either and SN1 or SN2 reaction, then you know that the reactant must have a halogen, the what it is reacting with must be a base. In SN2 reactions, the base must have a negative charge. When doing a SN2 reaction, the halogen will break off in the transition state, leaving the original reactant as an ion and leaving you with a negative Br and the still negative CN (because it hasn't connected yet). After the transition state, the base will attach itself to the reactant, at that point it will lose its negative charge and become neutral.
TL:DR
CN is still negative in the transition state because it hasn't connected yet. You can tell because in the transition state, the halogen is breaking off and leaving you with an anion. If the CN was connected in the transition state, then the transition state would be the end of the energy diagram because that is the product, but instead, the transition state is just the highest point of energy because it takes much more energy to separate atoms from their bonds than to attach them.
@@hunterhorsch5104 Oh okay. Thank you is this the case with every transition state in SN2 or only in this specific example?
@@hunterhorsch5104 Thanks for the explanation bro. Really cleared some things up
The cyanide ion has a negative charge on its carbon to begin with. Approaching the transition state of an SN2 reaction between methyl bromide and cyanide would create a partial bond between the two carbons, as well as a partial bond between the carbon and bromine. At this transition state, no atom is wholly negative. The carbon of CN would be partially negative as it begins to form its new bond, and the Br would be partially negative as it prepares to break its bond.
Thankiee
You're welcome!
A lot of love from India....
yay
:)
Have you heard about jee advanced exam?
I have but I don't teach for it
@@Leah4sci can you recommend books for it?
Hey gaurav mishra everyone mostly solve the questions from MS CHOUHAN or Himnashu Pandey for jee advanced