Thank you soo much for your videos, Leah. You have no idea how much your videos has helped me in Organic Chemistry. My lectures are a joke. I literally have no clue what goes on in my 1,000 student lecture. All the professor does is ask questions in class and expect us to read the textbook. I don't mind reading the textbook but your videos seriously do an amazing job reinforcing the information.
I can't go to any of my organic chem classes this semester, because of my work schedules. These videos help me a lot. Thanks, Leah you make internet a better place.
Thank you for your videos. They are very good for refreshing my rusty chemistry for when I continue to study this fall. I particulary enjoy your very practical approach to problem solving and the very clear explanations. Also, your visual style shows that you put a lot of effort into it. You are awesome :)
I'm really thankfull Leah, cause you are just amazing. I only watch your videos and learn consepts very fast. You are really talented in teaching. Lots of love from Norway
Hello madam, Your videos really help me understand the concepts very well.Though I am from India,I follow your videos very well.Thank you for creating such videos. Regards, Sahithya
Thank you for this video, you have explained very well. Could i ask the following questions? Why do we not write the concentration of nucleophile in the rate law equation? The rate of reaction doesn't depend on the concentration of a nucleophile, how have we come to this consideration? Kindly explain. Thank you Ma'am! ❤
When look at rate of reaction, if one step is infinitely slower than the rest, that becomes the rate determining step. As soon as a carbocation (slow step forms) the nucleophile instantly attacks making its rate irrelevant. As for how this was determined, in lab observations
Ok, my semantics are incorrect. at 4:02 we see the beginning of reaction. By looking at that, can we predict that the product will be OH attached to the most stable carbon?
If anyone doesn't grasp this, Go to Crash Course (Nucleophiles and Electrophiles) watch it and then come back here. You'll thank me later for not wasting ur time trying to understand this.
Glad to have you come back here afterwards! :) For more detailed help with questions like this, I recommend joining the organic chemistry study hall. Details: leah4sci.com/join or contact me through my website leah4sci.com/contact/
The information is good, but why not start the series with an explanation of what SN1/2 and E1/2 are to start with? I had no idea what those were expecting some explanation at least late in the first video not the 8th. Now that I know what the reactions are exactly I have to rewatch the first handful of videos. I like the information and how it's presented, but the order of the information needs to be changed. What things are should come before how things work. Like introducing parts of an engine and what they do then describe the process of the system.
I'm still having trouble understanding what you mean by not being able to form a negative charge under acidic conditions. Can you explain what you mean in another way?
In acidic conditioms, there are free protons (hydrogens) in solution. No negative charge can form like this because one of the many protons will quickly create a bond to prevent that from happening. In a basic solution there is no chance of an acid or positive charge from forming because bases seek out protons in an effort to balance thier charge. Sorry nobody ever responded to you.
Acidic solutions are full of positive charges. If a negative charge even thinks about forming it will immediately bump into a proton and grab it forming something neutral, aka not be allowed to form
it would be SN1, I know it is your question of 2 years ago XD . Your chain is tertiair which would never be SN2 only SN1 or E1, your solvent is Polair Protic (CH3OH) which would give SN1 or E1, you are going to react with a very strong base which is hydroxide , which will give you rather a substitution than an elimination if you were to heat it you would get an E1 reaction.
HBr is a very strong acid and will not form in water. Instead you'll have the H+ picked up by water for a weaker hydronium and Br- off in solution surrounded and stabilized by water molecules
Thank you soo much for your videos, Leah. You have no idea how much your videos has helped me in Organic Chemistry. My lectures are a joke. I literally have no clue what goes on in my 1,000 student lecture. All the professor does is ask questions in class and expect us to read the textbook. I don't mind reading the textbook but your videos seriously do an amazing job reinforcing the information.
You're so very welcome! I prefer learning through videos, too, so I totally get what you're saying. Thanks for your kind words!
Have struggled with alot of videos only to just bump in your tutorial which has just simplified a 2hr lecture in minutes💪💪
Wow, so glad that it quickly helped you!
I can't go to any of my organic chem classes this semester, because of my work schedules. These videos help me a lot. Thanks, Leah you make internet a better place.
I'm glad it helped!
Thank you for your videos. They are very good for refreshing my rusty chemistry for when I continue to study this fall. I particulary enjoy your very practical approach to problem solving and the very clear explanations. Also, your visual style shows that you put a lot of effort into it. You are awesome :)
Glad to hear that! You are welcome!
Thank you for the help! You're the reason why I am able to take my organic exams with a smile!
yay! that's great! I'm glad the video helped. You are very much welcome.
Video very thorough! I encourage people to pause the video often and rewind to really understand the information.
Thanks! Glad you like it!
finally I get it
thanks for d video
chem. can be a real headache when one is messed up wid d mechanisms
deepshidha98
You're very welcome!
You truly explain things well. Thank you for posting these videos!
Thanks for your kind words :) Glad I could help
you are the best teacher so far i have seen
Thanks so much, happy to help!
You deserve a Nobel peace prize. Seriously
Too funny! Glad I could help :)
I'm really thankfull Leah, cause you are just amazing. I only watch your videos and learn consepts very fast. You are really talented in teaching. Lots of love from Norway
Aww thanks so much for your kind words!
You’re amazing. Thank you.
You're so welcome, and thanks for your kind words!
Hello madam,
Your videos really help me understand the concepts very well.Though I am from India,I follow your videos very well.Thank you for creating such videos.
Regards,
Sahithya
You're so very welcome!
Leah the genius, i wish you were my instructor.
Aww thanks!
thank you for this, love u!!!
You're so welcome!
Thank u so much it was amazing...
You're very welcome!
you are great, I love this channel.
Thank you for your kind words!
Thankyou - great explanation!
You're so very welcome!
Thank you for this video, you have explained very well. Could i ask the following questions?
Why do we not write the concentration of nucleophile in the rate law equation?
The rate of reaction doesn't depend on the concentration of a nucleophile, how have we come to this consideration? Kindly explain. Thank you Ma'am! ❤
When look at rate of reaction, if one step is infinitely slower than the rest, that becomes the rate determining step. As soon as a carbocation (slow step forms) the nucleophile instantly attacks making its rate irrelevant. As for how this was determined, in lab observations
Thank you so much! You explain things so well!
You're welcome!
and ma'am keep uploading more videos like this
v really appreciate yr efforts
You're so very welcome!
thank u from Swiss ❤
You're welcome 😊
You are the best
Aww thanks!
Fabulous, thank you!
You're so very welcome!
thank u orgowizard....!!!!
You're so very welcome!
You are truly a lifesaver
Glad I could help!
Thank you so much
You're welcome!
Ok, my semantics are incorrect. at 4:02 we see the beginning of reaction. By looking at that, can we predict that the product will be OH attached to the most stable carbon?
Thanku
You're welcome!
Thank you so much!!ma'am
You're welcome!
If anyone doesn't grasp this, Go to Crash Course (Nucleophiles and Electrophiles) watch it and then come back here.
You'll thank me later for not wasting ur time trying to understand this.
Glad to have you come back here afterwards! :)
For more detailed help with questions like this, I recommend joining the organic chemistry study hall. Details: leah4sci.com/join or contact me through my website leah4sci.com/contact/
Thanks
You're welcome
Thanks alot
You're welcome
Cant believe I only just found this...
Glad you found it, make sure to check out my other resources to help you! leah4sci.com/syllabus
The information is good, but why not start the series with an explanation of what SN1/2 and E1/2 are to start with? I had no idea what those were expecting some explanation at least late in the first video not the 8th. Now that I know what the reactions are exactly I have to rewatch the first handful of videos. I like the information and how it's presented, but the order of the information needs to be changed. What things are should come before how things work. Like introducing parts of an engine and what they do then describe the process of the system.
thanku so much, that's why i'm not understanding
This video is part of a series. To start at the beginning, go to Leah4Sci.com/sne
I'm still having trouble understanding what you mean by not being able to form a negative charge under acidic conditions. Can you explain what you mean in another way?
In acidic conditioms, there are free protons (hydrogens) in solution. No negative charge can form like this because one of the many protons will quickly create a bond to prevent that from happening. In a basic solution there is no chance of an acid or positive charge from forming because bases seek out protons in an effort to balance thier charge. Sorry nobody ever responded to you.
Acidic solutions are full of positive charges. If a negative charge even thinks about forming it will immediately bump into a proton and grab it forming something neutral, aka not be allowed to form
I'm wondering what would happen if you were to try the reaction of t-butyl bromide with hydroxide in methanol. Would that be SN1, 2, or E1, 2?
it would be SN1, I know it is your question of 2 years ago XD . Your chain is tertiair which would never be SN2 only SN1 or E1, your solvent is Polair Protic (CH3OH) which would give SN1 or E1, you are going to react with a very strong base which is hydroxide , which will give you rather a substitution than an elimination if you were to heat it you would get an E1 reaction.
Hydroxide is a strong base and so it would not be an SN1 reaction. Instead you're looking at E2 in this case
Is it possible to form HBr here? Instead of a water molecule forming the hydronium ion?
At which specific point in the video? (Timestamp please)
As a side product,7:59?
HBr is a very strong acid and will not form in water. Instead you'll have the H+ picked up by water for a weaker hydronium and Br- off in solution surrounded and stabilized by water molecules
you are the best
Thank you! I'm glad you're finding the videos useful.