I wish you had a full MCAT course, you explain stuff so well that I wish I found your videos earlier not 3 days out from my exam. I hope you would make more videos, you are a great teacher.
dude seriously, I have had the hardest time understanding this during my college career and decided today that I would figure it out. You explained it to me in 4 minutes and now I feel confident about it for the MCAT. Thank you. I hope you have success and growth on your channel. You're like the organic chemistry tutor channel. Good quality and easy explanations.
These are truly amazing videos; so helpful and concise! Hopefully, there will be more MCAT content videos covering wider range of topics in the future. Thank you for the resource!
Thanks for pointing that out. As much as I proofread I always feel like a couple errors always slip through the cracks. I’ll add a note in the description with the correction and will put this video on my list to be remade.
1:45 why couldn't you use the C=O to identify? C=0 is around 1700cm-1 with a narrow peak. Why did you use the -OH group in the carboxylic acid instead of the C=O
The C=O group in a carboxylic acid (CA) looks pretty much the same as an aldehyde or ketone the OH in a CA on the other hand is morphologically different and allows you to definitely identify a CA. This is probably beyond the scope of the MCAT though since they don’t tend to ask questions this specific. As well in the context of that specific question the only difference between the aldehyde and the CA is the OH group. So looking at the appearance of an OH peak would be key to making sure the reaction occurred.
It is a ketone. I made a mistake on this video someone else pointed it out and I will eventually remake this video and fix that error. I have notated the error in the video’s description. I should also probably pin a comment to notify people too. I debated taking the video down, but opted to leave it up as I have gotten feedback from my private tutoring students that the videos is still so helpful despite the error that they think I should leave it up.
For the last problem. why cant the answer include B. The OH could deprotonate in solution, resulting in an O-. Then the molecule would resonate in the ring structure to yield C=O and take the internal double bonds from 3 to 2
OHs can differ a little bit based on what is around them so their stretch can shift a bit. 3400-3200 is within the normal range of OH stretches so I) would still be correct. Depending on the source you reference you might see anything from 3200-3550 or 3200-3400 or 3300-3500. The MCAT isn’t trying to trick you and tends to test bigger concepts rather than minute details. So if see something that looks really close to the stretches you know 99% of the time it is that specific stretch.
Is there something confusing about the IR spectra graph at this time point? If so let me know and I will do my best to clear up any confusion you have.
Thanks for your videos! I'm a big fan. I was trying to understand why it was single bond, followed by triple and then double. According to this khan academy video, the first high wavenumber region is because of being bonded to a hydrogen (due to smaller mass). Then it's followed by triple, double, and then single bond (non-hydrogen) regions. ua-cam.com/video/MlMwMqM_Wsg/v-deo.html
I wish you had a full MCAT course, you explain stuff so well that I wish I found your videos earlier not 3 days out from my exam. I hope you would make more videos, you are a great teacher.
i like how you talk fast, clearly, and articulate everything so well. literally the best mcat youtube out there!
Where have you been my entire collegiate career?? Thank you so much for this, it comes across so easy when you explain it!
Glad the videos are helpful and easy to understand!
dude seriously, I have had the hardest time understanding this during my college career and decided today that I would figure it out. You explained it to me in 4 minutes and now I feel confident about it for the MCAT. Thank you. I hope you have success and growth on your channel. You're like the organic chemistry tutor channel. Good quality and easy explanations.
This is brilliant. Took a monotonous memorization chore to a concept I can form strategies around. Thank you!
These are truly amazing videos; so helpful and concise! Hopefully, there will be more MCAT content videos covering wider range of topics in the future. Thank you for the resource!
I have been struggling with IR spec since I covered chem.... this is so helpful
Good video. Only thing to mention is the aldehyde that you showed for the first example problem is a ketone.
Thanks for pointing that out. As much as I proofread I always feel like a couple errors always slip through the cracks. I’ll add a note in the description with the correction and will put this video on my list to be remade.
I love you. Please make an mcat course. Literally saved me. XO
I was trying the memorise the specific IR values and this made it SO much easier to learn, thank you!!
SIMPLY AMAZING wow thank you so much!!!
Glad you found the video helpful!
1:45 why couldn't you use the C=O to identify? C=0 is around 1700cm-1 with a narrow peak. Why did you use the -OH group in the carboxylic acid instead of the C=O
The C=O group in a carboxylic acid (CA) looks pretty much the same as an aldehyde or ketone the OH in a CA on the other hand is morphologically different and allows you to definitely identify a CA. This is probably beyond the scope of the MCAT though since they don’t tend to ask questions this specific. As well in the context of that specific question the only difference between the aldehyde and the CA is the OH group. So looking at the appearance of an OH peak would be key to making sure the reaction occurred.
@@EightfoldMCAT That makes much more sense. Thank you for taking the time to respond!
this is gold.
Glad the video was helpful!
Very helpful, thank you
Glad you liked the video :)
1:36 isn't that a ketone? Not an aldehyde?
It is a ketone. I made a mistake on this video someone else pointed it out and I will eventually remake this video and fix that error. I have notated the error in the video’s description. I should also probably pin a comment to notify people too. I debated taking the video down, but opted to leave it up as I have gotten feedback from my private tutoring students that the videos is still so helpful despite the error that they think I should leave it up.
@@EightfoldMCAT Oh okay. Got it! No worries. I should have read the description first!
when im a doctor, im donating good sums of money to your channels
first hurdle is mcat tho
Thank you this was helpful
this is so amazing
Thank you!
For the last problem. why cant the answer include B. The OH could deprotonate in solution, resulting in an O-. Then the molecule would resonate in the ring structure to yield C=O and take the internal double bonds from 3 to 2
In the last problem, it says I) 3400-3200 corresponding to O-H. I thought 3300-3500 corresponds to O-H. Wouldn’t that make answer choice I) incorrect?
OHs can differ a little bit based on what is around them so their stretch can shift a bit. 3400-3200 is within the normal range of OH stretches so I) would still be correct. Depending on the source you reference you might see anything from 3200-3550 or 3200-3400 or 3300-3500.
The MCAT isn’t trying to trick you and tends to test bigger concepts rather than minute details. So if see something that looks really close to the stretches you know 99% of the time it is that specific stretch.
@@EightfoldMCAT thank you so much! These videos are great and desperately needed in my case
Glad to help. Thanks for the question.
0:20???
Is there something confusing about the IR spectra graph at this time point? If so let me know and I will do my best to clear up any confusion you have.
@@EightfoldMCAT I just didn't understand what you said there is all.
I luv u
Thanks for your videos! I'm a big fan.
I was trying to understand why it was single bond, followed by triple and then double. According to this khan academy video, the first high wavenumber region is because of being bonded to a hydrogen (due to smaller mass). Then it's followed by triple, double, and then single bond (non-hydrogen) regions. ua-cam.com/video/MlMwMqM_Wsg/v-deo.html
Lee Frank Martin Elizabeth Robinson Christopher