No, this is because the 2 carbons you are talking about actually became part of the main chain. Please check that part again. We always look for the longest continuous chain for step 1. Hope that helps!
Hi Kyle, butanone does not exist. Why? Butanone would be a ketone (because it ends with one) but a ketone cannot have the double bond oxygen on the side (on oxygen 1). Hope that makes sense :)
Hello, that is not correct. Remember you are already saying it is a pent so you have already counted the 5 carbons in your main chain. So they cannot be part of the branches as well.
Hello :) We do have to give the -1 for this one. Sometimes the -1 - is not necessary but this is only for things like aldehydes, carboxylic acids for example where the functional group is always on carbon 1.
Mistake at 5:20 we must keep the OH on carbon number 2. Otherwise it becomes a positional isomer as well.
He is the king of physics and maths 😭🫂❤️❤️❤️
Omg I'm even 13 and i understood that, bcz you're awesome! Clear Explanations, thanks :))
Hello! I saw your channel :) I wish you the very best and cannot wait to hear about the incredible things you will do one day :)
@@kevinmathscience oh thank you 😊😊😻
Kevin you are truly my fighter😅😅
Tomorrow is going to cook me
Hi Kevin @ question 5.1.3 What if i made a straight chain ? Would that be wrong?
Hi Tanya, that would not work as your molecule would not be a TERTIARY alcohol.
Kevin for 5.3.2 isn't the IUPAC name 3-ethyl-pent-1-yne because there's 2 carbons under the side chain
No, this is because the 2 carbons you are talking about actually became part of the main chain. Please check that part again. We always look for the longest continuous chain for step 1. Hope that helps!
ohh I see it now, thanks
Hey Kevin, for molecule B how did you know it was a ketone not an aldehyde?
For 5.2.2, could a straight chain isomer also be butanone?
Hi Kyle, butanone does not exist. Why? Butanone would be a ketone (because it ends with one) but a ketone cannot have the double bond oxygen on the side (on oxygen 1). Hope that makes sense :)
CAN YOU PLEASE A VIDEO EXPLAINING BOILING AND SMELTING POINT PLEASE SIR
Kevin is just too good
when it comes to the functional group can we draw it instead of naming it or would that be considered to a different question
Hi, Kevin. For 5.2.3, is it wrong to say 3,3-dimethylpent-1-yne
Hello, that is not correct. Remember you are already saying it is a pent so you have already counted the 5 carbons in your main chain. So they cannot be part of the branches as well.
No Kevin, Thank you❤️✨
Haha glad it helped :)
Kevin, I will say that in the last example the -1- is unnecessary when giving the IUPAC name, but overall great vid:)
Hello :) We do have to give the -1 for this one. Sometimes the -1 - is not necessary but this is only for things like aldehydes, carboxylic acids for example where the functional group is always on carbon 1.
Um literary confused about 5.3.1 how do you know triple bond kelivn y not double bonds
I thought Compound A is an Aldehyde or am i wrong??
Excellent!! 😂
your 5.1.2 is not right
Yes it is
on the side
Cooked
Cooked