You don’t understand how helpful these videos are. I’ve been using your videos throughout this year and your consistently helping me get As in each test I’ve done. Now I have my chemistry mock exam in 2 days, hopefully that goes well too.
Hi sir, I’m in year 12 and I’m really hoping by the time I’m in year 13 you have also made all the year 2 videos also as these videos are a huge help 😅
I can only imagine the effort needed to produce this quality of videos and explanations, not just for this video but for the entire channel. Thank you very much for putting them on youtube for free.
Hi sir, hope you are well 😊 is there a trick to spotting these isomers?? I can always only spot like 2 for each molecule, I can never get the hydrogens to add up 😭😭
@@miawalker300 gotcha. Well my method is always to work downwards. Start with the longest unbranched chain. Then move one C to an internal position. See how many different places you can put the C. Name them roughly 1-methyl, 2-methyl... this trains you to think about other possible Isomers. 3-methyl?? This is often the same as one methyl (for chain of 5 or 6) so may not be an isomer. Then take a 2nd C off and repeat. Naming them is hugely useful because it stops you using the same one twice. I wouldn't necessarily worry about H if you've simply been asked how many isomers
In order to check your six, I suggest you name the ones you've drawn. The compare it to my list Hexane 2-methylpentane 3-methtlpentane 2,2-dimethylbutane 2,3-dimethylbutane Let me know what you get. I suspect you'll find that you've got 2 that are the same, maybe one is twisted, or you're numbering from different ends
hallo man i have spotted a mistake in your video at 6:06 you fergot to Connect a Carbon atom with a h atom. but your video is assem but a Little too long.
Hey, yes good spot! Don't do that in the exam!! Thanks for the feedback, glad it's useful 😊 I agree, it's a bit long - I've got the pace up more for the more recent videos!
Thank you 😀 C5H10 could be an alkene or a cycloalkane. So when deciding how many structural Isomers there are we need to consider all the different alkenes there can be, as well as all the different cycloalkanes there can be
........Br | C / | \ H | H H3C- C - CH3 | H Hello sir I hope you're doing well. I wanted to ask if this (2-Bromomethylpropane?) could be a position isomer for C4H9Br other than the ones you've mentioned @22:35? Also, thank you so much for such a succinct and immaculate explanation!
Thanks for your kind words! And good creativity with drawing the isomer! You've actually drawn 1-bromomethylpropane though. If you swap the H at the bottom for one of the methyl groups on the same C it's more clear. The atoms/groups on a carbon are constantly moving so the fact you got the chain turning at a right angle doesn't prevent it from being a chain of 3, and the bromine is on an end carbon. Well done for thinking deeply about this 😃
@@chemistrytutor It took me a while to draw the isomer here 😭. Thank you so much for your corrections sir. I completely overlooked the IUPAC rule of starting the count w the carbon attached to the substituent. Thank you sir for clarifying my doubts!
Hi, at around 27:00, why couldn't we have 1-methylbut-1-ene, where the methyl group is above, below or to the side of the first carbon? Thanks for the video too sir.
Good thinking... but when you have a hydrocarbon, you can't have 1-methyl anything actually. That would make it a longer chain by 1 that you've drawn curved. So 1-methylbut-1-ene as you named it, is actually pent-2-ene, just drawn twisted at the end
Yes definitely you can use my videos! The vast majority of the content is the same for all exam boards. The main differences are how they structure the exams. The best way to get an A* is lots of exam paper practice. There are only so many ways they can ask questions about chemistry. Once you've seen all the ones so far, there won't be surprises
Hi sir, at 17:01 (the isomers written in yellow,) why would it not be a chain isomer as it has a different carbon skeleton; the chains are reorganised differently bearing in mind that a cycloalkane is not a functional group? Is this correct?
Good question! I think from the point of view of functional groups, the key thing to consider is that alkenes have the C=C bond and cyclic Alkanes don't. Therefore they're different functional groups. Different FG is always more significant in terms of classification rather than different chains
Pleaseee can you do a a full transition metals video? Your videos are the ONLY chemistry resource that actually make me understand it so thank you for that, and currently we’re doing transition metals in class but my teacher is rushing through it as exams are soon. :/
That's really lovely to hear the videos are useful! I haven't got a TM full video scheduled for the near future. I'll add it to the list. I'll be releasing at least one exam question walkthrough soon. Invariably they end up being pretty detailed with their explanations. I've made these previously: th-cam.com/video/S_PdwSjT5Us/w-d-xo.html And: th-cam.com/video/tT44pTzruYc/w-d-xo.html
Good question. The answer is no, because what you've drawn is actually 3-methylpentane, that you've curled up/down at the two ends. You can never have 1-methyl anything, in fact, for the same reason.
Hi sir, at 27:08 wouldnt it still be 2-methylbut-1-ene since your always starting from the lowest number, ie from ‘C’ on the right in that case? Thanks
Hi, good question. It's 3-methylbut-1-ene because once you've started numbering from one end of the carbon you have to just keep counting up as you work along the chain. Since the "ene" is more important we number that as the 1 end and so counting upwards the methyl carbon has to considered to be on the 3rd carbon
Great question, but no, when you switch from a regular alkane to a cyclic alkane you lose 2 H atoms. So they aren't Isomers anymore. Thanks for commenting!
For the one on the middle on the right? Good thinking but you can only number from one of the ends and then progress up as you move towards the other end. We give the alkene highest priority so that makes it -1-ene. Then since we have designated one carbon as carbon number 1 then the methyl carbon must be on carbon number 3
Yes, definitely. I teach AQA and I always make my videos with aqa in mind. However 95% or more of exam board content is common across all exam boards. So they're good for everyone
In 26:58I think you made a mistake in naming the alkene C5H10; there's a 2 methyl but-1-ene but there can't be a 3 methyl but-1- ene cus we are meant to number from the lowest side so rather it'll still be a 2 methylbut-1-ene
Good thinking but you can only number from one of the ends and then progress up as you move towards the other end. We give the alkene highest priority so that makes it -1-ene. Then since we have designated one carbon as carbon number 1 then the methyl carbon must be on carbon number 3
That's a great question, and I can see why you'd suggest it. In fact 3-methylbut-2-ene doesn't exist. It's identical to 2-methylbut-2-ene in terms of structure. So we name it in the way that keeps the numbers as small as possible 😊
You can never have a 'branch' on the end of a chain. So with a propane chain of three carbons, the 1 and 3 position are actually ends. Both of those positions turns propane into butane
At 30 minutes? There is only one ethyl group coming from the ring of 3. Each ethyl group is C2H5, so diethyl cyclopropane is C7H14 To have cylcopropane and only 5 Carbon, the largest size of two branchss you could have is dimethyl
For organic Chemistry you will need E/Z. Cis/trans has some use in the transition metals topic, but it's normally necessary to learn it. It does depend on what exam board you're doing. For AQA for instance the only mention of cis/trans on the specification is cis platin an anticancer drug. Whereas E/Z isomerism has its own section in the alkenes topic
It's because of how I framed the question. I asked which of the drawn formulae was an isomer of the straight chain I drew of butane. So the only Isomers possible *have* to be branched
@@mayaresam4355 if you draw that out you've got a chain of 4 that is bent and you've actually got 2-methtyl butane Always look for the longest chain rather than the horizontal looking part
Good question. No we can't. If you draw that out as you think it would look, try naming it from the other end instead. You'll see its 2-methylbut-2-ene
Honestly thankyou so much, I was so stuck on the differences between cis/trans and E/Z for so long that I was lost, but your video was so concise and clear! Thankyou so much!
It depends when your exams are, and what type of exam you're working towards. Official external exams... do all of them and then go looking for more questions for challenge 👍
That's really great to hear! So glad they're useful! This document of links may be helpful? drive.google.com/file/d/1s3I5prjbJRR1U1lcKpXO9TQpQMaVoP66/view?usp=drivesdk
Since either of the double bond carbons could be numbered as "2" then the one with the methyl group attached is automatically numbered as 2. If you drew 3-methylbut-2-ene and 2-methylbut-2-ene you'd see that they're both the same. One is the other just rotated
Your video was soo helpful but I was a little confused at about 22:00 in the video. Why couldn't you position the bromine at the 3rd or 4th carbon? Why can it only be positioned at the 1st and 2nd carbon?
Glad it's useful! You can put it in those positions. BUT, once it's placed in either of those positions, you number from whichever end of the chain makes the numbers the smallest. So if it was on what looks like position 4, for instance, you could turn the molecule over or simply start numbering from the opposite end
That's correct yes. Carbon has 4 outer electrons so needs another 4 to make the 8, hence 4 covalent bonds. Oxygen in group 6, only needs 2 more electrons, so 2 covalent bonds
Sometimes, e.g. for structural and skeletal formulae. But if you've been asked for the displayed formula, you need to show every bond, including the one between the O and H
That's because the group you want to call methyl isn't a branch, the chain is just twisted at the end. It's all part of the main chain of hexane. For something that is only made of C and H, you can't actually have 1-methyl
At 29:09 what is the square cycloalkane called? Is there a methyl group there and if there is how would you know which carbon atom it’s coming out of to name it? Also (sorry I know I’m asking a lot) but for changing the functional group can certain molecules only be one other type of molecule. Like can carboxylic acids only be changed to esters so can butanoic acid only change it’s functional group to become an ester? Your video is saving my life in a level Chem thank you so much
A ring of 4 would be cyclobutane. With a methyl group sticking out, it would be methyl cyclobutane. When there is only one branch on a cyclic compound there is no need for a number. The carbon the single branch is on is assumed to be carbon number 1
Also, in terms of 1 FG turning into another FG, yes. Carboxylic acids and Esters are the only Single FG isomer. However, you could technically have a molecule with 2 separate FG being an isomer. So a botanic acid could be an isomer of 2-hydroxybutanal with a C=O group and an OH group on 2 different carbon atoms. Not normally that complicated for a levels though
Good question: 2-ethyl butane in fact doesnt exist, its actually 3-methyl pentane. If you've got a picture of it drawn out you should see that what you've done is bent the hydrocarbon chain and what you were thinking of as the ethyl group is actually the beginning of the pentane chain
Hey Sir, regarding ionic bonds. In NaCl, how exactly is the Cl atom stable after gaining an electron from Na (by doing this, its outermost 3rd shell goes from 7 to 8 electrons..BUT..dont you need 18 electrons in 3rd shell to be stable?
Great question! The 3d sub-shell has much more energy than 3p, so it's almost like it's part of the 4th shell not the 3rd. Which is why you fill the 4s before 3d. P block atoms in general don't need to have a full d sub-shell (from the outermost energy level) to be considered stable.
Hi sir, at 21:57, you were talking about isomer of C4H9Br, you named the first isomer to be 1-bromo butane, I was just thinking, isn't alkane group having the molecular formula of CnH2n+2 ??? But from the compound given, the molecular formula of CH is an alkali, which is CnH2n+1 with br, please how do you explain this, I'm getting confused
Good question. The answer is that 1 bromo butane isn't an alkane, it is a halogenoalkane. Halogenoalkanes don't have the same general formula, it varies based on the type of halogen present, and the number of them present as well (it can be more than one). You can still refer to any branches present as 'alkyl' groups (methyl/ethyl etc), but fundamentally its still a Halogenoalkane
@@ezedavis7760 I always try to respond to comments on videos to help people out. I've got loads of other a level videos you can check out as well. If there was a particular topic you'd be interested in then let me know and I'll see what I can do. I have an idea that I may do streamed tuition at some point, but not just yet
Thank you for this video sir but i am a bit confused on 41:50 as i undertsand e/z isomerism but some other resources say that you use the atomic number(proton number) which is 6 for carbon and some say use the mass number which is 12 for carbon so i think thats where i am lost
You're correct, there's often confusion between them. I think it's because both normally go together (larger Ar also has a larger atomic number mostly). Strictly speaking, Atomic number is correct. Most a level questions don't require you to make that distinction. Normally you're drawing or Naming the different isomers
I really enjoy watching your videos as these explain the topics far better than my teachers and in my recent mock i used your videos to get an outstanding grade. Keep it up!
You don’t understand how helpful these videos are. I’ve been using your videos throughout this year and your consistently helping me get As in each test I’ve done. Now I have my chemistry mock exam in 2 days, hopefully that goes well too.
That's brilliant to know thank you!
And well done on such consistent high scores!
Best of luck for your mocks 😀
how did it go
A* in that mock exam mate, I have now got a levels in roughly 3 months so wish me luck!@@ali_saleemzz7217
🤍
Introduction - 00:44
Chain Isomers - 4:18
Position Isomers - 8:18
Functional group Isomers - 11:56
Questions on Structural Isomers - 15:56
Sterioisomers - 30:16
E/Z Isomers - 32:14
Questions on Sterioisomers - 39:51
Thanks for this!
ze zame zide
😀
Why is this being recommended to me on my vacation 😭😭
@@donmujtaba happy holiday!
Hi sir, I’m in year 12 and I’m really hoping by the time I’m in year 13 you have also made all the year 2 videos also as these videos are a huge help 😅
That's definitely my plan!
Thanks for the feedback!
I can only imagine the effort needed to produce this quality of videos and explanations, not just for this video but for the entire channel. Thank you very much for putting them on youtube for free.
You're kind to take the time to give this feedback! I really appreciate it 😀
these videos are a massive help, especially with the example question for us to try. thank you for your hard work in putting these up !
thank you thank you thank you! I really love the way you explain everything, so clear and easy to understand!!
That's really lovely feedback, thank you 😊
I'm pleased you found it useful
To be very honest I had problems with isomerism but this video was very explanatory. Thank you very much for this lesson .
You're really welcome. It's great for me to know it's useful 🙂
Zee zame zide was crazy but it helps💯
Whatever helps you remember 👍👍
Hi sir, hope you are well 😊 is there a trick to spotting these isomers?? I can always only spot like 2 for each molecule, I can never get the hydrogens to add up 😭😭
There are definitely strategies that can help. Was it a particular type of structural isomer you struggled with? Chain etc?
@@chemistrytutor mainly chain isomers
@@miawalker300 gotcha. Well my method is always to work downwards. Start with the longest unbranched chain. Then move one C to an internal position. See how many different places you can put the C. Name them roughly 1-methyl, 2-methyl... this trains you to think about other possible Isomers. 3-methyl?? This is often the same as one methyl (for chain of 5 or 6) so may not be an isomer. Then take a 2nd C off and repeat. Naming them is hugely useful because it stops you using the same one twice. I wouldn't necessarily worry about H if you've simply been asked how many isomers
@@chemistrytutor thankyou so much you are such a lifesaver honestly, you're literally the only reason I've understood organic chemistry so far 😂😂
@@miawalker300 that's very kind of you to say. I'm glad it's helping. Well done for sticking at it!
At 24:48 shouldnt there be 6 isomers?
i dont know if im fully correct but i got 6 isomers and i did check again after you got 5 isomers.
In order to check your six, I suggest you name the ones you've drawn. The compare it to my list
Hexane
2-methylpentane
3-methtlpentane
2,2-dimethylbutane
2,3-dimethylbutane
Let me know what you get. I suspect you'll find that you've got 2 that are the same, maybe one is twisted, or you're numbering from different ends
I'm really glad to have come across this video. Thank you so much sir
I'm really pleased it's useful! I've got other videos you might also like. Let me know if you can't find what you need!
Hi I was taught that when looking at priority groups for CIP rules you look at atomic number NOT atomic mass?
You're 100% correct!! Well spotted. That's an important error to clarify, thank you!
Loveeeeeeeeee it
Good video
@@vent5125 thanks 😃
THANK YOU SO MUCH can you please make videos for A2 organic :)
Its on my list 😀
I have some question walkthrough videos already about aromatic and biochemistry
hallo man i have spotted a mistake in your video at 6:06 you fergot to Connect a Carbon atom with a h atom. but your video is assem but a Little too long.
Hey, yes good spot! Don't do that in the exam!!
Thanks for the feedback, glad it's useful 😊
I agree, it's a bit long - I've got the pace up more for the more recent videos!
Hi for the 28:45 isn’t the C5H10 is alkene ? If it’s alkene can we use cycloalkane for the structural drawing? However, it’s really a useful video 🥹
Thank you 😀
C5H10 could be an alkene or a cycloalkane. So when deciding how many structural Isomers there are we need to consider all the different alkenes there can be, as well as all the different cycloalkanes there can be
........Br
|
C
/ | \
H | H
H3C- C - CH3
|
H
Hello sir I hope you're doing well. I wanted to ask if this (2-Bromomethylpropane?) could be a position isomer for C4H9Br other than the ones you've mentioned @22:35?
Also, thank you so much for such a succinct and immaculate explanation!
Thanks for your kind words! And good creativity with drawing the isomer!
You've actually drawn 1-bromomethylpropane though. If you swap the H at the bottom for one of the methyl groups on the same C it's more clear. The atoms/groups on a carbon are constantly moving so the fact you got the chain turning at a right angle doesn't prevent it from being a chain of 3, and the bromine is on an end carbon. Well done for thinking deeply about this 😃
@@chemistrytutor It took me a while to draw the isomer here 😭.
Thank you so much for your corrections sir. I completely overlooked the IUPAC rule of starting the count w the carbon attached to the substituent. Thank you sir for clarifying my doubts!
@@dancindwight you're very welcome! Your determination does you credit! 👏
Hi, at around 27:00, why couldn't we have 1-methylbut-1-ene, where the methyl group is above, below or to the side of the first carbon? Thanks for the video too sir.
Good thinking... but when you have a hydrocarbon, you can't have 1-methyl anything actually. That would make it a longer chain by 1 that you've drawn curved. So 1-methylbut-1-ene as you named it, is actually pent-2-ene, just drawn twisted at the end
at 30:00 i think you have named cyclo alkanes wrongly. btw this vid was very useful..
Glad it was useful!
In what way do you mean about the naming... which one of the 5 cyclo compounds do you mean?
I'm doing OCR (A) A Level Chemistry, can I follow your videos? Can I get an A*?
Yes definitely you can use my videos! The vast majority of the content is the same for all exam boards. The main differences are how they structure the exams.
The best way to get an A* is lots of exam paper practice. There are only so many ways they can ask questions about chemistry. Once you've seen all the ones so far, there won't be surprises
@@chemistrytutor Thanks you and I really appreciate the effort you put to make these videos.
is there a way you can also include optical isomerism cause it is now there in my syllabus if you don't mind pls Lovely content too
Separate video 😃
th-cam.com/video/Fi1zlk6ZuOs/w-d-xo.html
Hi sir, at 17:01 (the isomers written in yellow,) why would it not be a chain isomer as it has a different carbon skeleton; the chains are reorganised differently bearing in mind that a cycloalkane is not a functional group? Is this correct?
Good question! I think from the point of view of functional groups, the key thing to consider is that alkenes have the C=C bond and cyclic Alkanes don't. Therefore they're different functional groups. Different FG is always more significant in terms of classification rather than different chains
so many students are saved cuz of u thnx 👍🏾👍🏾
Thanks, that's very kind
Pleaseee can you do a a full transition metals video? Your videos are the ONLY chemistry resource that actually make me understand it so thank you for that, and currently we’re doing transition metals in class but my teacher is rushing through it as exams are soon. :/
That's really lovely to hear the videos are useful!
I haven't got a TM full video scheduled for the near future. I'll add it to the list. I'll be releasing at least one exam question walkthrough soon. Invariably they end up being pretty detailed with their explanations.
I've made these previously:
th-cam.com/video/S_PdwSjT5Us/w-d-xo.html
And:
th-cam.com/video/tT44pTzruYc/w-d-xo.html
@@chemistrytutor thank you so much!
@@sahib3035 😀
Hello sir!
at 24:38 can we have 1,2,3 trimethylpropane as one other isomer
Good question. The answer is no, because what you've drawn is actually 3-methylpentane, that you've curled up/down at the two ends. You can never have 1-methyl anything, in fact, for the same reason.
@@chemistrytutor Thanks a lot!
Saved me 🙏🙏🙏
I'm glad they've helped ☺️
Hi sir, at 27:08 wouldnt it still be 2-methylbut-1-ene since your always starting from the lowest number, ie from ‘C’ on the right in that case? Thanks
Hi, good question. It's 3-methylbut-1-ene because once you've started numbering from one end of the carbon you have to just keep counting up as you work along the chain. Since the "ene" is more important we number that as the 1 end and so counting upwards the methyl carbon has to considered to be on the 3rd carbon
Thank you sir it is useful for my exam
And it is easy topic now
That's great to hear, thank you! And well done 👏
@@chemistrytutor thank you sir
U forgot a hydrogen for the chain isomers
Great spot 👍
tysm! you really helped
😀 you're welcome
Hi sir, the question at 21:57,, didnt you foreget 1-bromocyclobutane as functional group isomers ?
Great question, but no, when you switch from a regular alkane to a cyclic alkane you lose 2 H atoms. So they aren't Isomers anymore.
Thanks for commenting!
23:47 why did you CH3 on the second carbon atom instead of 1st carbon atom?
If you put a methyl group on the first carbon then it would not be a methyl group. It would be a longer carbon skeleton 😊
you are th best
Thank you 😊
Very helpful video👍
@Jenny20302 thank you 😊
In 27:07, shouldn't the naming be 2-mehtylbut-1-ene since we start counting from the side which has the least number of carbons?
For the one on the middle on the right? Good thinking but you can only number from one of the ends and then progress up as you move towards the other end. We give the alkene highest priority so that makes it -1-ene. Then since we have designated one carbon as carbon number 1 then the methyl carbon must be on carbon number 3
@@chemistrytutor Got it! Thank you so much sir! These videos truly are a life saver.
@rooooooooooootib thanks for your kind feedback 😀
is this aqa?
Yes, definitely. I teach AQA and I always make my videos with aqa in mind. However 95% or more of exam board content is common across all exam boards. So they're good for everyone
For c5h10, is 3-methyl but-1-ene correct?
Yes, you can have 3- methylbut-1-ene. This keeps the double bond functional group ad the highest priority and therefore it gets the 1 👌
@@chemistrytutor so then there'll be a total of 11 isomers?
@@zainabbi4992 it's still 10 Isomers for C5H10. 3-methylbut-1-ene was one of the ones I included in my 5 possible alkenes
@@chemistrytutor oh yes i overlooked, what about 3 methyl but 2 ene?
@zainabbi4992 number it from the other end. I think I drew out this example and then crossed it out as a sneaky repetition
Thank you very much!
So glad to have found this video and channel
Lovely to know its useful!
Thank you very much
I'm glad it's useful!! 😃
In 26:58I think you made a mistake in naming the alkene C5H10; there's a 2 methyl but-1-ene but there can't be a 3 methyl but-1- ene cus we are meant to number from the lowest side so rather it'll still be a 2 methylbut-1-ene
Good thinking but you can only number from one of the ends and then progress up as you move towards the other end. We give the alkene highest priority so that makes it -1-ene. Then since we have designated one carbon as carbon number 1 then the methyl carbon must be on carbon number 3
Hi Sir, regarding 28:00, wouldn't another isomer be 3-methylbut-2-ene? Because there are no duplicates. Thanks in advance!
That's a great question, and I can see why you'd suggest it. In fact 3-methylbut-2-ene doesn't exist. It's identical to 2-methylbut-2-ene in terms of structure. So we name it in the way that keeps the numbers as small as possible 😊
34:15
Great😂
If it helps, it's a good thing
Sir at 24:40 can you have 2,3-dimethylpropane , 1,3-dimethylpropane
You can never have a 'branch' on the end of a chain. So with a propane chain of three carbons, the 1 and 3 position are actually ends. Both of those positions turns propane into butane
really nice, clear video - thank you! (:
Thank you 😊
3:00 wouldn't it be: diethylcyclopropane?
And, for the rules of higher or lower priority we use atomic numbers not the mass number.
At 30 minutes? There is only one ethyl group coming from the ring of 3. Each ethyl group is C2H5, so diethyl cyclopropane is C7H14
To have cylcopropane and only 5 Carbon, the largest size of two branchss you could have is dimethyl
This vedio has helped me to calculate the number of isomers especially that include ring isomers like C6H12
Thank u sir from Pakistan 🇵🇰
@@livetofullest786 that's great to hear! Thanks for your kind feedback 😀
Do we also need to know cis/trans isomers? Is there a video on that? My teacher went through them even though I'm year 12.
For organic Chemistry you will need E/Z. Cis/trans has some use in the transition metals topic, but it's normally necessary to learn it. It does depend on what exam board you're doing. For AQA for instance the only mention of cis/trans on the specification is cis platin an anticancer drug. Whereas E/Z isomerism has its own section in the alkenes topic
exactly
👌
20:11 how did you know that the structural formula in the question has a branch. Y not just a chain of 4 , which would give the second answer
It's because of how I framed the question. I asked which of the drawn formulae was an isomer of the straight chain I drew of butane. So the only Isomers possible *have* to be branched
Find the video very help. Noticed at 24:56 we could have 6 isomers as 2-ethylbutane could be structured and have the same molecular formula
Good thought, but, actually 2-ethylbutane is the same as 3-methylpentane, one is just twisted more. 😃
@@chemistrytutor What about 2-propylpropane??
@@mayaresam4355 if you draw that out you've got a chain of 4 that is bent and you've actually got 2-methtyl butane
Always look for the longest chain rather than the horizontal looking part
27:50 can we also have 3-methyl-but-2-ene?
Good question. No we can't. If you draw that out as you think it would look, try naming it from the other end instead. You'll see its 2-methylbut-2-ene
ohhh thanks @@chemistrytutor
@@SammyS909 👌
Thank you for the video! At 45:16 what are the names of E and Z isomers?
On the right? E- and Z- oct-2-ene
Thank you so much sir this was really helpful❤
Thanks for the kind feedback 😀
Amazing video, thank you for your effort and tome
So nice of you! Glad it was useful 😀
Honestly thankyou so much, I was so stuck on the differences between cis/trans and E/Z for so long that I was lost, but your video was so concise and clear! Thankyou so much!
Thanks for your kind words 😊
I'm really glad the video is useful!
Hello sir, how many past papers do you reckon I should do?
It depends when your exams are, and what type of exam you're working towards. Official external exams... do all of them and then go looking for more questions for challenge 👍
Thank you
No problem! Happy it's useful 😊
Thank you so much sir!! I have no idea about this chapter till i came across your videos. Thank you so much again!!!
That's really great to hear! So glad they're useful!
This document of links may be helpful?
drive.google.com/file/d/1s3I5prjbJRR1U1lcKpXO9TQpQMaVoP66/view?usp=drivesdk
at 28:24, can 3-methylbut-2-ene also not be an isomer?
Since either of the double bond carbons could be numbered as "2" then the one with the methyl group attached is automatically numbered as 2. If you drew 3-methylbut-2-ene and 2-methylbut-2-ene you'd see that they're both the same. One is the other just rotated
Your video was soo helpful but I was a little confused at about 22:00 in the video. Why couldn't you position the bromine at the 3rd or 4th carbon? Why can it only be positioned at the 1st and 2nd carbon?
Glad it's useful!
You can put it in those positions. BUT, once it's placed in either of those positions, you number from whichever end of the chain makes the numbers the smallest. So if it was on what looks like position 4, for instance, you could turn the molecule over or simply start numbering from the opposite end
thank youuu@@chemistrytutor
@@littlefry6691 😀
Sir at 14:22 the oxygen only has 2 bonds around it does that mean that only carbon is suppose to have 4 bonds around it?
That's correct yes. Carbon has 4 outer electrons so needs another 4 to make the 8, hence 4 covalent bonds.
Oxygen in group 6, only needs 2 more electrons, so 2 covalent bonds
Thank you Sir ❤
Can we write it like this OH instead ft O-H
Sometimes, e.g. for structural and skeletal formulae. But if you've been asked for the displayed formula, you need to show every bond, including the one between the O and H
hi sir... Have u covered hemolytic and hemolytic fission?
Hi there, I discuss homolytic fission at length in this video...
th-cam.com/video/-P-YWVK-v4w/w-d-xo.html
You literally saved my exams thanks so much 🙏
Thank you 😊
Good luck!
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That's lovely to know, thanks 😊
Isn't there another isomer for C6H14 2-ethyl 2methyl propane ?
2 ethyl propane is actually methyl butane that's been twisted.
2 ethyl 2 methyl propane is in fact 2,2-dimethyl butane 😀
@@chemistrytutor thanks so much have a chem exam tomorrow, your videos are helping me out alot
@@breytonkruger7214 thats great to know! Best of luck for the exam! 😀
Thank up.
God bless. 🙏🏼
Thank you! Glad it's useful 😊
8:05 why is this not 1-methylpentane?
That's because the group you want to call methyl isn't a branch, the chain is just twisted at the end. It's all part of the main chain of hexane. For something that is only made of C and H, you can't actually have 1-methyl
Always think 'what's the longest continuous chain?' Including turns and twistS
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Thank you very much, can you give me this summary?
I'm not sure what you mean?
@@chemistrytutor Im just want your file that you explain from it
This video helps a lot, thankk you alwayss. I cannot imagine your effort on the quality of explaination. thanks a lot again
Thank you very much for your kind feedback 😀
At 29:09 what is the square cycloalkane called? Is there a methyl group there and if there is how would you know which carbon atom it’s coming out of to name it?
Also (sorry I know I’m asking a lot) but for changing the functional group can certain molecules only be one other type of molecule. Like can carboxylic acids only be changed to esters so can butanoic acid only change it’s functional group to become an ester?
Your video is saving my life in a level Chem thank you so much
A ring of 4 would be cyclobutane.
With a methyl group sticking out, it would be methyl cyclobutane. When there is only one branch on a cyclic compound there is no need for a number. The carbon the single branch is on is assumed to be carbon number 1
Also, in terms of 1 FG turning into another FG, yes. Carboxylic acids and Esters are the only Single FG isomer.
However, you could technically have a molecule with 2 separate FG being an isomer. So a botanic acid could be an isomer of 2-hydroxybutanal with a C=O group and an OH group on 2 different carbon atoms.
Not normally that complicated for a levels though
You are a god! That's all I have to say. A GOD!
Thanks so much! It's really great to know it's useful!
why can't 2-ethyl butane be an structural isomer for c6h14
Good question: 2-ethyl butane in fact doesnt exist, its actually 3-methyl pentane. If you've got a picture of it drawn out you should see that what you've done is bent the hydrocarbon chain and what you were thinking of as the ethyl group is actually the beginning of the pentane chain
😍 Thank You For Your Efforts To Make this video 🤩
You're very welcome 😀
you explain it in a very detailed way that anyone could understand 🤯
Excellent! I'm really glad it's useful 😀
Hey Sir, regarding ionic bonds.
In NaCl, how exactly is the Cl atom stable after gaining an electron from Na (by doing this, its outermost 3rd shell goes from 7 to 8 electrons..BUT..dont you need 18 electrons in 3rd shell to be stable?
Great question! The 3d sub-shell has much more energy than 3p, so it's almost like it's part of the 4th shell not the 3rd. Which is why you fill the 4s before 3d.
P block atoms in general don't need to have a full d sub-shell (from the outermost energy level) to be considered stable.
@@chemistrytutor ; see this is where i need some brushing up..thanks for that!
@@kennyx8482 no worries 👍
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😃 thanks
Hi sir, at 21:57, you were talking about isomer of C4H9Br, you named the first isomer to be 1-bromo butane, I was just thinking, isn't alkane group having the molecular formula of CnH2n+2 ???
But from the compound given, the molecular formula of CH is an alkali, which is CnH2n+1 with br, please how do you explain this, I'm getting confused
Good question. The answer is that 1 bromo butane isn't an alkane, it is a halogenoalkane. Halogenoalkanes don't have the same general formula, it varies based on the type of halogen present, and the number of them present as well (it can be more than one).
You can still refer to any branches present as 'alkyl' groups (methyl/ethyl etc), but fundamentally its still a Halogenoalkane
@@chemistrytutor thanks, i get it now... Please how can I get to reach you?? I'm having problems with chemistry way too much
@@ezedavis7760 I always try to respond to comments on videos to help people out. I've got loads of other a level videos you can check out as well. If there was a particular topic you'd be interested in then let me know and I'll see what I can do.
I have an idea that I may do streamed tuition at some point, but not just yet
where can I find A2 isomerism?
Optical Isomerism is the next video I will be making. Out this week 😃
It's out now 😃
th-cam.com/video/Fi1zlk6ZuOs/w-d-xo.html
@@chemistrytutor thank you so much!!!
@@alizakhalid472 you're welcome 😊
Let me know what you think 😃
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😂 if it helps you remember it, perfect
Thank you for this video sir but i am a bit confused on 41:50 as i undertsand e/z isomerism but some other resources say that you use the atomic number(proton number) which is 6 for carbon and some say use the mass number which is 12 for carbon so i think thats where i am lost
You're correct, there's often confusion between them. I think it's because both normally go together (larger Ar also has a larger atomic number mostly). Strictly speaking, Atomic number is correct. Most a level questions don't require you to make that distinction. Normally you're drawing or Naming the different isomers
@@chemistrytutor ahh i understand now Thank you so much sir for taking the time to explain!
I really enjoy watching your videos as these explain the topics far better than my teachers and in my recent mock i used your videos to get an outstanding grade. Keep it up!
That's really great to hear! Thanks for your kind words, and well done!