I watched this video in the morning and I had my test today. I aced it and credit goes to you. I am really going to share this. Please keep doing thiss. Ur a lifesaver!!!
A good question from @instafeed asking what causes the C-H bond break... The answer is that the chlorine free radical breaks the bond. It, along with an H plus 1 Electron from the C-H bond form HCl, leaving the 'CH3 with its unpaired electron. Free radicals are very reactive which is why this can happen, but this reaction specifically happens because the 2 products are slightly more stable than the 2 reactants. That being said, this reaction along with most organic reactions is actually Reversible. Which makes sense, since the CH3 free radical will also be unstable and reactive and so could reverse the process
15:53 I still can't get it right. Do we add the ch3ch3 to get C2h6 and then reacted with a cl radical? 😣 but even then I don't get what you got sir as a I finish my propagation.
Good question, but no, it's slightly different to that. When you add the 2 propagation steps together to get the overall equation the things you cross out are the things that don't change. If you have a CH3 radical on both sides then they haven't been changed overall by this reaction. It's the same with the Cl free radical. It would be like a recipe telling you to get an ingredient out before cooking and then it later telling you to put it back unused. So you cross out the free radicals because overall they don't change, not for any other reason
23:35 Sir I don't understand what you meant in that part. Does it mean that these are separate reactions that could then undergo free radical substitution? If yes when would questions like this come up in an exam? Would they specify that they would like us to further react the products of the reaction with chlorine? (Before we work out the 3 stages of free radical substitution)
They would definitely tell you what you were making and starting with. I just wanted to make clear that when you do this, it doesn't necessarily only substitute once. Most common though is methane to chloromethane
19:30 Sir does it matter which hydrogen we replace? Would it be correct to write CH2CH3Cl + HCl ? Also sir does it matter the order of the reactants and products in the propagation step? For example at 14:00 could I write: CH4 + 'Cl ----> 'CH3 + HCl (first reaction) Or would it have to be in the exact same order you wrote? I'm often confused about this when writing chemical reactions in general.
1) for the structural formula you need to make sure of two things. First, that the carbon atom only has 4 bonds (your second one has 5). Second, that the H that is replaced leads to the formation of the specified product. 2) the chemicals on each side of the arrow can be in any order
This lesson was a hell of a disaster for me! But this simple clarification you, did save me! THANK YOU LOADS!!!!!!!!!!!!!! I read your details and it said your students wanted you to start this channel. Huge THANKS to the students who came up with the idea. Your tutor is saving many of us kids from being grounded because of bad grades. 😁 PLEASE KEEP ON GOING! You are the best
I watched this video in morning and I had my test today. I aced and the credit goes to you. Your a lifesaver please continue makimg these videos. Is it posisble if you do exam questions videos too ? But Thank you soo much. I am going to share this with all my peers.
The Cl2 molecule during propagation reacts with a methyl radical, and during termination it is formed. Someone said somewhere else that this Cl2 molecule formed is now stable and will 'no longer take part in the reaction'; I thought it could still undergo the reaction with the methyl radical like in the propagation step, or does it not (why not, is it only due to possibly low concentrations of the methyl radical at that point?) Sry if this is weirdly written
It is just as reactive as the chlorine molecule that reacted with the methyl radical in the first place. I wonder if the source you mentioned meant something different? Namely, the Cl2 molecule produced is nowhere near as reactive as the Cl radical and so is stable and can't react with a CH4 like Cl radicals can?
@@chemistrytutor ruclips.net/video/xMPQKzRQDzo/видео.html this is what he mentioned; do you know if 'no longer' (linked) here means definitely or generally during that moment? I get how it probably won't react as a molecule and that it's not reactive like the Cl radical like you suggested, but I wanted to clarify whether it could still be pulled back into reactions with other radicals because he didn't seem to say either. Thanks
@@55lorrw I think it's more of a throwaway statement rather than something important. It's been said to clarify the difference in reactivity now. There is no good reason why the Br2 formed in this way is any different to one that's never reacted yet. There is no difference between them
I cant remember which bit you're referring to, but... The colour of bromine water is due to the dissolved bromine. If that bromine reacts then the water will decolourise. The Br2 will most commonly act as an electrophile, but can behave in different ways, e.g. in FR substitution. If the Br2 reacts and is used up, then it won't have colour anymore.
@@chemistrytutor Thanks! P.s. a little bit confused but does your channel cover all the A Level Chemistry syllabus? And what exam boards is it applicable to?
@@maryamsadiq8780 All exam boards are 95% similar. The main differences are exam structure and the question style. I've covered all of y12 with Explained videos, and Y13 is more exam walkthroughs. Here is a doc with links to all the videos... drive.google.com/file/d/1s3I5prjbJRR1U1lcKpXO9TQpQMaVoP66/view?usp=drivesdk
@@chemistrytutor Thank you so so much! I need help regarding my exam prep. I have A Level Chemistry P4 (CAIE International) in 1 week. I haven't studied properly. My target is an A. Can I still get it? What do I do at this point? I haven't practiced past year questions as well and need to catch up on the core syllabus too! What do I do?!?
@@maryamsadiq8780 I think practice questions are the best thing now. Great revision generally, bit they will highlight where you need to watch videos and hit the books etc. It's not too late!
In the second reaction during propagation, why would the Cl be a radical in the products, as surely one is given to the methyl radical, and the other is just left on its own?
I'm not totally clear what you mean. The chlorine must be a free radical in the second product because of the number of electrons it has. Methyl free radical has 7 electrons in the outer energy level, so it has an unpaired electron (free radical) and needs 1 more to make 8. The Cl - Cl bond contains 2 electrons, this breaks and momentarily each chlorine takes 1 from the bond meaning they have 7 electrons so both have an unpaired electron and so both are (temporarily) free radicals. Pretty much instantly, one of the chlorine forms a new covalent bond using its unpaired electron and the unpaired electron in the methyl free radical, so both become stable and now have a share of a total of 8 electrons. The other chlorine still only has 7, and so exists as a free radical
I'm glad you like the video. Glad the pronunciation is also entertaining as well, to add to the experience 😃 I have to say I'm intrigued to know the way you're used to hearing it said...
You're pronouncing the "gen" part of the word as if it rhymes with "gene". Every other chemist I know pronounces it to rhyme with "jen"; hal o jen o al kane. It matters not 😂
I watched this video in the morning and I had my test today. I aced it and credit goes to you. I am really going to share this. Please keep doing thiss. Ur a lifesaver!!!
That's amazing well done you! I'll take 5% credit. Thank you for the lovely feedback 😊
A good question from @instafeed asking what causes the C-H bond break...
The answer is that the chlorine free radical breaks the bond. It, along with an H plus 1 Electron from the C-H bond form HCl, leaving the 'CH3 with its unpaired electron.
Free radicals are very reactive which is why this can happen, but this reaction specifically happens because the 2 products are slightly more stable than the 2 reactants.
That being said, this reaction along with most organic reactions is actually Reversible. Which makes sense, since the CH3 free radical will also be unstable and reactive and so could reverse the process
15:53 I still can't get it right. Do we add the ch3ch3 to get C2h6 and then reacted with a cl radical? 😣 but even then I don't get what you got sir as a I finish my propagation.
Good question, but no, it's slightly different to that. When you add the 2 propagation steps together to get the overall equation the things you cross out are the things that don't change. If you have a CH3 radical on both sides then they haven't been changed overall by this reaction. It's the same with the Cl free radical. It would be like a recipe telling you to get an ingredient out before cooking and then it later telling you to put it back unused. So you cross out the free radicals because overall they don't change, not for any other reason
23:35 Sir I don't understand what you meant in that part. Does it mean that these are separate reactions that could then undergo free radical substitution?
If yes when would questions like this come up in an exam? Would they specify that they would like us to further react the products of the reaction with chlorine? (Before we work out the 3 stages of free radical substitution)
They would definitely tell you what you were making and starting with.
I just wanted to make clear that when you do this, it doesn't necessarily only substitute once.
Most common though is methane to chloromethane
19:30 Sir does it matter which hydrogen we replace? Would it be correct to write CH2CH3Cl + HCl ?
Also sir does it matter the order of the reactants and products in the propagation step? For example at 14:00 could I write:
CH4 + 'Cl ----> 'CH3 + HCl (first reaction)
Or would it have to be in the exact same order you wrote?
I'm often confused about this when writing chemical reactions in general.
1) for the structural formula you need to make sure of two things. First, that the carbon atom only has 4 bonds (your second one has 5). Second, that the H that is replaced leads to the formation of the specified product.
2) the chemicals on each side of the arrow can be in any order
@@chemistrytutor Thank you so much sir! I finally understand. You're the best.
@@NishatMazumder-s4y happy to help 😊
This lesson was a hell of a disaster for me! But this simple clarification you, did save me!
THANK YOU LOADS!!!!!!!!!!!!!!
I read your details and it said your students wanted you to start this channel. Huge THANKS to the students who came up with the idea. Your tutor is saving many of us kids from being grounded because of bad grades. 😁
PLEASE KEEP ON GOING! You are the best
That's really lovely to hear!
Thanks so much for taking the time to send the feedback. It really means a lot to know it's useful 😃
@@chemistrytutor Any possibility to organize a live session? Via zoom or so?
@@miyumithisewnithennakoon8141 that's not a bad shout. Never done live RUclips. I'll Certainly give it a go near to the real exams!
@@chemistrytutor Im willing to organize a private lesson with you. Cab you let me know if it is possible and if so your fee?
@@miyumithisewnithennakoon8141 feel free to email me... thechemistrytutor123@gmail.com
your are saving my upcoming exams
thank you loads
I watched this video in morning and I had my test today. I aced and the credit goes to you. Your a lifesaver please continue makimg these videos. Is it posisble if you do exam questions videos too ? But Thank you soo much. I am going to share this with all my peers.
I'll definitely be looking to do exam question videos over the next few weeks. I have done some already, but mostly for A2 Physical Chemistry
ruclips.net/video/uPVnWDqppic/видео.html
23:00 Sir where is the Cl2 coming from in all 3 reactions (as a reactant) Is it the chlorine that hasn't reacted yet?
Yes that's correct 😃
27:19 was this removed from the 2024 syllabus? I'm checking it and its not there.. but they also didn't say it was removed?
Which syllabus is it? It's in most of them. For instance in aqa it's in the chlorination of alkanes section
@@chemistrytutor Oh mine is CIE international syllabus 9701
@@chemistrytutor Its the CIE international syllabus 9701
@@glo457 is it not in 13.2? 1d and 2a
@@chemistrytutor Oh i was referring to CFC's and the ozone layer information.. That bit seems to be removed
The Cl2 molecule during propagation reacts with a methyl radical, and during termination it is formed.
Someone said somewhere else that this Cl2 molecule formed is now stable and will 'no longer take part in the reaction'; I thought it could still undergo the reaction with the methyl radical like in the propagation step, or does it not (why not, is it only due to possibly low concentrations of the methyl radical at that point?)
Sry if this is weirdly written
It is just as reactive as the chlorine molecule that reacted with the methyl radical in the first place. I wonder if the source you mentioned meant something different? Namely, the Cl2 molecule produced is nowhere near as reactive as the Cl radical and so is stable and can't react with a CH4 like Cl radicals can?
@@chemistrytutor ruclips.net/video/xMPQKzRQDzo/видео.html this is what he mentioned; do you know if 'no longer' (linked) here means definitely or generally during that moment? I get how it probably won't react as a molecule and that it's not reactive like the Cl radical like you suggested, but I wanted to clarify whether it could still be pulled back into reactions with other radicals because he didn't seem to say either. Thanks
@@55lorrw I think it's more of a throwaway statement rather than something important. It's been said to clarify the difference in reactivity now. There is no good reason why the Br2 formed in this way is any different to one that's never reacted yet. There is no difference between them
@@chemistrytutor yeah haha that makes sense, thank you
Why does the bromine water decolourise if there is no alkene that was produced since only a halogenoalkane is made.
I cant remember which bit you're referring to, but...
The colour of bromine water is due to the dissolved bromine. If that bromine reacts then the water will decolourise.
The Br2 will most commonly act as an electrophile, but can behave in different ways, e.g. in FR substitution. If the Br2 reacts and is used up, then it won't have colour anymore.
lovely
@@gamuchirainmago3328 thanks 😊
Greetings!
Is it still valid for 2023 exams of Cambridge International A Level?
Yes definitely! 😀
@@chemistrytutor Thanks!
P.s. a little bit confused but does your channel cover all the A Level Chemistry syllabus? And what exam boards is it applicable to?
@@maryamsadiq8780 All exam boards are 95% similar. The main differences are exam structure and the question style.
I've covered all of y12 with Explained videos, and Y13 is more exam walkthroughs.
Here is a doc with links to all the videos...
drive.google.com/file/d/1s3I5prjbJRR1U1lcKpXO9TQpQMaVoP66/view?usp=drivesdk
@@chemistrytutor Thank you so so much! I need help regarding my exam prep. I have A Level Chemistry P4 (CAIE International) in 1 week. I haven't studied properly. My target is an A. Can I still get it? What do I do at this point? I haven't practiced past year questions as well and need to catch up on the core syllabus too! What do I do?!?
@@maryamsadiq8780 I think practice questions are the best thing now. Great revision generally, bit they will highlight where you need to watch videos and hit the books etc.
It's not too late!
In the second reaction during propagation, why would the Cl be a radical in the products, as surely one is given to the methyl radical, and the other is just left on its own?
I'm not totally clear what you mean. The chlorine must be a free radical in the second product because of the number of electrons it has.
Methyl free radical has 7 electrons in the outer energy level, so it has an unpaired electron (free radical) and needs 1 more to make 8.
The Cl - Cl bond contains 2 electrons, this breaks and momentarily each chlorine takes 1 from the bond meaning they have 7 electrons so both have an unpaired electron and so both are (temporarily) free radicals.
Pretty much instantly, one of the chlorine forms a new covalent bond using its unpaired electron and the unpaired electron in the methyl free radical, so both become stable and now have a share of a total of 8 electrons.
The other chlorine still only has 7, and so exists as a free radical
Omg thank u so much this is all amazing!!!!
does this follow AQA spec?
@tazhehe9919 yes indeed. I teach aqa so I made it for my students.
Its suitable for other courses too, but definitely aqa
Thank you so much
You're most welcome 😀
Amazinggggggg
Love it... but have never heard halogenoalkanes pronounced like that before...😂
I'm glad you like the video. Glad the pronunciation is also entertaining as well, to add to the experience 😃
I have to say I'm intrigued to know the way you're used to hearing it said...
You're pronouncing the "gen" part of the word as if it rhymes with "gene". Every other chemist I know pronounces it to rhyme with "jen"; hal o jen o al kane. It matters not 😂
@@andrewwhitehead2002 😀 interesting!
@@andrewwhitehead2002 He couldn't care less + hes the teacher and you're watching his video so pipe down
ARE U IRISH
Irish great grand parent... do I sound Irish?
@@chemistrytutor a little and u teach so good i rlly APPERICATE dat
@@iluvdawgs00 thanks 😊