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carbocation exam question help please( electrophilic addition)

the question is asking for the structure of the alterntive carbocation formed in reaction between hydrogen bromide and propene. I don't understand why the answer shows that the plus sign is on the right hand side of the carbon in CH3CH2CH2. why is it it using ch3ch2ch2 that's not an alkene or the final product?

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Reply 1
Original post by usernamenew
the question is asking for the structure of the alterntive carbocation formed in reaction between hydrogen bromide and propene. I don't understand why the answer shows that the plus sign is on the right hand side of the carbon in CH3CH2CH2. why is it it using ch3ch2ch2 that's not an alkene or the final product?


hope u can see the qjuetion and markscheme, i understand the mechanismfor part i but not the last part ii, i don't know how the structure is showing primary carbocation?
(edited 5 years ago)
Original post by usernamenew
why is it it using ch3ch2ch2 that's not an alkene or the final product?


Because that's the carbocation.
Original post by usernamenew
the question is asking for the structure of the alterntive carbocation formed in reaction between hydrogen bromide and propene. I don't understand why the answer shows that the plus sign is on the right hand side of the carbon in CH3CH2CH2. why is it it using ch3ch2ch2 that's not an alkene or the final product?


The carbocation can form either on the middle carbon or the carbon on the edge of propene. The final product should either be CH3CHBrCH3 or CH3CH2CH2Br
Reply 4
Original post by Forecast
Because that's the carbocation.


i dont get that, is the carbocation the product? im confused where does the ch3ch2ch2 come from if its not propene
Original post by usernamenew
i dont get that, is the carbocation the product? im confused where does the ch3ch2ch2 come from if its not propene


The carbocation is the intermediate ...its what happens in the process of changing from an alkene to halogenoalkane..the carbon gets a positive charge and attracts the bromine
Reply 6
Original post by HAFKHA$123
The carbocation is the intermediate ...its what happens in the process of changing from an alkene to halogenoalkane..the carbon gets a positive charge and attracts the bromine


thanks but i dont get how ch3ch2ch2 is the secondary carbocation if the plus sign is on the right carbon
Reply 7
Original post by Anonymouspsych
The carbocation can form either on the middle carbon or the carbon on the edge of propene. The final product should either be CH3CHBrCH3 or CH3CH2CH2Br


thanks, how do u know it can also be on the middle carbon, why does the carbocation not use the Br too because thats the product. and how is the carbocation a secondary if the plus sign is on the righthand side carbon
Reply 8
Ch3Ch2Ch2+ is the primary carbocation. The extra H comes from the addition of H from HBr, breaking the double bond to the left, so the positive charge goes on the terminal carbon.

The mark scheme is saying that you give marks for the secondary carbocation in (ii) if Ch3Ch2Ch2+ (primary) was used in part (i)
Reply 9
Original post by _NMcC_
Ch3Ch2Ch2+ is the primary carbocation. The extra H comes from the addition of H from HBr, breaking the double bond to the left, so the positive charge goes on the terminal carbon.

The mark scheme is saying that you give marks for the secondary carbocation in (ii) if Ch3Ch2Ch2+ (primary) was used in part (i)


ah thank u so much for the explanation, that really helped, i misread the markscheme thinking that was the secondary carbocation.
do u mind helping me on another question its on polymers
Original post by usernamenew
thanks, how do u know it can also be on the middle carbon, why does the carbocation not use the Br too because thats the product. and how is the carbocation a secondary if the plus sign is on the righthand side carbon


The carbocation is not the final product, it is an intermediate species through which the mechanism proceeds to get to the final product. Think of the H-Br molecule, the H will have a delta positive charge and Br will have a delta negative charge as it is more electronegative. So first, it is the H which will bond to the alkene and then the Br will take the electrons from the covalent bond to become a bromide ion which will then bond to the carbocation (essentially at this point the Br- acts as a nucleophile).

It can be on the middle carbon or one of the carbons on the edge because major and minor products can form. If the plus sign is in the middle carbon, then it is a secondary carbocation because you you two R groups on either side of the carbon. But if the plus sign is on the rhs carbon, then it is a primary carbocation.
Original post by Anonymouspsych
The carbocation is not the final product, it is an intermediate species through which the mechanism proceeds to get to the final product. Think of the H-Br molecule, the H will have a delta positive charge and Br will have a delta negative charge as it is more electronegative. So first, it is the H which will bond to the alkene and then the Br will take the electrons from the covalent bond to become a bromide ion which will then bond to the carbocation (essentially at this point the Br- acts as a nucleophile).

It can be on the middle carbon or one of the carbons on the edge because major and minor products can form. If the plus sign is in the middle carbon, then it is a secondary carbocation because you you two R groups on either side of the carbon. But if the plus sign is on the rhs carbon, then it is a primary carbocation.


thank u so much!!:h:
Original post by usernamenew
thank u so much!!:h:


Original post by Anonymouspsych
The carbocation is not the final product, it is an intermediate species through which the mechanism proceeds to get to the final product. Think of the H-Br molecule, the H will have a delta positive charge and Br will have a delta negative charge as it is more electronegative. So first, it is the H which will bond to the alkene and then the Br will take the electrons from the covalent bond to become a bromide ion which will then bond to the carbocation (essentially at this point the Br- acts as a nucleophile).

It can be on the middle carbon or one of the carbons on the edge because major and minor products can form. If the plus sign is in the middle carbon, then it is a secondary carbocation because you you two R groups on either side of the carbon. But if the plus sign is on the rhs carbon, then it is a primary carbocation.


could u please explain why in this case the the major products secondary carbocation is B? B only has 1 alkyl, ive drawn the mechanism out but i till dont get it
Reply 13
Original post by usernamenew
could u please explain why in this case the the major products secondary carbocation is B? B only has 1 alkyl, ive drawn the mechanism out but i till dont get it


The answer is C because.

Can't be A and B as contain a carbocation and a halogen, that is not how electrophilic addition works as the Hydrogen is always attacked first since Bromine is more delta negative (electrophilic).

D is a primary carbocation. This will appear in very small quantities and is the minor intermediate (consequently the minor primary halogenoalkane forms from this).

Due to electronic inductive vectors from the R groups that stabilise positive charges, i.e Electron density is pushed from 2 R groups to the positive charge. The Secondary carbocation is made more thermodynamically stable in solution.

The primary carbocation only has one inductive vector from 1 R group and so is less thermodynamically stable and less likely to form in the first place.

Basically a case of 2 'stability' vectors winning over 1 'stability' vector.

The secondary carbocation is the major intermediate which forms the secondary halogenoalkane product. Therefore C is the answer.

If you had something like 2 methyl propene, then by the same logic. The tertiary carbocation would be the most stable (3 charge vectors) and would form in huge orders of magnitude greater than the primary carbocation, therefore 2 bromo - 2 methyl propane would be the major product of that.
(edited 5 years ago)
Original post by _NMcC_
The answer is C because.

Can't be A and B as contain a carbocation and a halogen, that is not how electrophilic addition works as the Hydrogen is always attacked first since Bromine is more delta negative (electrophilic).

D is a primary carbocation. This will appear in very small quantities and is the minor intermediate (consequently the minor primary halogenoalkane forms from this).

Due to electronic inductive vectors from the R groups that stabilise positive charges, i.e Electron density is pushed from 2 R groups to the positive charge. The Secondary carbocation is made more thermodynamically stable in solution.

The primary carbocation only has one inductive vector from 1 R group and so is less thermodynamically stable and less likely to form in the first place.

Basically a case of 2 'stability' vectors winning over 1 'stability' vector.

The secondary carbocation is the major intermediate which forms the secondary halogenoalkane product. Therefore C is the answer.

If you had something like 2 methyl propene, then by the same logic. The tertiary carbocation would be the most stable (3 charge vectors) and would form in huge orders of magnitude greater than the primary carbocation, therefore 2 bromo - 2 methyl propane would be the major product of that.



thank u very much, i got c too and i was wondering what wrong
could u help me with this question please i wrote the major product as CH3CHBRCH2CH3 is that right if the reaction was between but-1-ene and hbr?
the answer in thhe markscheme says its C
Reply 15
Original post by usernamenew
thank u very much, i got c too and i was wondering what wrong
could u help me with this question please i wrote the major product as CH3CHBRCH2CH3 is that right if the reaction was between but-1-ene and hbr?
the answer in thhe markscheme says its C


Here they're not just using any old hydrogen, but they are "marking" it by using a different isotope of Hydrogen - Deuterium. So in the question, they also want to know where the Deuterium ends up in the organic molecule. In other words, they want to know where the Hydrogen that was originally in the Hydrogen Bromide ends up.

If you follow the mechanism, and realise the major intermediary is the secondary carbocation, you should see that the D ends up on Carbon #1, and the Br on Carbon #2. So the correct answer is the third option.

Spoiler

Original post by have
Here they're not just using any old hydrogen, but they are "marking" it by using a different isotope of Hydrogen - Deuterium. So in the question, they also want to know where the Deuterium ends up in the organic molecule. In other words, they want to know where the Hydrogen that was originally in the Hydrogen Bromide ends up.

If you follow the mechanism, and realise the major intermediary is the secondary carbocation, you should see that the D ends up on Carbon #1, and the Br on Carbon #2. So the correct answer is the third option.

Spoiler



haha thank u thats why i was confused lol. so what i wrote initially was wrong CH3CHBRCH2CH3 if it was just hbr and but-1-ene?
Reply 17
Original post by usernamenew
haha thank u thats why i was confused lol. so what i wrote initially was wrong CH3CHBRCH2CH3 if it was just hbr and but-1-ene?


That's right if it was just HBR with But-1-ene
Original post by have
That's right if it was just HBR with But-1-ene

ok thank u, one final question lol
im a bit confused as to why sometimes when naming compounds u dont always mention the ethyl group and instead use the methyl?
i wouldve named this compound 3-ethyl-4-methylhex3-ene
but the markscheme says its 3,4dimethylhex-3-ene why is the ethyl group ignored? isnt ch2ch3 an ethyl?
Reply 19
Original post by usernamenew
ok thank u, one final question lol
im a bit confused as to why sometimes when naming compounds u dont always mention the ethyl group and instead use the methyl?
i wouldve named this compound 3-ethyl-4-methylhex3-ene
but the markscheme says its 3,4dimethylhex-3-ene why is the ethyl group ignored? isnt ch2ch3 an ethyl?


nothing's been ignored? The correct name is 3,4-dimethylhex-3-ene.
I don't know where you got your name from?

Maybe it'll help if you drew the complete displayed formula.

Spoiler

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