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Chapter 11: Amines

Re: Chapter 11: Amines

Postby goldstanda3269 » Sun Dec 17, 2017 3:01 am

karencowlishaw wrote:Q3
How do you know when the electrons in the resonance structure will move from the double bond to N or the lone pair of electrons will move from N into the ring? I’m finding it hard to track the direction of movement of electrons in resonance structures, is there a way to remember this?

1) Atoms don't move in resonance structures.
2) You can only move electrons in π bonds or lone pairs (that are in p orbitals; don't worry, if the lone pair is not in a p orbital ACER will make this point very clear before the question).
3) The overall charge of the molecule must remain the same.
4) The bonding framework of a molecule must remain intact (you are not breaking the molecule or adding to it).

Consider browsing through the various resonance structures in Chapter 5 in the GS book followed by the part of the videos on Aromatics that explore resonance if you have not seen that already.
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Re: Chapter 11: Amines

Postby brad.willo2670 » Thu Jun 20, 2019 4:46 pm

Q9: 9) Pyridine is a much stronger base than pyrolle but still does not form aliphatic (non-aromatic) hydrocarbons. Why?

Part of explanation: There are 6 sp2 hybridized orbitals in pyridine (1 each for the 5 carbons in the ring and 1 for the nitrogen) thus each double bond is actually composed of the following 2 bonds: 1 sigma which in this case is sp2-sp2 and 1 pi bond which is p-p; because nitrogen only has 3 bonds, it must have a lone pair of electrons (ORG 11.1.1). In pyridine the lone pair is not delocalised (it is in fact perpendicular to the aromatic p system) so that means it is available for bonding.

Am I right in thinking that the lone pair in pyridine is not delocalised because it inhabits the empty sp2 orbital? N in this situation (being bonded 3 times, but only to 2 atoms) would have 3 sp2 and 1 p orbitals.
Therefore, because of the double bond only 2 sp2 are used and 1 p orbital (for double) bond, rather than all 3 sp2 orbitals being used. This means that the lone pair in the sp2 orbital is available for bonding (ie as a base to accept a proton) and does not contribute to aromaticity?

Alternatively if N were bonded to 3 different atoms (as it is in Pyrolle) then all 3 sp2 orbitals are occupied leaving only the p orbital which could then be delocalised as part of the aromatic pi system?

Any light shed on this would be much appreciated
Thanks!
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Re: Chapter 11: Amines

Postby brad.willo2670 » Thu Jun 20, 2019 4:58 pm

Q8: Electrophilic substitution occurs predominantly at which position for pyridine?

In this situation it appears that N is acting as a meta director (electron withdrawing group) - is this because it is only bonded to two atoms? Given that it has a lone pair and does not have a more electronegative atom attached to it I would suspect it would function as an electron donating group?

Also as regards the diagram in the explanation it appears the Electrophile adds to the Para (4) position despite the passage describing the Meta (3) position as the most likely point for E+ substitution. Have I misinterpreted the diagram?

Thanks in advance
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Re: Chapter 11: Amines

Postby horbowiec.7527 » Thu Jul 15, 2021 6:38 pm

This is regarding ORG CHM Chapter 11 Worked Solution Q6:
Answer choice says A: Less then 3 molecules. This includes 2: "primary amine with itself (top middle molecule) and H of primary amine with O of formaldehyde (bottom middle molecule), that's all!"

How about bottom left, 2 molecules of water between each other? And bottom right molecule primary amine with H in water? Surely that's correct too? I chose those 4 molecules as correct, so answer C...

Regards,
Ania
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Re: Chapter 11: Amines

Postby goldstanda3269 » Fri Jul 16, 2021 3:27 am

horbowiec.7527 wrote:This is regarding ORG CHM Chapter 11 Worked Solution Q6:
Answer choice says A: Less then 3 molecules. This includes 2: "primary amine with itself (top middle molecule) and H of primary amine with O of formaldehyde (bottom middle molecule), that's all!"

How about bottom left, 2 molecules of water between each other? And bottom right molecule primary amine with H in water? Surely that's correct too? I chose those 4 molecules as correct, so answer C...


It seems that you fell into a little trap. If it were water in the 2 instances that you described, then you would be correct. However, it is formaldehyde: the H's are attached to C and thus cannot engage in H-bonding (we need the H to be directly attached to any of the 3 atoms: FON).
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Re: Chapter 11: Amines

Postby horbowiec.7527 » Fri Jul 16, 2021 9:19 am

I absolutely did, thanks!

Regards,
Ania
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Re: Chapter 11: Amines (Q24,Q25 Master Series)

Postby rockypatel8090 » Mon Aug 02, 2021 10:28 pm

Can some please explain me the these questions? Cannot understand enough from worked solutions.
Having hard time understanding 1-aza-cope, 2-aza-cope, etc. Thanks :)
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Re: Chapter 11: Amines (Q24,Q25 Master Series)

Postby goldstanda3269 » Tue Aug 03, 2021 5:34 pm

rockypatel8090 wrote:Can some please explain me the these questions? Cannot understand enough from worked solutions.
Having hard time understanding 1-aza-cope, 2-aza-cope, etc. Thanks :)


Of course, you are not expected to understand everything since it is an advanced topic. You will definitely be confronted with advanced topics during the real exam, the point is to just use the information provided to be informed enough to discover the answer.

Also, note that this is not the first time this topic has arisen in GAMSAT-level practice questions:

Chapter 4: Questions 38–39
Chapter 6: Questions 26-27
Chapter 7: Question 40


rockypatel8090 wrote:Having hard time understanding 1-aza-cope, 2-aza-cope, etc. Thanks :)


Observing the structures would suggest that 1-aza means that nitrogen is in position 1, 2-aza: N in position 2, 3-aza: N in position 3: no use to commit that to memory but it is useful to practice your powers of observation with new material.

We are told that these reactions break and re-make a single bond (sigma) while in the environment of a double bond (pi).

Consider going back over the previous sigmatropic rearrangement questions (referenced above) and then if you have questions about this unit, please be as specific as possible with your question.

Also, we re-wrote the Worked Solution for this question hoping that it would be more clear to you by adding images from the passage. It will be live within 24 hours.
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Re: Chapter 11: Amines

Postby rockypatel8090 » Tue Aug 03, 2021 8:48 pm

Thank you!!
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Re: Chapter 11: Amines Q19

Postby skogias6733 » Wed Jan 12, 2022 8:03 pm

Hi,

I cant see how is methadone is disobeying rule c) and missing the secondary carbon (CH) in the linker?

In meperidine the first blue dot is one of the carbons of the teriary amine, (the carbon attached to Nitrogen) and the second blue dot is one of the carbons making up the quaternary carbon,

So in methadone why cant the first blue dot also be one of the carbons of the tertiary amine? (attached to Nitrogen)

I am Not sure what i am missing.


Thanks in advance
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Re: Chapter 11: Amines

Postby goldstanda3269 » Wed Jan 12, 2022 8:58 pm

skogias6733 wrote:I cant see how is methadone is disobeying rule c) and missing the secondary carbon (CH) in the linker?

In meperidine the first blue dot is one of the carbons of the teriary amine, (the carbon attached to Nitrogen) and the second blue dot is one of the carbons making up the quaternary carbon,

So in methadone why cant the first blue dot also be one of the carbons of the tertiary amine? (attached to Nitrogen)


Good question!

In the image, methadone only has 1 blue dot (CH2), because the carbon above it is a tertiary carbon (CH not CH2). The rule requires 2 CH2s in a row as a linker and methadone does not have that.
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Re: Chapter 11: Amines

Postby skogias6733 » Thu Jan 13, 2022 12:19 pm

ohhhh right i was concentrating on the C's not the Hs thank you=)
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Re: Chapter 11: Amines Q25

Postby skogias6733 » Fri Jan 14, 2022 12:27 pm

Hi,

Am i able to please get some further explanation on this question?
I can see why its not C or D but i am a little confused between why it is A and not B.
Dont they both have the same number of double bonds as the starting product?

The answers say:
A: The end of the allyl is expected to sigma bond with the ring and the double bond moves over one position, similar to azo-Cope. We start and end with the same number of double bonds. Correct.

B: New sigma bond (good) but the double bond never moved (not good).

How did the double bond never move in B?? In the reactant the double bond is at the end of the allyl group and in B it appears that it has moved over one. Whereas in A it appears that the double bond has not moved over as it is still on the end?

Could you please point out what i am missing? Thanks!!
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