You forgot to take in account of a hydride shift when you did the alkylation. So you'd have a secondary carbo cation when adding on the meta position.
yeah, there will be a hydride shift since the secondary carbo cation is more stable. In that case, there is a mixture primary and secondary, it is not the best way to do it , I have to say by having primary halide. But on the exam, if professor really picky on it ... I would do an extra acyl group step instead.
Yes, the MgBr is the idea. This is how I plan to go about (1) add SO3/H2SO4 (2) add excess Br2 -this is to add br and each meta position (3) Mg/ether (4) attacks an aldehyde on one side and propyl bromide on the other side. (note: Noate: On the step 2 we add Br on two meta position related to SO3 on the ring (5)Use H2O to get rid of SO3 (6) then use NBS to add Br on the allylic position on propyl bromide. (7 )reduce the C=O on the aldehyde
Thanks so much! I'm glad you did electrophilic substitution.
One question: could you (instead of using NBS), add a propanal acyl group, then substitute the ethyl at the meta position, then reduce (instead of doing it at the end) the ketone and use a sulfonic acid (like PTSA) to make the OH a leaving group, then use Sn2 with bromine? This requires one more step, but I guess the real question is: what is the yield for NBS vs. another method...
You can use the propanal acyl group and reduce the ketone later. SN2 work when Nu: attack the alkyl halide (br) and the halide leaves the carbon ... for here ... I am not quite clear on your mechanism. - btw I just add something on the video. Technically we cannot get any yield by adding alkylation on the a ring the way I did it - becuase there is a withdrawing group on the ring. Freidel craft reaction doesn't work on withdrawing group!
Don't take it personally but don't they teach you nomenclature in Chem classes? Or there's only 'this guy/that guy' in your books?
shelteringshade 11 months ago
A very good way to learn about Benzene
TheUbaid143 11 months ago
You have some pretty coldy hands :3
I would love to hug em. n.n
tino2kp 1 year ago
nice
balochiz 1 year ago
where did the CH3 come from on the acylation? My book has it turning into acylium ions. with a double and tripple bond.
ss4vegeta1 2 years ago
You forgot to take in account of a hydride shift when you did the alkylation. So you'd have a secondary carbo cation when adding on the meta position.
kirbykarter 2 years ago
yeah, there will be a hydride shift since the secondary carbo cation is more stable. In that case, there is a mixture primary and secondary, it is not the best way to do it , I have to say by having primary halide. But on the exam, if professor really picky on it ... I would do an extra acyl group step instead.
YourFormulaSheet 2 years ago
Yes, the MgBr is the idea. This is how I plan to go about (1) add SO3/H2SO4 (2) add excess Br2 -this is to add br and each meta position (3) Mg/ether (4) attacks an aldehyde on one side and propyl bromide on the other side. (note: Noate: On the step 2 we add Br on two meta position related to SO3 on the ring (5)Use H2O to get rid of SO3 (6) then use NBS to add Br on the allylic position on propyl bromide. (7 )reduce the C=O on the aldehyde
YourFormulaSheet 2 years ago
Thanks so much! I'm glad you did electrophilic substitution.
One question: could you (instead of using NBS), add a propanal acyl group, then substitute the ethyl at the meta position, then reduce (instead of doing it at the end) the ketone and use a sulfonic acid (like PTSA) to make the OH a leaving group, then use Sn2 with bromine? This requires one more step, but I guess the real question is: what is the yield for NBS vs. another method...
One again, thank you so much :D.
lapo3399 3 years ago
You can use the propanal acyl group and reduce the ketone later. SN2 work when Nu: attack the alkyl halide (br) and the halide leaves the carbon ... for here ... I am not quite clear on your mechanism. - btw I just add something on the video. Technically we cannot get any yield by adding alkylation on the a ring the way I did it - becuase there is a withdrawing group on the ring. Freidel craft reaction doesn't work on withdrawing group!
YourFormulaSheet 3 years ago
so are you saying that instead of using freidel craft reaction, we should just use grinard reaction MgBr-alkyl group?
fh1mahfanzai 2 years ago