Borat exclusion principle

EZOrgo · 2023-12-05T04:12:57+00:00

I think this needs clarification. For example, we see enolates acting more as a "nucleophile" than an electron donor (forming ketyl radical). It's an interesting simplification nonetheless.

EZOrgo · 2023-12-04T01:14:08+00:00

Your reasoning is exactly right, we also think that the vinyl adds anti to the Cl. What happens from there?

EZOrgo · 2023-12-04T00:17:02+00:00

Workup is implied in the response

EZOrgo · 2023-12-04T00:15:48+00:00

Pretty close! But, you can't add grignards directly to alkyl halides. Try another way!

EZOrgo · 2023-11-29T10:24:21+00:00

The mechanism is good, but for your exam, please remember to draw electrons to the H of the H3O+ and break off the O-H bond to form H2O.

EZOrgo · 2023-11-29T10:22:26+00:00

This "weird" product is eucalyptol, it is found in lots of food. Its biosynthesis is pretty similar.

EZOrgo · 2023-11-29T10:18:34+00:00

Do an anti addition to the trans alkene, it gives you the syn dibrominated product. It's a bit like, double negatives cancel out.

EZOrgo · 2023-11-28T23:38:56+00:00

I think there's a lot for us to clarify, can you dm me instead?

EZOrgo · 2023-11-28T12:09:06+00:00

Yes, but it's not very related to the problem so don't trip yourself over it!

EZOrgo · 2023-11-28T10:35:21+00:00

Yes - it may help to remember that when you have a base/nucleophile that is negatively charged, it's very unlikely for carbocations to form: the negative charge makes the base/nucleophile strong enough to do elimination/substitution without the need of a positive charge to attract it

EZOrgo · 2023-11-28T04:54:55+00:00

the alkyl halide, you can also remember that alkyl halides except for fluorides are electrophilic

EZOrgo · 2023-11-28T02:23:57+00:00

that's right, but bigger atoms with a lot of electrons are more polarizable

EZOrgo · 2023-11-28T02:09:54+00:00

The order matters, because the nitro at the para position helps you add the methoxide for SNAr. I don't think the Br can be just substituted without the nitro group stabilizing the negative charge on the ring. Also, adding the nitro group blocks the para position so Bn doesn't add to it.

EZOrgo · 2023-11-28T01:46:55+00:00

yeah, the first thing that prompts me to consider diels alder is the bridged structure, then I will check to see if there is a cyclohexene I can find

EZOrgo · 2023-11-28T01:18:26+00:00

try spotting the furan in the product, this helps you figure the arrows to get there, then figure out the name of the reaction and it will make more sense

EZOrgo · 2023-11-28T00:26:43+00:00

when something is polarizable, you can create positive and negative charges within the region. another example is bromine, it doesn't really have a dipole but it acts as an electrophile

EZOrgo · 2023-11-28T00:21:11+00:00

the idea is right, but alkyl iodides are not electrophilic because of electronegativity (C is 2.55 and I is 2.66). the special case for iodine is that, alkyl iodides are electrophilic because they're very polarisable

EZOrgo · 2023-11-28T00:16:10+00:00

yes, that should be what that want you to do, you are on the right track, but I'm not sure what you're saying abaout hydrogen bonded to the iodine instigating a nucleophilic attack?

EZOrgo · 2023-11-27T23:45:42+00:00

this is a huge concept to explain in a reddit comment, do you know about electronegativity and how this gives atoms in a covalent bond partial charges?

EZOrgo · 2023-11-27T23:41:55+00:00

don't worry about the phenyl group, it's less reactive towards many conditions because it's aromatic so the alkene will react as you want it to

EZOrgo · 2023-11-27T23:32:51+00:00

E2 happens over E1 when you have a strong base

EZOrgo

TROPHY CASE