Hey /sci. Undergrad here

Pretty sure it's 1bromo1phenylethan2al

1(1isopropyl3chloropentane),4methylcyclohexane? Idk how to name those long ones, I'm just an ochem 1 student

steric hindrance from different groups nigger tits.

>what is chemdraw?

or just name the goddamned things yourself like a fucking human bean.

2e,5e)-4-OXOHEPTA-2,5-DIENOIC ACID

3-bromo-3-phenylpropANAL

cant tell, is that iodine attached to the six carbon rigng?

it's clearly a propanal,
>ethan2al
the notion of ...-2-al is a problem, b/c the aldehyde terminates the chain, that said it is possible to name an aldehyde group as a substituent using the name 'formyl' for the group.

but the chemistry (in the first two steps) in OP's pic is electrophilic aromatic substitution

Your right it definitely is propanal. But wouldn't then the phenyl group be highest priority? Or at least the carbon with the most substituted groups be 1?So 1bromo1phenylpropan3al?

If you name it as an aldehyde, then the aldehyde carbon is usually the 1 position.

You could name it as an alkane:
1-bromo-1-phenyl-3-oxopropane.

I don't remember which of the two is preferred in IUPAC.

Ah gotcha, I've never named anything with oxo. So if you name a substituent on the main carbon chain that substituent becomes the 1 group?

>So if you name a substituent on the main carbon chain that substituent becomes the 1 group?
No, not generally. But it will often be the case for groups which necessarily terminate the carbon chain (aldehyde, carboxylic acid, etc).

The answer is resonance. Resonance is the movement of electrons through a pi system. Steric factors do play a role in where groups go, but the OP/M directing theory is strictly based on electronic effects.

Keep in mind that in EAS the aromatic ring is the nucleophile.

Withdrawing groups are withdrawing because they can accept a pair of electrons from the aromatic pi system. Look at the nitro group, there exists a resonance form where a pair of electrons comes out of the ring, forms a double bond with N and a pair of electrons in the NO double bond goes to the O. Now, draw the resonance forms moving the positive charge around the ring, you will see that the positive charge only exists at the ortho and para positions. The OP positions have been "deactivated" because they cannot act as nucleophiles now.

Now look at an activating group, like an amino group in aniline. This time, the nitrogen has a lone pair that it can donate to the ring. Draw the resonance form of the lone pair on nitrogen forming a double bond with the ring, and a pair of electrons in the ring moving to the ortho position, forming a negative charge there. Now draw the resonance forms with the lone pair moving around the ring, you will see that only the ortho and para positions get the negative charge. The OP positions are "activated" and can act as nucleophiles.

OP/M directing can also happen via induction and it works in the same way, you just can't draw resonance forms for it. To rationalize this, you must realize that partial positive charges in molecules are always balanced by partial negatives on adjacent atoms. Apply the same logic do EW/ED substituents via induction.

NAS is also a thing. In this case everything is reversed and you perform nucleophilic attacks on the "deactivated", electron poor positions of the ring. But the chemistry is totally different now, because a hydride must leave instead of a proton.

>anal
gfffffssss