[susdujcrd]

well, I know the preferred mechanism will be Sn2, but I'm not sure "where" will the hydroxide attack.
From what I've learned, these types of reactions are always backside attacks, but here there is too much steric hindrance
for the ion to attack from "inside" the ring, so it will be forced to react from the same "side" of the bromine.
Am I right?

[discodermolide]

well, I know the preferred mechanism will be Sn2, but I'm not sure "where" will the hydroxide attack.
From what I've learned, these types of reactions are always backside attacks, but here there is too much steric hindrance
for the ion to attack from "inside" the ring, so it will be forced to react from the same "side" of the bromine.
Am I right?

I think you will get ring flip to the other cyclohexane which will the be attacked by the OH^- from the bottom face, as shown in the picture.

[Guitarmaniac86]

Small nucleophiles will tend to attack axially because they can better avoid 1,3 diaxial interactions so they can approach from above axially. Larger nucleophiles will tend to attack equitorially if this means there is less steric hindrance (1,3 diaxial interactions again) to do so, and only if the ring is locked in a specific conformation with a t-butyl group or sufficiently large group on the ring.

As discodermolide said, you will most likely get ring flip to a more stable conformation, though I think the hydroxide would attack axially, and end up "pointing out from the plane of the page" to avoid steric interactions with the methyl group.

[discodermolide]

Small nucleophiles will tend to attack axially because they can better avoid 1,3 diaxial interactions so they can approach from above axially. Larger nucleophiles will tend to attack equitorially if this means there is less steric hindrance (1,3 diaxial interactions again) to do so, and only if the ring is locked in a specific conformation with a t-butyl group or sufficiently large group on the ring.

As discodermolide said, you will most likely get ring flip to a more stable conformation, though I think the hydroxide would attack axially, and end up "pointing out from the plane of the page" to avoid steric interactions with the methyl group.

The drawing I posted shows "axial attack" of the OH group i.e from "underneath". The steric crowding between the methyl group and the OH group will be less than that of the bromine.

[Guitarmaniac86]

Small nucleophiles will tend to attack axially because they can better avoid 1,3 diaxial interactions so they can approach from above axially. Larger nucleophiles will tend to attack equitorially if this means there is less steric hindrance (1,3 diaxial interactions again) to do so, and only if the ring is locked in a specific conformation with a t-butyl group or sufficiently large group on the ring.

As discodermolide said, you will most likely get ring flip to a more stable conformation, though I think the hydroxide would attack axially, and end up "pointing out from the plane of the page" to avoid steric interactions with the methyl group.

The drawing I posted shows "axial attack" of the OH group i.e from "underneath". The steric crowding between the methyl group and the OH group will be less than that of the bromine.

Sorry if I implied it didnt show axial attack, I was agreeing with you. I saw the OP and was in reply mode, but you posted before I did, and as I was typing my response, I was trying to re-word what I had written. Basically, I should have just not replied when you beat me to the punch.

[discodermolide]

Small nucleophiles will tend to attack axially because they can better avoid 1,3 diaxial interactions so they can approach from above axially. Larger nucleophiles will tend to attack equitorially if this means there is less steric hindrance (1,3 diaxial interactions again) to do so, and only if the ring is locked in a specific conformation with a t-butyl group or sufficiently large group on the ring.

As discodermolide said, you will most likely get ring flip to a more stable conformation, though I think the hydroxide would attack axially, and end up "pointing out from the plane of the page" to avoid steric interactions with the methyl group.

The drawing I posted shows "axial attack" of the OH group i.e from "underneath". The steric crowding between the methyl group and the OH group will be less than that of the bromine.

Sorry if I implied it didnt show axial attack, I was agreeing with you. I saw the OP and was in reply mode, but you posted before I did, and as I was typing my response, I was trying to re-word what I had written. Basically, I should have just not replied when you beat me to the punch.

No problems, the more perspectives on a question the better

[susdujcrd]

thank you guys, excellent explanations.

[orgopete]

well, I know the preferred mechanism will be Sn2, but I'm not sure "where" will the hydroxide attack.
From what I've learned, these types of reactions are always backside attacks, but here there is too much steric hindrance
for the ion to attack from "inside" the ring, so it will be forced to react from the same "side" of the bromine.
Am I right?

The ring flips are correct. I am dubious of an SN2 reaction unless you are explicitly asked to write an SN2 product. A rule of thumb I like to recall from Brown, Foote, and Iverson for a reaction like this is elimination is the preferred product with any base with a pKa greater than 12.

[susdujcrd]

well, I know the preferred mechanism will be Sn2, but I'm not sure "where" will the hydroxide attack.
From what I've learned, these types of reactions are always backside attacks, but here there is too much steric hindrance
for the ion to attack from "inside" the ring, so it will be forced to react from the same "side" of the bromine.
Am I right?

unless you are explicitly asked to write an SN2 product.

That is the case. It was before we learned about elimination reactions