nvm I understood, you can find ksp of insoluble compounds using dissociation equation

How do you find the Ksp of an extremely insoluble compound? Like for a soluble compound you would write the dissociation equation and use the molar solubility provided & the ratios in the balanced equation to calculate Ksp, but for an insoluble compound is it still accurate to write the dissociation equation and calculate it from there, if it's barely going to dissociate? I don't remember if we covered this in class, and I searched up the definiton of Ksp, and it said it calculates the solubility of a "sparingly soluble compound," is that true?

Quote from: amimi on November 08, 2025, 07:31:55 PMThank you! So for those four structures I incorrectly said were racemic, I would circle no instead and place everything on lines instead of wedges or dashes?

Yes

Thank you! So for those four structures I incorrectly said were racemic, I would circle no instead and place everything on lines instead of wedges or dashes?

All products are correct for regiochemistry.You have used Marks and antimarks correcetly.However you are doing error in identification of racemeic mixture.
In hydroiodination (HI) you are not getting any chiral centre.Check Carbon with methy and iodine is connected to same groups in the ring.Similarly in oxymercuration and demercuration , acid catalyzed hydration and reduction of pi bond with H2 in the presenece of metal catalyst.

Hello! I was wondering if you could check these problems. Thank you so much!
 

step by step.

Protonation of the alkene (electrophilic addition)
• In H₂SO₄/MeOH, the alkene π bond is the nucleophile and grabs H⁺.
• Proton adds in the Markovnikov direction to give the more stable carbocation (a cyclobutyl-carbinyl cation).

Fast ring expansion (carbocation rearrangement)
• A C–C bond from the adjacent cyclobutane migrates (1,2-shift) to the positively charged carbon.
• This converts the strained 4-membered ring into a 5-membered ring and places the positive charge on the new cyclopentyl carbon.
• Driving force: relief of cyclobutane ring strain and formation of a more stabilized carbocation (the classic cyclobutyl-carbinyl ⇌ cyclopentyl rearrangement).

Nucleophilic attack by methanol
• MeOH (the solvent, present in large excess and far more nucleophilic here than HSO₄⁻) attacks the planar carbocation to form a C–O bond.
• The oxygen bears a positive charge in this "oxonium" intermediate.

Deprotonation (regeneration of the acid catalyst)
• Another MeOH molecule acts as a base and removes the extra proton from oxygen.
• This neutralizes the oxonium, giving the ether product and regenerating H⁺ (acid-catalyzed addition overall).

Net reaction: "hydro-alkoxylation" of the alkene by MeOH under acid catalysis to give a methyl dicyclopentyl ether.
• The key feature is the carbocation rearrangement (ring expansion) that occurs before nucleophilic capture; without it you'd retain a 4-membered ring, but the system preferentially expands to a 5-membered ring because it is both less strained and better at stabilizing the cation.
• Stereochemistry: attack occurs on a planar carbocation, so if a stereocenter were formed it would come from either face; here the final product (a monosubstituted cyclopentane) is not stereogenic.

Hello! I would appreciate some help with these questions, thank you so much!!

Hi Sorry for late reply
Just saw this post.
Order is D B A C
First Go with Atom
O is more EN than N so it is less basic than N .It means D is least basic as O is holding its LP tightly.
Then B because it has so many EN atoms (3 Cl)  on it.
C is most basic because O is EDG here and making ring electron rich so deloaclization of LPs of N atom rons is less.

Hi! I wanted to be sure of my reasoning for this question. I thought B would be the weakest base, since it is most stabilized due to the presence of resonance, induction, and the negative charge being on N. However, structure B has resonance, in addition to the negative being on the more electronegative oxygen.