[webassignbuddy]

Would the answer be C and D (the 2 ions that form the acidic salt)?

[webassignbuddy]

For Part B, the answer would be X and Y right, because when both molecules are deprotonated, aromaticity is created, thus making both molecules stable immediately which is favored, as opposed to the molecule in Z.

For Part C, the answer would be proton, right?

For Part D, would molecule X and Y look like this after the reaction?

[webassignbuddy]

Well then...

[webassignbuddy]

Anyone? 
Please help

[orgopete]

Start with the basic salt, I think you may find it easier. Then look at X, Y, and Z. If the salt is basic, which will react with it? For example, if X was HCl, and the base was NaOH, how would they react? One of the choices of X, Y, or Z will behave as a weak acid.

Then go back and find the other pair and how might it react with X, Y, or Z (it will be a different reaction).

[webassignbuddy]

Start with the basic salt, I think you may find it easier. Then look at X, Y, and Z. If the salt is basic, which will react with it? For example, if X was HCl, and the base was NaOH, how would they react? One of the choices of X, Y, or Z will behave as a weak acid.

Then go back and find the other pair and how might it react with X, Y, or Z (it will be a different reaction).

The basic salt would be Na⁺NH₂^-.

1. Concept:

pKa = refers to the dissociation of the molecules

• The LARGER the pKa value is = MORE dissociation of the molecules and therefore MORE H+ ions (the EASIER it is for the base to pluck off/deprotonate an H from the acid) = the WEAKER the acid.
• The SMALLER the pKa value is = LESS dissociation of the molecules and therefore LESS H+ ions (the HARDER it is for the base to pluck off/deprotonate an H from the acid) = the STRONGER the acid.
  

2. For Example: A base NaOH can react with an acid HCl to form NaCl and H₂O when OH- (from the base) deprotonates the acid HCl.

• So the concept in this problem is: This basic salt/the base NaNH₂ will react with one of the molecules X, Y, or Z that acts as a weak acid.
  

3. Back to the problem

• NaNH₂ = strong base, with a LARGE pKa (Na+ is the spectator ion, NH₂^- will do the deprotonating)
• Cyclopentadiene (molecule Y) = acid, with a pKa of 16 (from teachers notes)
  

4. Reaction:

NH₂^- (from the basic salt/base) will deprotonate cyclopentadiene's hydrogen (letter d). Cyclopentadiene would then become aromatic like the example below.



So, now that it's aromatic, the system goes DOWN in energy.
And the product = NH₃, and cyclopentadienyl anion with a pKa of 17.

So Y would be the answer to Part E and the answer to Part F would be "proton", right?

[webassignbuddy]

What about Parts A, B, C, and D?

I got the answer to B, which is Molecule X and the answer to C is hydride ion. But what is the acidic salt?

This is due in 44 min

[orgopete]

Okay, you have the basic salt (although amide should have been the base rather than isopropoxide). Sooooo, if A and C are the basic salt, do you think B and D might be the acidic salt? You should be able to work through it in a similar fashion. Just be careful with your electrons.