[jlax31]

which hydrogen (underlined) is more acidic?

I believe it is the one on the right with resonance and proximity explaining why the h would be more easily lost. any help?

[Dan]

I believe it is the one on the right with resonance and proximity explaining why the h would be more easily lost. any help?

Can you explain your answer in more detail? Resonance is the key, but I do not see why you chose the indanone on the right on that basis.

[Rookie_OC_Lover]

My guess :
Right one.
to know why, imagine the compound has lost its Hydrogen.
for the right structure it takes electron movements on 4 carbons to establish a more stable state. from the N to the carbon in Ketone group.
for the one on left, it takes 5 carbons to travel.

[Dan]

for the right structure it takes electron movements on 4 carbons to establish a more stable state. from the N to the carbon in Ketone group.

Really? Are you saying that the nitrogen anion is conjugated to the meta electron withdrawing ketone? Have you tried drawing the curly arrows on paper?

Question: which is more acidic out of these two?

[jlax31]

I believe it is the one on the right with resonance and proximity explaining why the h would be more easily lost. any help?

Can you explain your answer in more detail? Resonance is the key, but I do not see why you chose the indanone on the right on that basis.

Sure, I will give it a shot. When the carbonyl group is in the position on the right it pulls electron density out through induction (e- O atom) and resonance. I guess my answer would be that the polarity of this carbonyl group might pull the H off the Nitrogen? I believe there is more to it though where resonance and possibly a more stable cation/anion is the answer.

[Dan]

When the carbonyl group is in the position on the right it pulls electron density out through induction (e- O atom) and resonance.

This is where you are mistaken. The N lone pair is conjugated only in the o/p positions.

[jlax31]

When the carbonyl group is in the position on the right it pulls electron density out through induction (e- O atom) and resonance.

This is where you are mistaken. The N lone pair is conjugated only in the o/p positions.

Ok, the electron pair delocalizes the structure, makes it a weaker base than usual, and is conjugated only in the o/p positions. So, if the lone pair is in the para position on the left, would that some how affect the acidity of the hydrogen? The cyclopentanone is a substituent connected to the para position? Still confused..

[Dan]

Do this one first:

Question: which is more acidic out of these two?

1. Draw the conjugate bases (anions) of both structures
2. Determine which of the two can delocalise the -ve chage all the way through to O.
3. The more stable conjugate base is the one with the -ve charge shared between N and O.
4. So which of the acids is more acidic?

[opsomath]

I endorse Dan's problem that he posed as an analogy to the two-ring system. Try doing that one first, it's simpler. It's easy to get confused when there are multiple rings.

Don't just stare at it. Get a pencil and a blank piece of paper and draw curly arrows moving electrons around. For organic chemistry, your fingers have to teach it to your brain by drawing lots and lots of structures.