Using resonance, select the most acidic protons in the hexane-2,4-dione molecule. the most acidic proton is the one that, upon removal, will yield the most stable conjugate base.
check all that apply.
[hexane-2,4-dione molecule with protons numbered as follows according to iupac nomenclature: protons on c1 are numbered 1 through 3, protons on c3 are numbered 4 and 5, protons on c5 are numbered 6 and 7, protons on c6 are numbered 8 through 10.]

check all that apply.

protons 1, 2, or 3
protons 4 or 5
protons 6 or 7
protons 8, 9, or 10

Protons 4 or 5 are most acidic.

Explanation:

As mentioned there are four types of hydrogen in Hexane-2,4-dione.

The numbering are shown in the figure for your better understanding.

As shown in the figure if we remove the proton from carbon number 3 (protons number 4 or 5), it will result in formation of a carbanion. This carbanion is stabilized due to presence of -I group oxygen and there is possibility of resonance (as shown).

Thus protons 4 or 5 are most acidic.

answer: -

reaction:

na + cl ⇌ nacl

balanced chemical equation:

na + 2cl ⇌ nacl2

sodium + clorine goes to form sodium chloride (table salt).

the product is nacl2.

and the reactants are na and cl.

(note: if you don’t know what ⇌ means, read below: -

⇌(double head arrow) means reversible reaction).