Question

on 5-methyl-1,4-hexadiene, if a acid catalyzed reaction were to occur, where would the hydronium ion attack?

Answer 1

there are two double bonds in the structure so where would it attack

Answer 2

would it be better to attack at carbon 4 because it can be stabilized by resonance?

Answer 3

In an acid-catalyzed reaction on 5-methyl-1,4-hexadiene, the hydronium ion (H3O+) would likely attack the most nucleophilic carbon atom. In this case, the carbon atoms are numbered as follows:

CH3 ─ CH2 ─ CH ─ CH2 ─ CH═CH2
1 2 3 4

The hydronium ion (H3O+) would most likely attack carbon atom number 3 since it is a terminal carbon (thus more exposed) and also next to a double bond (making it more nucleophilic).

Answer 4

Apologies for the oversight. Since there are two double bonds in the structure, the hydronium ion can potentially attack at two different carbon atoms.

In the case of 5-methyl-1,4-hexadiene, the hydronium ion (H3O+) can attack either at carbon atom number 2 or carbon atom number 4. These positions represent the conjugated double bonds, which are more electron-rich and therefore more susceptible to nucleophilic attack by the hydronium ion.

Answer 5

Yes, attacking at carbon atom number 4 would indeed be preferred as it can be stabilized by resonance. When the hydronium ion attacks carbon 4, the resulting carbocation intermediate can undergo resonance stabilization with the adjacent double bond.

The resonance stabilization involves delocalization of the positive charge across the double bonds, distributing the charge over a larger area. This resonance stabilization increases the stability of the intermediate and makes the attack at carbon 4 more favorable.