Organic Chemistry---Question about carbocation rearrangement in a dehydration reaction?

Question

I'm thinking of treating 1-(1,4-cyclodienyl)-1- propanol with concentrated H2SO4 and heat to get 1-(1-propenyl)-1,4- cyclodiene. Will carbocation rearrangement cause my double bond in the propene side chain to form instead at the carbon in the ring, or will the double bond that's already in the ring block the rearrangement from getting to the most stable carbon?

It's a diene ring with a 3 carbon chain attached; the alcohol is atached to the chain

Answer 1

You didn't get the names right. Propanol has only 3 C atoms but your compound has a 4th one too.

Answer 2

-A1: Picture #1 appears to be (3S,4E)-3-chloro-1,4-hexadiene. -A2: The dehydration occurs via an E1 mechanism and gives the most highly substituted (trisubstituted) alkene as the major product because it is more stable so it forms faster. The dehydrohalogenation reaction occurs via an E2 mechanism. In that mechanism, the H and Br are lost in a single step in an anti periplanar fashion. In the given starting material, it is impossible to arrange the molecule so that the H on the same C as the CH3 and the Br are are arranged anti and periplanar because that H and Br are cis to each other. However, the Br and one of the H's on carbon 3 can be arranged anti and periplanar, so loss of that H and Br leads to the observed product even though it is not the most highly substituted product. -A3: The Zaitsev product is the more highly substituted alkene formed by the dehydration. -A4: Dehydration- An unshared electron pair on the alcohol O moves to the acid catalyst and picks up an H+. The resulting ion loses a molecule of water by moving the C-O bond onto the O to give a carbocation. A water molecule acts as a base by having an electron pair on the O remove the H+ from the ring carbon with the CH3 on it and allowing the C-H bonding electrons move between the carbocation carbon and the adjacent ring carbon to give a pi bond. Dehydrohalogenation- Begin by carrying out a cyclohexane "ring flip" if required to get the Br in an axial position. The OH- in solution begins to bond to the C-H hydrogen that is trans to the Br and is also axial, but in the opposite direction. Simultaneously the C-H bond moves between the two ring carbons and the C-Br bond breaks so that the electrons in it move entirely onto the Br to give Br- -A5: I suspect that the question should say 3-methyl-2-butanol. If so, the same dehydration mechanism as was described above will give 2-methyl-2-butene as the major product. Note that the numbering of the carbon chain is reversed for the alkene product from what it was for the alcohol.