How do you write out the mechanism for the reaction of #"C"_5"H"_11"OH"# with hot sulfuric acid to determine which option gives the greatest number of isomeric alkenes?

Well, let's draw these out and see what they look like.

Based on what they look like, we can determine which products are possible. Let's write one mechanism for each one.

Note that a protonated hydroxyl group is a better leaving group than a regular hydroxyl group. Hence, it will be protonated before leaving.

DEHYDRATION OF PRIMARY ALCOHOLS YIELDS TERMINAL ALKENES

1) This alcohol is a primary alcohol, so there is only one possible dehydration product: the terminal alkene. The #E2# mechanism becomes:

No other isomers are possible for this starting reactant. It doesn't matter how carbons 3-5 rotate their bonds; it's still the same compound. Also, the #E1# mechanism can occur as well. I just didn't draw it because it is shown in the fourth alcohol possibility.

DEHYDRATION OF SECONDARY ALCOHOLS MAY YIELD TWO OR THREE DIFFERENT ALKENES

2) For this secondary alcohol, the major product is formed according to Zaitsev's rule, which says that the #\mathbf(beta)#-carbon with the fewest hydrogens donates the hydrogen to the base that performs the elimination. The #E2# mechanism becomes:

Path 1 gives the two major products. Path 2 gives the minor product.

The major products include one cis and one trans isomer, while the minor product has no more than one isomer, being a terminal alkene. Also, the #E1# mechanism can occur as well. I just didn't draw it because it is shown in the fourth alcohol possibility.

SYMMETRY LIMITS THE NUMBER OF DIFFERENT ALKENES POSSIBLE

3) This secondary alcohol is symmetrical, so there is no possibility of more than two alkenes: the cis and trans geometric isomers. The #E2# mechanism becomes:

If you try eliminating the other #beta#-proton and then flip the product horizontally, it's the same molecule as the products drawn out in the mechanism directly above. Ergo, there are only two. Not three. Not four. Two. Also, the #E1# mechanism can occur as well. I just didn't draw it because it is shown in the fourth alcohol possibility.

DEHYDRATION OF TERTIARY ALCOHOLS MAY INVOLVE 1,2-ALKYL OR HYDRIDE SHIFTS

4) This one is interesting; not the answer, but interesting.

Once the #"OH"_2^(+)# leaves, there is no hydrogen to pluck off of the tertiary carbon, so the methyl group must shift (a 1,2-methyl shift) to the right in order to generate a more stable tertiary carbocation. As a result, this molecule is more likely to form a carbocation, and experiences more #E1# reactions than #E2#.

After the proton elimination, we have the E and Z geometric isomers of the final alkene.

Thus, we have overall:

Evidently, option 2 produces 3 isomeric alkenes.