Question

question 7 (10 points) listen what product is formed in the hydration reaction shown below?
chemical structure + h₂o $xrightarrow{h_2so_4}$
four options with chemical structures
none of the answers predict the correct product of the reaction.

Explanation:

Step1: Identify Reaction Type

The reaction is an acid - catalyzed hydration of an alkene. The general mechanism follows Markovnikov's rule, where the hydrogen from \(H_2O\) (in the presence of \(H_2SO_4\) as an acid catalyst) adds to the less substituted carbon of the double bond, and the -OH group adds to the more substituted carbon.

First, let's analyze the starting alkene. The alkene has a double bond with one carbon having two methyl groups (more substituted) and the other carbon having an ethyl group and a hydrogen (less substituted).

Step2: Apply Markovnikov's Rule

According to Markovnikov's rule, during acid - catalyzed hydration, the proton (\(H^+\)) from the acid (or from the protonated water) will add to the carbon of the double bond with the most hydrogens (less substituted carbon). Then, the water molecule (or \(OH^-\)) will add to the more substituted carbon.

The starting alkene: Let's draw the structure of the starting alkene. The double bond is between a carbon (let's call it C1) with two methyl groups and a carbon (C2) with an ethyl group and a hydrogen. So, C2 is less substituted (has one H), C1 is more substituted (has no H, two methyl groups).

When \(H^+\) (from \(H_2SO_4\) - catalyzed hydration, the \(H_2O\) is protonated to \(H_3O^+\), which donates \(H^+\)) adds to C2 (less substituted), forming a carbocation on C1 (more substituted). Then, the \(H_2O\) (or \(OH^-\)) attacks the carbocation on C1, forming an alcohol.

Now, let's look at the options:

• The first option: The structure has the -OH group on the more substituted carbon (the carbon with two methyl groups), which is consistent with Markovnikov's rule. Let's check the carbon skeleton. The starting alkene has a 5 - carbon skeleton (ethyl group + two methyl groups on the double bond carbons, so total carbons: 3 (from ethyl: \(CH_3CH_2\)) + 2 (from the two methyls on the other carbon) = 5? Wait, the starting alkene: let's count the carbons. The left - hand carbon of the double bond is connected to an ethyl group (\(CH_3CH_2\)) and the right - hand carbon is connected to two methyl groups (\(CH_3\)) and a methyl? Wait, the starting alkene structure: the double bond is between a carbon with \(CH_3CH_2\) - and a carbon with two \(CH_3\) - groups (so the right - hand carbon of the double bond has two methyls, so it's a tetrasubstituted? Wait, no, the double bond: one carbon has \(CH_3CH_2\) - and \(H\) (wait, no, the drawing: the left carbon of the double bond has an ethyl group (two carbons) and the right carbon has two methyl groups (so the double bond is between a carbon with \(CH_3CH_2\) - and a carbon with \((CH_3)_2C=\) - so the double bond is \(CH_3CH_2 - C = C(CH_3)_2\).

So, the double bond is between \(C - C(CH_3)_2\) (where the first C is \(CH_3CH_2 - C\)). So, the carbons of the double bond: the left C (let's call it C3) has \(CH_3CH_2\) - and is bonded to the double bond, the right C (C4) has two \(CH_3\) - groups. So, when we add \(H_2O\) in acid, \(H^+\) adds to C3 (since it has a hydrogen, while C4 has no hydrogens), forming a carbocation on C4. Then, \(H_2O\) attacks C4, forming an alcohol on C4. So the product should have the -OH group on the carbon with two methyl groups (C4), which is the first option.

The other options: the second option has two -OH groups, which is not possible in a simple hydration (it would be a diol, but this is a simple alkene hydration, not a dihydroxylation). The third option has the -OH on a different carbon, not the most substituted. The fourth option has the -OH on a less substituted carbon, violating…

Answer:

The first option (the structure with -OH on the carbon with two methyl groups)