Question

what product is formed in the hydration reaction shown below?
chemical structure + h₂o $xrightarrow{h_2so_4}$
four options with chemical structures, and
one of the answers predict the correct product of the reaction.\

Explanation:

Step1: Recall Markovnikov's Rule

For the hydration of alkenes (reaction of an alkene with \(H_2O\) in the presence of an acid like \(H_2SO_4\)), Markovnikov's rule states that the hydrogen atom from \(H_2O\) (which protonates the alkene first) adds to the carbon of the double bond with more hydrogen atoms, and the hydroxyl group (\(-OH\)) adds to the carbon with fewer hydrogen atoms (more substituted carbon).

Step2: Analyze the Reactant Alkene

The reactant alkene has a double bond. Let's identify the carbons of the double bond. One carbon of the double bond is bonded to two methyl groups (so it's a tertiary - like carbon in terms of substitution around the double bond) and the other carbon is bonded to an ethyl group and a hydrogen? Wait, no, let's draw the reactant structure: the alkene is \(CH_3CH_2C(CH_3)=C(CH_3)_2\)? Wait, no, looking at the structure, the alkene has a double bond where one carbon is connected to two methyl groups and the other carbon is connected to an ethyl group and a hydrogen? Wait, actually, the alkene is \(CH_3CH_2C(CH_3)=C(CH_3)_2\)? Wait, no, the left - hand carbon of the double bond: let's count the substituents. The double bond is between two carbons. One carbon (let's call it C1) has two methyl groups (\(CH_3\)) attached, and the other carbon (C2) has an ethyl group (\(CH_3CH_2\)) and a hydrogen? Wait, no, the structure is: the alkene is \(CH_3CH_2 - C(CH_3)=C(CH_3)_2\). So when we protonate the alkene, the proton (\(H^+\)) from \(H_2O\) (with \(H_2SO_4\) as catalyst) will add to the less substituted carbon of the double bond. Wait, the carbon with more hydrogens: the carbon in the double bond that has a hydrogen? Wait, no, the two carbons in the double bond: C1: \(C(CH_3)=C(CH_3)_2\)? Wait, no, the left carbon of the double bond is \(CH_3CH_2 - C(CH_3)=\), and the right carbon is \(=C(CH_3)_2\). So the left carbon of the double bond (C - 1 of the double bond) has one hydrogen? Wait, no, let's do the substitution count. The double - bonded carbons: for the carbon on the left of the double bond (let's say carbon A), it is bonded to \(CH_3CH_2 -\), \( - CH_3\), and the double bond. For the carbon on the right of the double bond (carbon B), it is bonded to two \( - CH_3\) groups and the double bond. So carbon A has one hydrogen (since \(CH_3CH_2 -\) is \(C_2H_5 -\), \( - CH_3\) is \(C_1H_3 -\), so the total number of hydrogens on carbon A: in \(CH_3CH_2 -\), the carbon adjacent to the double bond has two hydrogens? Wait, maybe a better way: the alkene is \(CH_3CH_2C(CH_3)=C(CH_3)_2\). So the double bond is between \(C(CH_3)\) (attached to \(CH_3CH_2 -\)) and \(C(CH_3)_2\). So the carbon with the \(CH_3CH_2 -\) group (let's call it C1) has one hydrogen (because \(C(CH_3)\) has three bonds: \(CH_3CH_2 -\), \( - CH_3\), and the double bond, so the remaining bond is to a hydrogen). The carbon with two \( - CH_3\) groups (C2) has no hydrogens (since it's bonded to two \( - CH_3\) groups and the double bond, so four bonds: two \( - CH_3\), double bond, so no hydrogens). According to Markovnikov's rule, the \(H^+\) (from \(H_2O\) protonated by \(H_2SO_4\)) will add to the carbon with more hydrogens, which is C1 (since it has one hydrogen, C2 has zero). Then the \(OH^-\) (from \(H_2O\) after losing \(H^+\)) will add to C2 (the more substituted carbon).

Step3: Determine the Product Structure

After adding \(H\) to C1 and \(OH\) to C2, the product will have the \(OH\) group on the carbon that was originally the more substituted carbon of the double bond (the one with two methyl groups). Let's look at the options:

- The first option: The structure has the \(OH\) group on the carbon that is bonded to two methyl groups (the more substituted carbon from the double bond). Let's check the structure: the reactant alkene, after hydration, the product should have the \(OH\) on the carbon that was part of the double bond with two methyl groups. The first option's structure: the carbon with \(OH\) is bonded to two methyl groups and another carbon, which matches the Markovnikov addition.

- The second option has two \(OH\) groups, which is not possible in a simple hydration (hydration of an alkene gives an alcohol, not a diol in this case with \(H_2SO_4\) as catalyst for Markovnikov hydration; diols are formed via other methods like osmium tetroxide oxidation).

- The third option has the \(OH\) on a different carbon, not the more substituted one.

- The fourth option has the \(OH\) on a carbon that was not part of the double - bond's more substituted carbon.

Answer:

The first option (the structure with the \(OH\) group on the carbon bonded to two methyl groups)