Reagents such as \(\mathrm{HCl}\) , HBr, and $\mathrm{HOH}\left(\mathrm{H}_{2} \mathrm{O}\right)$ can add across carbon-carbon double and triple bonds, with \(\mathrm{H}\) forming a bond to one of the carbon atoms in the multiple bond and \(\mathrm{Cl}\) , Br, or OH forming a bond to the other carbon atom in the multiple bond. In some cases, two products are possible. For the major organic product, the addition occurs so that the hydrogen atom in the reagent attaches to the carbon atom in the multiple bond that already has the greater number of hydrogen atoms bonded to it. With this rule in mind, draw the structure of the major product in each of the following reactions.

First, we need to identify the reagent being used and the multiple bonds present in the reactant molecule. Pay attention to whether the reactant contains double or triple carbon-carbon bonds. It's also important to identify the carbon atoms involved in the multiple bonds.

Based on Markovnikov's rule, the hydrogen atom in the reagent should attach to the carbon atom in the multiple bond that already has the greater number of hydrogen atoms bonded to it. For this step, you should determine which carbon in the multiple bonds will have the hydrogen atom added to it. Ensure you follow this rule for each specified reaction.

Since we now know where the hydrogen atom will bond, we can complete the other part of the reagent bonding to the other carbon atom. In this step, the remaining Cl, Br, or OH in the reagent will bond to the other carbon atom involved in the multiple bond, thus forming the final product. We can't provide accurate products without the specific reactions. However, below are some examples of how the solutions should be presented: Example 1: Reactant: CH3CH=CH2 + HCl Step 1: The reagent is HCl, and the reactant has a double bond between C2 and C3. Step 2: Applying Markovnikov's rule, the hydrogen atom in HCl should bond to C2, as it has more hydrogen atoms bonded to it (2) compared to C3 (1). Step 3: The remaining Cl in the reagent will bond to C3. So, the major product will be CH3-CH(Cl)-CH3. Example 2: Reactant: CH3CH=CHCH3 + HOH Step 1: The reagent is HOH, and the reactant has a double bond between C2 and C3. Step 2: Applying Markovnikov's rule, the hydrogen atom in HOH should bond to C2, as it has more hydrogen atoms bonded to it (2) compared to C3 (1). Step 3: The remaining OH in the reagent will bond to C3. So, the major product will be CH3-CH(OH)-CH2-CH3.