a) There are several reagents that can be used to effect addition to a double bond including: acid and water, oxymercuration–demercuration reagents, and hydroboration–oxidation reagents. Inspect the final product and select all the reasons why hydroboration–oxidation reagents were chosen to effect the following transformation instead of the other reagents. Oxymercuration–demercuration reagents give the Markovnikov product. The reaction requires anti-Markovnikov product without sigmatropic rearrangement. Hydroboration–oxidation reagents yield the Markovnikov product of addition. Addition with acid and water as reagents gives the Markovnikov product. The reaction requires sigmatropic rearrangement. Hydroboration–oxidation reagents yield the anti-Markovnikov product of addition. Addition with acid and water as reagents avoids sigmatropic rearrangements. The reaction requires the Markovnikov product without sigmatropic rearrangement.
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Oxymercuration-demercuration reagents give the Markovnikov product, but the reaction requires the anti-Markovnikov product. So, this option is not suitable. Show more…
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Reagents such as HCI, HBr, and HOH (H_O) can add across carbon-carbon double and triple bonds, with H forming a bond to one of the carbon atoms in the multiple bond and $\mathrm{Cl}, \mathrm{Br},$ 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. a. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}_{2}+\mathrm{H}_{2} \mathrm{O} \stackrel{\mathrm{H}^{2}}{\longrightarrow}$ b. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}_{2}+\mathrm{HBr} \longrightarrow$ c. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}+2 \mathrm{HBr} \longrightarrow$
Electrophilic addition of hypohalous acids to alkenes yields a 1,2-haloalcohol called a halohydrin. Halohydrin formation, however, does not result from the addition of HOBr, for example. Instead, the addition is done indirectly by reaction of the alkene with Br2 in the presence of water. The reaction also works with Cl2 to give chlorohydrins instead of bromohydrins. The reaction proceeds through a cyclic intermediate known as a bromonium ion. In the second step of the reaction, water is the nucleophile and reacts with the bromonium ion to yield the product. The reaction can also proceed using an alcohol as the solvent instead of water, to give a 1,2-haloether. Draw curved arrows to show the movement of electrons in this step of the mechanism.
Madhur L.
Reagents such as $\mathrm{HCl}, \mathrm{HBr}$, 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 $\mathrm{Cl}, \mathrm{Br}$, or $\mathrm{OH}$ 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. a. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow$ b. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CH}_{2}+\mathrm{HBr} \longrightarrow$ c. $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CH}+2 \mathrm{HBr} \longrightarrow$
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