Question

If, instead of \(\mathrm{CCl}_4\), water is used as the solvent in a halogen addition reaction, a halohydrin is formed. The student proposed that such a reaction would follow Mechanism A with water replacing bromine as the nucleophile. If this hypothesis is correct, which of the following is the most likely product for the addition of bromine to propene in water? A. 1,2-dibromopropane B. 1,3-dibromo-2-propanol C. 1-bromo-2-propanol D. 2-bromo-1-propanol

Short answer

C. 1-bromo-2-propanol

Step-by-step solution

- Understand the Reaction Context

The reaction in question is the addition of bromine to propene in water, forming a halohydrin. Normally, bromine would add to propene to form a dihalide, but in the presence of water, a halohydrin is formed where one of the halogen atoms is replaced by a hydroxyl group (OH).

- Determine the Mechanism

The mechanism involves the formation of a bromonium ion intermediate after bromine adds to the double bond of propene. This intermediate is then attacked by water, which acts as the nucleophile, opening the bromonium ion and adding to the carbon atoms.

- Identify the Bromonium Ion Formation

The first step in the mechanism is the formation of a bromonium ion where the bromine molecule adds across the double bond of propene, forming a three-membered ring intermediate with Br connected to both carbons of the former double bond.

- Nucleophile Attack

Water, acting as the nucleophile, attacks the most substituted carbon of the bromonium ion. This results in the opening of the ring and the addition of the OH group at the more substituted position (Markovnikov addition). The bromine attaches to the less substituted carbon.

- Identify the Most Likely Product

Given the mechanism described, the major product will have the bromine on the less substituted carbon and the hydroxyl group on the more substituted carbon. For propene, this means attaching Br to the terminal carbon (C1) and OH to the middle carbon (C2), resulting in 1-bromo-2-propanol.

- Choose the Correct Option

Among the given options, identify which compound matches this structure. The choices are: A. 1,2-dibromopropane B. 1,3-dibromo-2-propanol C. 1-bromo-2-propanol D. 2-bromo-1-propanol. Option C. 1-bromo-2-propanol matches the structure where Br is on the first carbon and OH is on the second carbon.

Key concepts

bromonium ion

In organic chemistry, understanding intermediates is key to grasping reaction mechanisms. One such important intermediate is the bromonium ion. This ion forms during the addition of bromine (Br₂) to alkenes such as propene. Initially, the bromine molecule electrophilically adds to the double bond, creating a positively charged bromonium ion.
This ion is a three-membered ring where the bromine atom is simultaneously bonded to both carbons of the former double bond. The stability of the bromonium ion comes from the pseudo-triangular, strained ring it forms, which although not very stable, is crucial for the progression of the reaction.

nucleophilic attack

After forming a bromonium ion, the next step involves a nucleophilic attack. In simple terms, a nucleophile is a substance that donates an electron pair to an electrophile. Water (H₂O), which is rich in lone pairs of electrons, acts as the nucleophile in our reaction. This process involves the water molecule attacking the bromonium ion to open the strained ring.
Importantly, the water molecule usually targets the more substituted carbon atom of the bromonium ion because it is more positively charged and thus more receptive to an electron pair (this will connect us to the Markovnikov rule shortly). This attack results in the ring opening, creating a new carbon-oxygen bond, and eventually forming the halohydrin.

Markovnikov addition

The third critical concept is Markovnikov addition—a rule that guides the outcome of many electrophilic addition reactions. Essentially, Markovnikov's rule states that for the addition of HX (where X is a halogen) to an alkene, the hydrogen (H) will attach to the carbon with more hydrogen atoms initially, and the X group attaches to the carbon with fewer hydrogen atoms. In our case of halohydrin formation in the presence of water, this translates to a situation where the hydroxyl group (OH) from water attaches to the more substituted carbon.
For the addition of bromine to propene in water, this means that the OH group binds to the secondary carbon (C2), while the bromine attaches to the primary carbon (C1). This ensures that the resultant product follows Markovnikov's rule, leading to the formation of 1-bromo-2-propanol—aligning with the major product identified in the reaction mechanism outlined earlier.