Question

Consider the following reactions sequence.

Find the number of bromine atoms in major product B.

Answer 1

3

Answer 2

The starting material is 2H-chromene.

Step 1: Reaction with HBr (excess) and heat. Under these conditions, two reactions occur:

1. Electrophilic addition to the double bond: The double bond in the dihydropyran ring reacts with HBr. Protonation occurs at C3, leading to a carbocation at C4, which is resonance-stabilized by the benzene ring. Bromide then attacks the carbocation at C4. This yields 4-bromo-3,4-dihydro-2H-chromene.
2. Ether cleavage: With excess HBr and heat, the ether linkage is also cleaved. The oxygen atom is protonated, and bromide ion attacks the more substituted carbon atom (or the one that forms the more stable carbocation). In this case, the attack at C2 cleaves the C2-O bond.

Considering the combined effect, the double bond is saturated with HBr, and the ether linkage is cleaved. The cleavage of the ether linkage at C2-O will lead to the formation of a phenol and a brominated propyl chain. The double bond between C3 and C4 will add HBr.

A more likely pathway considering the conditions (excess HBr, heat) is that the ether cleavage happens first, or concurrently. When the ether bond cleaves, it forms a phenolic hydroxyl group and a 3-carbon chain with a double bond. The double bond then undergoes electrophilic addition of HBr.

Let's consider the structure of 2H-chromene. The double bond is between C3 and C4. When treated with HBr (excess, heat): The ether linkage is cleaved. The oxygen gets protonated, and Br- attacks C2. This opens the ring. The resulting intermediate is 2-(2-hydroxyphenyl)prop-2-ene. The double bond then reacts with HBr. Addition of HBr to the double bond gives 2-(2-hydroxyphenyl)-2-bromopropane. This is incorrect.

Let's re-evaluate the ether cleavage. The C-O bond cleavage under acidic conditions (HBr) will occur at the more substituted carbon or the one that can form a more stable carbocation. In 2H-chromene, the oxygen is connected to the benzene ring and C2. The double bond is between C3 and C4.

A more accurate mechanism for the reaction of 2H-chromene with excess HBr and heat involves cleavage of the ether bond and addition of HBr to the double bond. The ether cleavage typically occurs by protonation of oxygen, followed by nucleophilic attack by bromide. The attack can occur at C2 or the benzylic position. Given the presence of the double bond, addition of HBr to the double bond is also expected.

A common outcome for cyclic ethers with double bonds under strong acidic conditions with nucleophiles is ring opening and addition. For 2H-chromene, reaction with HBr (excess, heat) leads to cleavage of the ether bond and addition of HBr to the double bond. The ether linkage cleaves to form a phenol. The double bond reacts with HBr. The most accepted pathway is:

1. Protonation of the oxygen atom.
2. Attack by Br- at C2, leading to ring opening and formation of 2-bromophenol and allyl alcohol (prop-2-en-1-ol).
3. The allyl alcohol then reacts with excess HBr to undergo addition across the double bond, forming 1,2-dibromopropane. So, the intermediate product (A) is a mixture of 2-bromophenol and 1,2-dibromopropane.

Step 2: Reaction of (A) with Br2/H2O (excess). The reaction mixture contains 2-bromophenol and 1,2-dibromopropane.

1. Reaction with 1,2-dibromopropane: Alkenes react with Br2/H2O to form bromohydrins. However, 1,2-dibromopropane is already a dibrominated alkane. The addition of Br2 to an alkene in water leads to vicinal bromohydrins. If the starting material was propene, it would form 2-bromo-1-propanol or 1-bromo-2-propanol. Since we have 1,2-dibromopropane, this part of the mixture might not react significantly in the second step under these conditions, or it might undergo further substitution reactions. However, the question asks for the major product B, implying a single dominant pathway.

Let's reconsider the first step product. If the ether cleavage happens such that the double bond is preserved in the chain, and HBr adds to it. Let's assume the ether cleavage leads to a 2-hydroxyphenyl group and a 3-carbon chain. The structure of 2H-chromene is:

      O
     / \
    /   \
   C-----C
  / \   / \
 C   C=C   C
 |   |   |
 C---C---C

Where the benzene ring is fused to the C4a-C8a bond. Let's number the heterocycle: O is 1, C2, C3, C4, C4a (fused). Double bond between C3 and C4. Reaction with HBr (excess, heat): Ether cleavage at C2-O bond. Attack by Br- at C2. This opens the ring to form a 2-hydroxyphenyl group attached to a 3-carbon chain. The original double bond was between C3 and C4. After ring opening, this becomes a double bond in the chain. The structure would be: 2-hydroxyphenyl-CH2-CH=CH2. This is incorrect.

Let's assume the ether cleavage occurs differently. Consider the reaction of 2H-chromene with HBr. The double bond reacts with HBr first. Addition of HBr to the double bond. H+ adds to C3, carbocation at C4 (resonance stabilized). Br- attacks C4. Product: 4-bromo-3,4-dihydro-2H-chromene. Now, with excess HBr and heat, the ether linkage is cleaved. Protonation of O, attack by Br- at C2. This opens the ring to form 2-bromo-1-(2-hydroxyphenyl)propan-1-ol. This is still not right.

Let's assume the first step leads to saturation of the double bond and cleavage of the ether. The most reasonable outcome for the first step (HBr, excess, heat) is the cleavage of the ether bond and addition of HBr to the double bond. The ether cleavage of chromene with HBr leads to 2-bromophenol and allyl bromide. This is a known reaction. So, intermediate (A) is 2-bromophenol and allyl bromide.

Step 2: Reaction with Br2/H2O (excess). We have 2-bromophenol and allyl bromide.

1. Reaction of 2-bromophenol with Br2/H2O: Phenols are highly activated towards electrophilic aromatic substitution. With excess Br2/H2O, bromination will occur at the ortho and para positions relative to the hydroxyl group. Since there is already a bromine at the ortho position (C2), bromination will occur at C4 and C6. This will lead to 2,4,6-tribromophenol.
2. Reaction of allyl bromide with Br2/H2O: Allyl bromide is CH2=CH-CH2Br. The double bond will react with Br2/H2O to form a bromohydrin. CH2=CH-CH2Br + Br2 + H2O → CH2Br-CH(OH)-CH2Br (from attack of Br+ and H2O on the double bond, followed by Br- attack). This gives 1,2-dibromo-3-bromopropan-1-ol or similar.

The question asks for the major product B. Electrophilic aromatic substitution on activated rings like phenols is generally faster than addition to alkenes or substitution on alkyl halides. Therefore, the major product will likely arise from the reaction of 2-bromophenol.

The product from 2-bromophenol is 2,4,6-tribromophenol. Let's count the bromine atoms in 2,4,6-tribromophenol. There are 3 bromine atoms.

Let's double check the first step. Reaction of 2H-chromene with HBr (excess, heat). The ether linkage is cleaved. The C-O bond breaks. The oxygen atom is protonated, and bromide ion attacks. The most likely cleavage is at the C2-O bond, leading to 2-bromophenol and prop-2-en-1-ol. Then prop-2-en-1-ol reacts with excess HBr to give 1,2-dibromopropane. So, intermediate (A) is a mixture of 2-bromophenol and 1,2-dibromopropane.

Now, reacting this mixture with Br2/H2O (excess).

• 2-bromophenol + Br2/H2O (excess) → 2,4,6-tribromophenol (3 bromine atoms).
• 1,2-dibromopropane + Br2/H2O (excess). This is a saturated halide. It might undergo substitution or elimination reactions, but addition to double bonds is the primary reaction with Br2/H2O.

The major product B is formed from the more reactive species, which is the phenol. The major product B is 2,4,6-tribromophenol. Number of bromine atoms in 2,4,6-tribromophenol = 3.

Alternative interpretation of Step 1: If HBr adds to the double bond first, forming 4-bromo-3,4-dihydro-2H-chromene. Then ether cleavage occurs. Protonation of O, attack by Br- at C2. This opens the ring to form 2-bromo-1-(2-hydroxyphenyl)propan-1-ol. This molecule has 2 bromine atoms. Then reaction with Br2/H2O. The phenol ring will be brominated. The hydroxyl group is activating. So, the phenol part will get brominated at ortho and para positions. The product would be a brominated derivative of 2-bromo-1-(2-hydroxyphenyl)propan-1-ol. The benzene ring has positions 2, 4, 6 available for bromination. The product would have 2 bromines on the ring and 1 bromine on the side chain, total 3 bromines.

Both pathways lead to a product with 3 bromine atoms.

Final Check: Starting material: 2H-chromene. Reagent 1: HBr (excess, heat). Expected outcome: Ether cleavage and addition to double bond. Commonly accepted outcome for chromene with HBr/heat is cleavage to 2-bromophenol and allyl bromide, followed by addition of HBr to allyl bromide to give 1,2-dibromopropane. So, mixture of 2-bromophenol and 1,2-dibromopropane.

Reagent 2: Br2/H2O (excess). Reaction of 2-bromophenol with excess Br2/H2O gives 2,4,6-tribromophenol. This has 3 bromine atoms. Reaction of 1,2-dibromopropane with Br2/H2O. This would likely lead to further bromination or substitution, but the phenol reaction is much faster.

Therefore, the major product B is 2,4,6-tribromophenol. Number of bromine atoms = 3.

Let's consider the structure of 2H-chromene. It is a benzodihydropyran. The reaction with HBr (excess, heat) leads to cleavage of the ether bond and addition of HBr to the double bond. The ether bond cleaves to form a phenol. The double bond is saturated. The most common outcome is cleavage to 2-bromophenol and prop-2-en-1-ol, which then reacts with HBr to form 1,2-dibromopropane. So, compound A is a mixture of 2-bromophenol and 1,2-dibromopropane.

Compound B is formed by reacting A with Br2/H2O (excess). 2-bromophenol reacts with excess Br2/H2O to give 2,4,6-tribromophenol. This molecule has 3 bromine atoms. The reaction of 1,2-dibromopropane with Br2/H2O is less likely to produce the major product compared to the highly activated phenol.

Final answer is based on the number of bromine atoms in 2,4,6-tribromophenol. Number of bromine atoms = 3.