What is the product for the following three-step reaction sequence? Br2 + CH3OH → CH3ONa + heat → BrCCl + IBr + O
The Correct Answer and Explanation is:
To determine the correct product from the three-step reaction sequence provided, we need to analyze each reaction step carefully:
Given Reaction Sequence:
- Br₂, hν
This is a radical bromination reaction, specifically allylic or benzylic bromination due to light (hν). The molecule is 1-methyl naphthalene, and the benzylic position (the CH₃ group) is particularly reactive. A bromine radical substitutes a benzylic hydrogen, forming 1-(bromomethyl)naphthalene. - CH₃OH, CH₃ONa, heat
This is a nucleophilic substitution reaction under basic conditions, favoring E2 elimination due to heat. The CH₃ONa acts as a strong base, abstracting a β-hydrogen to eliminate HBr and form a double bond between the benzylic position and the ring — resulting in an exocyclic alkene (i.e., a vinyl group attached to the ring). - Br₂, CCl₄
This is a halogen addition across a double bond. The double bond formed in step 2 reacts with Br₂ to form a vicinal dibromide (anti-addition). This leads to the addition of two Br atoms across the double bond in a trans (anti) fashion.
Now, let’s analyze the options:
We look for a final product that:
- Has two Br atoms on the exocyclic double bond carbons,
- Shows anti stereochemistry (i.e., Br atoms on opposite faces),
- Retains the original naphthalene ring structure.
Option V fits this perfectly:
- It has two Br atoms on the exocyclic carbon positions,
- It shows correct trans stereochemistry,
- The rest of the structure is unaltered.
Correct Answer:
E) V
This sequence showcases typical benzylic bromination, E2 elimination to form an alkene, and anti-addition of Br₂ across the resulting double bond — a classic example of multi-step synthesis in organic chemistry.
