Question:
Identify the major product for the reaction:
Identify the major product for the reaction:
Show Hint
Phenol + Br\(_2\) at low temperature in CHCl\(_3\) → para substitution; dehalogenation with Na gives hydroquinone.
Updated On: Jul 18, 2026
- 1
- 2
- 3
- 4
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Electrophilic bromination of phenol.
Phenol undergoes bromination at ortho and para positions due to activating –OH group. Low temperature in chloroform favors selective para substitution.
Step 2: Formation of tribromophenol intermediate.
At 273K, bromination primarily yields para-bromophenol. Further treatment with Na in dry ether removes bromine atoms via reduction (Wurtz-type or similar) forming hydroquinone.
Step 3: Identify reaction pathway.
Sequence: selective bromination → dehalogenation with sodium → 1,4-dihydroxybenzene (hydroquinone).
Step 4: Check positions.
Para substitution ensures hydroxyl groups at 1,4 positions. Ortho positions may also react but sterically less favorable; para dominates at low temperature.
Step 5: Eliminate incorrect options.
Phenol ether, naphthalene derivative, or biphenyl are not formed in this reaction sequence.
Step 6: Final conclusion.
\[ \boxed{\text{Hydroquinone (1,4-dihydroxybenzene)}} \]
Phenol undergoes bromination at ortho and para positions due to activating –OH group. Low temperature in chloroform favors selective para substitution.
Step 2: Formation of tribromophenol intermediate.
At 273K, bromination primarily yields para-bromophenol. Further treatment with Na in dry ether removes bromine atoms via reduction (Wurtz-type or similar) forming hydroquinone.
Step 3: Identify reaction pathway.
Sequence: selective bromination → dehalogenation with sodium → 1,4-dihydroxybenzene (hydroquinone).
Step 4: Check positions.
Para substitution ensures hydroxyl groups at 1,4 positions. Ortho positions may also react but sterically less favorable; para dominates at low temperature.
Step 5: Eliminate incorrect options.
Phenol ether, naphthalene derivative, or biphenyl are not formed in this reaction sequence.
Step 6: Final conclusion.
\[ \boxed{\text{Hydroquinone (1,4-dihydroxybenzene)}} \]
Was this answer helpful?
0
0
Top AP EAPCET Chemistry Questions
- The four quantum numbers of the valence electron of potassium are :
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
- In which of the following the election gain enthalpy of elements is correctly arranged?
- The radii of $ \text{F,}{{\text{F}}^{-}}\text{O} $ and $ {{\text{O}}^{2-}} $ are in the order
- Which of the following has the highest first ionization potential?
View More Questions
Top AP EAPCET Ethers Questions
Better results for the preparation of ethers X and Y can be obtained from reactant pairs respectively
Identify the product/s formed in the following reaction
- What is the IUPAC name of the product Y formed in the given sequence of reactions? \[ \text{Isobutane} \xrightarrow{\text{KMnO}_4} \text{X} \xrightarrow{\text{(i) Na, (ii) }\text{CH}_3-\text{Br}} \text{Y} \]
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions