Organic Compounds Containing Oxygen: Chemistry | JEE Main
An organic compound 'A' with emperical formula C6H6O gives sooty flame on burning. Its reaction with bromine solution in low polarity solvent results in high yield of B. B is
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Hint 1 of 3
Given the empirical formula C6H6O and that the compound burns with a sooty flame, what is compound A?
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Step-by-step solutionView
Correct answer
Compound A is phenol, which undergoes monobromination in low-polarity solvents like CS2 or CHCl3 to yield 4-bromophenol (p-bromophenol) as the major product.
Option analysis
Why each option works or fails
A ·
Believing that bromine always yields 2,4,6-tribromophenol regardless of the solvent polarity. Recall that polysubstitution to 2,4,6-tribromophenol occurs in highly polar solvents like water (bromine water), whereas non-polar solvents suppress phenoxide formation and lead to monobromination.
B ·
Assuming that ortho-bromophenol is favored over para-bromophenol due to proximity effects. Steric hindrance from the hydroxyl group directs the incoming electrophile predominantly to the less hindered para position, making 4-bromophenol the major product.
C ·
None. This is the correct major product, 4-bromophenol. Phenol reacts with Br2 in low-polarity solvent (e.g., CS2 or CHCl3) at low temperature to yield p-bromophenol (4-bromophenol) predominantly.
D ·
Confusing phenol's activating ortho/para directing effect with meta-direction, or mistaking the product for 3-bromophenol. The -OH group is a strongly activating +M ortho/para director, so substitution at the meta position does not occur.
Reviewed route
Solution
StepWorking
01identify
Empirical formula C6H6O gives a sooty flame, indicating a high degree of unsaturation typical of an aromatic compound. This identifies 'A' as phenol (C6H5OH). The reaction is electrophilic aromatic substitution (monobromination) using Br2 in a low polarity solvent such as CS2 or CHCl3 at low temperature.
02mechanism
In non-polar solvents like CS2, the ionization of phenol to the highly activating phenoxide ion is suppressed. The −OH group moderately activates the ring, directing incoming electrophiles to ortho and para positions.
03product
Due to steric hindrance at the ortho position, the para isomer undergoes less repulsion and forms as the major product with high yield. Hence, 'B' is 4-bromophenol (p-bromophenol).
✓verify
In contrast, aqueous bromine (extBr2/H2O) would yield 2,4,6-tribromophenol. The condition specifies low polarity solvent, confirming 4-bromophenol as the major product.
Hints that build this answer step by step
Given the empirical formula C6H6O and that the compound burns with a sooty flame, what is compound A?
Phenol (C6H5OH)
How does reacting phenol with Br2 in a low-polarity solvent (like CS2) differ from using aqueous bromine water?
Low-polarity solvent prevents ionization of phenol into phenoxide, resulting in monobromination instead of tribromination.
Between the ortho and para positions, which monobrominated isomer forms in higher yield?
4-Bromophenol (para-isomer), because of lower steric hindrance compared to the ortho position.
Why doesn't phenol undergo tribromination in CS2 or CHCl3?
In non-polar solvents, phenol does not ionize into phenoxide ion, which is far more strongly activating. Therefore, the ring is only moderately activated and monobromination takes place.