The Correct Option is C
Solution and Explanation
Reaction of \( C_6H_5Br + CH_3Br \) with \( \text{NaOH (alc.)} \) to Form B:
In the first step, the reaction of bromobenzene (\( C_6H_5Br \)) with methyl bromide (\( CH_3Br \)) in the presence of alcoholic NaOH undergoes a nucleophilic substitution to form B, which is ortho-bromophenol.
Reaction of B with \( \text{HBr (ether)} \) to Form C:
In the second step, treating ortho-bromophenol with \( \text{HBr} \) in ether leads to the formation of C, which is para-bromophenol.
Identifying the Relationship Between A and C:
Compound A (bromobenzene) and compound C (para-bromophenol) are position isomers because they differ in the position of the bromine and hydroxyl groups on the benzene ring.
Conclusion:
The structures of B and C are as shown in Option (3), and A and C are position isomers.
Top Questions on Electrochemistry
- What is the standard electrode potential of a half-reaction in which electrons are transferred from \(Ag^+\) to \(Ag\)?
- VITEEE - 2025
- Chemistry
- Electrochemistry
- The standard electrode potential for $\text{Zn}^{2+}/\text{Zn}$ is $-0.76$ V and for $\text{Cu}^{2+}/\text{Cu}$ is $+0.34$ V. The EMF of the cell
$$ \text{Zn}|\text{Zn}^{2+}||\text{Cu}^{2+}|\text{Cu} $$ is:- BITSAT - 2025
- Chemistry
- Electrochemistry
- The standard reduction potentials are: \[ \mathrm{Cu^{2+} + 2e^- \rightarrow Cu} \quad E^\circ = +0.34\,V \] \[ \mathrm{Zn^{2+} + 2e^- \rightarrow Zn} \quad E^\circ = -0.76\,V \] Calculate the standard cell potential for the reaction: \[ \mathrm{Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu} \]
- BITSAT - 2025
- Chemistry
- Electrochemistry
- In the electrolysis of molten \( \mathrm{NaCl} \), what is produced at the cathode?
- BITSAT - 2025
- Chemistry
- Electrochemistry
Consider the following half cell reaction $ \text{Cr}_2\text{O}_7^{2-} (\text{aq}) + 6\text{e}^- + 14\text{H}^+ (\text{aq}) \longrightarrow 2\text{Cr}^{3+} (\text{aq}) + 7\text{H}_2\text{O}(1) $
The reaction was conducted with the ratio of $\frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}]} = 10^{-6}$
The pH value at which the EMF of the half cell will become zero is ____ (nearest integer value)
[Given : standard half cell reduction potential $\text{E}^\circ_{\text{Cr}_2\text{O}_7^{2-}, \text{H}^+/\text{Cr}^{3+}} = 1.33\text{V}, \quad \frac{2.303\text{RT}}{\text{F}} = 0.059\text{V}$- JEE Main - 2025
- Chemistry
- Electrochemistry
Questions Asked in JEE Main exam
- If the equation of the parabola with vertex $\left(\frac{3}{2},\, 3\right)$ and the directrix $x + 2y = 0$ is
$$ ax^2 + by^2 - cxy - 30x - 60y + 225 = 0, $$ then $\alpha + \beta + \gamma$ is equal to:- JEE Main - 2025
- Calculus
- In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are $ t_1 $ and $ t_2 $ (s), respectively. The ratio $ t_1 / t_2 $ will be:
- JEE Main - 2025
- Chemical Kinetics
- HA $ (aq) \rightleftharpoons H^+ (aq) + A^- (aq) $ The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is 0.20 °C. The dissociation constant for the acid is Given: $ K_f(H_2O) = 1.8 \, \text{K kg mol}^{-1} $, molality ≡ molarity
- JEE Main - 2025
- Colligative Properties
20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol$^{-1}$)
- JEE Main - 2025
- Stoichiometry and Stoichiometric Calculations
- Which of the following binary mixture does not show the behavior of minimum boiling azeotropes?
- JEE Main - 2025
- Solutions