MHT CET · Chemistry · Alcohols Phenols and Ethers
Which among the following obtained as major product \(\mathrm{x}\) in the reaction stated below?

- A 2, 4, 6-Trinitro anisole
- B 4-Nitro anisole
- C 2-Nitro anisole
- D 3-Nitro anisole
Answer & Solution
Correct Answer
(B) 4-Nitro anisole
Step-by-step Solution
Detailed explanation
See the Complete Solution
Get step-by-step explanations for this and 2.5 Lakh+ more JEE, NEET & CET questions.
- Unlock all solutions
- Practice the full chapter
- Track accuracy across PYQs
4.8 rated on Google Play · 14,000+ reviews
More questions from Chemistry
- Oxidation state of iodine in \(\mathrm{I}_3^{-}\)isMHT CET 2021 Easy
- Which among the following is haloalkyne?MHT CET 2023 Easy
- What is the SI unit for electrochemical equivalent?MHT CET 2020 Easy
- What is the value of pH of a NaOH solution that dissociates \(2 \%\) in its 0.01 M solution?MHT CET 2025 Medium
- Dissolution of \(1.5 \mathrm{~g}\) of a non-volatile solute (mol. wt. \(=60\) ) in \(250 \mathrm{~g}\) of a solvent reduces its freezing point by \(0.01{ }^{\circ} \mathrm{C}\). Find the molal depression constant of the solvent.MHT CET 2010 Medium
- What is change in internal energy if a system gains \(\mathrm{xJ}\) of heat and yJ work is done on it?MHT CET 2022 Easy
More PYQs from MHT CET
- Which from following mixtures obeys Raoult's law?MHT CET 2025 Easy
- The logical statement
\((\sim(\sim \mathrm{p} \vee \mathrm{q}) \vee(\mathrm{p} \wedge \mathrm{r})) \wedge(\sim \mathrm{q} \wedge \mathrm{r})\) is equivalent toMHT CET 2023 Easy - An infinitely long straight conductor is bent into the shape as shown in the figure. It carries a current I ampere and the radius of the circular loop is r metre. Then the magnetic induction B at the centre of the circular part is:
MHT CET 2024 Easy - Which of the following did Meselson and Stahl use in the experiment to prove semiconservative DNA replication?MHT CET 2019 Hard
- Athelet's foot is caused by a ___________ .MHT CET 2017 Easy
- If \(A\) and \(B\) are non-singular matrices of order 2 such that \((\mathrm{AB})^{-1}=\frac{1}{6}\left[\begin{array}{cc}-7 & -3 \\ 2 & 3\end{array}\right]\) and \(\mathrm{A}^{-1}=\frac{1}{3}\left[\begin{array}{cc}4 & 3 \\ -1 & 0\end{array}\right]\) then \(\mathrm{B}^{-1}=\)MHT CET 2025 Medium