Considering the following reaction sequence,
the correct option(s) is(are)
the correct option(s) is(are)
The Correct Option is A, B, C
Approach Solution - 1
→ P may be → H2/Pd, ethanol; Sn/HCl
→ R may be → NaNO2/HCl; HNO2
→ U may be → (i) H3PO2, (ii) KMnO4 – KOH, Δ or (i) CH3 – CH2 – OH, (ii) KMnO4 – KOH, Δ
So the correct options are A,B and C.
Approach Solution -2
1. From \( \text{C}_6\text{H}_5\text{NO}_2 \) (Nitrobenzene) to Q:
The first step involves the reduction of nitrobenzene (\( \text{C}_6\text{H}_5\text{NO}_2 \)) to an amine derivative.
The reagent for reducing the nitro group to an amine group is \( \text{H}_2/\text{Pd} \), ethanol, which matches with option A and B for P → H₂/Pd, ethanol.
2. From Q to R:
The next step involves the conversion of the amine group to a diazonium salt using \( \text{NaNO}_2/\text{HCl} \).
This corresponds to option A and option B for R → NaNO₂/HCl.
3. From R to S:
The reaction involves the substitution of the diazonium group with a chlorine atom to form a chlorobenzene derivative.
This step is facilitated by CuCl, as shown in option C for S → \( \text{C}_6\text{H}_4\text{Cl} \) (Chlorobenzene).
4. From S to U:
The next step involves the oxidation of the chlorobenzene derivative into a carboxyl group, forming benzoic acid.
The reducing agent for this reaction is KMnO₄ - KOH, heat, which corresponds to U in option A and option C.
5. From U to T:
The final product, benzoic acid, is formed by the reaction of the compound under oxidation conditions.
This matches with the final step T: \( \text{C}_6\text{H}_5\text{COOH} \) in option A and option C.
Final Answer:
The correct matches are:
- P → H₂/Pd, ethanol: Reduction of nitrobenzene to aniline
- R → NaNO₂/HCl: Formation of a diazonium salt
- S → \( \text{C}_6\text{H}_4\text{Cl} \): Chlorination of the diazonium salt
- U → KMnO₄ - KOH, heat: Oxidation to benzoic acid
Correct Option:
The correct options are A, B, C.
Top Questions on Chemical Reactions of Alcohols Phenols and Ethers
Calculate the potential for half-cell containing 0.01 M K\(_2\)Cr\(_2\)O\(_7\)(aq), 0.01 M Cr\(^{3+}\)(aq), and 1.0 x 10\(^{-4}\) M H\(^+\)(aq).
- CBSE CLASS XII - 2025
- Chemistry
- Chemical Reactions of Alcohols Phenols and Ethers
- Phenol can be distinguished from propanol by using the reagent
- KCET - 2025
- Chemistry
- Chemical Reactions of Alcohols Phenols and Ethers
- Find out the final product C
- JEE Main - 2024
- Chemistry
- Chemical Reactions of Alcohols Phenols and Ethers
- An aqueous solution of hydrazine (\(N_2H_4\)) is electrochemically oxidized by \(O_2\), thereby releasing chemical energy in the form of electrical energy. One of the products generated from the electrochemical reaction is \(N_2(g)\). Choose the correct statement(s) about the above process:
- JEE Advanced - 2024
- Chemistry
- Chemical Reactions of Alcohols Phenols and Ethers
- According to Bohr's model, the highest kinetic energy is associated with the electron in the:
- JEE Advanced - 2024
- Chemistry
- Chemical Reactions of Alcohols Phenols and Ethers
Questions Asked in JEE Advanced exam
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity):
- JEE Advanced - 2025
- Waves and Oscillations
- Consider the vectors $$ \vec{x} = \hat{i} + 2\hat{j} + 3\hat{k},\quad \vec{y} = 2\hat{i} + 3\hat{j} + \hat{k},\quad \vec{z} = 3\hat{i} + \hat{j} + 2\hat{k}. $$ For two distinct positive real numbers $ \alpha $ and $ \beta $, define $$ \vec{X} = \alpha \vec{x} + \beta \vec{y} - \vec{z},\quad \vec{Y} = \alpha \vec{y} + \beta \vec{z} - \vec{x},\quad \vec{Z} = \alpha \vec{z} + \beta \vec{x} - \vec{y}. $$ If the vectors $ \vec{X}, \vec{Y}, \vec{Z} $ lie in a plane, then the value of $ \alpha + \beta - 3 $ is ________.
- If $$ \alpha = \int_{\frac{1}{2}}^{2} \frac{\tan^{-1} x}{2x^2 - 3x + 2} \, dx, $$ then the value of $ \sqrt{7} \tan \left( \frac{2\alpha \sqrt{7}}{\pi} \right) $ is.
(Here, the inverse trigonometric function $ \tan^{-1} x $ assumes values in $ \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) $.)- JEE Advanced - 2025
- Integral Calculus
Let $ a_0, a_1, ..., a_{23} $ be real numbers such that $$ \left(1 + \frac{2}{5}x \right)^{23} = \sum_{i=0}^{23} a_i x^i $$ for every real number $ x $. Let $ a_r $ be the largest among the numbers $ a_j $ for $ 0 \leq j \leq 23 $. Then the value of $ r $ is ________.
- JEE Advanced - 2025
- binomial expansion formula
- The total number of real solutions of the equation $$ \theta = \tan^{-1}(2 \tan \theta) - \frac{1}{2} \sin^{-1} \left( \frac{6 \tan \theta}{9 + \tan^2 \theta} \right) $$ is
(Here, the inverse trigonometric functions $ \sin^{-1} x $ and $ \tan^{-1} x $ assume values in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and $(-\frac{\pi}{2}, \frac{\pi}{2})$, respectively.)- JEE Advanced - 2025
- Inverse Trigonometric Functions
Concepts Used:
Chemical Properties - Alcohols, Phenols and Ethers
Alcohols, phenols, and ethers are organic compounds that have distinct chemical properties.
Alcohols are characterized by the presence of the hydroxyl (-OH) functional group, which makes them polar and capable of forming hydrogen bonds. They are typically classified as primary, secondary, or tertiary, depending on the number of alkyl groups attached to the carbon atom bearing the hydroxyl group. Alcohols undergo various chemical reactions, including oxidation, dehydration, and esterification.
Phenols are organic compounds that contain an -OH group attached to an aromatic ring. They are weaker acids than carboxylic acids but stronger acids than alcohols due to the resonance stabilization of the phenoxide ion. Phenols undergo various chemical reactions, including electrophilic substitution and oxidation.
Read More: Classification of Alcohol, Phenols and Ethers
Ethers are organic compounds that contain an oxygen atom bonded to two alkyl or aryl groups. They are characterized by their low boiling points and are often used as solvents. Ethers undergo various chemical reactions, including cleavage of the C-O bond and oxidation.
In summary, alcohols, phenols, and ethers have distinct chemical properties due to the presence of the hydroxyl or ether functional group. Understanding these properties is important for understanding their reactivity and potential applications in various fields, including chemistry, biology, and industry.