Which of the following compounds is least likely to give effervescence of mathrmCO_2 in presence of aq. mathrmNaHCO_3 ?

Solution & Explanation

### Core Logic For a compound to release mathrmCO_2 upon reaction with aqueous mathrmNaHCO_3, its acidity must be strictly greater than that of carbonic acid (mathrmH_2CO_3). Let's evaluate the acidity of the options: 1. **Picric Acid** (2,4,6-trinitrophenol - Option A): Extremely acidic (pK_a approx 0.38) due to three strong electron-withdrawing nitro groups. Reacts with mathrmNaHCO_3 easily. 2. **4-Nitrobenzoic Acid** (Option B): Carboxylic acids generally have pK_a approx 4text--5. More acidic than carbonic acid (pK_a approx 6.3). Reacts with mathrmNaHCO_3. 3. **Anilinium Chloride** (Option 3): A salt of a strong acid and weak base. The anilinium ion is quite acidic (pK_a approx 4.6) and easily decomposes mathrmNaHCO_3. 4. **4-Nitrophenol** (Option D): Only one nitro group is present. Its acidity (pK_a approx 7.15) is weaker than carbonic acid. Hence, it does not react with sodium bicarbonate to yield effervescence of mathrmCO_2. ### Pattern Recognition Acid-bicarbonate test shortcut: - All carboxylic acids and picric acid give a positive bicarbonate test. - Normal phenols and mono/di-nitrophenols are too weak to decompose bicarbonate. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Alcohols, Phenols and Ethers Class 12 Chemistry: Carboxylic Acids

More Alcohols, Phenols and Ethers Previous-Year Questions — Page 3

Q66 jee_main_2024_30_january_evening Preparation of Alcohols
Product A and B formed in the following set of reactions are:
Preparation of Alcohols diagram for Q66 - JEE Main 2024 Evening
The diagram shows a reaction pathway for synthesizing product A and B.
  • A. textA = mathrmCH_2OH text, B = mathrmCH_2OH
  • B. textA = mathrmCH_3OH text, B = mathrmCH_3OH
  • C. textA = mathrmCH_2OH text, B = mathrmCH_3OH
  • D. textA = mathrmCH_3OH text, B = mathrmCH_3OH

Solution

### Core Logic The given substrate is an alkene attached to a benzene ring (styrene derivative). Reaction 1: Acid-catalyzed hydration (H^+ / H_2O) This proceeds via Markovnikov's rule. The proton attacks the alkene to form the most stable carbocation (benzyllic and secondary). Water then attacks this carbocation to yield the alcohol at the substituted position (product A: Ph-CH(OH)-CH_3). Reaction 2: Hydroboration-Oxidation (B_2H_6, then H_2O_2/NaOH) This proceeds via anti-Markovnikov addition of water across the double bond without carbocation rearrangement. The OH group adds to the less substituted carbon atom of the alkene (product B: Ph-CH_2-CH_2-OH).
Reaction pathway for products A and B diagram for Q66 - JEE Main 2024 Evening
The diagram shows a reaction pathway for synthesizing product A and B.
### Pattern Recognition H^+/H_2O = Markovnikov hydration (carbocation intermediate). B_2H_6 / H_2O_2, OH^- = Anti-Markovnikov hydration (concerted, no rearrangement). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Alcohols Phenols and Ethers Class 11 Chemistry: Hydrocarbons
Q69 jee_main_2024_31_jan_morning Chemical Reactions of Alcohols
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R: Assertion A: Alcohols react both as nucleophiles and electrophiles. Reason R: Alcohols react with active metals such as sodium, potassium and aluminum to yield corresponding alkoxides and liberate hydrogen. In the light of the above statements, choose the correct answer from the options given below:
  • A. textA is false but R is true.
  • B. textA is true but R is false.
  • C. textBoth A and R are true and R is the correct explanation of A.
  • D. textBoth A and R are true but R is NOT the correct explanation of A.

Solution

### Core Logic Assertion (A) is correct. Alcohols can act as nucleophiles (in reactions involving O-H bond cleavage) and as electrophiles (in reactions involving C-O bond cleavage after protonation). Reason (R) is correct. Alcohols react with active metals to form alkoxides and H_2 gas. However, Reason (R) only explains the acidic nature of the O-H bond (which relates to nucleophilic behavior when broken) but does not explain the electrophilic behavior (C-O cleavage). Thus, (R) is not the correct explanation for (A). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Alcohols, Phenols and Ethers
Q74 jee_main_2024_31_jan_morning Acidity of Phenols
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R: Assertion A: pK_a value of phenol is 10.0 while that of ethanol is 15.9. Reason R: Ethanol is stronger acid than phenol. In the light of the above statements, choose the correct answer from the options given below:
  • A. textA is true but R is false.
  • B. textA is false but R is true.
  • C. textBoth A and R are true and R is the correct explanation of A.
  • D. textBoth A and R are true but R is NOT the correct explanation of A.

Solution

### Core Logic Assertion A is true: Phenol has a pK_a of around 10.0, and ethanol has a pK_a of around 15.9. Reason R is false: A lower pK_a value implies higher acidity. Therefore, phenol is a stronger acid than ethanol. This is because the conjugate base of phenol (phenoxide ion) is resonance-stabilized, whereas the ethoxide ion is destabilized by the +I effect of the ethyl group. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Alcohols, Phenols and Ethers
Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)