A. H₃C-CH(CH₃)-OH$H_{3}C-CH(CH_{3})-OH$Displays reactant and product chemical structures for matching with their respective reagents.
(I) (i) Oleum; (ii) NaOH, Δ$\Delta$; (iii) H⁺$H^{+}$
B. CH₃COOH CH₃CH₂OH$CH_{3}COOH \longrightarrow CH_{3}CH_{2}OH$
(II) (i) O₂$O_{2}$; (ii) H₂O/H⁺$H_{2}O/H^{+}$
C. CH₃CH₂CH₂OH CH₃-CH(OH)-CH₃$CH_{3}CH_{2}CH_{2}OH \longrightarrow CH_{3}-CH(OH)-CH_{3}$
(III) (i) CH₃OH, H⁺$CH_{3}OH, H^{+}$; (ii) H₂$H_{2}$, catalyst
D. Displays reactant and product chemical structures for matching with their respective reagents.
(IV) (i) conc. H₂SO₄$H_{2}SO_{4}$, Δ$\Delta$; (ii) H⁺/H₂O$H^{+}/H_{2}O$
Choose the correct answer from the options given below :
A.A-I, B-III, C-IV, D-II
B.A-II, B-IV, C-III, D-I
C.A-II, B-III, C-I, D-IV
D.A-II, B-III, C-IV, D-I
Solution & Explanation
Core Logic
A. Cumene hydroperoxide reaction: Cumene is treated with O₂$O_2$ followed by hydrolysis (H₂O/H^+$H_2O/H^+$) to form Phenol.
Displays reactant and product chemical structures for matching with their respective reagents.
B. Reduction of Carboxylic Acid: CH₃COOH$CH_3COOH$ is reduced using catalysts or transformed via intermediate esters to CH₃CH₂OH$CH_3CH_2OH$.
CH₃COOH [(ii) H₂, catalyst](i) CH₃OH, H^+ CH₃CH₂OH$CH_3COOH \xrightarrow[\text{(ii) } H_2, \text{catalyst}]{\text{(i) } CH_3OH, H^+} CH_3CH_2OH$
C. Conversion of 1^°$1^\circ$ alcohol to 2^°$2^\circ$ alcohol involves Dehydration followed by Hydration:
CH₃CH₂CH₂-OH [Dehydration]conc. H₂SO₄ / Δ CH₃CH=CH₂ H^+ / H₂O CH₃CH(OH)CH₃$CH_3CH_2CH_2-OH \xrightarrow[\text{Dehydration}]{\text{conc. } H_2SO_4 / \Delta} CH_3CH=CH_2 \xrightarrow{H^+ / H_2O} CH_3CH(OH)CH_3$
D. Benzene sulphonic acid to Phenol:
Displays reactant and product chemical structures for matching with their respective reagents.
Reagents used: (i) Oleum, (ii) NaOH, Δ$\Delta$, (iii) H^+$H^+$.
Step 1: Final Match
A arrow$\rightarrow$ II
B arrow$\rightarrow$ III
C arrow$\rightarrow$ IV
D arrow$\rightarrow$ I
Pattern Recognition
Standard organic conversions: Phenol from cumene requires O₂$O_2$ and acid. Phenol from benzene uses oleum and NaOH fuse. Dehydration/hydration shifts OH position.
Displays a multi-step organic conversion starting from propanol to product Z.
Step 1: Final Conclusion
Therefore, X = POCl₃$\mathrm{X} = \mathrm{POCl}_3$ and Z = CH₃CH₂CH₂Br$\mathrm{Z} = \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{Br}$.
Pattern Recognition
PCl₅$PCl_5$ yields POCl₃$POCl_3$, whereas PCl₃$PCl_3$ yields H₃PO₃$H_3PO_3$. HBr$HBr$ with peroxide is Kharasch effect (Anti-Markovnikov addition), placing the Bromine on the less hindered terminal carbon.
Chapter Mix
Class 12 Chemistry: Alcohols, Phenols and Ethers
Class 12 Chemistry: Haloalkanes and Haloarenes
Q74neet_2026_03_may_morningChemical Tests for Phenols
The functional group that can be identified through phthalein dye test is :
A. Phenolic
B. Alcohol
C. Aldehyde
D. Carboxylic acid
Solution
Core Logic
Phthalein Dye Test: Phenol on heating with phthalic anhydride in the presence of concentrated sulphuric acid forms a colourless condensation compound called phenolphthalein.
On further reaction with a base (like NaOH), it yields a deep pink (or red) coloured solution due to the formation of its sodium salt. This test is highly specific to the phenolic group.
Step 1: Final Conclusion
The functional group identified is Phenolic.
Pattern Recognition
Phthalein dye test = Phenol. Tollens'/Fehling's = Aldehyde. Lucas = Alcohols. Bicarbonate test = Carboxylic Acid.
Chapter Mix
Class 12 Chemistry: Alcohols, Phenols and Ethers
Class 12 Chemistry: Practical Chemistry
More Alcohols, Phenols and Ethers Questions — neet_2026_03_may_morning
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
