The functional group that can be identified through phthalein dye test is :

Year	2024	Total
Questions	3	3

Year

2024

Total

Questions

Solution & Explanation

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 Previous-Year Questions

Q46 neet_2026_03_may_morning Preparation Of Phenols And Alcohols

Match List I with List II

List-I	List-II
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. $CH₃COOH CH₃CH₂OH$	(II) (i) $O₂$ $O_{2}$; (ii) $H₂O/H⁺$ $H_{2}O/H^{+}$
C. $CH₃CH₂CH₂OH CH₃-CH(OH)-CH₃$	(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

Core Logic

A. Cumene hydroperoxide reaction: Cumene is treated with $O₂$ $O_2$ followed by hydrolysis ( $H₂O/H^+$ $H_2O/H^+$) to form Phenol.

Solution for Q46 - Reaction Mechanism — 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 $CH₃CH₂OH$ $CH_3CH_2OH$. $CH₃COOH [(ii) H₂, catalyst](i) CH₃OH, H^+ CH₃CH₂OH$

C. Conversion of $1^°$ $1^\circ$ alcohol to $2^°$ $2^\circ$ alcohol involves Dehydration followed by Hydration: $CH₃CH₂CH₂-OH [Dehydration]conc. H₂SO₄ / Δ CH₃CH=CH₂ H^+ / H₂O CH₃CH(OH)CH₃$

D. Benzene sulphonic acid to Phenol:

Reagents used: (i) Oleum, (ii) NaOH,

Δ

, (iii)

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.

Chapter Mix

Class 12 Chemistry: Alcohols, Phenols and Ethers

Q52 neet_2026_03_may_morning Reactions of Alcohols

In the following reaction sequence, X and Z respectively are :

Reactions of Alcohols diagram for Q52 - NEET 2026 Code 12 — Displays a multi-step organic conversion starting from propanol to product Z.

A. $X = POCl₃; Z = CH₃ - CH(Br) - CH₃$
B. $X = H₃PO₃; Z = CH₃CH₂CH₂ - Br$
C. $X = H₃PO₃; Z = CH₃ - CH(Br) - CH₃$
D. $X = POCl₃; Z = CH₃CH₂CH₂ - Br$

Solution

Core Logic

Step 1: Reaction of Alcohols with $PCl₅$ $PCl_5$ $CH₃CH₂CH₂-OH + PCl₅ arrow CH₃CH₂CH₂Cl + POCl₃ + HCl$ Thus, the by-product $X$ $X$ is $POCl₃$ $POCl_3$.

Step 2: Elimination (Dehydrohalogenation) $CH₃CH₂CH₂Cl alc. KOH, Δ CH₃CH=CH₂ (Y)$ (Propene)

Step 3: Anti-Markovnikov Addition (Peroxide effect) $CH₃CH=CH₂ HBr, (CH₃COO)₂ CH₃CH₂CH₂-Br (Z)$ (1-Bromopropane)

Solution for Q52 - Organic Sequence — 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 $Z = CH₃CH₂CH₂Br$ .

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

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The functional group that can be identified through phthalein dye test is :

Solution & Explanation

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Alcohols, Phenols and Ethers Previous-Year Questions

Solution

Core Logic

Step 1: Final Match

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

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The functional group that can be identified through phthalein dye test is :

Solution & Explanation

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Alcohols, Phenols and Ethers Previous-Year Questions

Solution

Core Logic

Step 1: Final Match

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

More Alcohols, Phenols and Ethers Questions — neet_2026_03_may_morning

Select Your Target Exam

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