NEET Previous Year Questions (2014-2025): Alcohols, Phenols & Ethers | Chemistry Class 12 PDF Download

Possible structures include:

• Four-membered rings (oxetanes) with different substitution patterns.
• Three-membered rings (epoxides) with different substitution patterns.

• Three-membered ring (epoxides): 6 isomers (including stereoisomers).
• Four-membered ring (oxetanes): 4 isomers (including stereoisomers).
• Total number of isomers = 6 (epoxides) + 4 (oxetanes) = 10.

Ans: (b)

Ans: (d)
Tertiary alcohols react instantaneously with Lucas reagent and gives immediate turbidity. In case of tertiary alcohols, they form halides easily with Lucas reagent (conc. HCl and ZnCl₂)

Ans: (a)
In o-nitrophenol intramolecular H-bonding is present.

Ans: (a)

Ans: (c)
Step 1: Methanol (CH₃OH) reacts with SOCl₂

• SOCl₂ (thionyl chloride) is a reagent that is used to convert alcohols to alkyl chlorides.
• CH₃OH + SOCl₂ → CH₃Cl (methyl chloride) + SO₂ + HCl
  This reaction replaces the hydroxyl group (-OH) with a chlorine atom (-Cl), forming CH₃Cl (methyl chloride).

Step 2: Reaction of CH₃Cl with KCN in ethanol

• KCN (potassium cyanide) reacts with CH₃Cl (methyl chloride) to replace the chlorine atom with a cyano group (-CN), forming CH₃CN (acetonitrile).
• CH₃Cl + KCN → CH₃CN + KCl
  This reaction results in the formation of acetonitrile (CH₃CN).

Step 3: Reduction of acetonitrile (CH₃CN) with Na/H₂

• Na/H₂ (sodium in hydrogen) is a strong reducing agent that reduces nitriles to primary amines.
• CH₃CN + Na/H₂ → CH₃CH₂NH₂ (ethylamine).
  This step reduces acetonitrile to ethylamine (CH₃CH₂NH₂).

Step 4: Reaction of ethylamine with NaNO₂ and HCl

• NaNO₂ (sodium nitrite) reacts with ethylamine (CH₃CH₂NH₂) in the presence of HCl (hydrochloric acid) to form a diazonium salt.
• The reaction proceeds as follows:
  CH₃CH₂NH₂ + NaNO₂ + HCl → CH₃CH₂N₂⁺Cl⁻ (ethyl diazonium chloride)
• However, ethyl diazonium chloride is unstable and undergoes hydrolysis when treated with water.

Step 5: Hydrolysis of the diazonium salt

• The diazonium salt formed in the previous step undergoes hydrolysis to form ethanol (CH₃CH₂OH).
• CH₃CH₂N₂⁺Cl⁻ + H₂O → CH₃CH₂OH + N₂ (nitrogen gas).
• The major product of the reaction is ethanol (CH₃CH₂OH).

Thus, the correct answer is (c) CH₃CH₂OH (ethanol).

​Ans: (a)
Step 1: Reaction of Ethanol (CH₃CH₂OH) with Phosphorus (P) and Iodine (I₂)

• When ethanol (CH₃CH₂OH) reacts with phosphorus (P) and iodine (I₂), it undergoes dehydrogenation and iodination to form 1-iodoethane (CH₃CH₂I).
• The reaction proceeds as follows:
  CH₃CH₂OH + I₂ + P → CH₃CH₂I (1-iodoethane).

Step 2: Reaction of 1-Iodoethane (CH₃CH₂I) with Magnesium (Mg) in dry ether

• Magnesium in dry ether reacts with 1-iodoethane to form an ethyl magnesium halide (Grignard reagent). This Grignard reagent is CH₃CH₂MgI.
• CH₃CH₂I + Mg → CH₃CH₂MgI (ethyl magnesium iodide).

Step 3: Reaction of Ethyl Magnesium Iodide with Formaldehyde (HCHO)

• The Grignard reagent (CH₃CH₂MgI) reacts with formaldehyde (HCHO) to form an intermediate alcohol, specifically propan-2-ol (secondary alcohol).
• CH₃CH₂MgI + HCHO → CH₃CH₂CH(OH)CH₃ (propan-2-ol).

Step 4: Hydrolysis of the Intermediate (Propan-2-ol)

• The final product D is obtained after hydrolysis of the intermediate. Hydrolysis converts the alcohol to a carboxylic acid. This will give propionic acid (C₂H₅COOH), which is the major product of this reaction.
• CH₃CH₂CH(OH)CH₃ + H₂O → C₂H₅COOH (propionic acid).

So, the correct answer is (d) Propionic acid, which is formed after the Grignard reaction and hydrolysis of the intermediate.

Ans: (d)

Ans: (b)

Ans: (a)
Due to presence of conjugation in product.

Ans: (c)

• (A) CrO₃ - H₂SO₄: This is a strong oxidizing agent, known as Jones reagent, which can oxidize alcohols (primary alcohols specifically) to carboxylic acids.
• (B) K₂Cr₂O₇ + H₂SO₄: This is another powerful oxidizing agent (chromic acid), which can also oxidize alcohols to carboxylic acids.
• (C) KMnO₄ + KOH/H₃O⁺: Potassium permanganate (KMnO₄) is a strong oxidizing agent that can oxidize alcohols to carboxylic acids.
• (D) Cu, 573 K: This reagent is used for the dehydrogenation of alcohols, typically producing aldehydes or ketones, not carboxylic acids.
• (E) CrO₃ + (CH₃CO)₂O: This is an esterification reaction (formation of esters), not related to the conversion of alcohols to carboxylic acids.

Hence, (A), (B), and (C) are the correct reagents for converting alcohols to carboxylic acids.

Ans: (a)
In the given reactions, we have two steps, and the task is to identify the reagent "X" that converts the carboxylic acid (R-COOH) to the alcohol (R-CH₂OH) and also participates in the transformation of the alkene (R-CH=CH₂) into an alcohol (R-CH₂-CH₂-OH).

Let's break down the reactions:

First reaction:

R-COOH → X → R-CH₂OH
In this reaction, R-COOH (a carboxylic acid) is converted into R-CH₂OH (an alcohol). The most common reagent to reduce a carboxylic acid to an alcohol is borane (B₂H₆) or LiAlH₄.

Second reaction:
R-CH=CH₂ → X → R-CH₂-CH₂-OH
Here, R-CH=CH₂ (an alkene) is reduced to R-CH₂-CH₂-OH (an alcohol). The typical reagent that reduces alkenes to alcohols is hydrogen (H₂) in the presence of Pd (palladium), which is a well-known catalyst for hydrogenation reactions.

Conclusion:

• In the first reaction, B₂H₆ (diborane) is the reagent that reduces carboxylic acids to alcohols.
• In the second reaction, H₂/Pd is the reagent that reduces alkenes to alcohols.

Thus, 'X' is B₂H₆.
Answer: (a) B₂H₆

Ans: (a)
Starting compound: The starting compound is a β-keto ester (a compound containing both a carbonyl group and an ester group).

• The structure has a ketone (C=O) group attached to the second carbon and an ester (C=O) group attached to the first carbon.

First step (NaBH₄):

• Sodium borohydride (NaBH₄) is a mild reducing agent. It selectively reduces ketones but does not affect esters.

• NaBH₄ will reduce the ketone group to a secondary alcohol, resulting in the intermediate product where the carbonyl group is replaced by an -OH group.

Second step (H₂SO₄, Δ):

• The ester group undergoes decarboxylation under acidic conditions (using sulfuric acid, H₂SO₄, and heat).

• This reaction removes the CO₂ group, leading to the formation of a cyclohexene structure with a double bond between two adjacent carbons.

Major product:
The major product is the cyclic alkene formed after the decarboxylation step, which is the 1-cyclohexene structure.
Thus, the correct answer is Option 1, which shows the major product as 1-cyclohexene.

Ans: (d)

Ans: (d)

At STP condition substitution at sp² carbon atom is not feasible.

Ans: (a)

Statement I is correct, Statement II incorrect.

Ans: (a)
Primary, secondary and tertiary alcohols can be differentiated by their reaction with (HCl + anhy ZnCl₂) Lucas reagent

Ans: (c)

Ans: (c)
This is Reimer–Tiemann reaction.

Ans: (d)
Lucas reagent [HCl + ZnCl₂] can be used for
1° / 2° / 3° alcohol but if ZnCl₂ is absent then only 3° alcohol can be used because it is most reactive due to 3° carbocation. 

Ans: (b)
The reaction is called Williamson synthesis

Ans: (a)
Acid reacts with sodium hydrogen carbonate as follows :
Among all the given options ortho-nitrophenol is weaker acid than HCO₃ hence, it does not react with NaHCO₃.