Alcohols

Alcohols with same molecular weight ate expected to have almost same boiling point, however two more factors other than molecular weight are important; they are namely H-bonding and surface area of molecule. Both these factors are least in alcohols and maximum in alcohols. Hence, alcohols have least boiling point while alcohols have maximum boiling point.

Dehydration of alcohols involves the formation of carbocation intermediates. More the stability of carbocation, more will be the ease of alcohol towards dehydration.
The order of stability of carbocations is

The order of ease of dehydration of alcohols is

Step 1: Understanding the structure of pinacol.
Pinacol is a vicinal diol (two hydroxyl groups on adjacent carbon atoms) with the structure 2,3-dimethylbutane-2,3-diol. The two hydroxyl groups are attached to carbon atoms 2 and 3.
Step 2: Analyzing the options.
(A) 2,3-dimethyl butane-1,4-diol: This name is incorrect as the hydroxyl groups are not positioned at carbon atoms 1 and 4.
(B) 2,2-dimethyl butane-1,3-diol: This is not correct, as the positions of the methyl groups and hydroxyl groups do not match the structure of pinacol.
(C) 2,3-dimethyl butane-2,3-diol: This is the correct IUPAC name, as it accurately describes the structure of pinacol, with two hydroxyl groups on adjacent carbons 2 and 3.
(D) 2,3-dimethyl butane-1,3-diol: This is incorrect because the hydroxyl groups are not on carbons 1 and 3.
Step 3: Conclusion.
The correct answer is (C) 2,3-dimethyl butane-2,3-diol, as it correctly represents the IUPAC name for pinacol.

Step 1: Understanding the reaction.
The given reaction involves the hydration of an alkene (but-1-ene) to form an alcohol (butan-2-ol). This reaction typically proceeds through the acid-catalyzed hydration of alkenes.
Step 2: Analyzing the options.
(A) H O : Hydrogen peroxide is not suitable for this transformation as it does not catalyze the hydration of alkenes.
(B) KOH aqueous: Potassium hydroxide in aqueous solution typically induces elimination reactions, not hydration.
(C) (i) cold Conc. H SO , (ii) H O: This is the correct answer. Cold concentrated sulfuric acid adds to the alkene, and hydrolysis with water results in the formation of butan-2-ol.
(D) B H : Diborane is used in hydroboration-oxidation reactions, but this is not the correct reagent for this specific transformation.
Step 3: Conclusion.
The correct answer is (C) (i) cold Conc. H SO , (ii) H O.

Step 1: Understanding Lucas reagent.
Lucas reagent is a solution of concentrated hydrochloric acid (HCl) and zinc chloride (ZnCl ). It is used to classify alcohols based on their reactivity.
Step 2: Analyzing the options.
(A) Concentrated carbolic acid with zinc chloride: This is not Lucas reagent.
(B) Concentrated hydrochloric acid with zinc chloride: This is the correct Lucas reagent, used for testing alcohols.
(C) Concentrated sulphuric acid with zinc sulphate: This is not Lucas reagent.
(D) Concentrated nitric acid with zinc nitrate: This is not Lucas reagent either.
Step 3: Conclusion.
The correct answer is (B) Concentrated hydrochloric acid with zinc chloride.

Step 1: Recall reaction with sodium metal.
Sodium metal reacts with compounds containing active hydrogen atoms (such as –OH or –COOH groups), liberating hydrogen gas.

Step 2: Analyse each compound.
(A) Ethanol ( ): Contains an –OH group, reacts with sodium to form sodium ethoxide and hydrogen gas.
(B) Phenol ( ): Contains phenolic –OH group, reacts with sodium to give sodium phenoxide.
(C) Dimethyl ether ( ): Does not contain an –OH group or acidic hydrogen, hence does not react with sodium.
(D) Acetic acid ( ): Contains –COOH group, reacts vigorously with sodium.

Step 3: Conclusion.
Dimethyl ether does not react with sodium metal because it lacks an active hydrogen atom.

Step 1: Identify the functional group.
The given compound contains an ether linkage .

Step 2: Select the parent chain.
The longer carbon chain attached to oxygen is 2-methylpropane. Hence, it is selected as the parent hydrocarbon.

Step 3: Identify the alkoxy substituent.
The smaller group attached through oxygen is an ethoxy group .

Step 4: Assign position numbers.
The ethoxy group is attached at carbon number 1 of the parent chain.

Step 5: Final name.
The correct IUPAC name is 1-ethoxy-2-methylpropane.

Step 1: Understanding acid catalysed hydration.
Acid catalysed hydration of alkenes involves the addition of water across the double bond in the presence of an acid catalyst such as . This reaction generally follows Markovnikov’s rule.

Step 2: Application to ethene.
When ethene undergoes acid catalysed hydration, water adds across the double bond to form ethanol as the product.

Step 3: Analysis of other options.
Methanol is not prepared by hydration of alkenes. Propan-1-ol and butan-1-ol are not obtained directly by acid catalysed hydration due to rearrangement and Markovnikov addition leading to secondary alcohols instead.

Step 4: Conclusion.
Therefore, the correct answer is (A) Ethanol.

Step 1: Analyzing the Reaction Sequence.
In the given sequence, the reaction starts with butan-2-ol. - In the first step, when butan-2-ol reacts with phosphorus trichloride (PCl ), it undergoes chlorination to form 2-chlorobutane (X). - In the second step, when 2-chlorobutane reacts with alcoholic potassium hydroxide (alc. KOH), it undergoes elimination to form But-2-ene (Y). - In the final step, when But-2-ene is treated with sulfuric acid (H SO ) and then heated, it undergoes dehydration to form But-2-ene (Z). Hence, Z is But-2-ene.

Step 2: Analyzing the options.
(A) Butan-1-ol: Incorrect. Butan-1-ol is not formed in this reaction sequence.
(B) 2-chlorobutane: Incorrect. 2-chlorobutane is formed in the first step of the reaction but is not the final product.
(C) Butan-2-ol: Incorrect. Butan-2-ol is the starting material, not the final product.
(D) But-2-ene: Correct. But-2-ene is the final product after the elimination and dehydration steps.

Step 3: Conclusion.
The correct answer is But-2-ene, corresponding to option (D).

Step 1: First reaction with .
Ethanol reacts with thionyl chloride in presence of pyridine to form ethyl chloride (A).

Step 2: Reaction with sodium ethoxide.
Ethyl chloride reacts with sodium ethoxide to form diethyl ether (B) via Williamson ether synthesis.

Step 3: Acidic hydrolysis under heat and pressure.
Diethyl ether on treatment with dilute sulphuric acid under heat and pressure undergoes cleavage to form ethanol.

Step 4: Conclusion.
Hence, the final product (C) formed is Ethanol.

Step 1: Understanding the reaction.
When 2-methylpropan-2-ol (tert-butyl alcohol) is treated with alumina (Al O ) at 423 K, it undergoes dehydration to form an alkene. The product is isobutylene (2-methylpropene).

Step 2: Analyzing the options.
(A) Propanone: This is a ketone and not formed in this dehydration reaction.
(B) Propene: This is an alkene, but the product from 2-methylpropan-2-ol will be isobutylene, not propene.
(C) Propanoic acid: This is a carboxylic acid and is not the product of this reaction.
(D) Isobutylene: Correct — 2-methylpropan-2-ol undergoes dehydration to form isobutylene.

Step 3: Conclusion.
The correct answer is (D) Isobutylene, as it is the product of the dehydration of 2-methylpropan-2-ol.

Step 1: Understanding dry ice.
Dry ice is solid carbon dioxide (CO ). It exists as a molecular solid, where individual CO molecules are held together by intermolecular forces, not ionic or covalent bonds.

Step 2: Analyzing the options.
(A) covalent solid: This is incorrect. Dry ice is not a covalent solid as the individual molecules are not covalently bonded to each other.
(B) ionic solid: This is incorrect. Dry ice is not composed of ions held together by electrostatic forces.
(C) molecular solid: Correct — Dry ice is a molecular solid, where CO molecules are held together by Van der Waals forces.
(D) metallic solid: This is incorrect. Dry ice does not exhibit metallic bonding.

Step 3: Conclusion.
The correct answer is (C) molecular solid.

Step 1: Identifying the preparation method.
Acid-catalyzed hydration of alkenes typically results in the formation of alcohols where the hydroxyl group (-OH) adds to the more substituted carbon. This leads to the formation of the Markovnikov product.
Step 2: Analyzing the options.
(A) Ethanol: Ethanol is formed by acid-catalyzed hydration of ethene.
(B) Propan-2-ol: Propan-2-ol is formed by acid-catalyzed hydration of propene.
(C) Propan-1-ol: Incorrect — Propan-1-ol is not the product of acid-catalyzed hydration of alkenes, as it would be the anti-Markovnikov product, which requires a different mechanism.
(D) 2-Methyl propan-2-ol: This is formed by acid-catalyzed hydration of isobutene.
Step 3: Conclusion.
The correct answer is (C) Propan-1-ol.

Pyridinium chlorochromate (PCC) is the most selective reagent for the oxidation of primary alcohols to aldehydes. It avoids further oxidation to carboxylic acids, which other reagents like KMnO4 and K2Cr2O7 tend to do. Therefore, PCC is ideal for the selective oxidation of alcohols to aldehydes without over-oxidation.

A primary alcohol is one in which the carbon atom attached to the –OH group is connected to only one other carbon atom. In option (A), CH CH CH OH (1-propanol), the –OH group is attached to the terminal carbon which is bonded to only one carbon atom, making it a primary alcohol. Option (B) is a secondary alcohol (2-propanol), and option (C) is a tertiary alcohol (tert-butanol).

Concept:
Solubility in water depends on hydrogen bonding with water molecules. Step 1: Analyze hydrogen bonding ability.

• Alcohols → strong H-bonding (O-H group)
• Amines → moderate H-bonding (N-H)
• Alkanes → no H-bonding

Step 2: Compare polarity.
Step 3: Conclusion. Correct order is option (A).

Pyridinium chlorochromate (PCC) is a selective reagent that is used to oxidize primary alcohols to aldehydes, without overoxidizing them to carboxylic acids, making it the ideal choice in this case.

Concept: Alcohols contain an group, where the hydrogen atom is slightly acidic. When alcohol reacts with an active metal such as sodium, the metal replaces the hydrogen of the hydroxyl group. This reaction forms a sodium alkoxide and releases hydrogen gas.
Step 1: {Write the general reaction.} For a general alcohol : where

• = sodium alkoxide
• = hydrogen gas

Step 2: {Example reaction.} For ethanol: Thus, ethanol forms sodium ethoxide and hydrogen gas.
Step 3: {Identify the products.} Therefore, the reaction between alcohol and sodium produces