Class 12 Chemistry Chapter 7 Alcohols, Phenols and Ethers

Key Concept:Reaction Mechanism, Real-world Application

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description] Ethers undergo cleavage reactions in the presence of strong acids like hydrogen halides. The order of reactivity for these halides is HI > HBr > HCl. This high reactivity is due to the weak bond between hydrogen and iodine, allowing the HI molecule to easily donate a proton and facilitate the ether cleavage process, forming alkyl iodides. However, the reason given relates to the historical replacement of diethyl ether as an anesthetic, which is not primarily due to its reactivity with HI but rather due to factors such as slow induction and recovery times, flammability, and irritant properties. Therefore, while both statements are true on their own, the reason provided does not correctly explain the assertion regarding ether cleavage efficiency.

Key Concept:Industrial Relevance

c) It yields both phenol and acetone, which are valuable products.
[Solution Description] The cumene process is highly efficient as it produces both phenol and acetone, which are valuable products in the chemical industry. This dual production makes the process economically favorable compared to other methods that might produce phenol alone.

Key Concept:Stereochemistry, Reaction Pathway

c) Racemic mixture
[Solution Description] During the reduction of a chiral ketone using $NaBH_{4}$, the reaction tends to proceed through the addition of hydride ions to the planar carbonyl group evenly from both sides. This results in the formation of a racemic mixture of alcohols, as there is no preference for either side of attack unless specific catalysts or chiral auxiliaries are used.

Key Concept:Basic Naming

b) Methanol
[Solution Description] The compound $CH_3OH$ consists of one carbon atom with an OH group attached, which is known as methanol. In IUPAC nomenclature for alcohols, we replace the ‘e’ in methane with ‘ol’, resulting in the name Methanol.

Key Concept:Reaction Comparison

c) Tertiary alcohol
[Solution Description] The Lucas Test uses $ZnCl_2$ and HCl to distinguish between primary, secondary, and tertiary alcohols based on their reactivity. Tertiary alcohols react rapidly at room temperature due to the stable carbocation formation, while secondary alcohols react more slowly, and primary alcohols are the slowest.

Key Concept:Basic Concept

c) Sulfuric acid
[Solution Description] The acid-catalysed hydration of alkenes involves using an acid to facilitate the reaction between an alkene and water, forming alcohols. Commonly used acids include sulfuric acid $(\text{H}_2\text{SO}_4)$ or phosphoric acid $(\text{H}_3\text{PO}_4)$.

Key Concept:Identify Alcohol Type

d) $CH_3CH(OH)CH_3$
[Solution Description] To identify whether an alcohol is primary, secondary, or tertiary, we must look at the carbon atom bonded to the hydroxyl group $(-OH)$. In a secondary alcohol, this carbon is attached to two other carbon atoms.

– $CH_3CH(OH)CH_3$: The hydroxyl-bearing carbon is connected to two other carbons.
– $CH_3CH_2OH$: The hydroxyl-bearing carbon is attached to only one other carbon (primary).
– $(CH_3)_3COH$: The hydroxyl-bearing carbon is attached to three other carbons (tertiary).
– $CH_3OH$: Only one carbon is present, so it’s primary.

Therefore, option A represents a secondary alcohol.

Key Concept:Comparative Stability, Advanced Mechanism Analysis, Reaction Pathway Prediction

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description] The assertion states that tertiary alcohols are less prone to oxidation because they do not have hydrogen atoms attached to the carbon with the –OH group, which is true. During oxidation, an alcohol typically loses a hydrogen atom from the carbon bonded to the –OH group to form a carbonyl compound. Since tertiary alcohols lack such hydrogens, they resist oxidation. The reason mentions that tertiary alcohols undergo dehydration more easily than other alcohols due to the formation of stable carbocations. This statement is also true. Dehydration involves the removal of a water molecule, forming a carbocation intermediate; tertiary carbocations are stabilized by hyperconjugation and +I effect, making their formation easier, thus facilitating dehydration. Although both statements are individually correct, the reason does not directly explain why tertiary alcohols are less prone to oxidation. Therefore, while both are true, the reason is not the correct explanation for the assertion.

Key Concept:Common vs IUPAC

c) n-Propyl alcohol

[Solution Description] Propan-1-ol has three carbon atoms and a hydroxyl group at the first carbon. The common name for such alcohols is derived from their alkyl chain followed by ‘alcohol’. Hence, Propan-1-ol is commonly known as n-Propyl alcohol.

Key Concept:Common Names

a) Diethyl ether
[Solution Description] The common name method for naming ethers involves using the names of the alkyl/aryl groups followed by the word ‘ether’. For $C_2H_5OC_2H_5$, both groups are ethyl, so the common name is diethyl ether.

Key Concept:Benzylic Identification

c) $C_6H_5CH_2OH$
[Solution Description] A benzylic alcohol has the $-OH$ group attached to a carbon next to an aromatic ring $(sp^3)$ hybridized. The compound $C_6H_5CH_2OH$ matches this definition as it contains a benzene ring directly bonded to the $CH_2OH$ group.

Key Concept:Mechanistic Pathways, Comparative Reactivity

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The assertion states that anisole undergoes bromination to yield para-bromoanisole predominantly. This is true because the methoxy group $(-OCH_3)$ on anisole is an electron-donating group via resonance, which increases electron density at both the ortho and para positions of the benzene ring. However, steric hindrance makes the para position more favorable for substitution than the ortho position.

The reason explains the effect of the methoxy group. It correctly mentions that the methoxy group donates electrons through resonance, increasing electron density specifically at the ortho and para positions, thereby enhancing nucleophilicity.

Both statements are true, and the reason explains why para-bromoanisole forms dominantly during bromination. Therefore, the correct answer is (a).

Key Concept:Reaction Mechanism

c) Sodium methoxide
[Solution Description] Methanol $(CH_3OH)$ reacts with sodium $(Na)$ to form sodium methoxide $(CH_3ONa)$ and hydrogen gas $(H_2)$. The reaction involves the cleavage of the O–H bond in methanol, where sodium donates an electron to the hydrogen, forming hydrogen gas and leaving behind the methoxide ion that pairs with sodium to form sodium methoxide. The balanced chemical equation is:
$$2 \ CH_3OH + 2 \ Na \rightarrow 2 \ CH_3ONa + H_2 \uparrow$$

Key Concept:Hybridisation Identification

b) sp2
[Solution Description] In phenols, the -OH group is attached to an $sp^2$ hybridised carbon of an aromatic ring.

Key Concept:Product Identification

a) Ortho
[Solution Description] In the Reimer-Tiemann reaction, a formaldehyde group (-CHO) is introduced at the ortho position relative to the hydroxyl group on the benzene ring. This is a crucial aspect of this reaction’s mechanism.

Key Concept:Basic Reaction Steps

b) Protonation of alkene

[Solution Description] The acid-catalyzed hydration of alkenes begins with the protonation of the alkene’s double bond, forming a carbocation intermediate. This happens when a proton $(H^+)$ from an acid attacks the pi-electron cloud of the alkene.

Key Concept:Product Prediction

c) Ortho hydroxybenzoic acid
[Solution Description] In Kolbe’s reaction, phenol is first deprotonated by reacting with a strong base such as sodium hydroxide to form the phenoxide ion. This phenoxide ion then reacts with carbon dioxide $(CO_2)$ to produce ortho hydroxybenzoic acid, also known as salicylic acid. The carboxylation occurs primarily at the ortho position due to the directive influence of the phenoxide ion.

Key Concept:Reagent Selection

c) Chromium trioxide $(CrO_3)$
[Solution Description] To oxidize a secondary alcohol to a ketone, Chromic anhydride $(CrO_3)$ is typically employed. It effectively converts secondary alcohols into ketones without over-oxidation to other products.

Key Concept:Reaction Pathways

b) Sodium borohydride $(NaBH_4)$

[Solution Description] Alcohols can be prepared from aldehydes through a reduction reaction. One common method for this conversion is the use of sodium borohydride $(NaBH_4)$, which is a mild reducing agent. The reaction involves the addition of hydrogen to the carbonyl group, converting the aldehyde into an alcohol.

The balanced chemical equation for the reduction of an aldehyde (RCHO) to an alcohol $(RCH_2OH)$ using $NaBH_4$ is:

$\text{RCHO} + \text{NaBH}_4 \rightarrow \text{RCH}_2\text{OH} + \text{NaBO}_2$

Therefore, the correct reagent for this preparation is sodium borohydride $(NaBH_4)$.

Key Concept:Reaction with Zinc Dust

a) Benzene
[Solution Description] When phenol is heated with zinc dust, it undergoes reduction, where zinc acts as a reducing agent and converts phenol into benzene by removing the oxygen atom.

Key Concept:Reaction Pathway, Reaction Energetics

b) Phenolate ion
[Solution Description] The reaction pathway involves phenol reacting with sodium, initially forming an unstable phenolate ion as an intermediate. This step requires breaking the O-H bond in phenol and transferring a proton to form hydrogen gas. The formation of the phenolate ion is critical before stabilization by sodium to form sodium phenoxide. Evaluating the energy required for O-H bond dissociation helps determine the overall feasibility.

Key Concept:Classification

c) Ether
[Solution Description] The compound $C_6H_5OCH_3$ consists of a phenyl group $(C_6H_5-)$ and a methoxy group $(-OCH_3)$. This structure is characteristic of an ether, specifically anisole, where an oxygen atom connects two hydrocarbon groups.

Key Concept:Advanced Chemical Processes, Real-world Applications

a) $a = 1,b = 2,c = 1$
[Solution Description] To balance the reaction $$\text{CO} + 2\text{H}_2 \rightarrow \text{CH}_3\text{OH}$$, we need to ensure that the number of atoms of each element is equal on both sides. Let’s determine each coefficient: – For Carbon (C): 1 C atom in CO and 1 C atom in $$\text{CH}_{3}\text{OH}$$. Thus, $a = 1$. – For Hydrogen $$(H):2 \text{H}_2$$ molecules contributing 4 H atoms on left side and 4 H atoms in $$\text{CH}_3\text{OH}$$ on the right side. Thus, $b = 2$. – For Oxygen (O): 1 O atom in CO and 1 O atom in $$\text{CH}_3\text{OH}$$ Thus, $c = 1$. Therefore, the balanced equation is: $$\text{CO} + 2\text{H}_2 \rightarrow \text{CH}_3\text{OH}$$. Hence, $a = 1, b = 2, c = 1$.

Key Concept:Functional Group Identification

c) Hydroxyl

[Solution Description] Phenol contains an -OH group attached directly to an aromatic ring, which classifies it under the hydroxyl group.

Key Concept:Nucleophile Identification

b) $${I}^-$$
[Solution Description] In the SN2 mechanism of ether cleavage using HI, the iodide ion $${I}^-$$ acts as the nucleophile. It attacks the carbon atom bonded to the ether oxygen after it has been protonated.

Key Concept:Reagent Function

b) Acid source
[Solution Description] In the reaction where aniline is converted to benzene diazonium chloride, hydrochloric acid (HCl) acts as an acid source. It provides the necessary acidic environment for the reaction between sodium nitrite $(NaNO_2)$ and aniline to form diazonium salt. Specifically, it helps generate nitrous acid $(HNO_2)$ in situ from $NaNO_2$, which is essential for diazotization.

Key Concept:Nomenclature Basics

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The given ether has a propyl group $(C_3H_7)$ and an ethyl group $(C_2H_5)$. According to IUPAC nomenclature for ethers, the larger alkyl group is named as the parent hydrocarbon. Here, the propyl group is larger than the ethyl group; therefore, in IUPAC nomenclature, it should indeed be called ethyl propyl ether.

Key Concept:Basic Properties

c) Causes blindness
[Solution Description] Methanol is highly poisonous to humans because it metabolizes into formaldehyde and formic acid in the liver, which can lead to metabolic acidosis. Ingestion of small amounts can cause blindness due to damage to the optic nerve.

Key Concept:Reaction Pathway

b) Loss of water to form carbocation
[Solution Description] The mechanism for the dehydration of ethanol includes three steps. The second step is the formation of a carbocation from the protonated alcohol. This step is the slowest and thus the rate-determining step because it involves breaking a bond and forming a charged intermediate.

Key Concept:Halogenation Reaction

d) 2,4,6-Tribromophenol
[Solution Description] When phenol reacts with bromine water, the main product is 2,4,6-tribromophenol due to the strong activating effect of the -OH group, leading to substitution at all ortho and para positions.

Key Concept:Reaction Conditions

b) Room temperature
[Solution Description] Phenol reacts with sodium hydroxide at room temperature to form sodium phenoxide and water according to the reaction: $\text{C}_6\text{H}_5\text{OH} + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{ONa} + \text{H}_2\text{O}$.

Key Concept:Nomenclature Basics

a) Ethyl propyl ether
[Solution Description] To determine the IUPAC name of an ether, identify the alkyl groups on either side of the oxygen atom and arrange them alphabetically. For $C_3H_7OC_2H_5$, the alkyl groups are propyl $(C_3H_7)$ and ethyl $(C_2H_5)$. Arranged alphabetically, this forms ethyl propyl ether.

Key Concept:Reaction Conditions

a) To increase reaction rate
[Solution Description] High temperature is necessary for the conversion of benzene sulphonic acid to sodium phenoxide as it increases the reaction rate. This ensures that the nucleophilic substitution reaction where $NaOH$ displaces the $SO_3H$ group proceeds at a practical rate.

Key Concept:Reaction Pathway Analysis, Reaction Limitations

a) Leads to exclusive ortho substitution relative to the substituent
[Solution Description] An electron-withdrawing group on the benzene ring influences the regioselectivity of electrophilic aromatic substitution by deactivating positions meta to itself due to reduced electron density. During Kolbe’s reaction, if such a substituent is present, the newly introduced carboxyl group tends to orient itself ortho or para to the existing substituent where electron density remains higher compared to the meta position. Hence, the presence of an electron-withdrawing group directs the carboxylation preferentially to the ortho position relative to its location due to resonance stabilization effects.

Key Concept:Common vs IUPAC

c) Hydroquinone
[Solution Description] The compound 1,4-benzenediol has two hydroxyl groups on the benzene ring, positioned at carbons 1 and 4. The common name for this structure is Hydroquinone.

Key Concept:Basic Reaction

c) Sodium ethoxide

[Solution Description] When ethanol reacts with sodium, the hydrogen from the hydroxyl group $(-OH)$ is replaced by sodium, forming sodium ethoxide and releasing hydrogen gas. The reaction can be represented as: $2\,C_2H_5OH + 2\,Na \rightarrow 2\,C_2H_5ONa + H_2 \uparrow$ Therefore, sodium ethoxide is formed.

Key Concept:Carbocation Stability

b) Secondary
[Solution Description] During the acid-catalyzed hydration of propene, the formation of a secondary carbocation is favored over a primary one. This is because a secondary carbocation has greater stability due to hyperconjugation and inductive effects compared to a primary carbocation.

The reaction follows Markovnikov’s rule where the hydrogen adds to the carbon with more hydrogens, resulting in a secondary carbocation.

Key Concept:Reaction Condition Effects

b) Decreased yield
[Solution Description] The preparation of phenol from chlorobenzene involves high-pressure conditions (320 atm). Reducing the pressure to 200 atm would decrease the rate or yield of the reaction. At lower pressures, the energy provided may not be sufficient for effective breaking of bonds, thus leading to decreased reaction efficiency and yield.

Key Concept:Product Type

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The assertion involves the reaction of a Grignard reagent, which is represented as $RMgX$, where $R$ is an alkyl or aryl group, with methanal ($H_2CO$). This reaction typically proceeds through nucleophilic addition to the carbonyl carbon of methanal, forming an intermediate adduct. Upon hydrolysis, this adduct gives rise to a primary alcohol, thus confirming that the assertion is true. The reason given states that methanal is the simplest aldehyde and results in primary alcohol formation when reacted with Grignard reagents. This explanation correctly identifies the nature of methanal and its role in producing primary alcohols, making both the assertion and the reason true, with the reason serving as the correct explanation for the assertion.

Key Concept:Basic Naming

b) Ethanol
[Solution Description]
The compound $C_2H_5OH$ is an alcohol derived from the alkane ethane $(C_2H_6)$. To name this alcohol, we replace the ‘e’ in ethane with ‘ol,’ resulting in ethanol.

Key Concept:Comparative Analysis

b) Butanol because it can form hydrogen bonds
[Solution Description] Butanol can form hydrogen bonds due to its hydroxyl (-OH) group, which increases its boiling point compared to diethyl ether, which lacks hydrogen bonding capability and relies mainly on weaker van der Waals forces.

Key Concept:Alkyl Group Effect

c) Solubility decreases due to decreased hydrogen bonding capacity.
[Solution Description] As the size of the alkyl group in an alcohol increases, its hydrophobic character also increases, which reduces the overall ability of the molecule to form hydrogen bonds with water molecules. This leads to a decrease in solubility in water.

Key Concept:Acidity Order

a) Propanol < Ethanol < Methanol [Solution Description] Alcohols have a polar O-H bond that contributes to their acidity. However, due to the electron-releasing inductive effect of alkyl groups, which decreases the positive character of hydrogen, the acidity of alcohols decreases as the size of the alkyl group increases. Therefore, the order of acidity is Methanol $(CH_3OH) > Ethanol (C_2H_5OH) > Propanol (C_3H_7OH)$.

Key Concept:Ortho, Meta, Para

b) Meta
[Solution Description] In 3-methylphenol, the hydroxyl group is at position 1, and the methyl group is at position 3 on the benzene ring. This position is known as meta.

Key Concept:Solubility and Chemical Reactions, Solubility and Functional Groups

b) Reactivity decreases with larger alkyl groups
[Solution Description] The size of the alkyl group affects both solubility and steric hindrance. Smaller alkyl groups allow alcohols to be more soluble in water, thus they participate better in aqueous reactions like nucleophilic substitutions. Larger alkyl chains hinder solubility and increase steric hindrance, decreasing reaction rates in water-based media.

Key Concept:Industrial Applications

c) Paint industry

[Solution Description] Methanol is used in various industries due to its properties as a solvent, particularly in the chemical and pharmaceutical industries where it acts as an effective medium for reactions and extractions. The paint industry also widely utilizes methanol as a solvent because it helps in dissolving resins and pigments.

Key Concept:Comparative Acidity, Real-World Application

c) Resonance stabilization of the phenoxide ion in phenol.
[Solution Description] Phenol is more acidic than ethanol due to the resonance stabilization of the phenoxide ion that forms when phenol donates a proton. This resonance stabilization makes it easier for phenol to release a proton, increasing its acidity. Ethanol lacks such resonance stabilization in its conjugate base, making it less acidic.

Key Concept:Mechanism Insight

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] The assertion states that during the reduction of ketones, a hydride ion is formed which attacks the carbonyl carbon, leading to the reduction process. This statement is true as the reduction mechanism typically involves nucleophilic attack by a hydride ion on the electrophilic carbonyl carbon.

The reason explains that these hydride ions are obtained from reducing agents such as $NaBH_4$ and $LiAlH_4$. This is also true; both sodium borohydride and lithium aluminium hydride are known to release hydride ions which facilitate the reduction of carbonyl groups in aldehydes and ketones.

Since the presence of reducing agents like $NaBH_4$ or $LiAlH_4$ directly leads to the generation of hydride ions necessary for the mechanism described in the assertion, the reason correctly explains the assertion.

Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept:Reaction Mechanism

c) Donation of an electron by sodium
[Solution Description]
When ethanol $(C_2H_5OH)$ reacts with sodium, the reaction involves the cleavage of the $O-H$ bond in ethanol. The sodium metal donates an electron to the oxygen atom in the $O-H$ bond, which leads to the formation of sodium ethoxide and hydrogen gas as shown below:

$2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2 \uparrow$

The key step here is the donation of an electron by sodium to form the alkoxide ion.

Key Concept:Common Names

d) Ethyl Alcohol
[Solution Description] Ethanol is also referred to as ethyl alcohol and has the chemical formula $C_2H_5OH$. It is widely used both industrially and recreationally.

Key Concept:Mono-, Di-, Trihydric

c) Trihydric alcohol
[Solution Description] Alcohols are classified by the number of hydroxyl groups they have:
– Monohydric: 1 hydroxyl group
– Dihydric: 2 hydroxyl groups
– Trihydric: 3 hydroxyl groups
Therefore, an alcohol with three hydroxyl groups is called trihydric alcohol.

Key Concept:Identify Symmetrical

c) Diethyl ether $(\text{C}_2\text{H}_5\text{OC}_2\text{H}_5)$
[Solution Description] A symmetrical ether has two identical alkyl or aryl groups attached to the oxygen atom. Among the given options, diethyl ether $\text{(C}_2\text{H}_5\text{OC}_2\text{H}_5\text{)}$ has identical ethyl groups attached to the oxygen atom, making it a symmetrical ether.

Key Concept:Reaction Mechanism

b) With bromine water
[Solution Description] When phenol is treated with bromine water, it undergoes an electrophilic aromatic substitution reaction. The presence of the -OH group on phenol activates the ring and makes it more reactive towards halogenation. Bromine water acts as a polar solvent and facilitates the rapid replacement of hydrogen atoms at the ortho and para positions, leading to the formation of 2,4,6-tribromophenol.

Key Concept:Reagent Identification

c) Assertion is true, but Reason is false.

[Solution Description] Sodium borohydride $$(\text{NaBH}_4)$$ is a weaker reducing agent compared to lithium aluminium hydride $$(\text{LiAlH}_4)$$ and cannot effectively reduce carboxylic acids to primary alcohols. While $$\text{LiAlH}_4$$ is capable of reducing carboxylic acids due to its strong reducing capability, $$\text{NaBH}_4$$ is not suitable for this purpose. Thus, the assertion is true because $$\text{LiAlH}_4$$ is indeed used for reducing carboxylic acids to primary alcohols. However, the reason given is false as $$\text{NaBH}_4$$ does not have the capability to perform such reductions.

Key Concept:Comparative Analysis

d) Bromine water
[Solution Description] When phenol reacts with bromine water, the high reactivity due to the activation effect of the -OH group leads to electrophilic substitution at all three active positions (2, 4, and 6). This results in the formation of 2,4,6-tribromophenol.

Key Concept:Comparative Analysis, Advanced Bonding Concepts

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] Methanol $(CH_3OH)$ has a hydroxyl group $(-OH)$, allowing it to engage in hydrogen bonding with other methanol molecules. This hydrogen bonding increases the boiling point because more energy is needed to break these interactions during phase transition from liquid to gas. On the other hand, methyl ethyl ether $(C_2H_5OCH_3)$, being an ether, lacks a hydrogen bonded to oxygen or nitrogen necessary for hydrogen bond formation. Consequently, its interactions are primarily van der Waals forces, which require less energy to overcome compared to hydrogen bonds, resulting in a lower boiling point. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept:Common Names

b) Resorcinol
[Solution Description] The compound 1,3-dihydroxybenzene refers to the benzene ring with two hydroxyl groups at positions 1 and 3. This structure corresponds to the common name Resorcinol.

Key Concept:Comparative Analysis, Real-world Application

c) Methanol can form hydrogen bonds.

[Solution Description] Methanol exhibits hydrogen bonding due to the presence of the hydroxyl (-OH) group, which significantly elevates its boiling point. Dimethyl ether lacks such capability for extensive hydrogen bonding because it has an ether linkage without a direct hydrogen bond donor like hydroxyl. Instead, it only has dipole-dipole interactions, which are weaker than hydrogen bonds. Therefore, despite both having similar molecular weights, the stronger intermolecular forces in methanol account for its higher boiling point.

Key Concept:Reaction Conditions

c) With anhydrous $ \text{AlCl}_3 $
[Solution Description] Anisole can undergo Friedel-Crafts alkylation at the ortho and para positions when reacted with an alkyl halide in the presence of anhydrous AlCl_3 as a catalyst. The methoxy group activates these positions for electrophilic attack.

Key Concept:Reaction Conditions

b) Low temperature and absence of a strong base
[Solution Description] In the dehydration of alcohols, low temperatures generally favor ether formation because it allows for the bimolecular nucleophilic substitution reaction (SN2) without providing sufficient energy for the elimination $$\frac{E_1}{E_2}$$ pathways that form alkenes.

Key Concept:Common vs IUPAC Names

c) n-Propyl alcohol
[Solution Description] The common name for Propan-1-ol $(C_3H_8O)$ is n-propyl alcohol, as it indicates a straight-chain structure without branching.

Key Concept:Reaction Type

b) SN2
[Solution Description] Williamson synthesis proceeds via an SN2 mechanism, where the nucleophilic alkoxide ion attacks the electrophilic carbon in the primary alkyl halide, resulting in the formation of an ether.

Key Concept:Reaction Mechanism

c) Protonation of the alcohol
[Solution Description] 2-methyl-2-propanol is a tertiary alcohol. The reaction mechanism with HI for tertiary alcohols follows an SN1 pathway. The first step involves the protonation of the alcohol oxygen, converting it into a good leaving group as water, which leaves to form a stable carbocation.

Key Concept:Mechanistic Insight, Advanced Comparative Analysis

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description]

Hydroboration-oxidation involves the formation of trialkyl boranes through a concerted syn-addition mechanism where both boron and hydrogen add across the double bond simultaneously without forming carbocation intermediates. This method is indeed more stereospecific compared to oxymercuration-demercuration, which may involve the formation of a mercurinium ion intermediate leading to potential rearrangements or different stereochemistry. Thus, the trialkyl borane intermediates are generally more stable in terms of maintaining specific regioselectivity (anti-Markovnikov), but they can be deemed less stable concerning intrinsic molecular stability when compared to their oxymercuration counterparts, due to the absence of strong C-Hg bonds. The Reason accurately describes why the hydroboration pathway is favored for certain transformations despite this perceived instability since it benefits from lower steric hinderance and avoids carbocations.