Organic Chemistry Trivia: Alcohols, Phenols & Ethers Quiz

The process of converting alkyl halides into alcohols involves substitution reaction. In this reaction, the halogen atom of the alkyl halide is replaced by a hydroxyl group (-OH) to form an alcohol. This substitution reaction occurs through the nucleophilic attack of a hydroxide ion (OH-) on the carbon atom bonded to the halogen. The halogen is displaced and replaced by the hydroxyl group, resulting in the formation of an alcohol.

The correct answer is (iv) > (iii) > (i) > (ii). This is because the acidity of a compound is determined by the stability of its conjugate base. In this case, para-nitrophenol (iv) has the highest acidity because the nitro group stabilizes the negative charge on the phenoxide ion through resonance. Meta-nitrophenol (iii) is next in acidity because the nitro group can still stabilize the negative charge, although to a lesser extent. Phenol (i) is less acidic than the nitrophenols because it lacks any electron-withdrawing groups to stabilize the negative charge. Methylphenol (ii) is the least acidic because the methyl group does not have any electron-withdrawing effect.

Propene on hydroboration-oxidation undergoes anti-Markovnikov addition, where the boron atom adds to the less substituted carbon atom, resulting in the formation of an alcohol. In this case, the less substituted carbon atom is the terminal carbon, so the boron atom adds to it. After oxidation, the boron atom is replaced by a hydroxyl group (OH), resulting in the formation of CH3CH2CH2OH, which is the correct answer.

When sodium benzene sulfonate reacts with NaOH, it forms sodium phenoxide. This is because the sodium ion from NaOH replaces the hydrogen ion in the sulphonate group, resulting in the formation of sodium phenoxide. When this solution is acidified, the phenoxide ion accepts a proton from the acid, forming phenol. Therefore, the correct answer is Phenol.

The given answer suggests that compound IV is the most reactive, followed by compound III, compound I, and finally compound II. This order is determined by the reactivity of the compounds, with compound IV being the most reactive and compound II being the least reactive.

The strongest acid among the given compounds is CH3OH (methanol). Methanol is a weak acid but it is stronger than the other two compounds listed. C6H6 (benzene) is a nonpolar compound and does not have any acidic properties. C2H6 (ethane) is also a nonpolar compound and does not exhibit any acidic behavior. Methanol, on the other hand, can donate a proton (H+) and therefore has weak acidic properties.

The monochlorination of toluene in sunlight results in the formation of benzyl chloride. When benzyl chloride is hydrolyzed with aqueous sodium hydroxide (NaOH), it undergoes a nucleophilic substitution reaction to form benzyl alcohol. Therefore, the correct answer is benzyl alcohol.

The Reimer-Tiemann Reaction involves the use of dichlorocarbene as an electrophile. In this reaction, a phenol or an enolizable aldehyde or ketone reacts with chloroform in the presence of a base to form a carbonyl compound with a substituted hydroxyl group. The dichlorocarbene acts as an electrophile by attacking the carbon of the carbonyl group, leading to the formation of a new carbon-carbon bond. This reaction is commonly used for the synthesis of salicylaldehydes and salicylic acids.

In the dehydration of alcohols to alkenes, the initial step is the protonation of the alcohol molecule. This involves the transfer of a proton from the concentrated sulfuric acid (H2SO4) to the alcohol molecule, resulting in the formation of a protonated alcohol. This protonation makes the alcohol a good leaving group, allowing the formation of a carbocation. Therefore, the correct answer is the protonation of the alcohol molecule.

Bromide ions (Br⁻) can function as nucleophiles in electrophilic aromatic substitution reactions. In the presence of an oxidizing agent, they can brominate the aromatic ring specifically at positions that are ortho to the hydroxyl group and para to the methyl group. The hydroxyl group (-OH) on the aromatic ring is an activating and ortho/para-directing group. Similarly, the methyl group (-CH33​) is also an activating group and directs substitution to the ortho and para positions. As a result, the combination of these substituents leads to preferential bromination at positions that are both ortho to the hydroxyl group and para to the methyl group.

The structure of iso-butyl alcohol is a branched chain compound with a hydroxyl group (-OH) attached to the second carbon atom in the chain. It has four carbon atoms in total, with three of them forming a linear chain and the fourth carbon branching off from the second carbon. The presence of the hydroxyl group makes it an alcohol.

Anisole reacts with Br2 in ethanoic acid to predominantly give only p-bromoanisole. This is because the reaction follows an electrophilic aromatic substitution mechanism, where the bromine molecule acts as the electrophile. The bromine molecule attacks the benzene ring of anisole, and due to the presence of electron-donating methoxy group (-OCH3), the attack occurs at the para position, resulting in the formation of p-bromoanisole as the major product. The ortho position is less favored due to steric hindrance caused by the methoxy group. Therefore, the reaction predominantly produces only p-bromoanisole.