JEE Advanced (Single Correct MCQs): Alcohols, Phenols & Ethers - JEE MCQ

• Firstly, alcohols are the most basic among all oxygen compounds.
• Now secondly, it is the fact that alkyl groups have an effect that means it will tend to give its electron towards the oxygen atom on account of increase in electron density on the oxygen atom.
• It gives its one electron pair and behaves as a base.

• The order of reactivity of alcohol with Lucas reagent is tert. > sec. > pri.
• Lucas test is based on the difference in the three types of alcohols (having 6 or less carbon) towards Lucas reagent (a mixture of conc. hydrochloric acid and anhydrous zinc chloride) at room temperature.
• The tertiary alcohols produce turbidity immediately, the secondary alcohols give turbidity within 5 – 10 minutes, and the primary alcohols do not give turbidity at all, at room temperature.
• Hence 2-methylpropan-2-ol (a 3° alcohol) reacts fastest.

Compounds having   groups show positive iodoform test.

(pentanone-2) gives this test.

NOTE :
The –OH group in phenol, being activating group, facilitates substitution in the o- and p-positions.

Reactions involving cleavage of carbon-oxygen bond, (C – OH) follows the following order :
Tertiary > Secondary > Primary

Secondary alcohols oxidise to produce ketone

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NOTE : In absence of CS₂, polyhalogenation in o- and p-positions takes place.

Riemer-Tiemann reaction involves electrophilic
substitution on the highly reactive phenoxide ring.

Compound (CH₃)₄N⁺I^– is most reactive due to (i) better
leaving group, I^– and (ii) due to the fact that the methyl group, with positive N, is more electron deficient.
Hence this group is more reactive towards nucleophile, OH^–

Fehling solution is a weak oxidising agent which can oxidise aldehyde but not ketone.
Primary alcohols undergoes oxidation with alkaline KMnO₄, acidic dichromate and conc. H₂SO₄ to give acids, whereas with Cu they give aldehydes.

• This reaction is an example of Williamson's synthesis.
• C₂H₅O^– will abstract proton from phenol converting the latter into phenoxide ion.
• This would then make nucleophilic attack on the methylene carbon of alkyl iodide forming C₆H₅OC₂H₅.
• But if C₂H₅O^– is in excess, C₂H₅OC₂H₅ will be formed.
• C₂H₅O^– is a better nucleophile than C₆H₅O^– (phenoxide) ion because in the former the negative charge is localized over oxygen, while in the latter it is delocalized over the whole molecular framework.
• So, it is C₂H₅O^– ion that would make nucleophilic attack at ethyl iodide to give diethyl ether (Williamson’s synthesis).

Addition of water to 2-phenylpropene follows Markownikov’s rule.

Conc. HCl, HBr and conc. HCl + ZnCl₂ all are
nucleophiles, thus convert alcohols to alkyl halides.
However, conc. H₃PO₄ is a good dehydrating agent which converts an alcohol to an alkene.

• Among the given compounds, hydroxybenzene (IV) has least molar mass and therefore possess least boiling point.
• Among the three isomeric dihydroxybenzenes, 1,2-dihydroxybenzene (I) forms intramolecular H-bonding with the result it will not form intermolecular H-bonding leading to lowest boiling point.
• On the other hand 1,3-dihydroxybenzene (II) and 1,4-dihydroxybenzene (III) do not undergo chelation, hence they will involve extensive intermolecular H-bonding leading to higher boiling point.
• Further intermolecular hydrogen bonding is stronger in the p-isomer than the m-isomer hence former has highest b.p.
• Thus the decreasing order of boiling points is III > II > I > IV.

In dye test, phenolic — OH group is converted to  — O– which activates the ring towards electrophilic aromatic substitution