NEET Alcohols, Phenols and Ethers Previous Year Questions with Solutions

Calculate the double bond equivalent (DBE) for C₄H₈O:

• • Because the question restricts compounds strictly to "cyclic ethers", the 1 degree of unsaturation is entirely utilized by the ring structure.
• • Possible ring sizes containing oxygen: 5-membered, 4-membered, and 3-membered.

Systematically count structural and stereoisomers for oxolanes and oxetanes:

• • 4-Membered (Oxetanes):
• • - 2-Methyloxetane: Carbon-2 is bonded to -H, -CH₃, -O-, and -CH₂-. All four groups are distinct, so C2 is a stereocenter, producing a pair of enantiomers (R and S) ⟹ 2 isomers.
• • - 3-Methyloxetane: Has an internal plane of symmetry through C3 and oxygen, achiral ⟹ 1 isomer.
• • Total 4-membered = 2 + 1 = 3 isomers.

Examine substituted oxiranes (epoxides):

• • - 2-Ethyloxirane: C2 is a chiral center, producing (R) and (S) enantiomers ⟹ 2 isomers.
• • - 2,2-Dimethyloxirane: Both methyl groups reside on the same carbon, achiral ⟹ 1 isomer.
• • - 2,3-Dimethyloxirane: Contains two stereocenters at C2 and C3:
• • * (2R, 3S)-2,3-dimethyloxirane is the cis-isomer with an internal plane of symmetry (meso form) ⟹ 1 isomer.
• • * (2R, 3R) and (2S, 3S) are the chiral trans-enantiomers ⟹ 2 isomers.
• • * Oxiranes subtotal = 2 + 1 + 3 = 6 isomers.
• • Grand Total = 1 (THF) + 3 (Oxetanes) + 6 (Oxiranes) = 10 isomers.
• • Matches Option (3) 10 perfectly.

Lucas reagent consists of equimolar anhydrous ZnCl2 (a strong Lewis acid) and concentrated hydrochloric acid. The alcohol coordinates with ZnCl2, turning the poor leaving group -OH into a good leaving group [-O(ZnCl2)H]:

In an SN1 pathway, departure of the leaving group produces a carbocation intermediate. The activation energy is lowest for tertiary carbocations owing to inductive stabilization (+I) and hyperconjugation (9 α-hydrogens in the tert-butyl cation):

Rapid attack by chloride ion yields tert-butyl chloride, which is immiscible in aqueous solution, producing immediate cloudiness / turbidity:

Intramolecular hydrogen bonding requires the hydrogen donor (e.g. -O-H) and hydrogen acceptor (e.g. -N⁺(=O)O⁻) to be positioned within the same molecule such that a strain-free 5- or 6-membered pseudo-ring (chelate ring) can form:

In ortho-nitrophenol (2-nitrophenol), the spatial distance between the phenolic hydrogen and the nitro oxygen is approximately 1.8 to 2.1 Å, facilitating rapid intramolecular proton-chelate formation. This internal bonding prevents association with neighboring molecules, resulting in lower boiling point and steam volatility.

In m-nitrophenol and p-nitrophenol, the large geometric distance (> 4.5 Å) precludes intramolecular bonding; instead, they associate strongly via intermolecular hydrogen bonding, conferring high boiling points. In HF, each molecule forms intermolecular hydrogen bonds with adjacent molecules:

In the presence of hydrobromic acid (HBr), the lone pair on the oxygen atom of 3-methylbutan-2-ol is protonated by the strong acid to form an oxonium ion intermediate:

Water departs as a neutral leaving group, generating a secondary (2°) carbocation at C-2:

The adjacent C-3 carbon has a tertiary C-H bond. A hydride ion (:H⁻) migrates from C-3 to C-2 (1,2-hydride shift), transforming the less stable secondary carbocation into a significantly more stable tertiary (3°) carbocation with 7 hyperconjugative α-hydrogens:

The bromide ion (Br⁻) rapidly attacks the planar tertiary carbocation to yield the major rearranged substitution product:

In the presence of concentrated HI, the lone pair on the ether oxygen is protonated, yielding a positively charged oxonium ion:

The bond between oxygen and the phenyl ring has partial double bond character due to resonance of oxygen lone pairs with the aromatic ring, making it resistant to cleavage. Furthermore, nucleophilic attack on an sp² phenyl carbon is energetically disfavored:

The benzyl C(sp³)–O bond cleaves readily via SN1/SN2 pathway because the developing positive charge at the benzylic carbon is stabilized by resonance with the phenyl ring. Iodide ion attacks the benzylic carbon:

Product A is benzyl iodide (C₆H₅–CH₂I) and Product B is phenol (C₆H₅–OH):

Acid-catalysed dehydration of alcohols occurs via an E1 elimination pathway. Protonation of the hydroxyl group yields an alkyloxonium ion, which loses water in the rate-determining step to generate a carbocation intermediate:

The rate of dehydration is directly proportional to the thermodynamic stability of the intermediate carbocation. A nitro group (–NO₂) is an extremely powerful electron-withdrawing group via both inductive (–I) and resonance (–M) effects. Placing a positive charge near –NO₂ results in severe electrostatic destabilization:

Options (1), (2), and (4) all possess a strongly electron-withdrawing –NO₂ group on the carbon skeleton. In contrast, Option (3) possesses no –NO₂ group; its carbocation is stabilized by adjacent alkyl electron donation (hyperconjugation and +I effect):

Grignard reagents ($RMgX$) act as powerful carbon nucleophiles. The polar organomagnesium bond provides a nucleophilic ethyl group ($C_2H_5^-$) that attacks the electrophilic carbonyl carbon of acetone (propan-2-one):

Acidic aqueous hydrolysis of the magnesium alkoxide intermediate decomposes it into the corresponding tertiary alcohol and basic magnesium bromide:

Identify the principal functional group and the longest continuous carbon chain: - Longest carbon chain containing the $-OH$ group has $4$ carbons (butane). - Numbering starts from the right to give the principal $-OH$ group the lowest locant ($2$): - Carbon 1: $-CH_3$ - Carbon 2: $-C(CH_3)(OH)-$ - Carbon 3: $-CH_2-$ - Carbon 4: $-CH_3$ - The substituents are a methyl group at position 2 and an -ol group at position 2.

The polarized carbon-magnesium bond in methylmagnesium chloride ($$\text{CH}_3^{\delta-} - \text{MgCl}^{\delta+}$$) acts as a strong nucleophile. The carbanion ($$\text{CH}_3^-$$) attacks the electrophilic carbonyl carbon of acetone: $$\text{CH}_3-\text{CO}-\text{CH}_3 + \text{CH}_3\text{MgCl} \xrightarrow{\text{dry ether}} (\text{CH}_3)_3\text{C-OMgCl}$$ This forms a magnesium alkoxide addition complex.

Subsequent hydrolysis with water in mildly acidic medium cleaves the oxygen-magnesium bond: $$(\text{CH}_3)_3\text{C-OMgCl} + \text{H}_2\text{O} \to (\text{CH}_3)_3\text{C-OH} + \text{Mg(OH)Cl}$$ The resulting product is 2-methylpropan-2-ol, whose common IUPAC/trivial name is tertiary butyl alcohol (tert-butyl alcohol).

The hydroxyl (-OH) group is attached to a tertiary carbon that is bonded to three other methyl carbon atoms. Thus, it is an unambiguous 3° alcohol.

Isopropyl alcohol (2-propanol) is a 2° alcohol formed from acetaldehyde + methyl Grignard; sec-butyl alcohol (2-butanol) is formed from propanal + methyl Grignard or acetaldehyde + ethyl Grignard; isobutyl alcohol is a 1° alcohol.

In the presence of hydrogen iodide ($$\text{HI}$$), anisole (methoxybenzene) is first protonated at the ether oxygen to form a methylphenyl oxonium ion: $$\text{C}_6\text{H}_5-\text{O}-\text{CH}_3 + \text{HI} \rightleftharpoons \text{C}_6\text{H}_5-\text{O}^+(\text{H})-\text{CH}_3 + \text{I}^-$$

The oxonium ion has two different $$\text{C}-\text{O}$$ bonds: 1. The $$\text{C}_{\text{phenyl}}-\text{O}$$ bond: Due to resonance with the benzene ring, this bond has partial double bond character ($$sp^2$$ hybridized carbon) and a shorter bond length ($$136\text{ pm}$$), making it exceptionally strong and resistant to cleavage. 2. The $$\text{C}_{\text{methyl}}-\text{O}$$ bond: This is a weaker, single $$\sigma$$-bond ($$sp^3$$ hybridized carbon).

The nucleophilic iodide ion ($$\text{I}^-$$) selectively attacks the less sterically hindered methyl carbon via an $$S_N2$$ pathway, displacing phenol as the neutral leaving group: $$\text{I}^- + \text{CH}_3-\text{O}^+(\text{H})-\text{C}_6\text{H}_5 \xrightarrow{S_N2} \text{CH}_3\text{I} + \text{C}_6\text{H}_5\text{OH}$$ Phenol does not react further with HI because the hydroxyl group cannot be displaced by iodide from the $$\text{sp}^2$$ phenyl ring.

The cleavage products are unequivocally Phenol ($$\text{C}_6\text{H}_5\text{OH}$$) and Methyl iodide ($$\text{CH}_3\text{I}$$), corresponding to Option (1).