Alcohols, Phenols and Ethers Set-7

$CH_3CH_2CH_2CH_2OH$ (butan‑1‑ol)

$CH_3CH(OH)CH_2CH_3$ (butan‑2‑ol)

$(CH_3)_3COH$ (2‑methylpropan‑2‑ol, tert‑butyl alcohol)

$CH_3CH_2OCH_2CH_3$ (diethyl ether)

$C_6H_5CH_2OCH_3$ (benzyl methyl ether)

$C_6H_5CH_2OCH_2CH_3$ (benzyl ethyl ether)

$CH_3CH_2OCH_2CH_3$ (diethyl ether)

$C_6H_5OCH_2CH_3$ (ethoxybenzene / phenetole)

First O‑methylation of phenol to anisole (e.g. $CH_3I/K_2CO_3$), then nitration of anisole under controlled (mild) conditions ($HNO_3$, low temperature)

First nitration of phenol with $HNO_3/H_2SO_4$, then O‑methylation of the nitrophenol (e.g. $CH_3I/K_2CO_3$)

Convert phenol to a diazonium derivative and try to introduce a nitro group via Sandmeyer type sequence, then methylate

Direct methylation of phenol with methanol/H$_2$SO$_4$ followed by vigorous nitration at high temperature

Both A and R are true, but R is not the correct explanation for A.

A is true and R is false.

Both A and R are false.

Both A and R are true and R is the correct explanation for A.

No reaction — 2,6‑dimethylphenol is recovered unchanged due to steric hindrance

$4$‑Formyl‑2,6‑dimethylphenol (i.e., $2,6$‑dimethyl‑4‑hydroxybenzaldehyde)

$2$‑Formyl‑2,6‑dimethylphenol (an ortho‑formyl salicylaldehyde derivative)

$2,6$‑Dimethyl‑4‑hydroxybenzoic acid (product of oxidative carboxylation)

Methyl bromide, $CH_3Br$

2‑Bromopropane, $CH_3CHBrCH_3$

tert‑Butyl bromide, $(CH_3)_3CBr$

Isobutyl bromide, $(CH_3)_2CHCH_2Br$

Sodium benzoate, $C_6H_5COO^-\,Na^+$

Sodium phenoxide, $C_6H_5O^-\,Na^+$

Both sodium benzoate and sodium phenoxide

Neither — both remain in the organic layer

$o$‑Nitrophenol, because intramolecular hydrogen bonding of the $o$‑isomer stabilises its conjugate base

$o$‑Nitrophenol, because the nitro group at ortho exerts a stronger $-I$ effect due to proximity

Both have the same acidity because intramolecular hydrogen bonding in $o$‑isomer exactly compensates better resonance in $p$‑isomer

$p$‑Nitrophenol, because the nitro group at para stabilises the phenoxide ion by resonance more effectively than the combined effects in the ortho isomer

Sodium phenoxide ($C_6H_5O^-Na^+$) + methyl iodide → anisole ($C_6H_5OCH_3$)

Sodium phenoxide ($C_6H_5O^-Na^+$) + bromobenzene ($C_6H_5Br$) → diphenyl ether ($C_6H_5OC_6H_5$)

Sodium tert‑butoxide ($(CH_3)_3C\!O^-Na^+$) + methyl bromide → methyl tert‑butyl ether ($(CH_3)_3COCH_3$)

Sodium phenoxide ($C_6H_5O^-Na^+$) + benzyl chloride ($C_6H_5CH_2Cl$) → benzyl phenyl ether ($C_6H_5OCH_2C_6H_5$)

1.0 mol

1.5 mol

2.0 mol

3.0 mol

$NaH$

$NaOH$

$NH_3$

$CH_3COO^-$

$(CH_3)_2CHBr + CH_3OH$ via $S_N1$

$CH_3Br + CH_3CH(OH)CH_3$ via $S_N1$

$(CH_3)_2CHBr + (CH_3)_2CHBr$ (multiple alkyl bromides)

$CH_3Br + (CH_3)_2CHOH$ via $S_N2$ (attack at the methyl carbon)

$(CH_3)_3CBr$ with $NaOCH_3$ — tertiary halide with small alkoxide (favours elimination)

$CH_3Br$ with $NaOCH_3$ — methyl halide with methoxide (good $S_N2$)

$C_6H_5CH_2Br$ (benzyl bromide) with $NaOCH_3$ — benzylic halide (good $S_N2$)

$CH_3CH_2CH_2Br$ with $NaOCH_3$ — primary halide (good $S_N2$)

(i) $Br_2$ (aq) → (ii) work-up

(i) $HNO_3$ (nitration) → (ii) $Sn/HCl$ (reduction) → (iii) $Br_2$

(i) acetylation: $(CH_3CO)_2O/\text{pyridine}$ (protect OH as acetate) → (ii) bromination with $Br_2$ (in $CCl_4$) → (iii) hydrolysis ($\mathrm{NaOH_{(aq)}}$) to regenerate OH

(i) Friedel–Crafts alkylation ($AlCl_3$) → (ii) bromination

Anisole $C_6H_5-O-CH_3$ — cleaved faster because resonance weakens the C–O bond

Isopropyl methyl ether $CH_3-O-CH(CH_3)_2$ — cleaved faster because cleavage proceeds by $S_N2$ at the methyl (or $S_N1$ on the secondary centre), whereas aryl–O cleavage in anisole is difficult ($S_N2$ at the $sp^2$ C is impossible and aryl carbocation formation is unfavourable)

Both ethers are cleaved at similar rates by $HI$

Neither ether is cleaved by $HI$ under these conditions