Alcohols, Phenols and Ethers

Cumene (isopropylbenzene) $\xrightarrow[(ii)\ H_2O,\ H^+]{(i)\ O_2}$ Phenol, so (i) matches (i) $O_2$; (ii) $H_2O/H^+$. $CH_3COOH \xrightarrow[(ii)\ H_2,\ catalyst]{(i)\ CH_3OH,\ H^+} CH_3CH_2OH$ (esterification followed by hydrogenolysis), so (ii) matches (i) $CH_3OH, H^+$; (ii) $H_2$, catalyst. $CH_3CH_2CH_2OH \xrightarrow[(ii)\ H^+/H_2O]{(i)\ conc.\ H_2SO_4,\ \Delta} CH_3-CH(OH)-CH_3$ (dehydration to propene, then Markovnikov hydration), so (iii) matches (i) conc. $H_2SO_4, \Delta$; (ii) $H^+/H_2O$. Benzene $\xrightarrow[(ii)\ NaOH,\ \Delta;\ (iii)\ H^+]{(i)\ Oleum}$ Phenol (sulphonation, alkali fusion, acidification), so (iv) matches (i) oleum; (ii) NaOH, $\Delta$; (iii) $H^+$. Correct match: i-b, ii-c, iii-d, iv-a.

Phthalein dye test is used for the identification of the phenolic group.

Alcohols react with Lucas reagent (conc. HCl + anhyd. $ZnCl_2$) via a carbocation intermediate, forming an alkyl chloride (turbidity). A tertiary carbocation is more stable than secondary or primary, so tertiary alcohols react instantaneously. 2-methylpropan-2-ol (tert-butyl alcohol), $(CH_3)_3C-OH$, is tertiary, so it reacts instantaneously.

The terminal alkene must be converted to an aldehyde with the carbonyl at the terminal carbon (anti-Markovnikov): (i) $BH_3$ (ii) $H_2O_2/OH^-$: hydroboration-oxidation gives the primary alcohol $-CH_2-CH_2-CH_2-OH$ (anti-Markovnikov addition of water) (iii) PCC: mild oxidation of the primary alcohol stops at the aldehyde $-CH_2-CH_2-CHO$ So the reagents are (i) $BH_3$ (ii) $H_2O_2/OH^-$ (iii) PCC.

$3ROH + PCl_3 \rightarrow 3RCl + H_3PO_3$ (A) $ROH + PCl_5 \rightarrow RCl + HCl + POCl_3$ (B) So A = $H_3PO_3$ and B = $POCl_3$.

Hydroboration of propene is anti-Markovnikov, so boron attaches to the less hindered terminal carbon, giving $(CH_3CH_2CH_2)_3B$ (not $((CH_3)_2CH)_3B$) — so Statement I is incorrect. However, oxidation of a borane such as $((CH_3)_2CH)_3B$ with $H_2O_2/NaOH$ does proceed with retention of configuration to replace the C–B bond with a C–OH bond, giving propan-2-ol, so Statement II (as a statement of the oxidation step) is correct.

$CH_3CH_2OH\xrightarrow{P+I_2}CH_3CH_2I$ (A); $CH_3CH_2I\xrightarrow{Mg,\ dry\ ether}CH_3CH_2MgI$ (B, Grignard reagent); $CH_3CH_2MgI$ reacts with HCHO to give an addition product, which on hydrolysis with $H_2O$ gives $CH_3CH_2CH_2OH$ (D, n-propyl alcohol, major product).

Reaction follows SN1 pathway involving Benzylic carbocation. [add image]

Formation of conjugated diene in option (c) makes the given reactant most reactive towards dehydration in acidic conditions. [add image]

Strong oxidising agents such as $CrO_3-H_2SO_4$ (A), $K_2Cr_2O_7-H_2SO_4$ (B), and $KMnO_4+KOH/H_3O^+$ (C) can oxidise a primary alcohol all the way to a carboxylic acid. (D) Cu at 573 K only dehydrogenates the alcohol to an aldehyde, and (E) $CrO_3$ with acetic anhydride (Collins-type reagent, Sarrett's reagent variant) typically stops oxidation at the aldehyde stage.

$NaBH_4$ selectively reduces the more reactive ketone in the side chain to a secondary alcohol without touching the ring, giving a diol-like intermediate with $-CH_2-CH(OH)-CH_3$ on the side chain and the ring ketone reduced too, giving $-OH$ on the ring as well. On treatment with $H_2SO_4/\Delta$, acid-catalysed dehydration (E1) occurs at both positions, ultimately giving the major product with a cyclopentene ring conjugated to a $-CH=CH-CH_3$ side chain (the more substituted, conjugated/stable alkene being favoured, per Zaitsev's rule).

The acidic strength of phenol increases due to electron-withdrawing groups, so nitrophenols are more acidic than phenol – Statement I is correct.
Order of acidic strength: p-nitrophenol > o-nitrophenol > m-nitrophenol > phenol, so they do not have the same acidic strength – Statement II is incorrect.

Statement I is correct: 1°, 2° and 3° alcohols are distinguished in the Lucas test by the time taken for turbidity to appear with conc. HCl + anhydrous $ZnCl_2$. Statement II is incorrect: tertiary alcohols are the most reactive and give turbidity immediately; secondary alcohols take about 5 minutes and primary alcohols do not give turbidity at room temperature. Reactivity towards Lucas reagent: 3° > 2° > 1°

(a) Kolbe's reaction: sodium phenoxide + $CO_2$, then $H^+$ gives salicylic acid. (b) Reimer-Tiemann reaction ($CHCl_3$/aq. NaOH, then $H^+$) gives salicylaldehyde. (c) Phenol with zinc dust on heating gives benzene. (d) Oxidation with $Na_2Cr_2O_7/H_2SO_4$ gives benzoquinone.

Gabriel phthalimide synthesis gives primary amines; Kolbe's reaction (sodium phenoxide + $CO_2$) gives salicylic acid; Williamson synthesis ($R-X+R'ONa\rightarrow R'OR+NaX$) gives ethers; Etard reaction (toluene + $CrO_2Cl_2$ in $CS_2$, then $H_3O^+$) gives benzaldehyde.

Chlorobenzene does not react with NaOH at STP; it needs 623 K and 300 atm. Benzene sulphonation (oleum) followed by NaOH fusion and acidification, diazotisation of aniline followed by warming with water, and the cumene process all give phenol.

$CH_3COCH_3 \xrightarrow{C_2H_5MgBr,\ dry\ ether} (CH_3)_2C(OMgBr)C_2H_5 \xrightarrow{H_2O,\ H^+} (CH_3)_2C(OH)CH_2CH_3$ IUPAC name: 2-methylbutan-2-ol

$NaBH_4$ reduces aldehydes and ketones but does not reduce esters. So only the ring C=O is reduced to –OH, and the $-CH_2COOCH_3$ group remains unchanged.

$CH_3COCH_3 \xrightarrow{CH_3MgCl} (CH_3)_3C-OMgCl \xrightarrow{hydrolysis} (CH_3)_3C-OH$ The product is tert-butyl alcohol (2-methylpropan-2-ol).

$C_6H_5-O-CH_3 + HI \rightarrow C_6H_5OH + CH_3I$ The ether oxygen is protonated; $I^-$ attacks the methyl carbon ($S_N2$), because the aryl C–O bond has partial double bond character and phenyl cation formation is unfavourable. So the products are phenol and methyl iodide.

Electron-withdrawing groups (like $-NO_2$) stabilise the phenoxide ion by delocalising the negative charge (through resonance and inductive effect), increasing acidity. Electron-donating groups ($-CH_3$, $-OCH_3$, $-C_2H_5$) destabilise the phenoxide ion, decreasing acidity. So p-nitrophenol, with the strong EWG $-NO_2$, is the strongest acid among the given options.

The $-OH$ group in phenol is a strong activating, ortho/para-directing group (due to resonance donation of the lone pair into the ring), making phenol far more reactive towards electrophilic aromatic substitution than benzene, chlorobenzene (weakly deactivating) or nitrobenzene (strongly deactivating).

Hydroboration-oxidation follows anti-Markovnikov addition: boron attaches to the less substituted (terminal) carbon, and subsequent oxidation with $H_2O_2/OH^-$ replaces the C-B bond with a C-OH bond. So 1-butene gives the primary alcohol $CH_3CH_2CH_2CH_2OH$ (n-butanol).

In phenol ($Ph-O-H$), the lone pair of electrons on oxygen is involved in resonance with the benzene ring, so oxygen carries a partial positive charge. Hence an incoming proton cannot attack it easily, and phenol is the most difficult to protonate.

This is the cumene process for phenol. Cumene is oxidised by $O_2$ to cumene hydroperoxide (A), $C_6H_5-C(CH_3)_2-O-O-H$. On treatment with $H^+/H_2O$, cumene hydroperoxide gives phenol and acetone ($H_3C-CO-CH_3$). So intermediate A is cumene hydroperoxide.