Question

(i) Explain the mechanism of the formation of ethanol by acid catalyzed hydration of ethene. (ii) Explain hydroboration-oxidation reaction with an example. (2½+2½=5)
OR
What happens when? (Write chemical equations only) (i) Ethyl bromide reacts with sodium ethoxide. (ii) Phenol is heated with conc. HNO₃ in the presence of conc. H₂SO₄. (iii) Methoxybenzene is heated with acetyl chloride in the presence of anhydrous AlCl₃. (iv) Phenol reacts with chloroform in the presence of aqueous NaOH. (v) Ethyl methyl ether is heated with conc. HI. (1+1+1+1+1=5)

Hint:

Acid hydration is a three-step Markovnikov addition through a carbocation; hydroboration-oxidation is anti-Markovnikov giving the primary alcohol. For the equations, recall Williamson, nitration to picric acid, Friedel-Crafts acylation, Reimer-Tiemann, and ether cleavage by HI.

Answer 1

Option 1

(i) Acid catalyzed hydration of ethene to ethanol. Overall: \(CH_2=CH_2 + H_2O \xrightarrow{H^+} CH_3CH_2OH\). It follows Markovnikov addition through three steps.
Step 1: Protonation. The \(\pi\) electrons of ethene attack a proton supplied by the acid (\(H_3O^+\) from \(H_2SO_4\)), forming a carbocation: \(CH_2=CH_2 + H^+ \rightarrow CH_3\overset{+}{C}H_2\) (ethyl carbocation).
Step 2: Nucleophilic attack of water. A water molecule uses its lone pair to attack the positively charged carbon, giving a protonated alcohol (oxonium ion): \(CH_3\overset{+}{C}H_2 + H_2O \rightarrow CH_3CH_2\overset{+}{O}H_2\).
Step 3: Deprotonation. Another water molecule removes the extra proton, regenerating the catalyst and giving ethanol: \(CH_3CH_2\overset{+}{O}H_2 + H_2O \rightarrow CH_3CH_2OH + H_3O^+\).

(ii) Hydroboration-oxidation. An alkene adds diborane \((BH_3)_2\); the boron atom adds to the less substituted (terminal) carbon and hydrogen to the more substituted carbon, so the addition is anti-Markovnikov and syn. The resulting trialkylborane is then oxidised by alkaline hydrogen peroxide to the alcohol.
Step 1: Addition of borane. \(3CH_3CH=CH_2 + (BH_3)_2 \rightarrow 2(CH_3CH_2CH_2)_3B\) (tripropylborane; boron on terminal carbon).
Step 2: Oxidation. \((CH_3CH_2CH_2)_3B + 3H_2O_2 \xrightarrow{OH^-} 3CH_3CH_2CH_2OH + H_3BO_3\).
Net result: propene gives propan-1-ol, the anti-Markovnikov (primary) alcohol.

Option 2: Equations only

(i) Williamson ether synthesis: \(C_2H_5Br + NaOC_2H_5 \rightarrow C_2H_5OC_2H_5 + NaBr\) (diethyl ether).

(ii) Nitration of phenol to picric acid (2,4,6-trinitrophenol): \(C_6H_5OH + 3HNO_3 \xrightarrow{conc.\ H_2SO_4} C_6H_2(NO_2)_3OH + 3H_2O\).

(iii) Friedel-Crafts acylation of anisole (para product): \(CH_3OC_6H_5 + CH_3COCl \xrightarrow{anhyd.\ AlCl_3} p\text{-}CH_3OC_6H_4COCH_3 + HCl\).

(iv) Reimer-Tiemann reaction giving salicylaldehyde: \(C_6H_5OH + CHCl_3 + 3NaOH \rightarrow o\text{-}HOC_6H_4CHO + 3NaCl + 2H_2O\).

(v) Ether cleavage by HI (iodide forms at the smaller alkyl group): \(CH_3OC_2H_5 + HI \rightarrow CH_3I + C_2H_5OH\).

\[\boxed{\text{Hydration gives Markovnikov ethanol; hydroboration gives anti-Markovnikov alcohol.}}\]