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 (mechanistic reasoning)

(i) Why acid hydration gives ethanol. The reaction needs an acid because ethene itself is not attacked by neutral water; the acid first makes the double bond electrophilic.
Step 1: The alkene is a base and grabs \(H^+\), producing the ethyl cation \(CH_3\overset{+}{C}H_2\). With ethene both carbons are equivalent, so only one cation forms.
Step 2: Water, being nucleophilic, bonds to the electron-poor carbon to give the oxonium ion \(CH_3CH_2\overset{+}{O}H_2\).
Step 3: Loss of a proton to a second water molecule frees the neutral product and returns \(H^+\), so the acid is a true catalyst: \(CH_2=CH_2 + H_2O \xrightarrow{H^+} CH_3CH_2OH\). For higher alkenes the \(H^+\) adds so as to give the more stable carbocation, which is why the rule is Markovnikov.

(ii) Hydroboration-oxidation as an anti-Markovnikov route. Diborane exists as \((BH_3)_2\); the electron-deficient boron seeks the carbon with more hydrogens (terminal), so \(-BH_2\) sits on the end carbon and \(-H\) on the inner carbon in a single concerted (syn) step. Alkaline peroxide then swaps boron for \(-OH\) with retention, so the \(-OH\) lands where boron was.
Addition: \(3CH_3CH=CH_2 + (BH_3)_2 \rightarrow 2(CH_3CH_2CH_2)_3B\).
Oxidation: \((CH_3CH_2CH_2)_3B + 3H_2O_2 \xrightarrow{OH^-} 3CH_3CH_2CH_2OH + H_3BO_3\).
Because \(-OH\) ends on the terminal carbon, propene yields propan-1-ol, opposite to acid hydration.

Option 2 (named reaction for each)

(i) Williamson synthesis (alkoxide + primary halide, \(S_N2\)): \(C_2H_5Br + NaOC_2H_5 \rightarrow C_2H_5OC_2H_5 + NaBr\).
(ii) Nitration of phenol to picric acid: \(C_6H_5OH + 3HNO_3 \xrightarrow{conc.\ H_2SO_4} C_6H_2(NO_2)_3OH + 3H_2O\).
(iii) Friedel-Crafts acylation of anisole (\(-OCH_3\) directs para): \(CH_3OC_6H_5 + CH_3COCl \xrightarrow{AlCl_3} p\text{-}CH_3OC_6H_4COCH_3 + HCl\).
(iv) Reimer-Tiemann reaction (dichlorocarbene formylates phenol at ortho): \(C_6H_5OH + CHCl_3 + 3NaOH \rightarrow o\text{-}HOC_6H_4CHO + 3NaCl + 2H_2O\).
(v) Ether cleavage by HI (\(I^-\) attacks the less hindered methyl carbon): \(CH_3OC_2H_5 + HI \rightarrow CH_3I + C_2H_5OH\).

\(\boxed{\text{Markovnikov (hydration) vs anti-Markovnikov (hydroboration); five named phenol/ether reactions.}}\)