Question

(a) Write the reagents which are used in the given conversions :

(i) Phenol to 2, 4, 6 - tribromophenol

(ii) Propene to propan-1-ol

(iii) Butan-2-one to butan-2-ol

(b) Explain the mechanism of acid catalyzed hydration of alkene to form corresponding alcohol.

Hint:

Use bromine water, then B2H6/THF followed by H2O2/NaOH, then NaBH4. The hydration mechanism has three steps: protonation to a carbocation, attack by water, loss of a proton.

Answer 1

Step 1: Understanding the Question:
Part (a) asks for one reagent set for each of three conversions. Part (b) asks for the three-step mechanism of acid catalysed hydration of an alkene.

Step 2: Phenol to 2,4,6-tribromophenol (a)(i):
The -OH group makes the ring very reactive towards electrophiles. With bromine water, all three ortho and para positions are brominated and a white precipitate forms.
\[ \mathrm{C_6H_5OH + 3Br_2\ (aq) \rightarrow 2,4,6-Br_3C_6H_2OH + 3HBr} \]
Reagent: bromine water (\(\mathrm{Br_2/H_2O}\)).

Step 3: Propene to propan-1-ol (a)(ii):
Direct acid hydration of propene follows Markovnikov addition and gives propan-2-ol. To get the primary alcohol we need anti-Markovnikov addition, which is hydroboration-oxidation.
\[ \mathrm{CH_3CH{=}CH_2 \xrightarrow{(i)\ B_2H_6/THF} (CH_3CH_2CH_2)_3B \xrightarrow{(ii)\ H_2O_2,\ NaOH} CH_3CH_2CH_2OH} \]
Reagents: diborane (\(\mathrm{B_2H_6}\)) in THF, then \(\mathrm{H_2O_2}\) with NaOH.

Step 4: Butan-2-one to butan-2-ol (a)(iii):
A ketone is reduced to a secondary alcohol by a hydride reagent.
\[ \mathrm{CH_3COCH_2CH_3 \xrightarrow{NaBH_4} CH_3CH(OH)CH_2CH_3} \]
Reagent: \(\mathrm{NaBH_4}\) (or \(\mathrm{LiAlH_4}\), or \(\mathrm{H_2}\) with Ni, Pd or Pt catalyst).

Step 5: Mechanism of acid catalysed hydration, step 1 (b):
Take ethene as the example. The acid gives \(\mathrm{H_3O^{+}}\). The pi bond of the alkene picks up a proton and a carbocation forms. This is the slow step.
\[ \mathrm{CH_2{=}CH_2 + H_3O^{+} \rightarrow CH_3{-}CH_2^{+} + H_2O} \]

Step 6: Mechanism, step 2:
Water is a nucleophile. Its oxygen lone pair attacks the carbocation and forms a protonated alcohol.
\[ \mathrm{CH_3{-}CH_2^{+} + H_2O \rightarrow CH_3{-}CH_2{-}O^{+}H_2} \]

Step 7: Mechanism, step 3:
A water molecule takes a proton from the protonated alcohol. The alcohol is released and the acid catalyst is regenerated.
\[ \mathrm{CH_3CH_2{-}O^{+}H_2 + H_2O \rightarrow CH_3CH_2OH + H_3O^{+}} \]
For an unsymmetrical alkene, the proton adds to the carbon with more hydrogens (Markovnikov rule), so the more stable carbocation forms.

Final Answer:
(a)(i) Bromine water. (ii) \(\mathrm{B_2H_6/THF}\), then \(\mathrm{H_2O_2/NaOH}\). (iii) \(\mathrm{NaBH_4}\) or \(\mathrm{LiAlH_4}\). (b) Protonation of the alkene, attack of water on the carbocation, then loss of a proton.