Consider the following statements about the preparation of alcohols from alkenes.
(A) In the acid-catalysed hydration of an unsymmetrical alkene, the addition of water takes place in accordance with Markovnikov’s rule.
(B) In the acid-catalysed hydration of an alkene, the first step is the protonation of the alkene to form a carbocation by nucleophilic attack of H3O+.
(C) In hydroboration, the boron atom gets attached to the sp2 carbon carrying the greater number of hydrogen atoms.
(D) In hydroboration–oxidation, the trialkyl borane formed is oxidised to the alcohol by hydrogen peroxide in the presence of aqueous sulphuric acid.
Which of the statements given above are incorrect? Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (B) and (D) only
Alkenes give alcohols in two ways. In acid-catalysed hydration, water adds through a carbocation, by Markovnikov’s rule. In hydroboration–oxidation, diborane adds first and the trialkyl borane is then oxidised, so the –OH group ends up on the carbon that had more hydrogen atoms.
(A) In the acid-catalysed hydration of an unsymmetrical alkene, the addition of water takes place in accordance with Markovnikov’s rule. — Correct statement. Propene, for example, gives propan-2-ol:
\( \displaystyle \ce{CH3-CH=CH2 + H2O} \xrightarrow{\ce{H+}} \ce{CH3-CH(OH)-CH3} \)
(B) In the acid-catalysed hydration of an alkene, the first step is the protonation of the alkene to form a carbocation by nucleophilic attack of H3O+. — Incorrect statement. H3O+ gives a proton to the alkene, so step 1 is an electrophilic attack. The nucleophilic attack comes in step 2, when water attacks the carbocation.
(C) In hydroboration, the boron atom gets attached to the sp2 carbon carrying the greater number of hydrogen atoms. — Correct statement. The –OH group later takes the place of boron, so the alcohol looks as if water had added opposite to Markovnikov’s rule; propene gives propan-1-ol.
(D) In hydroboration–oxidation, the trialkyl borane formed is oxidised to the alcohol by hydrogen peroxide in the presence of aqueous sulphuric acid. — Incorrect statement. The oxidation with hydrogen peroxide is carried out in the presence of aqueous sodium hydroxide, not an acid:
\( \displaystyle \ce{CH3-CH=CH2} \xrightarrow[\text{(ii) } \ce{H2O2},\ \text{aq. } \ce{NaOH}]{\text{(i) } \ce{(BH3)2}} \ce{CH3-CH2-CH2-OH} \)
NCERT (2026-27 edition):
7.4.1 Preparation of Alcohols – “(i) By acid catalysed hydration: Alkenes react with water in the presence of acid as catalyst to form alcohols. In case of unsymmetrical alkenes, the addition reaction takes place in accordance with Markovnikov’s rule.”
“The mechanism of the reaction involves the following three steps: Step 1: Protonation of alkene to form carbocation by electrophilic attack of H3O+.” … “Step 2: Nucleophilic attack of water on carbocation.”
“(ii) By hydroboration–oxidation: Diborane (BH3)2 reacts with alkenes to give trialkyl boranes as addition product. This is oxidised to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.”
“The addition of borane to the double bond takes place in such a manner that the boron atom gets attached to the sp2 carbon carrying greater number of hydrogen atoms. The alcohol so formed looks as if it has been formed by the addition of water to the alkene in a way opposite to the Markovnikov’s rule.”