Assertion (A): The addition of water to but-1-ene in acidic medium gives butan-1-ol.
Reason (R): The addition of water to an alkene in acidic medium proceeds through the formation of a primary carbocation.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Both (A) and (R) are false.
Alkenes react with water in the presence of an acid to give alcohols. For an unsymmetrical alkene the addition follows Markovnikov’s rule: H goes to the double-bond carbon that already carries more H atoms, and –OH goes to the other carbon.
\( \displaystyle \ce{CH3CH2CH=CH2 + H2O} \xrightarrow{\ce{H^+}} \ce{CH3CH2CH(OH)CH3} \)
Assertion (A): The addition of water to but-1-ene in acidic medium gives butan-1-ol. — False. The –OH group goes to C2, so the product is butan-2-ol, not butan-1-ol.
Reason (R): The addition of water to an alkene in acidic medium proceeds through the formation of a primary carbocation. — False. The first step is protonation of the alkene by \( \ce{H3O^+} \). The proton adds to C1 (the \( \ce{CH2} \) end), giving the secondary carbocation at C2, which is more stable than the primary carbocation that would form at C1:
\( \displaystyle \ce{CH3CH2CH=CH2 + H3O^+ -> CH3CH2}\overset{+}{\mathrm{C}}\ce{HCH3 + H2O} \)
Water then attacks this carbocation and loss of a proton gives butan-2-ol.
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.”
“Mechanism
The mechanism of the reaction involves the following three steps: Step 1: Protonation of alkene to form carbocation by electrophilic attack of H3O+.”