Question

Dehydration of alcohol in acidic medium involves

• A. i, ii, iii and v
• B. i, ii, iv and v
• C. i, ii, iii and v
• D. i, ii, iii and iv

Answer 1

Answer: (B)

i, ii, iv and v

Answer 2

The dehydration of alcohols in acidic medium typically proceeds via an E1 mechanism, especially for secondary and tertiary alcohols. The key steps are:

1. Protonation: The oxygen atom of the alcohol molecule is protonated by the acid catalyst, converting the poor leaving group (-OH) into a good leaving group (-OH₂⁺). $R-OH + H^+ \rightleftharpoons R-OH_2^+$
2. Formation of Carbonium Ion (Carbocation): The protonated alcohol loses a molecule of water to form a carbocation. This is usually the rate-determining step. $R-OH_2^+ \rightarrow R^+ + H_2O$
3. Deprotonation to form alkene: A base (like water or the conjugate base of the acid) removes a proton from a carbon atom adjacent to the carbocation center, leading to the formation of a carbon-carbon double bond (alkene). $R-CH_2-CH_2^+ + B^- \rightarrow R-CH=CH_2 + BH$

Analyzing the given statements: i) Protonation: This is the initial step. (Correct) ii) Dehydration: This refers to the elimination of water, which occurs when the protonated alcohol loses H₂O to form the carbocation. (Correct) iii) Formation of carbanion: A carbanion is a carbon species with a negative charge. The intermediate formed is a carbocation, which is positively charged. (Incorrect) iv) Formation of carbonium ion: This is an alternative term for carbocation. (Correct) v) Deprotonation to form alkene: This is the final step where a proton is removed to form the alkene. (Correct)

Therefore, the dehydration of alcohol in an acidic medium involves protonation, dehydration (loss of water), formation of a carbonium ion (carbocation), and deprotonation to form the alkene. The correct steps are i, ii, iv, and v.