Which of the following alcohols is dehydrated by heating with 20% phosphoric acid at 358 K?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2-Methylpropan-2-ol
Dehydration with acid goes through a carbocation. Tertiary carbocations are the most stable and the easiest to form, so the ease of dehydration is tertiary > secondary > primary. The tertiary alcohol therefore needs the mildest conditions, and 20% H3PO4 at 358 K is the mildest of the conditions used:
Ethanol (primary) — concentrated H2SO4 at 443 K.
Propan-1-ol (primary) — a primary alcohol too, so it is harder to dehydrate than secondary and tertiary alcohols.
Propan-2-ol (secondary) — 85% H3PO4 at 440 K.
2-Methylpropan-2-ol (tertiary) — 20% H3PO4 at 358 K.
\( \displaystyle \ce{(CH3)3C-OH} \xrightarrow[\text{358 K}]{\text{20\% } \ce{H3PO4}} \ce{CH2=C(CH3)2 + H2O} \)
NCERT (2026-27 edition):
7.4.4 Chemical Reactions – “Ethanol undergoes dehydration by heating it with concentrated H2SO4 at 443 K.”
“Secondary and tertiary alcohols are dehydrated under milder conditions. For example”
\( \displaystyle \ce{CH3CH(OH)CH3} \xrightarrow[\text{440 K}]{\text{85\% } \ce{H3PO4}} \ce{CH3-CH=CH2 + H2O} \)
\( \displaystyle \ce{(CH3)3C-OH} \xrightarrow[\text{358 K}]{\text{20\% } \ce{H3PO4}} \ce{CH3-C(=CH2)-CH3 + H2O} \)
“Thus, the relative ease of dehydration of alcohols follows the following order: Tertiary > Secondary > Primary”
“Tertiary carbocations are more stable and therefore are easier to form than secondary and primary carbocations; tertiary alcohols are the easiest to dehydrate.”