Question

Write chemical equations of dehydration reactions of primary, secondary and tertiary alcohols and also write mechanism of dehydration reaction of primary alcohol.

Hint:

The ease of dehydration of alcohols follows the order: tertiary $>$ secondary $>$ primary, due to increasing carbocation stability. Always use concentrated H$_2$SO$_4$ or H$_3$PO$_4$ and heat.

Answer 1

Dehydration reaction: Alcohols on heating with concentrated H$_2$SO$_4$ or H$_3$PO$_4$ undergo dehydration to give alkenes.
(i) Primary alcohol:
Example: Ethanol
\[ CH_3CH_2OH \;\xrightarrow[\;\Delta\;]{conc.\;H_2SO_4} \; CH_2{=}CH_2 + H_2O \] (ii) Secondary alcohol:
Example: 2-propanol
\[ CH_3{-}CHOH{-}CH_3 \;\xrightarrow[\;\Delta\;]{conc.\;H_2SO_4} \; CH_3{-}CH{=}CH_2 + H_2O \] (iii) Tertiary alcohol:
Example: tert-butanol
\[ (CH_3)_3COH \;\xrightarrow[\;\Delta\;]{conc.\;H_2SO_4} \; (CH_3)_2C{=}CH_2 + H_2O \] Mechanism of dehydration of primary alcohol (Ethanol):

Step 1: Protonation of alcohol.
\[ CH_3CH_2OH + H^+ \;\longrightarrow\; CH_3CH_2OH_2^+ \]

Step 2: Formation of carbocation (slow step).
\[ CH_3CH_2OH_2^+ \;\longrightarrow\; CH_3CH_2^+ + H_2O \]

Step 3: Deprotonation to form alkene.
\[ CH_3CH_2^+ \;\longrightarrow\; CH_2{=}CH_2 + H^+ \] Thus, the product is ethene.
\[ \boxed{\text{Primary alcohols undergo dehydration via E1 (or E2 under strong conditions), giving alkenes.}} \]