Question

Which one among the following compounds will most readily be dehydrated under acidic condition?
 

(A)	(B)	(C)	(D)
CH3-CH2-CH2-OH	CH3-CH(CH3)-CH2-OH	CH3-CH2-CH(CH3)-OH	C(CH3)(CH3)-CH2-OH
(1° alcohol)	(2° alcohol)	(2° alcohol)	(3° alcohol)

• A. A
• B. B
• C. C
• D. D

Hint:

Tip about dehydration of alcohols Key Points:
Protonation of -OH creates better leaving group (H$_2$O)
Carbocation stability determines rate: \chemfig{C^+(-[::+60]R)(-[:-60]R)-R>\chemfig{R-CH^+-R>\chemfig{R-CH_2^+
Rearrangements possible for less stable carbocations

Answer 1

The Correct Option is D

Dehydration under acidic conditions follows the carbocation stability trend:

\(3^\circ>2^\circ>1^\circ\)

(A) Primary alcohol (slowest)

(B) Secondary alcohol

(C) Secondary alcohol

(D) Tertiary alcohol (fastest)

The tertiary alcohol in (D) forms the most stable carbocation intermediate, making it dehydrate most readily.