Arrange the following compounds in increasing order of dehydrohalogenation reaction using alcoholic KOH.
A. Sec-butylbromide
B. 1-Bromopropane
C. Ethyl bromide
D. 2-Bromo-2-methylpentane
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → C, B, A, D
Heating an alkyl halide with alcoholic KOH removes H from the β-carbon and X from the α-carbon, giving an alkene (β-elimination):
\( \displaystyle \ce{R-CH2-CH2-X + KOH} \xrightarrow[\text{heat}]{\text{ethanol}} \ce{R-CH=CH2 + KX + H2O} \)
The ease of dehydrohalogenation depends on two factors that act in the same direction:
Class of the halide: 3° > 2° > 1°. A halide with more alkyl groups on the α-carbon gives a more substituted, more stable alkene.
Stability of the alkene formed: a more substituted alkene is more stable and forms more easily (Saytzeff rule).
Matching the given compounds:
(C) Ethyl bromide, \( \ce{CH3CH2Br} \): 1°, gives ethene (no alkyl group on the double bond)
(B) 1-Bromopropane, \( \ce{CH3CH2CH2Br} \): 1°, gives propene (one alkyl group)
(A) sec-Butyl bromide, \( \ce{CH3CHBrCH2CH3} \): 2°, gives mainly but-2-ene
(D) 2-Bromo-2-methylpentane, \( \ce{(CH3)2CBrCH2CH2CH3} \): 3°, gives mainly 2-methylpent-2-ene
Shortcut: The tertiary halide D reacts fastest and must come last. Options (3) and (4) start with C, but only Option (3) ends with D.