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Preparation and reactions of alkenes
Concepts tested here
- Saytzeff and Markovnikov rules
All Questions
2011 AIPMT-PRE 1 question
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In the following reactions,
The major products (A) and (C) are respectively
(a) Acid-catalysed dehydration of 3-methylbutan-2-ol: protonation of –OH and loss of water gives a carbocation, which then loses a proton.
By Saytzeff's rule the more substituted alkene is the major product: A = $CH_3-C(CH_3)=CH-CH_3$ (2-methylbut-2-ene, trisubstituted).
The minor product is B = $CH_3-CH(CH_3)-CH=CH_2$ (3-methylbut-1-ene).
(b) In the dark and in the absence of peroxide, HBr adds to A by the ionic mechanism following Markovnikov's rule.
$H^+$ adds to the CH carbon so as to form the more stable tertiary carbocation $CH_3-\overset{+}{C}(CH_3)-CH_2-CH_3$; $Br^-$ then attacks it.
C = $CH_3-C(CH_3)(Br)-CH_2-CH_3$ (2-bromo-2-methylbutane, major); D = $CH_3-CH(CH_3)-CH(Br)-CH_3$ (minor).
So A is $CH_3-C(CH_3)=CH-CH_3$ and C is $CH_3-C(CH_3)(Br)-CH_2-CH_3$.