In the following reactions, (i) CH3- CH3 |CH- OH |CH-CH3 H+/heat (Major product)A + (Minor product)B (ii) A [in absence of peroxide]HBr, dark (Major product)C + (Minor product)D the major products (A) and (C) are respectively
- CH2= CH3 |C-CH2-CH3 and Br |CH2- CH3 |CH-CH2-CH3
- CH3- CH3 |C=CH-CH3 and CH3- Br | CH3 |C-CH2-CH3
- CH2= CH3 |C-CH2-CH3 and CH3- Br | CH3 |C-CH2-CH3
- CH3- CH3 |C=CH-CH3 and CH3- CH3 |CH- Br |CH-CH3
Answer
(2) CH3- CH3 |C=CH-CH3 and CH3- Br | CH3 |C-CH2-CH3
Solution
(i) Acid-catalysed dehydration of 3-methylbutan-2-ol:
Protonation and loss of H2O give the 2° carbocation (CH3)2CH- +CH-CH3.
1,2-Hydride shift gives the more stable 3° carbocation (CH3)2 +C-CH2-CH3.
Loss of H+ gives the more substituted alkene (Saytzeff rule) as the major product:
A = (CH3)2C=CH-CH3 (2-methylbut-2-ene)
(ii) Addition of HBr in the absence of peroxide follows Markovnikov's rule: H goes to the carbon with more H atoms and Br to the more substituted carbon.
C = (CH3)2CBr-CH2-CH3 (2-bromo-2-methylbutane)
Answer: (2)