When neopentyl alcohol is heated with an acid, it slowly converted into an 85 : 15 mixture of alkenes A and B, respectively. What are these alkenes?

Solution

Chemistry ma’am · 3 steps

Neopentyl cannot eliminate as it is, so a methyl shifts first and the tertiary cation decides the alkenes.

1. Stuck as it is. The acid protonates the OH of (CH₃)₃C-CH₂OH, and water starts to leave. The carbon next to CH₂ is quaternary and has no hydrogen, so a plain elimination is impossible. A primary carbocation would also be very unstable.
2. Methyl shift. As water leaves, a methyl group moves from the quaternary carbon to the CH₂ carbon. This gives the much more stable tertiary cation (CH₃)₂C⁺-CH₂CH₃.

   H lost from	Alkene formed	C=C substitution	Share
   the CH₂ group	2-methylbut-2-ene, (CH₃)₂C=CHCH₃	trisubstituted	85 (A)
   a CH₃ on the cationic carbon	2-methylbut-1-ene, CH₂=C(CH₃)CH₂CH₃	disubstituted	15 (B)

3. So. By Saytzeff's rule the more substituted 2-methylbut-2-ene is the major alkene A and 2-methylbut-1-ene is the minor alkene B. That is option (a).

The idea to keep. If the carbon next to the leaving group is quaternary, expect a 1,2-methyl shift to a tertiary cation. Then list that cation's beta hydrogens: the more substituted alkene is the major one.

Answer from the NTA final answer key. Solution by PrepWiser.