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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper II covers organic chemistry and spectroscopy.

UPSC Civil Services (Mains), Chemistry optional · 2019 · Paper II

Q2(c). Dehydration of 3,3-dimethylbutan-2-ol: methyl shift and Saytzeff alkene

  • 20 marks
  • Answered
Question

Write different olefins expected to be formed in the following dehydration reaction:
(CH₃)₃CCHOHCH₃ →
Provide a mechanism for their formation. Which product is expected to be most stable?

Answer in brief

Three olefins form: 2,3-dimethylbut-2-ene (major, via the 1,2-methyl shift to the tertiary cation), 2,3-dimethylbut-1-ene (minor, same cation) and 3,3-dimethylbut-1-ene (small, unrearranged). The tetrasubstituted 2,3-dimethylbut-2-ene is the most stable (Saytzeff product, maximum hyperconjugation).

Key points

  • Water or HSO₄⁻ removes a β-H from the cation; the C–H electrons form the π bond (E1).
  • From the tertiary cation two β-positions are available: the CH of the isopropyl group (gives the Saytzeff tetrasubstituted alkene) and a CH₃ (gives the less substituted alkene). The more stable alkene has the more stable,…
  • The unrearranged 3,3-dimethylbut-1-ene forms only from the short-lived secondary cation and is a minor product. All steps are reversible in hot acid, which further enriches the most stable alkene.

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

Introduction. 3,3-Dimethylbutan-2-ol (pinacolyl alcohol) is a secondary alcohol next to a quaternary carbon. In hot H₂SO₄ it dehydrates by E1: the protonated alcohol loses water to give a secondary carbocation, which can either lose a proton directly or first rearrange by a 1,2-methyl shift to a more stable tertiary cation.

Step 1: protonation and ionisation (rate-determining)

Scheme: -OH is converted into the good leaving group -OH₂⁺.

Step 2: 1,2-methyl shift

Scheme: A methyl moves with its electron pair; the charge moves to the more substituted carbon.

The secondary cation has no H on the quaternary neighbour, so the only direct elimination is from the CH₃ group. A much faster process is the migration of a methyl from the tert-butyl carbon to the cationic carbon (Wagner–Meerwein shift). The resulting cation is tertiary and is stabilised by hyperconjugation with more C–H bonds, so the rearrangement is downhill and most of the product comes from it.

Step 3: loss of a proton: three olefins

StructureOlefinFromAmount
-0.4q2c_A2,3-dimethylbut-2-ene (tetrasubstituted)3° cation, H from C-3 (CH)major 2pt] -0.4q2c_B2,3-dimethylbut-1-ene (1,1-disubstituted)3° cation, H from CH₃minor 2pt] -0.4q2c_C3,3-dimethylbut-1-ene (monosubstituted)2° cation, no shiftsmall
  1. Water or HSO₄⁻ removes a β-H from the cation; the C–H electrons form the π bond (E1).
  2. From the tertiary cation two β-positions are available: the CH of the isopropyl group (gives the Saytzeff tetrasubstituted alkene) and a CH₃ (gives the less substituted alkene). The more stable alkene has the more stable, alkene-like transition state, so the tetrasubstituted alkene dominates.
  3. The unrearranged 3,3-dimethylbut-1-ene forms only from the short-lived secondary cation and is a minor product. All steps are reversible in hot acid, which further enriches the most stable alkene.

Which olefin is most stable?

Alkene stability rises with the number of alkyl groups on the C=C, through hyperconjugation of α-C–H bonds with the π bond and the stronger sp²–sp³ C–C bonds. 2,3-Dimethylbut-2-ene has four methyls on the double bond (12 α-H) and the lowest heat of hydrogenation of the three isomers; 3,3-dimethylbut-1-ene has no α-H on the quaternary side and is the least stable.

Figure: Relative enthalpy of the three C₆H₁₂ isomers (schematic).

Conclusion. Three olefins form: 2,3-dimethylbut-2-ene (major, via the 1,2-methyl shift to the tertiary cation), 2,3-dimethylbut-1-ene (minor, same cation) and 3,3-dimethylbut-1-ene (small, unrearranged). The tetrasubstituted 2,3-dimethylbut-2-ene is the most stable (Saytzeff product, maximum hyperconjugation).

Application. The rearrangement proves the carbocation pathway of acid-catalysed dehydration (a primary or secondary alcohol next to a quaternary centre always gives rearranged alkenes); 2,3-dimethylbut-2-ene from this reaction is a common tetrasubstituted alkene in ozonolysis and epoxidation studies.

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Model answer to Q2(c), 2019 Paper II: solution page 10
Solution page 10 of 23

All parts of Question 2