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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 · 2020 · Paper II

Q2(c). Directed aldol reaction through an aza-enolate

  • 20 marks
  • Answered
Question

Complete the following reaction sequence by identifying the missing intermediates/products/reagents as the case may be. Explain the chemistry of each step by writing detailed mechanism involved. Also identify the name reaction involved:
[structure] + [structure] → A → (LDA, THF) B → (C₆H₅COCH₃) C → (H₂O) D

Answer in brief

A = CH₃CH=NᵗBu (imine); B = CH₂=CH-N(Li)ᵗBu (lithium aza-enolate); C = ᵗBuN=CHCH₂C(OLi)(CH₃)Ph; D = 3-hydroxy-3-phenylbutanal, which dehydrates to 3-phenylbut-2-enal in acid. The name reaction is the directed aldol reaction (Wittig's imine/aza-enolate method).

Key points

  • Introduction. A crossed aldol reaction between acetaldehyde (as the enol component) and acetophenone fails under ordinary base: acetaldehyde condenses with itself much faster. Converting the aldehyde first into an imine and…
  • Scheme: A = N-ethylidene-tert-butylamine (an aldimine), formed by addition–elimination.

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. A crossed aldol reaction between acetaldehyde (as the enol component) and acetophenone fails under ordinary base: acetaldehyde condenses with itself much faster. Converting the aldehyde first into an imine and then into its lithium aza-enolate makes a single, specific nucleophile. This is the directed aldol reaction (Wittig's aza-enolate method).

Step 1: imine A

Scheme: A = N-ethylidene-tert-butylamine (an aldimine), formed by addition–elimination.

The amine nitrogen adds to C=O to give a carbinolamine; proton transfer and loss of water (acid-catalysed, or with a drying agent) give the C=N bond. The primary amine is only a weak base, so the imine forms faster than the aldehyde can self-condense. The bulky tert-butyl group stops the imine reacting with itself.

Step 2: aza-enolate B

Scheme: LDA removes the α-H of the imine quantitatively; Li binds the nitrogen.

LDA is a strong, hindered, non-nucleophilic base. It removes a proton from the CH₃ next to C=N (the conjugate base is delocalised over C and N), and does not add to C=N. B is the lithium aza-enolate CH₂=CH-N(Li)ᵗBu, the nitrogen analogue of an enolate. All the imine is converted before the ketone is added, so no uncombined aldehyde remains.

Step 3: aldol addition to acetophenone, C

Scheme: C–C bond formation: the nucleophilic CH₂ of B attacks the ketone C=O while Li moves from N to O.

The ketone oxygen coordinates Li and the terminal carbon of the aza-enolate attacks the carbonyl carbon through a cyclic, six-membered (Zimmerman–Traxler type) transition state. The new C–C bond links the acetaldehyde α-carbon to the ketone carbon. C is ᵗBuN=CH-CH₂-C(OLi)(CH₃)Ph: the imine is regenerated and the lithium is now on the alkoxide oxygen. The imine of C is a weak electrophile, so it does not react further.

Step 4: hydrolysis, D

Scheme: Water protonates the alkoxide and hydrolyses the imine; acid and heat dehydrate the aldol to the conjugated enal.

Water protonates the alkoxide to the tertiary alcohol. Water then adds to the protonated imine (iminium) to give a carbinolamine that expels tert-butylamine (the reverse of step 1). D = 3-hydroxy-3-phenylbutanal, the crossed aldol. If the work-up is acidic and warm, the tertiary benzylic alcohol dehydrates (E1: the cation is tertiary and benzylic) to the conjugated (E)-3-phenylbut-2-enal.

Conclusion. A = CH₃CH=NᵗBu (imine); B = CH₂=CH-N(Li)ᵗBu (lithium aza-enolate); C = ᵗBuN=CHCH₂C(OLi)(CH₃)Ph; D = 3-hydroxy-3-phenylbutanal, which dehydrates to 3-phenylbut-2-enal in acid. The name reaction is the directed aldol reaction (Wittig's imine/aza-enolate method).

Application. Aza-enolates (and the related SAMP/RAMP hydrazone anions) let chemists use aldehydes as the enol partner in crossed aldol and alkylation reactions, which ordinary base-catalysed aldol conditions cannot do cleanly.

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Model answer to Q2(c), 2020 Paper II: solution page 8
Solution page 8 of 21

All parts of Question 2