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INChO — Indian National Chemistry Olympiad — Answer Writing

INChO asks for the reasoning that multiple choice hides: the arrow-pushing, the intermediate, the equilibrium worked to a number. The product alone is rarely the answer.

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION
INDIAN NATIONAL CHEMISTRY OLYMPIAD
Time: 3 hoursMaximum Marks: 60

Answer all questions. Show mechanisms with curved arrows where asked. Use of a non-programmable calculator is permitted.

  1. 1.Predict the major product when 2-bromo-2-methylbutane is treated with sodium ethoxide in ethanol, and explain the mechanism and regiochemistry.10
  2. 2.Calculate the pH of the buffer described above and the change on adding 0.01 mol of strong acid.10
  3. 3.Explain the splitting of d-orbitals in an octahedral field and predict the magnetic behaviour of the given complex.12
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Organic Chemistry10 marks

Predict the major product when 2-bromo-2-methylbutane is treated with sodium ethoxide in ethanol, and explain the mechanism and regiochemistry.

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The product is 2-methyl-2-butene. Sodium ethoxide is a strong base so elimination takes place by E2 mechanism. The more substituted alkene is formed as per Zaitsev rule. 2-bromo-2-methylbutane is a tertiary halide and the carbon carrying the bromine has three alkyl groups attached to it. Ethoxide can remove a proton from either of two different beta carbons, so two alkenes are possible. Removal from the methyl group would give 2-methyl-1-butene, the less substituted alkene, while removal from the CH2 group gives 2-methyl-2-butene. The more substituted alkene is more stable because of hyperconjugation and the greater number of alkyl groups on the double bond, and the transition state leading to it is lower in energy. Therefore 2-methyl-2-butene is the major product and 2-methyl-1-butene is formed only in a small amount.
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Q. No. 1Organic ChemistryMax. marks 10
Q1

The product is 2-methyl-2-butene. Sodium ethoxide is a strong base so elimination takes place by E2 mechanism. The more substituted alkene is formed as per Zaitsev rule.? 2-bromo-2-methylbutane is a tertiary halide and the carbon carrying the bromine has three alkyl groups attached to it. Ethoxide can remove a proton from either of two different beta carbons, so two alkenes are possible. Removal from the methyl group would give 2-methyl-1-butene, the less substituted alkene, while removal from the CH2 group gives 2-methyl-2-butene. The more substituted alkene is more stable because of hyperconjugation and the greater number of alkyl groups on the double bond, and the transition state leading to it is lower in energy. Therefore 2-methyl-2-butene is the major product and 2-methyl-1-butene is formed only in a small amount.

The concerted anti-periplanar transition state, described step by stepWhy E2 rather than E1 or SN2 on this substrate with this base
5/10
Page 1Marks for each expected point are written down the right margin
The pen marks on your page
StrongCorrect major product.
VagueRight rule, but a mechanism question wants why: hyperconjugative stabilisation of the more substituted alkene in the transition state.
What the examiner wanted and did not find
  • The concerted anti-periplanar transition state, described step by step
  • Why E2 rather than E1 or SN2 on this substrate with this base
  • The Hofmann contrast with a bulky base such as potassium tert-butoxide
Model answer for this question

Identify the substrate as tertiary and ethoxide as a strong, small base, so E2 dominates over E1 and SN2 is blocked sterically. Describe the concerted step: base removes a beta hydrogen anti-periplanar to the leaving bromide, with C-H bond breaking, pi bond forming and C-Br breaking in one transition state. Compare the two available beta positions and explain that the more substituted alkene is favoured by hyperconjugative stabilisation in the transition state, giving 2-methyl-2-butene. Add the contrast: a bulky base gives the Hofmann product.

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Evaluation reportIllustrative
Step marking
  • Major product identified2/2

    Correct.

  • Mechanism identified and justified2/3

    E2 named correctly; the tertiary substrate and strong base as the reasons are not stated.

  • Mechanism drawn — anti-periplanar arrangement and concerted arrows0/3

    Not drawn or described. The arrows are the mechanism.

  • Regiochemistry explained rather than named1/2

    Zaitsev named without the stabilisation argument, and the Hofmann alternative with a bulky base is not contrasted.

Total 5/10AverageAI estimate — not an official score

INChO at a glance

Conducting bodyHomi Bhabha Centre for Science Education (HBCSE)
ModeHandwritten solutions
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What INChO sets

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1Physical Chemistry
2Organic Chemistry
3Inorganic Chemistry
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