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Treatment of 2,4,6-tri-tert-butylphenol with bromine in cold acetic acid gives the compound C18H29BrO in quantitative yield. The infrared spectrum of this compound contains absorptions at 1630 and 1655 cm^-1. Its 1H NMR spectrum shows only three peaks (all singlets), at δ 1.2, 1.3, and 6.9, in the ratio 9:18:2. What is a reasonable structure for the compound?

User Egl
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Answer:

2,4,6-tri-tert-butylcyclohexa-2,5-dienone

Step-by-step explanation:

1. Information from the formulas

C₁₈H₃₀O ⟶ C₁₈H₂₉BrO

A Br has replaced an H.

2. Information from the reaction

These look like the conditions for an electrophilic aromatic substitution.

3. Possible mechanism (Fig. 1)

The product must be highly symmetrical, because there are so few NMR signals.

(i) The bromonium ion attacks at the para position, forming a resonance-stabilized carbocation intermediate.

(ii) A bromide ion attacks the H of the hydroxyl group to form 2,4,6-tri-tert-butylcyclohexa-2,5-dienone.

4. Confirmatory evidence (Fig. 2)

(a) Infrared

1630 cm⁻¹: C=O stretch

1655 cm⁻¹ : C=C stretch

(b)NMR

1.2 (9h, s): the 4-tert-butyl group

1.3 (18H, s): the 2- and 6- tert- butyl groups

6.9 (2H, s): the alkene H atoms

The ratio is 9:18:2.

Treatment of 2,4,6-tri-tert-butylphenol with bromine in cold acetic acid gives the-example-1
Treatment of 2,4,6-tri-tert-butylphenol with bromine in cold acetic acid gives the-example-2
User Toidv
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