See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the reaction type: BF3 is a Lewis acid (electrophile) because boron has an empty p-orbital and only 6 electrons around it, making it electron-deficient. Acetone is a Lewis base (nucleophile) because oxygen has lone pairs (non-bonding electron pairs) that can be donated. Step 2 - Answer A (Critical HOMO / nucleophile): The HOMO is the highest occupied molecular orbital that acts as the nucleophile. In this Lewis acid-base interaction, the nucleophile is acetone acting through its oxygen atom. The lone pairs (non-bonding electron pair orbitals) on the oxygen of acetone constitute the HOMO that donates electron density to BF3. Options (a) is wrong because boron in BF3 has no non-bonding orbital with electrons - it is electron-deficient. Option (b) sigma-orbital of acetone is not the donor orbital in this reaction. Option (c) pi-orbital of acetone could in principle act as a nucleophile, but the non-bonding lone pair on oxygen is higher in energy (better HOMO) and more accessible for donation to Lewis acids. Therefore, answer A = (d) non-bonding electron pair orbital on oxygen. Step 3 - Answer B (Critical LUMO / electrophile): The LUMO is the lowest unoccupied molecular orbital that acts as the electrophile. BF3 has an empty p-orbital on boron (perpendicular to the plane of the three B-F bonds) which is the LUMO that accepts the electron pair from oxygen. Option (b) sigma-orbital of BF3 is occupied. Option (c) pi* of acetone is unoccupied but it is not the primary electrophilic site in this reaction - BF3 is the electrophile. Option (d) non-bonding orbital on oxygen is occupied. Therefore, answer B = (a) p-orbital of BF3. Step 4 - Answer C (Correct product): When the lone pair on oxygen of acetone donates into the empty p-orbital of BF3, a coordinate (dative) bond forms between O and B. Oxygen donates two electrons to boron, so oxygen becomes positively charged (O+) and boron acquires a negative formal charge (BF3-). The acetone skeleton is preserved (CH3 groups remain, carbonyl carbon remains sp2). This gives the product: (CH3)2C=O(+)-BF3(-), which corresponds to option (a). Option (b) shows O with a negative charge, which is incorrect - oxygen donates electrons so it becomes positive. Option (c) shows a different carbon skeleton inconsistent with acetone. Option (d) shows BF3 bonded directly to carbon with negative oxygen, which would require breaking the C=O pi bond and is not the primary product of Lewis acid-base adduct formation. Therefore, answer C = (a). Therefore, the correct answer is {"A": "d", "B": "a", "C": "a"}.