See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: LiAlH4 reduces ketones by delivering a hydride (H-) to the carbonyl carbon. Each mole of LiAlH4 provides 4 hydride equivalents (since AlH4- has 4 H atoms). The overall stoichiometry must be worked out based on how many moles of ketone are reduced and how many hydrides are consumed. Step 1: Determine hydrides needed. Each ketone (cyclobutanone) requires 1 hydride for reduction to an alkoxide. With 4 moles of cyclobutanone, we need 4 hydrides total. Step 2: Determine moles of LiAlH4 needed. Each mole of LiAlH4 provides 4 hydride equivalents. Therefore, 4 ketones require 4 H- ions, which comes from 4/4 = 1 mole of LiAlH4. Step 3: Verify. 1 LiAlH4 provides 4 H-, enough to reduce 4 molecules of cyclobutanone. After aqueous workup (H3O+), the alkoxide intermediates are protonated to give 4 molecules of cyclobutanol. The Al and Li are released as salts. Step 4: Why other options fail: - (b) x=2: Would provide 8 H-, far more than needed for 4 ketones. - (c) x=3: Would provide 12 H-, excessive. - (d) x=4: Would provide 16 H-, excessive. Only x=1 provides exactly 4 hydrides for 4 ketone molecules. Therefore, the correct answer is A.