See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: A symmetrical ketone is reduced by LiAlH4 to give a secondary alcohol (B). This alcohol (B), upon treatment with H+ and heat (acid-catalyzed dehydration), gives diastereomers. Step 1 - Identify the symmetrical ketone requirement: A symmetrical ketone has identical groups on both sides of the carbonyl. - Option (a): acetone (CH3COCH3) is symmetrical but gives 2-propanol on reduction, which has no chiral center and cannot give diastereomers. - Option (b): methyl ethyl ketone (CH3COCH2CH3) is NOT symmetrical. - Option (c): diethyl ketone (CH3CH2COCH2CH3, pentan-3-one) IS symmetrical - both sides have ethyl groups. - Option (d): butanal is an aldehyde, not a ketone. Step 2 - Reduction of option (c) with LiAlH4: CH3CH2-C(=O)-CH2CH3 + LiAlH4 -> CH3CH2-CH(OH)-CH2CH3 = pentan-3-ol (B). This is a secondary alcohol with no chiral center (the carbon bearing OH has two identical ethyl groups), so it is achiral. Step 3 - Dehydration of pentan-3-ol with H+/heat: Loss of water gives pent-2-ene (CH3CH=CHCH2CH3). Pent-2-ene has two geometric isomers: cis-pent-2-ene and trans-pent-2-ene. These two alkenes are diastereomers (cis/trans isomers are a type of diastereomer since they are stereoisomers that are not enantiomers). Step 4 - Why other options fail: - (a) Acetone gives 2-propanol, dehydration gives propene which has no stereoisomers. - (b) Not symmetrical, eliminated. - (d) Not a ketone, it is an aldehyde. Thus, the symmetrical ketone (A) = diethyl ketone (pentan-3-one), which upon LiAlH4 reduction gives pentan-3-ol (B), which upon acid-catalyzed dehydration gives cis and trans pent-2-ene (diastereomers). Therefore, the correct answer is C.