See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Concept: A carbonyl compound reacts with PhMgBr (a Grignard reagent) to give a racemic mixture when the reaction produces a chiral center at which two different groups are present in equal amounts from both faces of attack. This occurs when the carbonyl carbon, upon addition of Ph group, generates a new stereocenter that is not already chiral (or the existing chirality does not bias one face), leading to equal R and S products. Step 1: Identify all carbonyl isomers of C5H10O. The molecular formula C5H10O with degree of unsaturation = (2×5+2-10)/2 = 1, consistent with one C=O group (aldehyde or ketone). No rings or additional degrees of unsaturation. Step 2: List aldehydes (C5H10O, RCHO where R = C4H9): 1. Pentanal: CH3CH2CH2CH2CHO 2. 2-Methylbutanal: CH3CH2CH(CH3)CHO 3. 3-Methylbutanal: (CH3)2CHCH2CHO 4. 2,2-Dimethylpropanal (pivaldehyde): (CH3)3CCHO Step 3: List ketones (C5H10O): 5. Pentan-2-one: CH3COCH2CH2CH3 6. Pentan-3-one: CH3CH2COCH2CH3 7. 3-Methylbutan-2-one: (CH3)2CHCOCH3 8. Cyclopentanone (cyclic ketone, C5H8O - wait, MW check: C5H8O has DoU=2, not 1, so cyclopentanone is C5H8O, not C5H10O) Step 4: Determine which react with PhMgBr to give a racemic mixture. Addition of PhMgBr to a carbonyl gives an alcohol. A racemic mixture results if the product alcohol has a chiral center and the two faces of the carbonyl are equally accessible (no existing chirality biases the attack). - Pentanal: PhMgBr adds to give 1-phenyl-1-pentanol. The C1 bears OH, Ph, H, and n-Bu — chiral center, no pre-existing chirality in starting material, gives racemic mixture. YES. - 2-Methylbutanal: Product is PhCH(OH)CH(CH3)CH2CH3. Two stereocenters formed (C1 new, C2 pre-existing in product from prochiral aldehyde carbon). The aldehyde carbon becomes chiral; C2 is also a stereocenter. Since C2 already has defined configuration in the starting aldehyde (2-methylbutanal is chiral itself if racemic, but as a pure compound it has one stereocenter), the two faces of attack are diastereotopic, giving diastereomers not a simple racemic mixture. NOT a simple racemic mixture. - 3-Methylbutanal: PhMgBr adds to CHO to give PhCH(OH)CH2CH(CH3)2. Product has one chiral center (C1), no other stereocenters. Gives racemic mixture. YES. - 2,2-Dimethylpropanal (pivaldehyde): Product is PhCH(OH)C(CH3)3. One chiral center at C1. Gives racemic mixture. YES. - Pentan-2-one: PhMgBr adds to give Ph-C(OH)(CH3)(CH2CH2CH3). One chiral center. Gives racemic mixture. YES. - Pentan-3-one: PhMgBr adds to give Ph-C(OH)(CH2CH3)(CH2CH3). The two ethyl groups are identical, so no chiral center formed. Does NOT give racemic mixture (achiral product). - 3-Methylbutan-2-one: PhMgBr adds to give Ph-C(OH)(CH3)(CH(CH3)2). One chiral center at the former carbonyl carbon. Gives racemic mixture. YES. Step 5: Count the carbonyl isomers giving racemic mixture: 1. Pentanal — YES 2. 3-Methylbutanal — YES 3. 2,2-Dimethylpropanal — YES 4. Pentan-2-one — YES 5. 3-Methylbutan-2-one — YES 2-Methylbutanal gives diastereomers (not purely racemic), and Pentan-3-one gives an achiral product. Therefore, the correct answer is 5.