What does treatment of an alkene with O3 followed by Zn/H2O (reductive ozonolysis) accomplish?
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Correct answer
Option analysis
Believing that ozonolysis with Zn/H2O oxidizes aldehydes all the way to carboxylic acids as in oxidative workup. Recognize that Zn/H2O is a reductive workup that stops alkene cleavage products at aldehydes or ketones, not carboxylic acids.
Correct option: reductive ozonolysis cleaves the double bond to yield the dialdehyde/keto-aldehyde intermediate, which under the given conditions converts to the cyclopentene carbaldehyde derivative via intramolecular aldol condensation. Identify the carbon skeleton formed by oxidative cleavage of the double bond and subsequent intramolecular aldol cyclization.
Assuming the reagent reduces the carbonyl group to an alcohol rather than carrying out cleavage and condensation. Remember that O3 followed by Zn/H2O cleaves C=C bonds selectively to carbonyls; it is not a hydride-reducing agent like NaBH4.
Stopping at the open-chain dicarbonyl intermediate 'A' rather than continuing to the final product 'B'. Check whether the question asks for intermediate 'A' or final product 'B', and carry out the intramolecular reaction to obtain 'B'.
The reaction sequence begins with methyl benzoate reacting with excess phenylmagnesium bromide (Grignard reagent). Aqueous workup then gives compound A (a tertiary alcohol, triphenylmethanol). Compound A undergoes acid-catalyzed dehydration using concentrated sulfuric acid with heat to yield alkene B.
Step 1: Phenylmagnesium bromide attacks the ester carbonyl of methyl benzoate (C6H5COOCH3). Elimination of the methoxide leaving group yields benzophenone (C6H5COC6H5). A second equivalent of PhMgBr attacks benzophenone to form a magnesium alkoxide, which upon protonation with H3O+ gives triphenylmethanol (Ph3C-OH) as compound A.
Step 2: Triphenylmethanol has three phenyl groups on the central carbon. It has no beta-hydrogens directly on an aliphatic chain. In concentrated with heat, protonation of the group occurs. Subsequent loss of water generates the stable triphenylmethyl carbocation (trityl carbocation). To eliminate to an alkene, an intramolecular electrophilic aromatic substitution occurs. This Friedel-Crafts-type cyclization or elimination yields 9-phenylfluorene or related structures. Based on the given options, methyl benzoate reacts with to give triphenylmethanol. Dehydration then yields the corresponding stable alkene framework.
Compound B corresponds to Option B [Option 2 in 1-based index], which is 1,1-diphenyl-2-phenylethene (or the matching conjugated alkene structure shown in figure 2).
Two equivalents of the Grignard reagent add to the ester. This generates a tertiary carbinol with three aryl groups. Dehydration under strong acid with heat leads directly to the product shown in option B.
What does treatment of an alkene with O3 followed by Zn/H2O (reductive ozonolysis) accomplish?
It cleaves the carbon-carbon double bond, converting each alkene carbon into a carbonyl group (aldehyde or ketone).What is the structure of intermediate 'A' after cleaving the ring double bond?
An open-chain dicarbonyl intermediate containing aldehyde/ketone functional groups.Which species corresponds to compound 'B' formed from intermediate 'A'?
The cyclized cyclopentene carbaldehyde product resulting from intramolecular condensation of intermediate 'A'.Quick checks
Because the ketone formed after the first addition is more electrophilic than the starting ester, so the second addition is faster than the first.
