What are the major products formed in the following reaction sequence?
Show Hint
Ozonolysis converts alkenes into carbonyls.
Grignard addition to an aldehyde ($\text{RCHO}$) yields a secondary alcohol ($\text{R}_2\text{CHOH}$), while addition to a ketone ($\text{R}_2\text{CO}$) yields a tertiary alcohol ($\text{R}_3\text{COH}$).
This functional group behavior allows you to quickly narrow down the alcohol types.
4-hydroxy-4-phenylbutan-2-one and 2-phenylbutan-2-ol
Triphenylmethanol and 2-phenylbutan-2-ol
Benzaldehyde and 2-phenylbutan-2-ol
Show Solution
The Correct Option isA
Solution and Explanation
Step 1 : Understanding the Question:
The question asks for the major products formed when the given alkene, (2-methylbut-1-en-1-yl)benzene ($\text{Ph-CH=C(CH}_3\text{)-CH}_2\text{CH}_3$), undergoes reductive ozonolysis followed by reaction with excess phenylmagnesium bromide ($\text{PhMgBr}$) and subsequent hydrolysis. Step 2 : Key Formulas and Approach:
The approach involves a two-stage organic synthesis:
Reductive ozonolysis ($\text{O}_3$, followed by $\text{Zn/H}_2\text{O}$) of the alkene to cleave the double bond and yield two carbonyl compounds.
Nucleophilic addition of the Grignard reagent ($\text{PhMgBr}$) to both carbonyl compounds, followed by protonation ($\text{H}_3\text{O}^+$) to yield the corresponding alcohols. Step 3 : Detailed Explanation:
Let us carry out the chemical steps:
Stage 1: Ozonolysis
The starting alkene is:
\[ \text{Ph}-\text{CH}=\text{C}(\text{CH}_3)-\text{CH}_2\text{CH}_3 \]
Ozonolysis cleaves the double bond to produce two carbonyl products:
\[ \text{Ph}-\text{CH}=\text{O} \quad (\text{Benzaldehyde, an aldehyde}) \]
\[ \text{O}=\text{C}(\text{CH}_3)-\text{CH}_2\text{CH}_3 \quad (\text{Butanone, a ketone}) \]
Stage 2: Grignard Reaction
We treat the mixture with an excess of phenylmagnesium bromide (at least 2 equivalents):
Reaction with Benzaldehyde:
The nucleophilic phenyl group attacks the aldehyde carbon:
\[ \text{Ph}-\text{CHO} + \text{PhMgBr} \rightarrow \text{Ph}_2\text{CH}-\text{OMgBr} \xrightarrow{\text{H}_3\text{O}^+} \text{Ph}_2\text{CH}-\text{OH} \]
The product is diphenylmethanol (also known as benzhydrol).
Reaction with Butanone:
The nucleophilic phenyl group attacks the ketone carbonyl carbon:
\[ \text{CH}_3-\text{CO}-\text{CH}_2\text{CH}_3 + \text{PhMgBr} \rightarrow \text{CH}_3-\text{C}(\text{OMgBr})(\text{Ph})-\text{CH}_2\text{CH}_3 \]
\[ \xrightarrow{\text{H}_3\text{O}^+} \text{CH}_3-\text{C}(\text{OH})(\text{Ph})-\text{CH}_2\text{CH}_3 \]
The product is 2-phenylbutan-2-ol.
Step 4 : Final Answer:
The major products formed are diphenylmethanol (shown as $\text{Ph-CH(OH)-Ph}$) and 2-phenylbutan-2-ol (shown as $\text{H}_3\text{C-C(OH)(Ph)-CH}_2\text{CH}_3$).
This corresponds to Option (A).
