When a multi-step synthesis question leads to an answer not present in the options, re-read the question carefully for possible misinterpretations. If none are found, consider a likely typo in one of the reagents or starting materials, and see if a small change leads to one of the given answers. Ozonolysis is a great tool for this, as the products directly reveal the structure of the parent alkene.
Step 1: Understanding the Concept:
This is a multi-step organic synthesis problem. It involves a Grignard reaction, followed by dehydration of an alcohol to form an alkene, and finally ozonolysis of the alkene. Step 2: Detailed Reaction Sequence:
- Step 1: Formation of A
The starting material is ethanal (CH₃CHO). It reacts with a Grignard reagent, methylmagnesium bromide (CH₃MgBr), followed by acidic hydrolysis (H₃O⁺). This is a standard method for synthesizing a secondary alcohol. The nucleophilic methyl group from the Grignard reagent attacks the electrophilic carbonyl carbon of ethanal.
\[ \text{CH}_3-\text{CHO} + \text{CH}_3\text{MgBr} \rightarrow \text{CH}_3-\text{CH}(\text{OMgBr})-\text{CH}_3 \xrightarrow{\text{H}_3\text{O}^+} \text{CH}_3-\text{CH(OH)}-\text{CH}_3 \]
So, compound A is propan-2-ol.
- Step 2: Formation of B
Propan-2-ol (A) is heated with 20% H₃PO₄ at 358 K. This is an acid-catalyzed dehydration of an alcohol. An E1 or E2 elimination reaction occurs, removing a water molecule to form an alkene.
\[ \text{CH}_3-\text{CH(OH)}-\text{CH}_3 \xrightarrow{20%\text{H}_3\text{PO}_4, 358\text{K}} \text{CH}_2=\text{CH}-\text{CH}_3 + \text{H}_2\text{O} \]
So, compound B is propene.
- Step 3: Formation of C and D
Propene (B) undergoes reductive ozonolysis. First, ozone (O₃) is added across the double bond to form an ozonide. This ozonide is then cleaved by a reducing agent (Zn/H₂O). Reductive ozonolysis cleaves the double bond and forms two carbonyl compounds.
\[ \text{CH}_2=\text{CH}-\text{CH}_3 \xrightarrow{\text{(i) O}_3} \text{Ozonide} \xrightarrow{\text{(ii) Zn/H}_2\text{O}} \text{HCHO} + \text{CH}_3\text{CHO} \]
Wait, let me retrace the steps.
Step 1: Ethanal (2C) + CH₃MgBr (1C) → 3C secondary alcohol. Propan-2-ol. Correct.
Step 2: Dehydration of propan-2-ol gives propene. Correct.
Step 3: Ozonolysis of propene (\(\text{CH}_3-\text{CH}=\text{CH}_2\)). The double bond breaks. The CH₂ part becomes formaldehyde (HCHO), and the CH₃-CH part becomes ethanal (CH₃CHO).
So, the products are formaldehyde and ethanal.
This does not match any of the options. Let me re-read the starting material in the image.
Ah, the starting material in the image is not ethanal. It is Propanone (acetone), (CH₃)₂C=O.
Let's restart the synthesis with propanone. Revised Reaction Sequence:
- Step 1: Formation of A (with Propanone)
Propanone (a ketone) reacts with CH₃MgBr followed by hydrolysis. This forms a tertiary alcohol.
\[ (\text{CH}_3)_2\text{C=O} + \text{CH}_3\text{MgBr} \rightarrow (\text{CH}_3)_3\text{C-OMgBr} \xrightarrow{\text{H}_3\text{O}^+} (\text{CH}_3)_3\text{C-OH} \]
So, compound A is 2-methylpropan-2-ol (tert-butyl alcohol).
- Step 2: Formation of B
Dehydration of 2-methylpropan-2-ol (a tertiary alcohol) occurs readily to form an alkene.
\[ (\text{CH}_3)_3\text{C-OH} \xrightarrow{20%\text{H}_3\text{PO}_4, 358\text{K}} (\text{CH}_3)_2\text{C=CH}_2 + \text{H}_2\text{O} \]
So, compound B is 2-methylpropene.
- Step 3: Formation of C and D
Reductive ozonolysis of 2-methylpropene.
\[ (\text{CH}_3)_2\text{C=CH}_2 \xrightarrow{\text{(i) O}_3, \text{(ii) Zn/H}_2\text{O}} (\text{CH}_3)_2\text{C=O} + \text{HCHO} \]
The double bond breaks. The \((\text{CH}_3)_2\text{C=}\) part becomes propanone (acetone). The \(=\text{CH}_2\) part becomes methanal (formaldehyde).
So, the products C and D are propanone and methanal.
Again, this does not match the options. Let me check the image and my interpretation one more time.
The starting material is a 3-carbon ketone, Propanone. This is correct.
Step 1 gives 2-methylpropan-2-ol. This is correct.
Step 2 gives 2-methylpropene. This is correct.
Step 3 gives Propanone and Methanal. This is correct.
The options are: (A) Ethanoic acid, ethanal, (B) Ethanol, Propanone, (C) Ethanal, Propanone, (D) Propanal, Propanone.
There seems to be a significant error in the question, as the logical products (Propanone and Methanal) are not listed in any option.
Let's reconsider the starting material. What if it was propanal (CH₃CH₂CHO)?
Step 1: Propanal + CH₃MgBr \(\rightarrow\) Butan-2-ol.
Step 2: Dehydration of butan-2-ol gives a mixture of But-1-ene and But-2-ene (major, by Saytzeff's rule). Let's assume major product But-2-ene.
Step 3: Ozonolysis of But-2-ene (\(\text{CH}_3\text{-CH=CH-CH}_3\)) gives two molecules of ethanal (CH₃CHO).
This also does not match the options.
Let's go back to the original starting material in the image: Propanone. What if the Grignard reagent was different? The image clearly shows CH₃MgBr.
What if the dehydration step gives a rearrangement? No, not for a tertiary carbocation.
Let's re-examine the correct answer, Option (C): Ethanal and Propanone.
To get Ethanal and Propanone from ozonolysis, the alkene (B) must have the structure: \((\text{CH}_3)_2\text{C=CH-CH}_3\) (2-methylbut-2-ene).
How can we get 2-methylbut-2-ene from step 1 and 2?
To get 2-methylbut-2-ene, we need to dehydrate 2-methylbutan-2-ol.
To get 2-methylbutan-2-ol, the Grignard reaction must be between Propanone and Ethylmagnesium bromide (CH₃CH₂MgBr), or between Butan-2-one and CH₃MgBr. The question clearly specifies CH₃MgBr.
So, there is a clear inconsistency in the question. The reaction as written leads to (Propanone + Methanal). The answer provided (Ethanal + Propanone) implies the alkene was 2-methylbut-2-ene.
Let's assume there's a typo in the Grignard reagent and it should have been CH₃CH₂MgBr (ethylmagnesium bromide).
\[ (\text{CH}_3)_2\text{C=O} + \text{CH}_3\text{CH}_2\text{MgBr} \xrightarrow{\text{H}_3\text{O}^+} (\text{CH}_3)_2\text{C(OH)CH}_2\text{CH}_3 \text{ (2-methylbutan-2-ol)} \]
Dehydration of this alcohol gives 2-methylbut-2-ene as the major product.
\[ (\text{CH}_3)_2\text{C=CHCH}_3 \xrightarrow{\text{Ozonolysis}} (\text{CH}_3)_2\text{C=O} \text{ (Propanone)} + \text{CH}_3\text{CHO} \text{ (Ethanal)} \]
This set of products matches option (C). This is the most likely intended question, with a typo in the Grignard reagent.
Step 3: Final Answer (assuming typo in reagent):
Assuming the Grignard reagent was intended to be ethylmagnesium bromide (CH₃CH₂MgBr) instead of methylmagnesium bromide:
- Step 1: Propanone + CH₃CH₂MgBr \(\rightarrow\) 2-methylbutan-2-ol.
- Step 2: Dehydration of 2-methylbutan-2-ol \(\rightarrow\) 2-methylbut-2-ene.
- Step 3: Reductive ozonolysis of 2-methylbut-2-ene cleaves the double bond to give propanone and ethanal.
- Thus, the products C and D are Ethanal and Propanone. This corresponds to option (C).
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