Question

In the following sequence of reactions, the final product D is : 

 

• A. H\(_3\)C-CH=CH-CH(OH)-CH\(_2\)-CH\(_2\)-CH\(_3\)
• B. H\(_3\)C-CH\(_2\)-CH\(_2\)-CH\(_2\)-CH\(_2\)-C-H with O double bond
• C. CH\(_3\)-CH\(_2\)-CH\(_2\)-CH\(_2\)-CH\(_2\)-C-CH\(_3\) with O double bond
• D. CH\(_3\)-CH=CH-CH\(_2\)-CH\(_2\)-CH\(_2\)-COOH

Hint:

In complex multi-step synthesis questions, if the forward path seems illogical or doesn't match the options, try working backward from the options. Knowing that CrO\(_3\) oxidizes a secondary alcohol to a ketone is a key hint. This allows you to deduce the structure of the precursor C.

Answer 1

The Correct Option is C

Note: There appears to be a typo in the starting material shown in the question's image. The structure given is `CH₃-C≡C-H` (propyne), which has 3 carbons. The product options are all C7 compounds. A logical reaction sequence leading to the correct answer (Heptan-2-one) requires a C5 starting alkyne, such as pent-1-yne. The following solution assumes the starting material was intended to be pent-1-yne.
Step 1: Understanding the Question:
We need to follow a four-step reaction sequence to determine the final product D.
Step 2: Key Formula or Approach:
We will analyze each step of the reaction:
Deprotonation of a terminal alkyne.
Nucleophilic attack of the acetylide on an electrophile to form an alcohol.
Hydrogenation of the alkyne to an alkane.
Oxidation of a secondary alcohol to a ketone.
Step 3: Detailed Explanation (assuming starting material is pent-1-yne):
Reactant: Pent-1-yne, CH\(_3\)CH\(_2\)CH\(_2\)C≡CH.
Step 1: Formation of Acetylide (A)
The terminal alkyne reacts with the strong base sodium amide (NaNH\(_2\)) to form a sodium acetylide salt. \[ \text{CH}_3\text{CH}_2\text{CH}_2\text{C≡CH} + \text{NaNH}_2 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{C≡C}^-\text{Na}^+ \text{ (A)} + \text{NH}_3 \] Step 2: Formation of Alkyne-alcohol (B)
The problem is poorly depicted, but to arrive at a C7 product, the C5 acetylide (A) must react with a C2 electrophile. A logical choice that leads to a secondary alcohol is acetaldehyde (CH\(_3\)CHO), followed by an acidic workup. \[ \text{A} + \text{CH}_3\text{CHO} \xrightarrow{\text{1. ether}} \xrightarrow{\text{2. H}_3\text{O}^+} \text{CH}_3\text{CH}_2\text{CH}_2\text{C≡C-CH(OH)CH}_3 \text{ (B)} \] Product B is hept-3-yn-2-ol.
Step 3: Hydrogenation (C)
The alkyne (B) is treated with H\(_2\) and a Pd-C catalyst. This catalyst performs complete hydrogenation of the triple bond to a single bond. \[ \text{B} \xrightarrow{\text{H}_2/\text{Pd-C}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{-CH(OH)CH}_3 \text{ (C)} \] Product C is heptan-2-ol.
Step 4: Oxidation (D)
The secondary alcohol (C) is oxidized by chromium trioxide (CrO\(_3\)), a strong oxidizing agent (part of the Jones reagent). Secondary alcohols are oxidized to ketones. \[ \text{C} \xrightarrow{\text{CrO}_3} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{-C(=O)CH}_3 \text{ (D)} \] The final product D is heptan-2-one.
Step 4: Final Answer:
The final product D is heptan-2-one, which corresponds to option (C).