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Alcohols Phenols and Ethers question

2022 · 24 Jun · Shift 2 · Q12
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Alcohols Phenols and Ethers question

2022 · 24 Jun · Shift 2 · Q12

JEE MainChemistryAlcohols Phenols and EthersMCQ+4 / −1
Hex-4-ene-2-ol on treatment with PCC gives 'A'. 'A' on reaction with sodium hypoiodite gives 'B', which on further heating with soda lime gives 'C'. The compound 'C' is :
  1. A
    2-pentene
  2. B
    proponaldehyde
  3. C
    2-butene
  4. D
    4-methylpent-2-ene
View written solutionFree

Correct answer: C

  1. Oxidation of hex-4-ene-2-ol with PCC

Given alcohol: hex-4-ene-2-ol

Its structure is:

CH3−CH(OH)−CH2−CH=CH−CH3\text{CH}_3-\text{CH(OH)}-\text{CH}_2-\text{CH}=\text{CH}-\text{CH}_3CH3​−CH(OH)−CH2​−CH=CH−CH3​

PCC oxidizes a secondary alcohol to a ketone without affecting the double bond.

So compound AAA is:

CH3−CO−CH2−CH=CH−CH3\text{CH}_3-\text{CO}-\text{CH}_2-\text{CH}=\text{CH}-\text{CH}_3CH3​−CO−CH2​−CH=CH−CH3​

This is hex-4-en-2-one.

  1. Reaction of AAA with sodium hypoiodite

Sodium hypoiodite gives the iodoform reaction with methyl ketones.

A methyl ketone has the form:

R−CO−CH3\text{R}-\text{CO}-\text{CH}_3R−CO−CH3​

Here,

CH3−CO−CH2−CH=CH−CH3\text{CH}_3-\text{CO}-\text{CH}_2-\text{CH}=\text{CH}-\text{CH}_3CH3​−CO−CH2​−CH=CH−CH3​

can be written as:

CH3−CO−R\text{CH}_3-\text{CO}-RCH3​−CO−R quad \text{where} \quad R=\text{CH}_2-\text{CH}=\text{CH}-\text{CH}_3

In the iodoform reaction: $$\text{CH}_3\text{CO}R \xrightarrow[\text{NaOH}]{\text{I}_2} R\text{COO}^-\text{Na}^+ + \text{CHI}_3$$ Therefore, compound $B$ is: $$\text{NaOOC}-\text{CH}_2-\text{CH}=\text{CH}-\text{CH}_3$$ This is **sodium pent-2-enoate** if numbered from the carboxyl carbon: $$\text{^-OOC-CH}_2-\text{CH}=\text{CH-CH}_3$$ --- 3. **Heating $B$ with soda lime** Soda lime causes **decarboxylation** of sodium carboxylates: $$\text{RCOONa} \xrightarrow[\Delta]{\text{soda lime}} \text{RH} + \text{Na}_2\text{CO}_3$$ For $$\text{^-OOC-CH}_2-\text{CH}=\text{CH-CH}_3$$ removing the carboxyl group gives: $$\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3$$ This is **but-2-ene (2-butene)**. --- 4. **Checking options** - **A: 2-pentene** → incorrect - **B: propionaldehyde** → incorrect - **C: 2-butene** → correct - **D: 4-methylpent-2-ene** → incorrect --- 5. **Final answer** $$\boxed{\text{2-butene}}$$ So, the correct option is **C**.
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