AIPMT – 2009 – NEETmcqs

Phenol in a set of reactions (as captured below) yielded a product Z.

$\text{Phenol}\ \xrightarrow{\text{Zn Dust}}\ \text{X}\ \xrightarrow[\text{Anhy. }{AlCl_3}]{CH_3Cl}\ Y$
$Y\ \xrightarrow[KMnO_4]{\text{Alkaline}}\ Z$
The structure of the product Z would be:

Toluene
Benzaldehyde
Benzoic acid
Benzene

Exam: AIPMT – 2009

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AIPMT - 2009

Alcohols: Preparation & Properties

The product ‘Z’ in the below-mentioned reaction is:

$CH_{2}=CH_{2}$
$CH_{3}CH_{2}-O-CH_{2}-CH_{3}$
$CH_{3}-CH_{2}-O-SO_{3}H$
$CH_{3}CH_{2}OH$

Explanation:

Hint: Product Y is an alkene.

PBr₃ causes substitution of the alcoholic group by –Br . Bromoethane then undergoes dehydrohalogenation by alc. KOH to form

ethene, an olefin that is converted back to ethanol by H₂SO₄ and H₂O as shown in the flowchart.

$\underset{ethanol}{C_{2} H_{5}} \overset{P B r_{3}}{\to} C_{2} H_{5} B r \to_{- K B r, - H_{2} O}^{A l c .}$ $\underset{ethanol}{C_{2} H_{5}} \overset{P B r_{3}}{\to} C_{2} H_{5} B r \to_{- K B r, - H_{2} O}^{A l c .}$ $K O H C H_{2} = C H_{2} \overset{H_{2} S O_{4}}{\to} C H_{3} - C H_{2} O S O_{3} H \to_{- H_{2} S O_{4}}^{H_{2} O / ∆} \underset{e t h a n o l}{C H_{3} C H_{2} O H}$ $K O H C H_{2} = C H_{2} \overset{H_{2} S O_{4}}{\to} C H_{3} - C H_{2} O S O_{3} H \to_{- H_{2} S O_{4}}^{H_{2} O / ∆} \underset{e t h a n o l}{C H_{3} C H_{2} O H}$

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Exam: AIPMT – 2009

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AIPMT - 2009