Question

An organic compound a having molecular formula c3h6 on treatment with aqueous h2so4 gives b which on treatment with hcl/zncl2 gives
c. the compound c on treatment with ethanolic koh gives back the compound
a. identify the compounds a,b,c.

Answer 1

here is your answer.........

Answer 2

Compound identification

Given:

Molecular formula of organic compound = $C_3H_6$

• From the given molecular formula, organic compound A is identified as Propene. Its molecular structure is $CH_3CH=CH_2$.
• When propene is treated with aqueous $H_2SO_4$, due to hydration, it leads to the formation of compound 'B' called Propan-2-al.
• The equation involved in the reaction is as follows:

$C H_{3} C H=C H_{2} \stackrel{A q \cdot H_{2} S O_{4}}{\longrightarrow} C H_{3} C H-O H-C H_{3}$

• When Propan-2-al is treated with HCl/$ZnCl_2$ it forms 2-Chloropentane according to the following reaction:

$C H_{3} C H-O H-C H_{3} \stackrel{Z n C l_{2} / H C l}{\longrightarrow} C H_{3} C H C l C H_{3}$

• Propan-2-al when treated with ethanolic KOH yields propene as per the following reaction:

$C H_{3} C H C l C H_{3} \stackrel{\text { Alcoholic } K{O H}}{\longrightarrow} C H_{3} C H=C H_{2}$

Conclusion:

The compound are identified as:

A - Propene

B - Propan-2-al

C - 2-Chloropentane

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