Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYMedium

Consider the following reaction sequence: $\text{Ethanol} \xrightarrow{PBr_3} X \xrightarrow{\text{Alc. KOH}} Y \xrightarrow[(ii) H_2O, \Delta]{(i) H_2SO_4, \text{room temp.}} Z$ The product Z is:

CH₂=CH₂

CH₃CH₂OCH₂CH₃

CH₃CH₂OSO₃H

CH₃CH₂OH

Step-by-Step Solution

Step 1 (Formation of Alkyl Halide): Ethanol ( $CH_3CH_2OH$ ) reacts with Phosphorus Tribromide ( $PBr_3$ ) to undergo nucleophilic substitution, forming Ethyl bromide ( $X$ ).

Reaction: $3CH_3CH_2OH + PBr_3 \rightarrow 3CH_3CH_2Br + H_3PO_3$ .

Step 2 (Elimination): Ethyl bromide ( $X$ ) reacts with Alcoholic KOH. This is a dehydrohalogenation reaction (elimination) which yields Ethene ( $Y$ ).

Reaction: $CH_3CH_2Br + KOH(alc) \rightarrow CH_2=CH_2 + KBr + H_2O$ .

Step 3 (Hydration): Ethene ( $Y$ ) reacts with concentrated $H_2SO_4$ at room temperature to form ethyl hydrogen sulphate ( $CH_3CH_2OSO_3H$ ). Upon subsequent hydrolysis with water and heat, it regenerates Ethanol ( $Z$ ). This is the standard laboratory method for converting alkenes to alcohols.

Reaction: $CH_2=CH_2 + H_2O \xrightarrow{H^+} CH_3CH_2OH$ .

Conclusion: The reaction sequence converts ethanol to ethene and then back to ethanol. Thus, product Z is Ethanol ( $CH_3CH_2OH$ ).

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started