JEE Challenger
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ChemMedMain 2026 • 5th Apr - Shift 1

Identify Reagent and Product in Alcohol Dehydration Reaction

Consider compounds A, B and C with following structural formulae

A=CH3−CH2−CH2−CH2−CH2−OHA = \text{CH}_3 - \text{CH}_2 - \text{CH}_2 - \text{CH}_2 - \text{CH}_2 - \text{OH} B=CH2=CH−CH2−CH2−CH3B = \text{CH}_2 = \text{CH} - \text{CH}_2 - \text{CH}_2 - \text{CH}_3 C=HO−CH2−CH2−CH(OH)−CH3C = \text{HO} - \text{CH}_2 - \text{CH}_2 - \text{CH(OH)} - \text{CH}_3

For the conversion of B from A, reagent (D) required is ________ and structural formula of product (E) obtained when C undergoes same reaction using excess reagent (D) is ________.

Options

A

DEConc. H2SO4CH2=CH−CH(OH)CH3\begin{array}{|c|c|}\hline \text{D} & \text{E} \\ \hline \text{Conc. H}_2\text{SO}_4 & \text{CH}_2 = \text{CH} - \text{CH(OH)CH}_3 \\ \hline\end{array}

B

DEPCCHO−CH2−CH2−CH=CH2\begin{array}{|c|c|}\hline \text{D} & \text{E} \\ \hline \text{PCC} & \text{HO} - \text{CH}_2 - \text{CH}_2 - \text{CH} = \text{CH}_2 \\ \hline\end{array}

C

DEPCCCH2=CH−CH=CH2\begin{array}{|c|c|}\hline \text{D} & \text{E} \\ \hline \text{PCC} & \text{CH}_2 = \text{CH} - \text{CH} = \text{CH}_2 \\ \hline\end{array}

D

DEConc. H2SO4 or H3PO4CH2=CH−CH=CH2\begin{array}{|c|c|}\hline \text{D} & \text{E} \\ \hline \text{Conc. H}_2\text{SO}_4\text{ or H}_3\text{PO}_4 & \text{CH}_2 = \text{CH} - \text{CH} = \text{CH}_2 \\ \hline\end{array}

Correct

Topics & Concepts

Alcohols, Phenols and EthersAlcohols and EthersElimination Reactions

Step-by-Step Solution

To determine the reagent (D) and the structural formula of product (E), let us analyze the reactions step-by-step:

1. Conversion of Compound A to Compound B

  • Compound A: Pentan-1-ol (CH3−CH2−CH2−CH2−CH2−OH\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{OH})
  • Compound B: Pent-1-ene (CH2=CH−CH2−CH2−CH3\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}_3)

The conversion of a primary alcohol (A\text{A}) into an alkene (B\text{B}) involves the elimination of a molecule of water (H2O\text{H}_2\text{O}). This dehydration reaction requires a strong mineral acid dehydrating agent, such as concentrated sulfuric acid (Conc. H2SO4\text{Conc. H}_2\text{SO}_4) or phosphoric acid (H3PO4\text{H}_3\text{PO}_4), along with heating:

CH3−CH2−CH2−CH2−CH2−OH→ΔConc. H2SO4 or H3PO4CH2=CH−CH2−CH2−CH3+H2O\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{OH} \xrightarrow[\Delta]{\text{Conc. H}_2\text{SO}_4\text{ or H}_3\text{PO}_4} \text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}_3 + \text{H}_2\text{O}

Thus, reagent (D) is Conc. H2SO4\text{Conc. H}_2\text{SO}_4 or H3PO4\text{H}_3\text{PO}_4.

2. Reaction of Compound C with Excess Reagent (D)

  • Compound C: Butane-1,3-diol (HO−CH2−CH2−CH(OH)−CH3\text{HO}-\text{CH}_2-\text{CH}_2-\text{CH(OH)}-\text{CH}_3)

When butane-1,3-diol undergoes acid-catalyzed dehydration using excess reagent (D) and heat:

  1. Both hydroxyl (−OH-\text{OH}) groups are protonated and removed as water molecules.
  2. The elimination of the two −OH-\text{OH} groups yields two double bonds to form a stable, conjugated diene, buta-1,3-diene:

HO−CH2−CH2−CH(OH)−CH3→Δ,−2H2OExcess DCH2=CH−CH=CH2\text{HO}-\text{CH}_2-\text{CH}_2-\text{CH(OH)}-\text{CH}_3 \xrightarrow[\Delta, -2\text{H}_2\text{O}]{\text{Excess D}} \text{CH}_2=\text{CH}-\text{CH}=\text{CH}_2

Thus, product (E) is CH2=CH−CH=CH2\text{CH}_2=\text{CH}-\text{CH}=\text{CH}_2.

Conclusion:

  • Reagent D: Conc. H2SO4 or H3PO4\text{Conc. H}_2\text{SO}_4\text{ or H}_3\text{PO}_4
  • Product E: CH2=CH−CH=CH2\text{CH}_2=\text{CH}-\text{CH}=\text{CH}_2

This corresponds to Option D.

Identify Reagent and Product in Alcohol Dehydration Reaction | Chemistry PYQ Solution - JEE Challenger