Consider compounds $\mathrm{A}, \mathrm{B}$ and C with following structural formulae $\mathrm{A}=\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{OH}$ $\mathrm{B}=\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{3}$ $\mathrm{C}=\mathrm{HO}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}_{3}$ 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 ____

(D) D: Conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$ or $\mathrm{H}_{3} \mathrm{PO}_{4}$, E: $\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}$

(A) D: Conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$, E: $\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}(\mathrm{OH}) / \mathrm{CH}_{3}$

(B) D: PCC, E: $\mathrm{HO}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}=\mathrm{CH}_{2}$

(C) D: PCC, E: $\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}$

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Alcohols Phenols and Ethers

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Consider compounds

A, B

and C with following structural formulae

A = CH_{3} - CH_{2} - CH_{2} - CH_{2} - CH_{2} - OH

B = CH_{2} = CH - CH_{2} - CH_{2} - CH_{3}

C = HO - CH_{2} - CH_{2} - CH (OH) - 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 ____

(A)D: Conc. $H_{2} SO_{4}$ , E: $CH_{2} = CH - CH (OH) / CH_{3}$
(B)D: PCC, E: $HO - CH_{2} - CH_{2} - CH = CH_{2}$
(C)D: PCC, E: $CH_{2} = CH - CH = CH_{2}$
(D)D: Conc. $H_{2} SO_{4}$ or $H_{3} PO_{4}$ , E: $CH_{2} = CH - CH = CH_{2}$ ✓ Correct

Answer: (D)

Solution

Turning 1-pentanol into 1-pentene is a dehydration, the removal of water from an alcohol, achieved by a hot acid catalyst such as concentrated

H_{2} S O_{4}

H_{3} P O_{4}

; the oxidant PCC would instead convert the alcohol to a carbonyl compound and is unsuitable. Applying the same acid in excess to butane-1,3-diol,

HOC H_{2} C H_{2} CH (OH) C H_{3}

, strips away both hydroxyl groups to create two double bonds, giving buta-1,3-diene. Reagent D is concentrated

H_{2} S O_{4}

H_{3} P O_{4}

and product E is

C H_{2} = CH - CH = C H_{2}

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Alcohols Phenols and Ethers

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