IIT-JEE 2012 · Paper 2 · Chemistry Q39

ChemistryMultiple correct+4Question 39 of 60

With reference to the scheme given, which of the given statement(s) about T, U, V and W is (are) correct?

T is soluble in hot aqueous NaOH

U is optically active

Molecular formula of W is

\ce{C10H18O4}

V gives effervescence on treatment with aqueous

\ce{NaHCO3}

Answer and worked solution

Official answer: A, C, D

T is the δ-lactone 4-methyltetrahydropyran-2-one.

U: $\ce{LiAlH4}$ reduces the lactone to the diol $\ce{HOCH2CH2CH(CH3)CH2CH2OH}$ (3-methylpentane-1,5-diol).
W: excess acetic anhydride acetylates both OH groups: $\ce{AcOCH2CH2CH(CH3)CH2CH2OAc}$ , $\ce{C10H18O4}$ .
V: $\ce{CrO3}/\ce{H+}$ oxidises both primary alcohols: 3-methylpentanedioic acid.

Statements.

(A) Hot NaOH hydrolyses the lactone to the soluble sodium 5-hydroxy-3-methylpentanoate. True.
(B) In U, C3 carries two identical $\ce{CH2CH2OH}$ arms, so it is not a stereocentre and U is achiral. False.
(C) True.
(D) V is a dicarboxylic acid and liberates $\ce{CO2}$ from $\ce{NaHCO3}$ . True.

(A), (C), (D)

IIT-JEE 2012 · Paper 2, Chemistry question 39