Reaction Scheme Involving Lactone Reduction and Subsequent Reactions

To determine the correct statements about compounds $\mathbf{T}, \mathbf{U}, \mathbf{V},$ and $\mathbf{W}$, we analyze each step in the given reaction sequence:

1. Analysis of Compound $\mathbf{T}$ and Option A

• Compound $\mathbf{T}$ is a $\delta$-lactone (a cyclic ester), specifically 4-methyltetrahydro-2H-pyran-2-one with molecular formula $\mathrm{C}_6\mathrm{H}_{10}\mathrm{O}_2$.
• When heated with aqueous $\mathrm{NaOH}$, lactones undergo base-catalyzed hydrolysis (saponification) to form the sodium salt of a hydroxy carboxylic acid: $\mathbf{T} + \mathrm{NaOH(aq)} \xrightarrow{\Delta} \mathrm{HO-CH_2-CH_2-CH(CH_3)-CH_2-COONa}$
• The ionic sodium hydroxycarboxylate salt is highly polar and water-soluble. Thus, $\mathbf{T}$ is soluble in hot aqueous $\mathrm{NaOH}$.
• Option A is correct.

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2. Reduction of $\mathbf{T}$ to $\mathbf{U}$ and Option B

• Reduction of the lactone $\mathbf{T}$ using lithium aluminium hydride ($\mathrm{LiAlH}_4$) cleaves the cyclic ester linkage to yield a primary diol $\mathbf{U}$: $\mathbf{T} \xrightarrow{\mathrm{LiAlH}_4} \mathbf{U} : \mathrm{HO-CH_2-CH_2-CH(CH_3)-CH_2-CH_2-OH}$
• Compound $\mathbf{U}$ is 3-methylpentane-1,5-diol.
• The central carbon ($\mathrm{C\text{-}3}$) is attached to a hydrogen atom, a methyl group ($\mathrm{-\!CH_3}$), and two identical $-\mathrm{CH_2CH_2OH}$ chain segments.
• Due to the presence of two identical substituents on $\mathrm{C\text{-}3}$, the molecule possesses a plane of symmetry, making it achiral and optically inactive.
• Option B is incorrect.

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3. Formation of $\mathbf{W}$ and Option C

• Treatment of diol $\mathbf{U}$ with excess acetic anhydride, $(\mathrm{CH}_3\mathrm{CO})_2\mathrm{O}$, acetylates both hydroxyl groups to give the diacetate ester $\mathbf{W}$: $\mathbf{U} \xrightarrow{\text{excess }(\mathrm{CH}_3\mathrm{CO})_2\mathrm{O}} \mathbf{W} : \mathrm{CH_3COO-CH_2-CH_2-CH(CH_3)-CH_2-CH_2-OCOCH_3}$
• Counting the atoms in $\mathbf{W}$:
  • Carbons: $6 \text{ (from diol)} + 4 \text{ (from two acetyl groups)} = 10$
  • Hydrogens: $18$
  • Oxygens: $4$
• Thus, the molecular formula of $\mathbf{W}$ is $\mathrm{C}_{10}\mathrm{H}_{18}\mathrm{O}_4$.
• Option C is correct.

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4. Oxidation of $\mathbf{U}$ to $\mathbf{V}$ and Option D

• Oxidation of the primary diol $\mathbf{U}$ with Jones reagent ($\mathrm{CrO}_3 / \mathrm{H}^+$) oxidizes both primary alcohol ($-\mathrm{CH_2OH}$) groups into carboxylic acid ($-\mathrm{COOH}$) groups: $\mathbf{U} \xrightarrow{\mathrm{CrO}_3/\mathrm{H}^+} \mathbf{V} : \mathrm{HOOC-CH_2-CH(CH_3)-CH_2-COOH}$
• Compound $\mathbf{V}$ is a dicarboxylic acid (3-methylpentanedioic acid).
• Carboxylic acids react with aqueous sodium bicarbonate ($\mathrm{NaHCO}_3$) to liberate carbon dioxide ($\mathrm{CO}_2$) gas, resulting in effervescence: $\mathrm{R-COOH} + \mathrm{NaHCO}_3 \rightarrow \mathrm{R-COONa} + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2\uparrow$
• Option D is correct.

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Conclusion

The correct options are A, C, and D.