Chemistry : Aldehydes, Ketones and Carboxylic Acids

Glucose gives silver mirror with ammoniacal silver nitrate because it has

• aldehyde group

• ester group

• ketone group

• alcoholic silver nitrate

Fats are ester of

• sugar

• glycerol

• tributyrine

• polypeptide

The final product of the following reaction is/are 

End product of the following reaction is

${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{COOH} \underset{\mathrm{redP}}{\overset{{\mathrm{Cl}}_2}{\to }} \mathrm{X} \stackrel{\mathrm{Alc} \mathrm{KOH}}{\to }$

• ${\mathrm{CH}}_3\mathrm{HOCHCOOH}$

• ${\mathrm{HOCH}}_2{\mathrm{CH}}_2\mathrm{COOH}$

• ${\mathrm{CH}}_2=\mathrm{CHCOOH}$

• ${\mathrm{ClCH}}_2\mathrm{HOCHCOOH}$

Which reaction, intermediate is formed during the condensation reaction between acetaldehyde and formaldehyde ?

• $\mathbf{:}\stackrel{-}{\mathrm{C}}{\mathrm{H}}_2\mathrm{CHO}$

• $\stackrel{+}{\mathrm{C}}{\mathrm{H}}_2\mathrm{CHO}$

• $\stackrel{+}{\mathrm{C}}{\mathrm{H}}_2\mathrm{OH}$

• $\mathbf{:}\stackrel{-}{\mathrm{C}}\mathrm{HCHO}$

Aspirin is an acelytation product of

• o-hydroxybenzoic acid

• o-hydroxybenzene

• m-hydroxybenzoic acid

• p-dihydroxybenzene

An ester (A) with molecular formula C₉H₁₀O₂ was treated with excess of CH₃MgBr and the complex so formed was treated with H₂SO₄ to give an olefin (B). Ozonolysis of (B) gave a ketone with molecular formula C₈H₈O which shows positive iodoform test. The structure of (A) is

• C₆H₅COOC₂H₅

• C₆H₅COOC₆H₅

• C₆H₅COOCH₃

• p.H₃CO-C₆H₄-COCH₃

In a solution, 0.02 M acetic acid is 4% dissociated. The [OH^-] in the solution is

• 8 × 10^-4

• 2 × 10^-14

• 8 × 10¹⁰

• 1.25 × 10^-11

Silver salt of carboxylic acid on reaction with R-X gives

• alkyl halide

• ester

• aldehyde

• alcohol

Pentan-3-one is not obtained from

• 2, 2-dichloro pentane

• 3, 3-dichloro pentane

• pentan-3-ol

• pent-2-yne