$A$ compound gives a positive test with $I_2/NaOH$ and is extracted from benzene by saturated $NaHSO_3$. It may be

  • A
    $CH_3(CH_2)_4CHO$
  • B
    $CH_3(CH_2)_3COCH_3$
  • C
    $CH_3CH_2COCH_2CH_3$
  • D
    $CH_3(CH_2)_4CH_2OH$

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The reaction shown above is a Baeyer-Villiger rearrangement of an asymmetric ketone with magnesium monoperoxyphthalate hexahydrate (in the drawing,$Mg^{+2}$ is omitted for clarity). Identify the major product.

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Acetaldehyde cannot give:

In the given reaction $C_6H_5CHO + X \xrightarrow{(i) Zn, (ii) HOH/NH_4Cl} C_6H_5-CH(OH)-CH_2-COOC_2H_5$,$[X]$ will be:

Consider the reactions and identify $A, X, Y$ and $Z$:
$X (C_2H_6O)$ $\xrightarrow{Cu, 573 \ K} A$ $\xrightarrow{[Ag(NH_3)_2]^+, ^-OH, \Delta} \text{Silver mirror}$
$A \xrightarrow{^-OH, \Delta} Y$
$A \xrightarrow{NH_2NHCONH_2} Z$

Which of the following statements is false?

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