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STD 12 -8.1. Aldehydes and ketones — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 -8.1. Aldehydes and ketones4 Marks
MCQ
Compound '$A$' (molecular formula ${C_3}{H_8}O)$ is treated with acidified potassium dichromate to form a product '$B$' (molecular formula ${C_3}{H_6}O).$ '$B$' forms a shining silver mirror on warming with ammoniacal silver nitrate. '$B$' when treated with an aqueous solution of ${H_2}NCONHN{H_2}.HCl$ and sodium acetate gives a product '$C$'. Identify the structure of '$C$'
  • ✓
    $C{H_3}C{H_2}CH = NNHCON{H_2}$
  • B
    $\begin{array}{*{20}{c}}
      {C{H_3} - CH = NNHCON{H_2}} \\ 
      {\,\,|} \\ 
      {\,\,\,\,\,\,\,\,C{H_3}} 
    \end{array}$
  • C
    $\begin{array}{*{20}{c}}
      {C{H_3}CH = NCONHN{H_2}} \\ 
      {\,\,|\,\,\,\,} \\ 
      {\,\,\,\,\,\,\,\,C{H_3}\,\,\,\,\,} 
    \end{array}$
  • D
    $C{H_3}C{H_2}CH - NCONHN{H_2}$
✓

Answer

Correct option: A.
$C{H_3}C{H_2}CH = NNHCON{H_2}$
a
(a) $\mathop {C{H_3} - C{H_2} - C{H_2}OH}\limits_{{\text{(A)}}}  \xrightarrow{{[O]}}\mathop {C{H_3} - C{H_2} - CHO}\limits_{{\text{(B)}}} $  $C{H_3} - C{H_2} - \mathop {\mathop C\limits^| }\limits^H  = O + {H_2}NNHCON{H_2}\xrightarrow{{HCl}}$ 

  $C{H_3}C{H_2}CH = N - NHCON{H_2}$

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