In the following reactions,products $A$ and $B$ are

  • A
    $A = \text{3-hydroxy-6,6-dimethylcyclohexanone}; B = \text{6,6-dimethylcyclohex-2-enone}$
  • B
    $A = \text{3-hydroxy-2,2-dimethylcyclohexanone}; B = \text{6,6-dimethylcyclohex-2-enone}$
  • C
    $A = \text{2-hydroxy-2,3-dimethylcyclobutanecarbaldehyde}; B = \text{2,3-dimethylcyclobut-2-enecarbaldehyde}$
  • D
    $A = \text{2-hydroxy-2,3-dimethylcyclobutanecarbaldehyde}; B = \text{2-methylene-3,3-dimethylcyclobutanecarbaldehyde}$

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Acetaldehyde cannot show:

$C_{7}H_{10}O$ reacts with $CH_{3}MgBr$ to give a compound $C_{8}H_{12}O$ which gives a positive iodoform test. Identify the structure of $A$.

The major product of the above reaction is $A$.

$(A)$ $\xrightarrow{NH_2OH} (B)$ $\xrightarrow{H_2SO_4} (C)$ $\xrightarrow{H_3O^{\oplus}} (D) + (E)$ $\xrightarrow{CHCl_3, KOH} CH_3-NC$ (carbylamine test). $(D)$ $\xrightarrow{SOCl_2} (F)$ $\xrightarrow{(i) PhMgBr \text{ (excess) } (ii) H^{\oplus}} (G)$ $\xrightarrow[\Delta]{H^{\oplus}} (H)$ $\xrightarrow{CH_2I_2, Zn/Cu} \text{1,1-diphenylcyclopropane}$. The molecular weight of compound $(A)$ is:

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The reaction of hydrazine $(NH_2NH_2)$ with aldehydes and ketones yields a compound with the general structure:

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