The starting material is propene $(CH_3-CH=CH_2)$. The reaction with $H_3O^+$ (acid-catalyzed hydration) follows Markovnikov's rule to give $A$ (propan$-2-$ol,$CH_3-CH(OH)-CH_3$). The reaction with $(i) BH_3/THF$ followed by $(ii) H_2O_2/OH^-$ (hydroboration-oxidation) follows anti-Markovnikov's rule to give $B$ (propan$-1-$ol,$CH_3-CH_2-CH_2OH$). What is the relationship between products $A$ and $B$?

  • A
    Functional isomer
  • B
    Metamers
  • C
    Position isomer
  • D
    Geometrical isomer

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Which of the following will show tautomerism?

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Assertion $A$: Enol form of acetone $[CH_{3}COCH_{3}]$ exists in $< 0.1 \%$ quantity. However,the enol form of acetylacetone $[CH_{3}COCH_{2}COCH_{3}]$ exists in approximately $15 \%$ quantity.
Reason $R$: Enol form of acetylacetone is stabilized by intramolecular hydrogen bonding,which is not possible in the enol form of acetone. Choose the correct statement:

The pair of metamers is :-

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