Question

Z forms 2,4-dinitrophenylhydrazone and gives yellow ppt with NaOH + $I_2$. What are X and Y?

• A. $BH_3, H_2O_2/OH^-; CrO_3$
• B. $BH_3, H_2O_2/OH^-; PCC$
• C. $H_2O/H^+ ; CrO_3$
• D. $H_2O/H^+ ; KMnO_4/OH^-, \Delta$

Hint:

{Iodoform Test:} Identifies $CH_3-CO-$ or $CH_3-CH(OH)-$ groups. {Hydroboration:} Anti-Markovnikov (1-ol). {Acid Hydration:} Markovnikov (2-ol).

Answer 1

The Correct Option is A

Step 1: Analyze Z:
- Forms 2,4-DNP hydrazone $\Rightarrow$ Z is an Aldehyde or Ketone. - Gives Yellow ppt with $NaOH + I_2$ (Iodoform test) $\Rightarrow$ Z contains a methyl ketone group ($CH_3-C=O$) or methyl carbinol group (but Z is carbonyl). So Z is a Methyl Ketone.
Step 2: Precursor of Z (Compound formed after X):
Let the intermediate be 'Alcohol A' ($C_6H_{14}O$). Since Z is a methyl ketone, Alcohol A must be a secondary alcohol with a $CH_3-CH(OH)-$ group. Oxidation of this alcohol by Y gives Z. Reagent Y needs to be an oxidizing agent ($CrO_3, PCC$, etc.).
Step 3: Reactant Alkene ($C_6H_{12}$) reaction with X:
Alkene $\xrightarrow{X}$ Alcohol A. Reaction X must produce a secondary alcohol with a terminal methyl group. Let's consider the alkene. Usually, Hex-1-ene ($CH_3(CH_2)_3CH=CH_2$). - Using $BH_3, H_2O_2/OH^-$ (Hydroboration-Oxidation): Anti-Markovnikov addition. Product: 1-Hexanol (Primary). Oxidation gives Hexanal. Hexanal gives DNP but Negative Iodoform test. (Incorrect). - Using $H_2O/H^+$ (Acid Hydration): Markovnikov addition. Product: 2-Hexanol ($CH_3(CH_2)_3CH(OH)CH_3$). - Oxidation of 2-Hexanol gives 2-Hexanone ($CH_3(CH_2)_3COCH_3$). - 2-Hexanone is a Methyl Ketone. It gives Positive DNP and Positive Iodoform test. (Correct).
Step 4: Identify Reagents:
X = $H_2O/H^+$ (Acid Hydration). Y = Oxidizing agent (e.g., $CrO_3$, PCC). Option 3 lists: $X = H_2O/H^+$ and $Y = CrO_3$. This fits perfectly.