CUET Preparation Today
The correct sequence of reagents to carry out the following conversion is:
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(i) $Sn/HCl$ (ii) $Conc.HNO_3/Conc.H_2SO_4$ (iii) $DIBAL-H$ (iv) $NaNO_2/HCl$ (v) $HBF_4$ (i) $Conc.HNO_3/Conc.H_2SO_4$ (ii) $Sn/HCl$ (iii) $DIBAL-H$ (iv) $NaNO_2/HCl$ (v) $HBF_4$ (i) $NaNO_2/HCl$ (ii) $DIBAL-H$ (iii) $Sn/HCl$ (iv) $Conc.HNO_3/Conc.H_2SO_4$ (v) $HBF_4$ (i) $NaNO_2/HCl$ (ii) $DIBAL-H$ (iii) $Sn/HCl$ (iv) $HBF_4$ (v) $Conc.HNO_3/Conc.H_2SO_4$ |
(i) $Conc.HNO_3/Conc.H_2SO_4$ (ii) $Sn/HCl$ (iii) $DIBAL-H$ (iv) $NaNO_2/HCl$ (v) $HBF_4$ |
The correct answer is Option (2) → (i) $Conc.HNO_3/Conc.H_2SO_4$ (ii) $Sn/HCl$ (iii) $DIBAL-H$ (iv) $NaNO_2/HCl$ (v) $HBF_4$ Initial compound: Benzonitrile ($-CN$) Target compound: m-Fluorobenzaldehyde Step 1: Nitration $-CN$ is strongly electron-withdrawing and meta directing. Treatment with Conc. $HNO_3$ / Conc. $H_2SO_4$ gives m-nitrobenzonitrile. Step 2: Reduction Reduce $-NO_2$ to $-NH_2$ using $Sn/HCl$. Step 3: Reduction of nitrile to aldehyde DIBAL-H selectively reduces $-CN$ to $-CHO$. Gives m-aminobenzaldehyde. Step 4: Diazotization Treat $-NH_2$ with $NaNO_2 / HCl$ at low temperature. Forms diazonium salt. Step 5: Balz–Schiemann reaction Treat diazonium salt with $HBF_4$. Fluorine replaces diazonium group. Final product: m-Fluorobenzaldehyde.
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