Question

Give the correct sequence of reagents used for the following conversion.

C—OCH, o C—H

• A. $\text{DIBAL}-\text{H},{\text{H}}_3{\text{O}}^{+}/\Delta ,{\text{NaBH}}_4$
• B. ${\text{H}}_3{\text{O}}^{+}/\Delta ,{\text{NaBH}}_4,\text{DIBAL}-\text{H}$
• C. $\text{DIBAL}-\text{H},{\text{NaBH}}_4,{\text{H}}_3{\text{O}}^{+}/\Delta$
• D. ${\text{NaBH}}_4,\text{DIBAL}-\text{H},{\text{H}}_3{\text{O}}^{+}/\Delta$

Answer 1

The correct option is D ${\text{NaBH}}_4,\text{DIBAL}-\text{H},{\text{H}}_3{\text{O}}^{+}/\Delta$

Ester will react with $H_3{O}^{+}$ to give carboxylic acid. So sequence (b) is not possible.
DIBAL-H can reduce ketone to alcohol and ester to aldehyde but further reaction with $NaB{H}_4$ reduce the aldehyde to alcohol so we get diol product. So sequence (a) is not possible.

In sequence (d), DIBAL-H will reduce ester to aldehyde and further reaction $H_3{O}^{+}$ it gives elimination product and then reacting with $NaB{H}_4$ , it reduce aldehyde to alcohol. So sequence (d) is not possible.

Sequence (c):