Which can reduce $R C O O H ⟶ R C H_{2} O H$ ?

N a B H_{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

N a / C_{2} H_{5} O H

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B H_{3} / T H F / H_{3} O^{+}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

H_{2}

/ catalyst

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $B H_{3} / T H F / H_{3} O^{+}$
$N a B H_{4}$ can reduce only aldehyole and ketones not acids.
$R - C O O H ⟶ R C H_{2} O H$ is done by $B H_{3} / T H F / H_{3} O$ .

Suggest Corrections

Similar questions

C H_{3} - C H = C H - C O O H \to C H_{3} - C H = C H - C H_{2} O H

R C O O H \to R C H_{2} O H

N a B H_{4} / H_{3} O^{+}

N a B H_{4} / H_{3} O^{+}

R C O O H \to R C H_{2} O H

N a B H_{4} / H_{3} O^{+}

N a B H_{4} / H_{3} O^{+}

Join BYJU'S Learning Program