Question

$\longrightarrow$

The above conversion is carried out

• A. KOD /D2O, HÅ/$\Delta$, LiAlH4
• B. HÅ/$\Delta$ / KOD, D2O, LiAlH4
• C. KOD/ D2O, LiAlH4, HÅ/$\Delta$
• D. LiAlH4, HÅ $\Delta$ ,KOD/H2O

Answer 1

Answer: (C)

KOD/ D2O, LiAlH4, HÅ/$\Delta$

Answer 2

$\xrightarrow { \mathrm { KOH } / \mathrm { D } _ { 2 } \mathrm { O } }$ $\xrightarrow { \mathrm { H } ^ { \oplus } / \Delta }$