Question

The end product (Y) of the reaction sequence:
$CH_3NHCH_3\xrightarrow[NaOH]{PhCOCl} (X) \xrightarrow[H_2O]{LiAlH_4}(Y)$

• A. $PhN \left( CH _{3}\right)_{2}$
• B. $PhCH _{2} N \left( CH _{3}\right)_{2}$
• C. $PhCH ( OH ) N \left( CH _{3}\right)_{2}$
• D. $PhCON \left( CH _{3}\right)_{2}$

Answer 1

Answer: (B)

$(-\overset{\underset{||}{O}}{C} - NH -)$ group is $N$-substituted amide, so by $LAH$ it is converted to $(- CH ,- NH )$ group.