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Chemistry
The reaction, ROH + H2CN2 in the presence of HBF4, gives the following product
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Q. The reaction, $ROH + H_2CN_2$ in the presence of $HBF_4$, gives the following product
VITEEE
VITEEE 2010
A
$ROCH_3$
B
$RCH_2OH$
C
$ROHCN_2N_2$
D
$RCH_2CH_3$
Solution:
$\ce{ROH + H2CN2 ->[HBF_4] ROCH_3 + N_2}$