In the given carbylamine reaction,the value of $x$ (moles of $KOH$ consumed) is:
$C_6H_5NH_2 + CHCl_3 + xKOH \rightarrow C_6H_5NC + 3KCl + 3H_2O$
$C_6H_5NH_2 + CHCl_3 + xKOH \rightarrow C_6H_5NC + 3KCl + 3H_2O$
- A$1$
- B$2$
- C$3$
- D$4$
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Medium
View SolutionReaction of $R-CONH_2$ with a mixture of $Br_2$ and $KOH$ gives $R-NH_2$ as the main product. The intermediates involved in this reaction are :-
$(A)$ $R-CONHBr$
$(B)$ $R-NH-Br$
$(C)$ $R-N=C=O$
$(D)$ $R-CONBr_2$
$(A)$ $R-CONHBr$
$(B)$ $R-NH-Br$
$(C)$ $R-N=C=O$
$(D)$ $R-CONBr_2$
Difficult
View SolutionThe major products $A$ and $B$ formed in the following reaction sequence are :
DifficultJEE MAIN 2021
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