MCQ
The reaction $C{H_2} = CH - C{H_3} + HBr \to C{H_3}CHBr - C{H_3}$ is
- ANucleophilic addition
- ✓Electrophilic addition
- CElectrophilic substitution
- DFree radical addition
$HBr \rightarrow H ^{+}+ Br ^{-}$
Therefore, this reaction show electrophilic addition reaction.
$CH _2= CH - CH _3+ HBr \longrightarrow CH _3-\stackrel{\oplus}{ C } H - CH _3 \stackrel{ Br ^{-}}{\longrightarrow} CH _3- CHBr - CH _3$
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$(a)$ $H _{2} C _{2} O _{4} \rightleftharpoons H ^{+}+ HC _{2} O _{4}^{-}$
$(b)$ $HC _{2} O _{4}^{-} \rightleftharpoons H ^{+}+ HC _{2} O _{4}^{2-}$
$(c)$ $H _{2} C _{2} O _{4} \rightleftharpoons 2 H ^{+}+ C _{2} O _{4}^{2-}$
The relationship between $K_{a_{1}}, K_{ a _{2}}$ and $K_{ a _{3}}$ is given as
| List-$I$ (Molecule) | List-$II$(Shape) |
| $A$ $\mathrm{NH}_3$ | $I$ Square pyramid |
| $B$ $\mathrm{BrF}_5$ | $II$ Tetrahedral |
| $C$ $\mathrm{PCl}_5$ | $III$ Trigonal pyramidal |
| $D$ $\mathrm{CH}_4$ | $IV$ Trigonal bipyramidal |
Choose the correct answer from the option below :
a $20\ L$ box containing $X_6$ and $X_3$ at equilibrium at $1500\ K$ . $K_p = 4 \times 10^{18}\ atm$ . Assuming ${P_{{x_3}}} > > {P_{{x_6}}}$ and total pressure is $10\ atm$ .The partial pressure of $X_6$ will be