MCQ
$P{H_4}I + NaOH$ forms
- ✓$P{H_3}$
- B$N{H_3}$
- C${P_4}{O_6}$
- D${P_4}{O_{10}}$
$PH _4 I + NaOH \rightarrow PH _3+ H _2 O + NaI$
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Which conformer of above compound is most stable across $C_2 - C_3$ ?
Statement $I$ : Bromination of phenol in solvent with low polarity such as $\mathrm{CHCl}_3$ or $\mathrm{CS}_2$ requires Lewis acid catalyst.
Statement $II$ : The lewis acid catalyst polarises the bromine to generate $\mathrm{Br}^{+}$.
In the light of the above statements, choose the correct answer from the options given below :

