Oxidation of allyl alcohol with a peracid gives a compound of molecular formula C₃H₆O₂, which contains an asymmetric carbon atom. The structure of the compound is

The number of alkene (s) which can produce 2-butanol by the successive treatment of (i) B₂H₆ in tetrahydrofuran solvent and (ii) alkaline H₂O₂ solution is

Methoxybenzene on treatment with HI produces

$$\mathop {{C_4}{H_{10}}O}\limits_{(N)} \mathop {\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{{H_2}S{O_4}}^{{K_2}C{r_2}{O_7}}} }\limits_{} {C_4}{H_8}O\mathop {\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{Warm}^{{I_2}/NaOH}} }\limits_{} CH{I_3}$$

Here, N is