From $\mathrm{B}_{2} \mathrm{H}_{6}$ all the following can be prepared except\

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Professor PC Thomas Classes · Answer

In diborone two of the 6 hydrogen atoms form part of a ring and they cannot be methylated\ $\mathrm{B}_{2} \mathrm{H}_{6} \xrightarrow{\mathrm{MCH}_{3}} \mathrm{~B}_{2} \mathrm{H}_{5} \mathrm{CH}_{3} \xrightarrow{\mathrm{MCH}_{3}} \mathrm{~B}_{2} \mathrm{H}_{4}\left(\mathrm{CH}_{3}ight)_{2}$\ $\xrightarrow{\mathrm{MCH}_{3}} \mathrm{~B}_{2} \mathrm{H}_{3}\left(\mathrm{CH}_{3}ight)_{3} \xrightarrow{\mathrm{MCH}_{3}} \mathrm{~B}_{2} \mathrm{H}_{2}\left(\mathrm{CH}_{3}ight)_{4}$\ $\mathrm{B}_{2} \mathrm{H}_{6}+6 \mathrm{H}_{2} \mathrm{O} ightarrow 2 \mathrm{H}_{3} \mathrm{BO}_{3}+6 \mathrm{H}_{2}$\ $2 \mathrm{NaH}+\mathrm{B}_{2} \mathrm{H}_{6} ightarrow 2 \mathrm{NaBH}_{4}$\ $3 \mathrm{~B}_{2} \mathrm{H}_{6}+6 \mathrm{NH}_{3} ightarrow 2 \mathrm{~B}_{3} \mathrm{~N}_{3} \mathrm{H}_{6}+12 \mathrm{H}_{2}$