First slide

Orthoboric acid

Question

$\large BC{l_3} + {H_2}\xrightarrow[{450^\circ C}]{{Cu - Al}}X + HCl\$
$\large X\xrightarrow{{methylation}}Z\$ . Z in the reaction is

Easy

A

CH₃BH₂
B

(CH₃)₄B₂H₂
C

(CH₃)₃B₂H₃
D

(CH₃)₆B₂

Solution

$BC{l_3} + 6{H_2}\xrightarrow[{{{450}^0}}]{{Cu - Al}}\mathop {\mathop {{B_2}{H_6}}\limits_{Diborane} }\limits_X + 6HCl$
${B_2}{H_6} + 4C{H_3}I\xrightarrow{{Methylation}}{B_2}{H_2}{[C{H_3}]_4} + 4HI$
OR
During methylation reaction only four terminal hydrogens are replaced by methyl group but not bridged hydrogens.

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