An alkene CH-3CH - CH-2 is treated with B-2H-6 in presence of H-2O-2- The final product formed is-CH-3CH-OH-CH-3-CH-3CH-2CH-2-3BCH-3CH-2CH-2OHCH-3CH-2CHO

Question

Toppr Admin · Accepted Answer

Hydroboration-oxidation reaction follows anti-Markovnikov's addition of $H - O H$ across $C = C$ to give alcohol.
Thus an alkene $C H_{3} C H = C H_{2}$ when treated with $B_{2} H_{6}$ in presence of $H_{2} O_{2}$ will yield the final product as $C H_{3} C H_{2} C H_{2} O H$

An alkene CH3CH=CH2 is treated with B2H6 in presence of H2O2. The final product formed is:CH3CH(OH)CH3(CH3CH2CH2)3BCH3CH2CH2OHCH3CH2CHO

Hydroboration-oxidation reaction follows anti-Markovnikov's addition of H−OH across C=C to give alcohol.Thus an alkene CH3CH=CH2 when treated with B2H6 in presence of H2O2 will yield the final product as CH3CH2CH2OH

An alkene $C H_{3} C H = C H_{2}$ is treated with $B_{2} H_{6}$ in presence of $H_{2} O_{2}$ . The final product formed is:
$C H_{3} C H (O H) C H_{3}$
$(C H_{3} C H_{2} C H_{2})_{3} B$
$C H_{3} C H_{2} C H_{2} O H$
$C H_{3} C H_{2} C H O$

Hydroboration-oxidation reaction follows anti-Markovnikov's addition of $H - O H$ across $C = C$ to give alcohol.
Thus an alkene $C H_{3} C H = C H_{2}$ when treated with $B_{2} H_{6}$ in presence of $H_{2} O_{2}$ will yield the final product as $C H_{3} C H_{2} C H_{2} O H$