Identify the correct reagents that would bring about the following transformation. [figure]
- (i) H2O/H+, (ii) CrO3
- (i) BH3, (ii) H2O2/OH , (iii) PCC
- (i) BH3, (ii) H2O2/OH , (iii) alk. KMnO4
- (i) H2O/H+, (ii) PCC
Answer
(2) (i) BH3, (ii) H2O2/OH , (iii) PCC
Solution
The product is an aldehyde with the C=O at the terminal carbon, so water must add in the anti-Markovnikov way and the primary alcohol must then be oxidised only up to the aldehyde.
(i) BH3, (ii) H2O2/OH (hydroboration–oxidation): C6H11CH2CH=CH2 C6H11CH2CH2CH2OH
(iii) PCC oxidises the 1 alcohol to the aldehyde: C6H11CH2CH2CH2OH C6H11CH2CH2CHO
H2O/H+ (Markovnikov) would give a 2 alcohol and hence a ketone; alk. KMnO4 would oxidise further to the acid.
Answer: (2)