Question

could you please explain the solutions for 26, 27 and 28

L-a la call NaBHA " vi - V HC-0 0-CH reme CH,OH + H+ (eg, H.SO.) [2 eg of CH3OH, H20 a pdt) N", remove H,

Answer 1

Solution: OH OH (26) The given reaction OH NUBH4 - CH3 Mechanism: Na Bни H- oct 0 -- H - Et OH 04 EtOH teto B H₂ RR ugg i H₃ BTH still reducing 1 molecule of NaBlu reduce 4 molecule of ketone. agent- 04 contain two chirad center. After reduction this compound Thus it is diastereomers. I : the Number of stereoisomer = 2 0 ne number of chiral center. ..2 .2 =4 and I contain four stereoisomer (mixture of enati enantiomers diasteromers. It contain one pair of diuste somers OH Olle OR Racemic Diasteromers. 27 2eg. LH3OH tec-o so- CH₂

Mechanismi DOR OYOCH Call HO-CH₂ DO step-1 Step 2 step-3 Co-CH3 CH3 - HO-CH₂ 15 -O Step 5 step- -H₂O step. 4 H₂C-9 0 -CH₂ Protonation step.1 Proton of carbonyl group in . It increase electrophilic nature of carbonyl group. step-2: I CH₂-OH attack on protonated carbong 9 Step.3: protonation of - OH group. Step-4- Elimination of H₂O molecule. 4 - CH3 stepes- Second molecule of Charoh attack on a step-6: loss of hydrogen atom to give ketul. It is a protection of kelone in acidic medium to give ketal. (28) ů Hoon Hachy Home Mechanism: IG CIL 40 Ho: OH the- "3 step-1 - CH3 step-2 step 3 OH -H2º -H 40 ) Clot Her CH₂ си, Step-u - CH3 step-s -CH3

step- & pr protonation of carbonyl group by H. It incre philic nature of carbony group. oh step-2: CHHO attack on protonated group. carbonyl step-3: protonation of -OH Step-4: Elimination of H₂O molecule. steps: See second hydroxy group of ethylene glycol attack on Do It is intramolecular ker:04 reaction. step-6: loss of hydrogen atom to It is protection of ketone with give cyclic ketul. give cyclic ketul. ethylene glycol in H to