Q.In the following sequence, acetaldehyde (CH3CHO) is treated with
You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
Start your 14-day free trial to unlock the full solution →Acetaldehyde + CH3MgBr (then H2O) gives propan-2-ol (A). Dehydration gives propene (B). Hydroboration-oxidation of propene adds -OH anti-Markovnikov, giving propan-1-ol (C). A (propan-2-ol) and C (propan-1-ol) are positional isomers, so option (ii) is correct.
Step 1 - forming A
A Grignard reagent adds to the carbonyl carbon: CH3MgBr adds a CH3 group to CH3CHO. After aqueous work-up:
CH3CHO + CH3MgBr --> CH3-CH(OMgBr)-CH3 --(H2O)--> CH3-CH(OH)-CH3
So A = propan-2-ol (a secondary alcohol).
Step 2 - forming B
Heating a secondary alcohol with H2SO4 dehydrates it (loss of water) to the alkene:
CH3-CH(OH)-CH3 --(H2SO4, heat)--> CH3-CH=CH2
So B = propene.
Step 3 - forming C
Hydroboration-oxidation adds water across the double bond with anti-Markovnikov orientation (-OH goes to the less substituted, terminal carbon):
CH3-CH=CH2 --(i) BH3;
(ii) H2O2/OH- --> CH3-CH2-CH2-OH
So C = propan-1-ol (a primary alcohol).
Relationship …
Unlock everything free for 14 days
- Full step-by-step solutions
- Concept-first explanations
- Methods, shortcuts & mistakes
- PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.