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NCERT Exemplar · Q9

Q.In the following sequence, acetaldehyde (CH3CHO) is treated with

(i) CH3MgBr and then
(ii) H2O to give A; A is heated with H2SO4 to give B; B then undergoes hydroboration-oxidation to give C. State the relationship between compounds A and C.
(i) Identical
(ii) Positional isomers
(iii) Functional isomers
(iv) Optical isomers
Haryana BsehMCQ· 1mImportance★★★★★
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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 …

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