Consider compounds A, B and C with following structural formulae A = CH3-CH2-CH2-CH2-CH2-OH B = CH2=CH-CH2-CH2-CH3 C = HO-CH2-CH2-CH(OH)-CH3 For the conversion of B from A , reagent (D) required is and structural formula of product (E) obtained when C undergoes same reaction using excess reagent (D) is .
- D: Conc. H2SO4; E: CH2=CH-CH(OH)CH3
- D: PCC; E: HO-CH2-CH2-CH=CH2
- D: PCC; E: CH2=CH-CH=CH2
- D: Conc. H2SO4 or H3PO4; E: CH2=CH-CH=CH2
Answer
(4) D: Conc. H2SO4 or H3PO4; E: CH2=CH-CH=CH2
Solution
A → B is the loss of a water molecule from an alcohol (pentan-1-ol → pent-1-ene), i.e. acid-catalysed dehydration ⇒ D = conc. H2SO4 or H3PO4 (PCC is an oxidising agent and would give an aldehyde).
C (butane-1,3-diol) with excess acid loses both –OH groups as two molecules of water:
HO-CH2-CH2-CH(OH)-CH3 [ ]conc. H2SO4 (excess) CH2=CH-CH=CH2 + 2H2O
⇒ E = buta-1,3-diene (conjugated diene, extra stable)
Answer: (4)