Q.An alkene (A), on ozonolysis, gives propanone and an aldehyde (B). When (B) is oxidised, (C) is obtained. (C), treated with Br2/P, gives (D), which on hydrolysis gives (E). When propanone is treated with HCN followed by hydrolysis, it also gives (E). Identify A, B, C, D and E.

Step 1. Start from the second, independent route to E: propanone (acetone), treated with HCN, gives the cyanohydrin (CH3)2C(OH)CN, which on hydrolysis gives E = 2-hydroxy-2-methylpropanoic acid, (CH3)2C(OH)COOH (Sections 12.5.A.1/12.10.2).

Step 2. E is also stated to come from D by hydrolysis; since HVZ-type hydrolysis simply swaps an alpha-bromo for an alpha-hydroxy group, D must be the alpha-bromo acid with the SAME skeleton as E: D = 2-bromo-2-methylpropanoic acid, (CH3)2C(Br)COOH.

Step 3. D is formed from C by Br2/P -- the Hell-Volhard-Zelinsky reaction (Section 12.12.D.1) -- which installs Br at the alpha-carbon of an acid; so C must be the PARENT acid of D with a plain alpha-hydrogen in place of Br: C = isobutyric acid (2-methylpropanoic acid), (CH3)2CH-COOH.

Step 4. C is obtained by oxidising B; since oxidation of an aldehyde gives the same-carbon-count acid (Section 12.5.B.1), B must be the matching aldehyde: B = isobutyraldehyde (2-methylpropanal), (CH3)2CH-CHO. …

Work BACKWARD from E, using the second (propanone + HCN, then hydrolysis) route to pin down E's exact structure first, then reverse each named step in turn (HVZ bromination reversed to a plain alpha-H acid, oxidation reversed to t …

• Trying to solve the chain forward from an assumed, unconstrained alkene A -- the problem is far more tractable solved backward from E, since E's structure is independently fixed by the propanone+HCN r …