Q.An ether (A), C5H12O, when heated with excess of hot HI produce two alkyl halides which on hydrolysis form compound (B)and (C), oxidation of (B) gave and acid (D), whereas oxidation of (C) gave a ketone (E). Deduce the structural formula of (A), (B), (C), (D) and (E). (Carried exactly as printed — the book runs "(B)and" together and prints "gave and acid" for "gave an acid".)
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Start your 14-day free trial to unlock the full solution →Step 1. Read off the structural clues. Ether A has formula C5H12O -- five carbons, one oxygen, fully saturated (degree of unsaturation zero, consistent with an acyclic ether). Excess hot HI cleaves it (section 11.5.3) into TWO alkyl iodides (both fragments converted all the way to iodides because HI is in EXCESS); hydrolysing these two iodides regenerates two alcohols, B and C. Oxidation of B gives an ACID (D) -- meaning B must be a PRIMARY alcohol (only primary alcohols oxidise through to a carboxylic acid, section 11.4.4). Oxidation of C gives a KETONE (E) -- meaning C must be a SECONDARY alcohol (secondary alcohols oxidise to ketones).
Step 2. Deduce the ether's two alkyl fragments. Since A is unsymmetrical (its two cleavage fragments, B and C, are structurally different -- one primary, one secondary), A must be a mixed ether R-O-R' where R gives a primary alcohol and R' gives a secondary alcohol, with a combined total of 5 carbons (C5H12O has 5 C). The simplest such split is a 1-carbon primary fragment (methyl) and a 4-carbon secondary fragment (sec-butyl, since a 4-carbon group whose attachment carbon has one H and one further CH3 branch gives a secondary alcohol on hydrolysis): A = CH3-O-CH(CH3)-CH2-CH3 (methyl sec-butyl ether, IUPAC 2-methoxybutane). Checking the formula: 1 (OCH3 carbon) + 4 (sec-butyl carbons) = 5 carbons; total H = 3 (OCH3) + 1 (CH) + 3 (branch CH3) + 2 (CH2) + 3 (terminal CH3) = 12 H; one O -- matches C5H12O exactly. …
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