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Compound (A) C_(6)H_(12)O_(2) on reduction with LiAlH_(4) yields two compounds B and C. The compound (B) on oxidation gave ( D), which on treatment with aqueous alkali and subsequent heating furnished E. The latter on catalytic hydrogenation gave ( C). Compound (D) on oxidation gave monobasic acid (molecular formula weight =60). Deduce the structure of (A), (B), (C), (D) and (E). |
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Answer» Solution :(i) E is MONOBASIC ACID (RCOOH) having molecular weight 6 and it is formed from D on oxidation. So E must be acetic acid and D must be acetaldehyde. (ii) (B) on oxidation gives `CH_(3)CHO`. So (B) must be alcohol `CH_(3)CH_(2)OH`. (iii) Acetaldehyde (D) on treating with aqueous alkali (NaOH) gives aldol which on heating gies 2-butenal (E). `CH_(3) -CHO overset(NaOH) to underset("Aldol")(CH_(3)CHOHCH_(2)CHO overset("heat")underset(-H_(2)O) to CH_(3)-underset("2-Butenal"(E))(CH = CH) CHO` (iv) Compound E on cattalytic hydrogenation gives butyl alcohol. `CH_(3)-CH=CHCHO overset(H_(2)) to CH_(3)CH_(2)-CH_(2)-underset("1-Butanol")(OH` (v) Hence compound (A) must be an ester. Ester (A) on reduction with `LiAlH_(4)` yeilds two alcohols (B) and (C). (A) is ethyl butyrate. `CH_(3)CH_(2)CH_(2)COOCH_(2)CH_(3) overset(LiAlH_(4)) to underset("Butyl alcohol")(CH_(3)CH_(2)OH) + underset("Ethyl alcohol")(CH_(3)CH_(2)CH_(2)CH_(2)OH)` 'A' can also be `CH_(3)COOCH_(2C) H_(2)CH_(2)CH_(3)`. This structure will be answering all the above reactions. |
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