Q.An alkene A on ozonolysis gives a mixture of Pentan-3-one and Ethanal. Write the structure and IUPAC name of 'A'.
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Start your 14-day free trial to unlock the full solution →Reconstructing the alkene from its ozonolysis products by joining the two carbonyl carbons back into a C=C bond gives A = 3-ethylpent-2-ene, CH3-CH=C(CH2CH3)(CH2CH3).
Ozonolysis breaks a C=C double bond and converts each of the two carbons into a C=O (carbonyl) carbon, so to find the original alkene we simply rejoin the two carbonyl carbons of the given products back into a double bond, removing the oxygen atoms.
Product 1: Pentan-3-one, CH3-CH2-CO-CH2-CH3. Its carbonyl carbon is attached to two ethyl (C2H5) groups.
Product 2: Ethanal, CH3-CHO. Its carbonyl carbon is attached to one H and one CH3 group.
Rejoining these two carbonyl carbons with a double bond (in place of the two C=O oxygens) gives:
(C2H5)2C = CH-CH3
Writing this as a continuous IUPAC chain: take one of the two ethyl groups as a continuation of the main chain (through the double bond), and the other ethyl group becomes a substituent. …
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