An alkene 'A' on ozonolysis gives a mixture of ethanal and…

Question 15 of 32intermediate🔧 ApplyShort Answer2 marks

An alkene 'A' on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of 'A'.

Correct Answer

Structure of 'A': $CH_3-CH=C(CH_2CH_3)_2$

IUPAC name of 'A': 3-ethylpent-2-ene

Exercise: EXERCISES | Q: 9.5 | (Page Page 33)

For More Understanding

Explanation

Ozonolysis of an alkene cleaves the carbon-carbon double bond to form carbonyl compounds. The given products are ethanal ( $CH_3CHO$ ) and pentan-3-one ( $CH_3CH_2COCH_2CH_3$ ). To find the original alkene, the oxygen atoms from the carbonyl groups of the products are removed, and the two carbonyl carbons are joined together with a double bond. This gives the structure $CH_3-CH=C(CH_2CH_3)_2$ . The longest carbon chain containing the double bond has five carbons, with an ethyl substituent at carbon 3, making the IUPAC name 3-ethylpent-2-ene.

Solution Steps

Step 1: Identify the ozonolysis products: ethanal ( $CH_3CHO$ ) and pentan-3-one ( $CH_3CH_2COCH_2CH_3$ ).
Step 2: Remove the oxygen atoms from the carbonyl carbons of both products.
Step 3: Join the two carbonyl carbons with a double bond to get the alkene structure: $CH_3-CH=C(CH_2CH_3)_2$ .
Step 4: Determine the IUPAC name by finding the longest chain of 5 carbons with a double bond at position 2 and an ethyl group at position 3, which is 3-ethylpent-2-ene.

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