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Chapter 5: Problem 41

CC1CCCC(=O)C1C=O 470 \(\stackrel{\mathrm{KOH}}{\longrightarrow}(\mathrm{X})\). The product \((\mathrm{X})\) is (a) CC1CCCC(=O)C1C (b) CC1CCCCC1C (c) CC1CCCC(O)C1C (d) CC1CCC=CC1C=O

Short Answer

Expert verified
Options: (a) CC1CCCC(=O)C1C (b) CC1CCCCC1C (c) CC1CCCC(O)C1C (d) CC1CCC=CC1C=O Answer: (d) CC1CCC=CC1C=O

Step by step solution

01

Identify the given molecule

The given molecule in the SMILES notation is: CC1CCCC(=O)C1C=O. It represents a cyclic compound that contains a six-membered ring with two carbonyl groups.
02

Understand the reaction

The given reaction involves the treatment of the given cyclic compound with potassium hydroxide (KOH). It is an elimination reaction where KOH acts as a strong base to eliminate hydrogen and one of the carbonyl groups while forming a double bond in the ring.
03

Predict the product

The most probable outcome of the reaction with KOH would be the elimination of a small molecule from one of the carbonyl groups (likely a water molecule) and formation of a double bond within the ring. This would lead to the formation of an alpha, beta-unsaturated carbonyl compound, retaining the other carbonyl group due to its relative non-reactivity.
04

Compare the product with the given options

Now, we should compare the product formed in the reaction with the given options in the question to find the correct answer. (a) CC1CCCC(=O)C1C - This option retains both carbonyl groups, however, it does not have any double bond in the ring resulting from the elimination of hydroxyl group. (b) CC1CCCCC1C - This option has neither carbonyl groups nor double bond in the ring. (c) CC1CCCC(O)C1C - This option has an alcohol rather than a carbonyl group which is incorrect. (d) CC1CCC=CC1C=O - This option seems to be the correct answer as it retains one carbonyl group and has a double bond within the ring which is formed due to the elimination of hydroxyl group by KOH. Hence, the correct answer is (d) CC1CCC=CC1C=O.

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