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Chapter 7: Problem 55

This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct. A compound containing a conjugated double bond, decolourises bromine water. On heating with \(\mathrm{I}_{2}\) and \(\mathrm{NaOH}\), a yellow precipitate is formed. If the compound contains five carbons including a methyl substituent, the compound may be (a) 3 -methyl but- 3 -en- 2 -one. (b) 3 -methyl but- 3 -enal (c) 3 -methyl but- 2 -enal (d) 2 -methyl but- 2 -enal

Short Answer

Expert verified
Answer: 3-methyl but-2-enal

Step by step solution

01

Identify the functional groups and double bond positions in the given choices

Inspect the 4 given compounds: (a) 3-methyl but-3-en-2-one: Functional group - ketone (\(\ce{-C=O}\)), and double bond at the third position. (b) 3-methyl but-3-enal: Functional group - aldehyde (\(\ce{-CHO}\)), and double bond at the third position. (c) 3-methyl but-2-enal: Functional group - aldehyde (\(\ce{-CHO}\)), and double bond at the second position. (d) 2-methyl but-2-enal: Functional group - aldehyde (\(\ce{-CHO}\)), and double bond at the second position.
02

Check decolorizing bromine water

A compound decolorizes bromine water due to the presence of unsaturated bonds, particularly about conjugated dienes. In our case, there is only one double bond in all the given compounds. However, the one that is conjugated with an electron-withdrawing group would show stronger bleaching power. (a) 3-methyl but-3-en-2-one: Conjugated with the ketone group. (b) 3-methyl but-3-enal: Conjugated with the aldehyde group. (c) 3-methyl but-2-enal: Conjugated with the aldehyde group. (d) 2-methyl but-2-enal: Not conjugated with the aldehyde group. Based on the conjugated systems, we can eliminate (d) 2-methyl but-2-enal.
03

Check formation of yellow precipitate

The yellow precipitate can be formed in the presence of an α-hydrogen (a hydrogen adjacent to a carbonyl group). This compound will form haloform (in this case, iodoform) with iodine and sodium hydroxide, which forms a yellow precipitate. (a) 3-methyl but-3-en-2-one: Contains one α-hydrogen (next to the ketone group). (b) 3-methyl but-3-enal: Contains no α-hydrogen (next to the aldehyde group). (c) 3-methyl but-2-enal: Contains two α-hydrogen (next to the aldehyde group). We can eliminate choice (b) 3-methyl but-3-enal, as it does not contain any α-hydrogen.
04

Choose the correct compound

After considering the functional groups, conjugation, decolorizing bromine water, and formation of a yellow precipitate, we have narrowed down the possible compounds to (a) 3-methyl but-3-en-2-one and (c) 3-methyl but-2-enal. Both compounds fit the criteria given in the problem, but only one can be chosen as the correct answer. Choice (c) 3-methyl but-2-enal satisfies all the given criteria, and it has a more pronounced conjugation between double bond and carbonyl group. Thus, the compound could be: (c) 3 -methyl but- 2 -enal

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