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Chapter 1: Problem 140

Identify the statement which is not correct among the following (a) \((+)\) and meso tartaric acids are diastereomers. (b) The enol percentage of acetyl acetone is more than that of acetone. (c) Meso butane- 2,3 -diol can easily be resolved to give \((+)\) and \((-)\) isomers. (d) \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\) is known as 3 -methyl-2-butanol.

Short Answer

Expert verified
a. (+) and meso tartaric acids are diastereomers. b. The enol percentage of acetyl acetone is more than that of acetone. c. Meso butane-2,3-diol can easily be resolved to give (+) and (-) isomers. d. $\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH})\mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}$ is known as 3-methyl-2-butanol. Answer: c. Meso butane-2,3-diol can easily be resolved to give (+) and (-) isomers.

Step by step solution

01

Statement (a)

\((+)\) and meso tartaric acids are diastereomers. Diastereomers are stereoisomers that are not related as mirror images, meaning they have at least one stereocenter being inverted and at least one staying the same. \((+)\) tartaric acid has a specific optical rotation, while meso tartaric acid is achiral (optically inactive) because it has a plane of symmetry. Thus, they are diastereomers.
02

Statement (b)

The enol percentage of acetyl acetone is more than that of acetone. The enol percentage refers to the presence of enol \(\mathrm{C=C-O-H}\) functional groups in a compound at equilibrium. Acetylacetone has a higher enol percentage than acetone because the presence of the enol form is stabilized by hydrogen bonding and also due to conjugation between the two carbonyl groups.
03

Statement (c)

Meso butane-2,3-diol can easily be resolved to give \((+)\) and \((-)\) isomers. Meso compounds are achiral due to the presence of a plane of symmetry, and they do not have enantiomers. Therefore, we cannot resolve a meso compound into \((+)\) and \((-)\) isomers. This statement is not correct.
04

Statement (d)

$\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}$ is known as 3-methyl-2-butanol. The IUPAC nomenclature rules dictate that we must name this compound accordingly. In this case, we have a 4-carbon chain, so the base name is butanol. The alcoholic group is on the second carbon, and there is a methyl group on the third carbon, making the name 3-methyl-2-butanol. This statement is correct.

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Most popular questions from this chapter

Identify the compounds having chiral centres among the following (i) 1-Bromopentane (ii) 1-Chloro-2-methylpentane (iii) 2 -Chloro- 2 -methylpentane (iv) Citric acid (v) 1-Chloro-2-bromobutane

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