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A molecule is achiral if it has a plane of symmetry in any conformation. Each of the following conformations does not have a plane of symmetry, but rotation around a carbon–carbon bond forms a conformation that does have a plane of symmetry. Draw this conformation for each molecule.

a.

b.

Short Answer

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a.

b.

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01

Plane of symmetry

The term "plane of symmetry" refers to an imaginary plane that divides a molecule into two equal/identical halves.

02

Achiral molecules

The molecules that are superimposable with their corresponding mirror images are termed achiral.

Achiral molecules possess a plane of symmetry.

03

Rotation around carbon-carbon bonds

a. A rotation around the carbon-carbon bond ( and ) yields a conformer of the molecule that has a plane of symmetry.

Conformer of a

b. The rotation about the carbon-carbon ( and ) bond by yields the conformer that possesses a plane of symmetry.

Conformer of b

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Most popular questions from this chapter

Which group in each pair is assigned the higher priority?

a. -CH3,-CH2CH3

b. -I,-Br

c. -H,-D

d. -CH2Br,-CH2CH2Br

e. localid="1648296449951" -CH2Cl,-CH2CH(CH3)2

f.-CH2OH,-CHO

Without drawing out the structures, label each pair of compounds as enantiomers or diastereomers.

a. (2R,3S)-hexane-2,3-diol and (2R,3R)-hexane-2,3-diol

b. (2R,3R)-hexane-2,3-diol and (2S,3S)-hexane-2,3-diol

c. (2R,3S,4R)-hexane-2,3,4-triol and (2S,3R,4R)-hexane-2,3,4-triol

Draw all the possible stereoisomers for each compound and label pairs of enantiomers and diastereomers.

a.

b.

Rank the following groups in order of decreasing priority.

  1. -F,-NH2,-CH3,-OH
  2. -CH3,-CH2CH3,-CH2CH2CH3,-(CH2)3CH3
  3. -NH2,-CH2NH2,-CH3,-CH2NHCH3
  4. -COOH,-CH2OH,-H,-CHO
  5. -Cl,-CH3,-SH,-OH
  6. -C≡CH,-CH(CH3)2,-CH2CH3,-CH=CH2

Amygdalin, a compound isolated from the pits of apricots, peaches, and wild cherries, has been used as an unsanctioned anticancer drug both within and outside of the United States. One hydrolysis product formed from amygdalin is mandelic acid, used in treating common skin problems caused by photo-aging and acne.

a. How many stereogenic centres are present in amygdalin? What is the maximum number of stereoisomers possible?

b. Draw both enantiomers of mandelic acid and label each stereogenic centre as R or S.

c. Pure (R)-mandelic acid has a specific rotation of –154. If a sample contains 60% of the R isomer and 40% of its enantiomer, what is [α]of this solution?

d. Calculate the ee of a solution of mandelic acid having [α] = +50. What is the percentage of each enantiomer present?

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