14

Suggested solutions for Chapter 14

PROBLEM 1 Are these molecules chiral? Draw diagrams to justify your answer. OH
HO2C
HO2C

Ph
O

OH

OH

O

OH
O

O

O

Purpose of the problem Reinforcement of the very important criterion for chirality. Make sure you understand the answer.

Suggested solution Only one thing matters – does the molecule have a plane of symmetry? We need to redraw some of them to see if they do. On no account look for chiral centres or carbon atoms with four different groups or anything else. Just look for a plane of symmetry. If the molecule has one, it isn’t chiral. The first compound has been drawn with carboxylic acids represented in two different ways. The two CO2H groups are in fact the same and the molecule has a plane of symmetry (shown by the dashed lines). It isn’t chiral.

Me

OH

OH

Me

OH

=

HO2C

O

OH

HO2C

CO2H

HO2C

CO2H

2

Solutions Manual to accompany Organic Chemistry 2e

The second compound is chiral but if you got this wrong don’t be dismayed. Making a model would help but there are only two plausible candidates planes of symmetry: the ring itself, in the plane of the page, and a plane at right angles to the ring. The molecule redrawn below with the tetrahedral centre displayed shows that the plane of the page isn’t a plane of symmetry as the CO2H is on one side and the H on the other, and neither is the plane perpendicular to the ring, as Ph is on one side and H on the other. No plane of symmetry: molecule is chiral. Ph

H

Ph

Ph

no plane of symmetry perpendicular to ring

no plane

CO2H

H

of symmetry
CO2H in plane of page

H

CO2H

The third compound is not chiral because of its high symmetry. All the CH2 groups are identical so the alcohol can be attached to any of them. The plane of symmetry (shown by the dotted lines) may be easier to see after redrawing, and will certainly be much easier to see if you make a model.

=

OH

OH

The fourth compound needs only the slightest redrawing to make it very clear that it is not chiral. The dashed line shows the plane of symmetry at right angles to the paper.

OH
O

O

■ Spiro compounds, which

contain two rings joined at a single atom, are discussed on p. 653 of the textbook.

The final acetal (which is a spiro compound) is drawn flat but the central carbon atom must in fact be tetrahedral so that the two rings are orthogonal. By drawing first one and then the other ring in the

Solutions for Chapter 14 – Stereochemistry

plane of the page it is easy to