A Few Things From the Organic Unit, Followed up

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Presentation transcript:

A Few Things From the Organic Unit, Followed up CHM 122 Week 12, I&II A Few Things From the Organic Unit, Followed up

Geometric Isomers in Organic Molecules CHM 122 Week 12, I&II Geometric Isomers in Organic Molecules The only type of geometric isomer “type” we will discuss is cis-trans isomerism With TM complexes, we had cis-trans isomerism in square planar and octahedral complexes Ligands were either 90° (cis) or 180° (trans) apart from one another Cis = “same side”; trans = “across from one another” In organic molecules, cis-trans isomers can occur in certain alkenes Alkene  a double bond is necessary (but not sufficient)

Example of a cis-trans isomer pair (see p. 920 in Tro and Table 20 Example of a cis-trans isomer pair (see p. 920 in Tro and Table 20.9 for properties) Cis = “same side of double bond” Trans = “opposite sides of double bond” (across the double bond)

The prior two compounds are only different because rotation about a db is hindered—a p bond would need to break p. 425 in Tro (Section 10.7; 1st semester)

When is cis-trans isomerism possible? Consider the following “generalized” alkene with substituents To have cis-trans isomers, both of the following must be true: A  B AND D  E Can’t be same A Can’t be same D C C B E Alternate view: If either A = B OR D = E, then NO cis-trans isomer pair. (Flipping vertically would yield same [superimposable] structure: not isomers) AND

Which of the following compounds could exhibit geometric isomerism? a) CH2=CHCH3 b) CH3CH=CHCH3 c) CCl2=CF2 d) CHCl=CF2 Answer: b HINT: Redraw structures as on prior slide to better “see” the four “things” attached! (See board) Copyright © 2011 Pearson Education, Inc.

Which of the following compounds could exhibit geometric isomerism? a) CH2=CHCH3 b) CH3CH=CHCH3 c) CCl2=CF2 d) CHCl=CF2 Answer: b Copyright © 2011 Pearson Education, Inc.

See doc camera examples from text

Optical Isomerism in molecules with a tetrahedral carbon Review from metal complexes unit (next slide)

Chirality Any molecule with a nonsuperimposable mirror image is said to be chiral A pair of (such) nonsuperimposable mirror images are called optical isomers or enantiomers Tro: Chemistry: A Molecular Approach, 2/e

Optical Activity Enantiomers have all the same physical properties except one – the direction they rotate the plane of plane-polarized light each one of the enantiomers will rotate the plane the same amount, but in opposite directions that’s why enantiomers are called “optical isomers” Tro: Chemistry: A Molecular Approach, 2/e

Optical Isomerism in molecules with a tetrahedral carbon Any carbon with 4 different substituents will be a chiral center (chirality center, technically) Thus, if a molecule has exactly one chiral carbon (center), it will have an optical isomer If any two (or more) of the substituents are identical, the center will not be chiral and the mirror image will be superimposable (not isomers) Tro: Chemistry: A Molecular Approach, 2/e

See board for way to “rotate” on paper. To tell if two mirror image structures are superimposable, rotate structure and compare See board for way to “rotate” on paper. Tro: Chemistry: A Molecular Approach, 2/e

Which of the following can exhibit optical isomerism? d) Answer: c Copyright © 2011 Pearson Education, Inc.

Which of the following can exhibit optical isomerism? Same  no optical isomer right C) Same  no optical isomer (left C) Same  no optical isomer a) b) c) d) Same  no optical isomer right C) Answer: c Same  no optical isomer (left C) None are the same on left! All three are same on right  will be an optical isomer (left C is a chiral(ity) center; exactly one in molecule) Copyright © 2011 Pearson Education, Inc.

See doc camera examples from text

Which functional groups are present in the molecule penicillin shown below? Alcohol Amine Ester Carboxylic acid All of the above Answer: e Copyright © 2011 Pearson Education, Inc.

Which functional groups are present in the molecule penicillin shown below? Alcohol Amine Ester Carboxylic acid All of the above Answer: e Copyright © 2011 Pearson Education, Inc.