TBR Organic chemistry Passage IV question 26

[Joy Bunny]

The question asks which compound has the LARGEST dipole moment and I got it down two choices:

C) 1,1-difluoro-2,2-dichloroethane
D) 1,1-difluoro-2,2-dichloropropane

The explanation given is that the methyl group in choice D donates its electrons to the electron poor central carbon placing a slightly positive charge on the methyl group and increasing the dipole moment.

I do not understand this explanation. Why exactly does the electron donating properties of the methyl group increase the dipole moment? The vectors are the same for both answer choices so how exactly does it increase the dipole? Should there be a vector that shows the electron donating properties of the methyl group?

I think of it like a game of tug-of-war and the vectors point in the same direction for each answer choice so what exactly am I missing here?

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[DrknoSDN]

The tug of war analogy works. With ethane both carbons are partially positive because of the attached electron withdrawing groups. It's has a smaller dipole because both carbons similar in how much positive charge they have.
When you add the methyl group to carbon 2 the 2 chlorine atoms are pulling electrons from carbon 2, but they can also pull from carbon 3 because it is nearby. There is less overall withdraw from carbon 2 because some of the electrons from 3 float over to 2 and stabilize the charge formation.

If you are thinking of it visually with vectors maybe think of electrons flowing toward the halogen atoms from anywhere they can. As the vector from carbon 3 (close enough) flows towards chloride, it would pass through carbon 2 making carbon 2 less positive. As C2 becomes more neutral, the difference between the positive C1 and the more neutral-ish C2 gets larger so the dipole increases.

 

[manohman]

Bumping this.

While I understand how due to the additional carbon that can donate electron density to the second carbon, the second carbon becomes less electropositive and so the difference in partial charge between C1 and C2 increases, wouldn't the difference in partial charge between the Chloride and Carbon 2 decrease to offset that increased difference between C1 and C2?

 

[DrknoSDN]

wouldn't the difference in partial charge between the Chloride and Carbon 2 decrease to offset that increased difference between C1 and C2?

>>>---(This answer is based on real world chemistry and not MCAT simplifications. The book answer is the correct simplification.)

Online Chemical Structure Analysis:
http://www.chemicalize.org/structure/#!mol=1,1-Dichloro-2,2-difluoroethane&source=fp
and
http://www.chemicalize.org/structure/#!mol=structureId:1592768914082&source=fp

If you look at the charge distribution of the two molecules you can see the dipole between C1 and C2 goes down only slightly. (from 0.27<->0.17 in the first molecule to 0.27<->0.18 in the second).

In the first molecule the electron withdrawing groups pull from their respective carbons which in turn pull from their attached hydrogens.
In the second molecule C2 loses the ability to pull from that hydrogen and instead pulls from the adjacent methyl carbon which then pulls from its 3 hydrogens.

The change in dipole comes from the loss of the C2 Hydrogen which carried a partial positive charge into the now adjacent C3 which has a partial negative (instead of the positive that was on the hydrogen).

The result is that the first molecule has relatively balanced charge distribution with the perimeter of the molecule being more negative than the central carbons. Adding the methyl group lets the hydrogens on the far side of the molecule to carry a partial positive while on the very opposite end the chlorine atoms have a partial negative.


The book answer is just ignoring the hydrogens and adding the partial charge present on any hydrogens to their parent carbon. You get the same result. Hope that doesn't make it more confusing. GL.