Organic Chemistry

PROBLEM-SOLVING STRATEGY

The para-nitroanilinium ion is units more acidic than the anilinium ion

( versus 4.58), but para-nitrobenzoic acid is only unit more acidic

than benzoic acid ( versus 4.20). Explain why the nitro substituent causes a

large change in in the one case and a small change in in the other.

Don’t expect to be able to solve this kind of problem simply by reading it. First, you

need to remember that the acidity of a compound depends on the stability of its conjugate

base (Sections 1.18 and 7.10). Next, draw structures of the conjugate bases in question and

compare their stabilities.

When a proton is lost from the para-nitroanilinium ion, the electrons that are left behind are

shared by five atoms. (Draw resonance contributors if you want to see which atoms share the

electrons.) When a proton is lost from para-nitrobenzoic acid, the electrons that are left

behind are shared by two atoms. In other words, loss of a proton leads to greater electron

delocalization in one base than in the other base. Because electron delocalization stabilizes a

compound, we now know why the addition of a nitro substituent has a greater effect on the

acidity of an anilinium ion than on benzoic acid. Now continue on to Problem 14.

PROBLEM 14

p-Nitrophenol has a of 7.14, whereas the of m-nitrophenol is 8.39. Explain.

16.6 The Ortho–Para Ratio

When a benzene ring with an ortho–para-directing substituent undergoes an elec-

trophilic aromatic substitution reaction, what percentage of the product is the ortho

isomer and what percentage is the para isomer? Solely on the basis of probability, one

would expect more of the ortho product because there are two ortho positions available

to the incoming electrophile and only one para position. The ortho position, however,

is sterically hindered, whereas the para position is not. Consequently, the para isomer

will be formed preferentially if either the substituent on the ring or the incoming elec-

trophile is large. The following nitration reactions illustrate the decrease in the

ortho–para ratio with an increase in the size of the alkyl substituent:

50%

p-ethylnitrobenzene

50%

o-ethylnitrobenzene

ethylbenzene

H 2 SO 4

++HNO 3

CH 2 CH 3 CH 2 CH 3 CH 2 CH 3

NO 2

NO 2

39%

p-nitrotoluene

61%

o-nitrotoluene

toluene

H 2 SO 4

++HNO 3

CH 3 CH 3 CH 3

NO 2

NO 2

pKa pKa

NNH 2

O

−O +

NCO

O O

−O +

−

pKa pKa

pKa=3.44

pKa=0.98 0.76 pKa

3.60 pKa

640 CHAPTER 16 Reactions of Substituted Benzenes