Carboxylic Acids Flashcards Preview

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Flashcards in Carboxylic Acids Deck (24)
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1

How do you account for the fact that the difference betweenthe first and second ioniztion constants decreases with increasing distance between the carboxyl groups?

Ex:

  • Oxalic Acid (HO2CCO2H) pK1: 1.2, pK2: 4.2
  • Succinic (HO2CCH2CH2CO2H), pK1: 4.2, pK2: 5.6

20.32

Inductive effects of funcitonal groups are transmitted through sigma bonds.

  • For oxalic acid, the EWG inductive effect of one carboxylic acid group decreases the acidity of the remaining group.
  • However, as the length of the carbon chain increases, the effect of one funcitonal group on another decreases. In this example the influcence of the second carboxylic acid group on the ionization of the first is barely felt by succinic acid.

2

How would you convert butanenitrile into the following compounds?

  1. 1-butanol
  2. butylamine
  3. 2-methyl-3hexanone

30.35

3

Prepare from benzene:

  1. m-chlorobenzoic acid
  2. p-bromobenzoic acid
  3. phenylacetic acid, C6H5CH2CO2H

 

20.36

4

Predict the product of the reaction of p-methylbenzoic acid with each of the following:

  • (a) LiAlH4, then H3O+
  • (b) N-Bromosuccinimide in CCl4
  • (c) CH3MgBr in ether, then H3O+
  • (d) KMnO4, H3O+

20.37

Note: for (c), the acidic proton reacts with the Grignard reagent to form methane, for no net reaction.

5

How would you carry out the following transformations?

20.39

*For (b): Only Grignard carboxylation can be used bc -CN brings about elimination of the tertiary bromide to form a double bond.

6

When do you use Grignard carboxylation vs. nitrile hydrolysis (when preparing carboxylic acids)?

20.40

Use nitrile hydrolysis:

  • in presence of acidic hydroxyl groups because Grignards will deprotonate

 

Use Grignards:

  • in presence of secondary halides (ie: bromide) because nitrile hydrolysis will make optically active product from optically active reagent

 

7

Synthesize 3-Methyl-2-hexanoic acid (E, Z mixture) from starting materials of 5 carbons or fewer.

8

What is the product and (reaction type: SN1, SN2, E1, E2) for a reaction of 2-chloro-2-methylpentane with NaCN?

Tertiary halides won't react Sn2 bc of bulkiness (sterics). E2 is the only thing that occurs.

9

Prepare the following: 

20.48

10

Propose a synthesis of the anti-inflammatory drug Fenclorac from

phenylcyclohexane.

11

The pKa’s of five p-substituted benzoic acids (YC6H4CO2H) follow.

  • Rank the corresponding substituted benzenes (YC6H5) in order of their increasing reactivity toward electrophilic aromatic substitution.
  • If benzoic acid has pKa = 4.19, which of the substituents are activators and which are deactivators?

Of all listed, Only -Si(CH3)3 is and activator

12

Transform the following: 

13

Identify the missing reagents a–f in the following scheme:

14

2-Bromo-6,6-dimethylcyclohexanone gives 2,2 dimethylcyclopentanecarboxylic acid on treatment with aqueous NaOH followed by acidification, a process called the Favorskii reaction. Propose a mechanism.

15

Propose the mechanism.

20.60

16

  1. Deprotonation
  2. Decarboxylation
  3. Protonation

17

Propose a mechanism

20.64

18

Summary of Reactions (MM Ch 20.a)

19

Summary of Reactions (MM Ch 20.b)

20

Propose a synthesis of ibuprofen, starting from benzene. 

21

Find the flaw in the following syntheses and then propse solutions for the errors.

  1. Adding magnesium to an alkyl halide produces a very strong, very basic nucleophile. 
    Adding another nucleophile like CN-  will not accomplish very much. A simple solution is to swap CO2 for the second step
  2. The protection step is a good idea as alcohols are incompatible with strongly basic 
    reagents like LAH and EtLi, but unfortunately we cannot stop an LAH reduction at the 
    aldehyde so adding an alkyl lithium would achieve little beynod deprotonation. Our only 
    reaction that does something like that is using DIBAL on an ester. Here we should be 
    successful with simple protection of the alcohol followed by treatment of the acid with 
    SOCl2 and then Et2CuLi. This will produce the ketone.
  3. The main issue here is that treatment of the resultant nitrile with acid will also dehydrate the tertiary alcohol. Base hydrolysis is preferred.
  4. This path to nitriles goes through an SN2 mechanism, which is not compatible with tertiary alkyl halides. A Grignard reagent is preferred for this type of substrate. Another issue is the presence of the aldehyde, which would not be compatible with either carboxylic acid forming paths (you would get at least some addition to the carbonyl). You could protect the aldehyde as an acetal and then do the Grignard path to the acid and then do the amide formation followed by deprotection. Although ketals are removed under aqueous acid conditions, they can be quite robust and will outlive some acidic workups.

22

 Show how to accomplish the following transformations.

23

 Show how to accomplish the following transformations.

24

Compare reducing a carboxylic acid with LAH vs. 

BH3•THF?

 

Reduction with borane is often preferred over reduction with LAH, because borane reacts selec- tively with a carboxylic acid moiety in the presence of another carbonyl group. As an example, if the following reaction were performed with LAH instead of borane, both carbonyl groups would be reduced.