Chapter 16: Aldehydes and Ketones (Carbonyl Compounds)


The Carbonyl Double Bond


  1. Both the carbon and oxygen atoms are hybridized sp2, so the system is planar.



  1. The three oxygen sp2 AO’s are involved as follows: The two unshared electorn pairs of oxygen occupy two of these AO’s, and the third is involved in sigma bond formation to the carbonyl carbon.



  1. The three sp2 AO’s on the carbonyl carbon are involved as follows: One of them is involved in sigma bonding to one of the oxygen sp2 AO’s, and the other two are involved in bonding to the R substituents.
  2. The 2pz AO’s on oxygen and the carbonyl carbon are involved in pi overlap, forming a pi bond.



  1. The pi BMO, formed by positive overlap of the 2p orbitals, has a larger concentration of electron density on oxygen than carbon, because the electrons in this orbital are drawn to the more electronegative atom, where they are more highly stabilized. This result is reversed in the vacant antibonding MO.



  1. As a consequence of the distribution in the BMO, the pi bond (as is the case also with the sigma bond) is highly polar, with the negative end of the dipole on oxygen and the positive end on carbon. We will see that this polarity, which is absent in a carbon-carbon pi bond, has the effect of strongly stabilizing the C=O moiety.



Resonance Treatment of the Carbonyl Pi Bond



1.    Note that the ionic structure (the one on the right side) has one less covalent bond, but this latter is replaced with an ionic bond (electrostatic bond).



2.    This structure is a relatively “good” one, therefore, and contributes extensively to the resonance hybrid, making this bond much more thermodynamically stable than the C=C pi bond, for which the corresponding ionic structure is much less favorable (negative charge is less stable on carbon than on oxygen).



3.    The carbonyl carbon therefore has extensive carbocation character, while the oxygen has extensive oxyanion character (or alkoxide character). The corresponding canonical structure for an alkene pi bond has carbanion character, which is much less energetically favorable than oxyanion character.




q      The bond dissociation energy, D, of an alkene type pi bond (C=C) is ca. 65 kcal/mol.


q      The bond dissociation energy, D, of a carbonyl type pi bond (C=O) is ca. 85 kca/mol.


q      The polarity of the carbonyl pi bond is directly responsible for the enhanced bond strength.


q      Very Important:  The carbonyl pi bond is stronger, it is also more reactive kinetically, because of the carbocation and alkoxide character, the same feature which makes this bond thermodynamically stronger. The consequence of this is that reactions at the carbonyl pi bond are often faster than those at the alkene pi bond, but the equilibrium constant is much less favorable (they do not necessarily proceed to the right).







Consider the following isomeric carbonyl compounds:


q      The equation for the heat of combustion of each is the same:


4H8O  +  11/2 O2 ® 4 CO­2  + 4 H2O


q      The heat of combustion of butanal is 8 kcal/mol greater than that of butanone. Therefore, butanone is 8 kcal/mole more stable, thermodynamically, than butanal. This is a reliable generalization for comparisions of aldehydes and ketones.



q      The basis for this stability order is carbocation character at the carbonyl carbon, and the ability of alkyl groups to stabilize carbocation character. The ketone has two alkyl substituents to stabilize this carbocation character, while the aldehyde has only one.


q      Methanal (formaldehyde), the simplest (one carbon) aldehyde, has no alkyl groups and is the least stable of the simple carbonyl compounds.


q      Typically, the least stable carbonyl compounds react most rapidly, i.e., kinetic rates inversely parallel thermodynamic stabilities. Thus, the reactivity order is ketones>aldehydes>methanal (formaldehyde).




ALDEHYDES.  Replace the terminal “-e” of the corresponding alkane having the same number of carbon atoms with “-al”.


q      The parent (main) chain must contain the aldehyde carbon.


q      The aldehyde carbon is always C1, so no locant number is needed for the aldehyde functionality.


q      Examples:


KETONES. Replace the terminal “-e” of the corresponding alkane with the ketone suffix “-one”.


q      The main chain must contain the ketone carbonyl carbon.


q      Numbering of the main chain begins with the end of the main chain nearest to the carbonyl carbon.


q      The position of the ketone carbon must be specified with a numeric locant (unless there is only one possible ketone having that main chain).


q      In case of ties in regard to the numbering of the main chain (that is, the carbonyl carbon is equidistant from both chain termini), numbering proceeds from the end of the main chain closest to the first substituent.


q      Examples:





1.    Reactions with Grignard reagents.



q      An analogous reaction, of course, applies to ketones.


q      The reaction is very fast, even at room temperature, because the carbonyl group has a highly electrophilic carbonyl carbon, which combines with the nucleophilic carbon of the Grignard reagent.


q      Recall that the C-Mg bond is polar, and the carbon has carbanion character. This is an important general strategy for forming carbon-carbon bonds: combine a reagent which has carbocation character with one which has carbanion character.


q      Synthetically, the reaction is used to make alcohols. The immediate product is the magnesium salt of the conjugate base of the alcohol, but acidic, aqueous workup provides the corresponding alcohol.


q      IMPORTANT: Note that aldehydes give secondary alcohols, and ketones yield tertiary alcohols. Addition of Grignard reagents to methanal uniquely gives primary alcohols.


q      Finally, note that Grignard reagents do not add to alkene pi bonds at all, even though these bonds are much weaker than the carbonyl pi bonds. This is essentially because the carbons of an alkene pi bond do not have carbocation character.


q      Exercise: Sketch three different possible syntheses of the tertiary alcohol shown below, using the Grignard method illustrated above. You may use any Grignard reagent and any carbonyl compound as starting materials.


q      Exercise: Sketch two different possible syntheses of the secondary alcohol shown below, using the Grignard method.


q      Sketch a possible synthesis of the primary alcohol shown below, using the Grignard method of synthesis.







q      The alkoxide character derives from the product (P)  character of the TS.

q      This alkoxide character  is further stabilized by covalent and ionic bonding to the magnesium ion. The Mg-O bond is a very stable one.

q      A final factor which makes this TS an especially favorable (low energy) one is the electrostatic attraction between the positively charge carbonyl carbon and the partially negatively charged carbon of the Grignard reagent. This derives from reactant (R) character in the TS (properly positioned reactants, of course).





q      By simply substituting carbon for oxygen in the transition state model for addition to a carbonyl group (along with generalized valencies to that carbon), we can obtain a TS model for the analogous but hypothetical addition of a Grignard reagent to an alkene.


q      We use the Method of Competing Transition States to compare the relative merits of these two reaction (relative rates).

o    Carbanion character is much less favorable than oxyanion character.


o    Mg-C bond character is less favorable than Mg-O bond character in the TS.


o    Electrostatic attraction is present in the carbonyl addition TS but is simply missing in the alkene addition TS, because the alkene pi bond is not polar.


o    Consequently, since all of these effects favor the TS for the addition of Grignard reagents to carbonyl compounds, it is much more favorable than than the TS for addition to an alkene.





q      The hydrogens of lithium aluminum hydride and sodium borohydride (see below) are hydridic, since they share a portion of the negative overall charge of the tetrahydroaluminate or tetrahydroborate anions. They are therefore nucleophilic.


q      Just as Grignard reagents provide nucleophilic carbon, complex metal hydride anions provide nucleophilic hydrogen. The nucleophilic additions of these hydride reagents to carbonyl compounds proceed mechanistically very much like the Grignard additions and also yield alcohols as final products, after aqueous workup.


q      The development of a TS model for the addition of LAH to an aldehyde is shown below:



q      Since a new carbon-carbon bond is not being added in this case, aldehydes yield primary alcohols and ketones yield secondary alcohols.


q      Exercise: Provide a synthetic sketch, starting with an appropriate organic halide and using LAH as a reagent, for the synthesis of the following two alcohols.You may use any appropriate aldehyde or ketone.



q      Explain why a tertiary alcohol cannot be synthesized using the hydride method.











Development of the Transition State Model for the Slow Step



q      The first step is termed “slow”, but not “rate determining” because it is at least partially reversible. Recall that a rate-determining step has a rate which is exactly equal to the rate of formation of the final product. This cannot be the case for a reversible or partially reversible step, because in this case some of the product of the first step doesn’t go on to the final product, but reverts to the starting materials (the reverse of step 1). Therefore the rate of final product formation is less than the rate of the first (forward) step. Nevertheless, this step is the slowest step of the reaction and its rate most strongly effects the rate of product formation. That is why we consider the TS for this step, in loose analogy to the TS of a rds.


q      Note: Hydroxide ion is re-formed in the second step, so the reaction is truly catalytic.


q      Hydroxide anion is the nucleophile, while the carbonyl carbon is the electrophilic site.


q      Both the electrostatic attraction between the carbonyl carbon and the negatively charged nucleophile in the proprerly oriented TS (reactant character) and the development of oxyanion character in the TS are favorable factors which cause this reaction to occur at a rapid rate, even though the C=O bond is a relatively strong one.


q      The first step, though fast, is reversible as a direct result  of the strength of the C=O bond. The latter bond strength also causes this equilibrium to lie on the side of the carbonyl compound (to the left in the equilibrium, as written).


q      Analogous nucleophilic additions to the much weaker pi bonds of simple alkenes do not occur at all (see analogy  to the Grignard and hydride reactions), since carbanion character is far less favorable than oxyanion character and there is no assisting electrostatic attraction in the case of addition to a non-polar alkene.


q      The problem with addition of water to alkenes is not the equilibrium position, but the rate (as reflected in the discussion of TS characters). The equilibrium for hydration of an alkene (as you may recall from the first semester) lies well to the right, in favor of the alcohol. Again, the reason for this is the relative weakness of the C=C pi bond. So in both rate and equilibrium, hydration of an alkene is in sharp contrast to hydration of a carbonyl compound.


q       The second step of the two-step reaction mechanism is a proton transfer from oxygen to oxygen; such reactions are extremely fast and reversible.


Mechanism of the Acid-Catalyzed (Electrophilic) Addition of Water to Carbonyl Compounds (AdE2)


q      Note that water is too weak a nucleophile to add to the neutral carbonyl group at a reasonable rate, but if the carbonyl group is protonated to provide the conjugate acid shown, the carbonyl carbon becomes even more electrophilic (it has more partial positive charge), and even water can add to this.


q      The conjugate acid of the carbonyl compound actually has two resonance structures, one indicating oxonium ion character and the other carbocation character.



q      Although the neutral carbonyl compound also has carbocation character, the structure which exhibits this character has charge separation (see below). Therefore this structure is of higher energy and contributes less to its resonance hybrid than the carbocation structure in the conjugate acid, which is not charge-separated.



q      It is important to again recognize that the slow step is the one in which the strong C=O bond is broken (viz., the 2nd step). The other steps are extremely rapid proton transfers from O to O.


q      The original acid, hydronium ion, is regenerated in the third step, so the reaction is catalytic.


q      Interestingly, acid catalyzed hydration is much faster than the corresponding acid-catalyzed hydration of an alkene (though the alkene pi bond is weaker). But again, the equilibrium is more favorable for alkene hydration than for carbonyl hydration.


q      In acid catalyzed hydration of an alkene, the rds is protonation of the pi bond. This requires breaking the pi bond. In acid catalyzed hydration of a carbonyl compound, this protonation is quite easy because no bond has to be broken, the unshared electron pair on oxygen is a non-bonding pair of electrons. Once this conjugate acid is formed, it is extremely reactive because it has such extensive carbocation character. The unshared pair of electrons on oxygen provides an easy way to facilitate addition to the pi bond.








q      The equilibrium between a ketone and water, on the one hand,  and its hydration product on the other strongly favors the reactant ketone, because of the strength of the C=O pi bond.  For example, the hydration of acetone, at equilibrium, forms only 0.1% of the hydrate; 99.9% of the acetone remains unhydrated.


q      In the case of an aldehyde, the equilibrium is somewhat more favorable for hydration than for a ketone, but the equilibrium still somewhat favors the aldehyde. In neither case does the reaction go to completion.


q      Of the simple aldehydes, only methanal is more or less completely hydrated in the equilibrium.


q      These relative extents of hydration of carbonyl compounds follow the order of thermodynamic stabilities discussed previously, viz., ketones are more stabilized than typical aldehydes than methanal. Thus, alkyl groups stabilize the carbonyl group (via carbocation character) more than they stabilize the hydrate.


q      Again, it is important to note that hydration of an alkene pi bond goes to completion, whereas the hydration of carbonyl compounds does not. This accords with the relative thermodynamic stabilities of the C=C and C=O pi bonds.




q      The addition of alcohols to carbonyl compounds follow exactly the same mechanisms as for addition of water.


q      The reactions, as in the case of water, can occur either by an acid-catalyzed or a base-catalyzed mechanism.


q      The product hemiacetal, instead of being a 1,1-diol, is a 1-alkoxy-1-ol, i.e., an ether-alcohol instead of a diol.


q      Hemiacetals are so-called because they are half way toward the formation of an acetal from a carbonyl comound, which we will consider momentarily. An acetal is a 1,1-diether (a 1,1-dialkoxy comound).


The Acid-Catalyzed Mechanism for Hemiacetal Formation.



q      Note that the only difference from acid-catalyzed hydration is that R’ replaces one of the hydrogens. The catalyst, instead of being hydronium ion is the conjugate acid of the alcohol, since the reaction would be carried out in the alcohol as the solvent.


q      Also, the alcohol replaces water is the nucleophile in step 2 and as the base in step three.


q      The hemiacetal product also has one OR’ group replace one of the alcohol functions.


The Base-Catalyzed Mechanism of Hemiacetal Formation



q      Note that the nucleophile in step 1 and the base in step 2 is not hydroxide ion, but an alkoxide anion.




q      Acetals are formed from hemiacetals in a strictly acid-catalyzed process. Base catalysis is ineffective in converting a hemiacetal to an acetal.


q      Although hemiacetal formation is an addition reaction (AdE or and, depending upon whether the mechanism is acid- or base-catalyzed, the conversion of a hemiacetal to an acetal is a substitution reaction.



q      The overall reaction is replacing an –OH group with an OR’ group. The mechanism is most similar to an SN1 substitution mechanism.









q      The formation of acetals usually starts with the carbonyl compound and proceeds via a two stage process, the first (addition)stage yielding the hemiacetal and the second (substitution) stage converting the hemiacetal to an acetal. Recall that, whereas base catalysis can accomplish the first stage, it cannot convert a hemiacetal to an acetal (we will return to discuss the reasons for this momentarily).



q      The overall reaction for the conversion of a carbonyl compound to an acetal (see below) requires two moles of alcohol.


q      If, instead, a diol is used to provide both R’O groups, a cyclic acetal is formed. The most commonly used diol is the very readily available ethylene glycol (1,2-ethanediol).



q      In comparing these two equilibria (one using two molecules of a monofunctional alcohol and one using one molecule of a difunctional alcohol), we can see that in the first case three molecules of reactant are converted to two molecules of product, but in the second case two molecules of reactant are converted to two molecules of product.



q      This is important because molecules are free to move about (translate) and this freedom contributes to their thermodynamic stability through the entropy factor. This freedom is referred to as translational entropy. When a reaction takes place which requires the loss of translational entropy, this is an unfavorable factor with respect to the free energy change and to the completion of the equilibrium. When 2 molecules of reactant are converted to 2 molecules of product, little or no loss of translational entropy occurs. But when 3 molecules of reactant are converted to 2 molecules of product, the translational entropy of one molecule is lost. This is a very significant factor with respect to the position of the equilibrium.


q      As a result , the equilibrium for the formation of a cyclic acetal is much more favorable than for an acylic acetal. In particular, the equilibrium for the formation of an aldehyde cyclic acetal is now favorable to the acetal, whereas it was not in the case of an acylic acetal. Consequently, the formation of a cyclic acetal in organic synthesis is much more convenient than forming an acylic acetal.


q      In the case of cyclic acetal formation, the step which is really making the difference is not the addition step, but the second, substitution step, because this step is essentially intramolecular. Intramolecular reactions are typically strongly favored by entropy (note that one molecule is converted to two molecules, thereby increasing translational freedom).





q      Recall that hydroxide ion is a rather poor leaving group, whereas water is a good leaving group.


q      Refer to step 2 of the mechanism for the conversion of a hemiacetal to an acetal, noting the analogy to an SN1 reaction, with water as the leaving group. In the absence of acid, the leaving group would have to be a hydroxide anion, which is a poor leaving group.


q      It is also important to note that the converse is alsotrue, that is, that an acetal cannot be converted to a hemiacetal in base.


q      Thus, the acetal function is especially synthetically useful because it is stable in basic solutions and contains only two relatively unreactive ether functionalities. On the other hand, it is readily converted back to the precursor carbonyl compound by acid-catalyzed hydrolysis. The mechanism is the reverse of acid-catalyzed acetal formation (start with the product and read backward in the mechanism).






q      The ability to smoothly convert reactive aldehyde or ketone functionality to a relatively inert acetal functionality is highly useful in organic synthesis. The acetal functionality is, in effect, a protected carbonyl group.


q      An excellent example of the use of the acetal protecting group strategy is the formation of a Grignard reagent from a bromoaldehyde, as shown below. If the aldehyde functionality were not protected, this Grignard reagent could not be formed in a synthetically useful manner, because it would quickly react with a molecule of the aldehyde. However, if the aldehyde functionality is first protected as an acetal function, the Grignard can be prepared quite nicely, and the protected Grignard reagent can then be used for whatever synthetic purpose may be desired. In the example shown, the Grignard is used to react with benzaldehyde to give, after hydrolysis, a hydroxyaldehyde.




q      Note that there are essentially three requirements for an effective protecting group.

o    It must be formed in high yield

o    In the protected form it must be unreactive

o    The protected functionality must be efficiently re-converted to the original functionality. This step is called “de-protection”.


q      The de-protection step is very facile in the case of an acetal, because in the course of acidic, aqueous workup it is quickly hydrolyzed to the carbonyl functionality.




q      In the case of the synthetic transformation below, the preferred starting material is a keto ester, and it is desired to selectively reduce the ester functionality to  an alcohol functionality, without affecting the ketone function (the target molecule retains the ketone function).



q      It is known that lithium aluminum hydride will efficiently reduce an ester function in the manner desired (see later), but we also know that it will reduce the ketone function, in this case to a secondary alcohol. This would not lead to the desired product, if both functionalities were reduced.



q      There is an even more serious problem. If we were to attempt to use only the stoichiometric amount of the hydride reagent necessary to selectively reduce one of the functionalities, the ketone would be the one preferentially reduced, because the ester function is resonance stabilized (see later), and therefore less reactive than the ketone. (By the way, LAH is highly reactive, and not very selective, anyway, but any selectivity would favor reaction with the ketone function.



q      We must therefore adopt a protecting group strategy, as shown below. Again, note that the acidic, aqueous workup required for LAH reductions is also that required to de-protect the acetal functionality.






I.              Additions of Conjugate Bases of Terminal Alkynes.


q      Since terminal alkynes are unusually acidic, their conjugate bases may be prepared quantitatively by using a sufficiently strong base. In this connection, sodium amide is an appropriately strong base, as we learned in the first semester (the pKa’s of terminal alkynes are ca. 25, whereas the pKa  of ammonia is ca. 38).


q      These anions are carbon centered anions (carbanions), and, like Grignard reagents, are highly nucleophilic. They therefore add in the same manner as do Grignard reagents to carbonyl compounds.



II.            Addition of Cyanide Ion.




q      The cyanide anion is another carbon centered nucleophile which can add readily to carbonyl functionalities. As in the case of Grignard reagents and alkynide anions, the special importance of cyanide anion as a nucleophile is that it forms a new carbon-carbon bond.


q      The reaction only requires a catalytic amount of the cyanide anion, since the intermediate alkoxide anion (the conjugate base of the product) can abstract a proton from HCN, generating more cynanide anion.



q      The CN functionality is called a nitrile group. Nitrile groups are at the same oxidation state as a carboxylic acid group or an ester group, and can be readily converted to either one of these


q      You may recall from the first semester’s studies of SN reactions, that the cyanide anion is a very strong nucleophile.





q      .The so-called “Wittig reagent” is another reagent containing nucleophilic carbon, and which can add readily to a carbonyl group. The reagent, as illustrated below, has extensive carbanion character, and thus is quite nucleophilic.


q      This reagent is rather unique in that it allows the formation of a new doubly bonded carbon. The net result is the conversion of an aldehyde on ketone to an alkene of choice. The overall reaction is illustrated below. It can be used to form mono-, di-, tri-, or tetrasubstituted alkenes. A mixture of cis and trans alkene isomers are usually formed.



q      It is very important to note that the conversion of a carbonyl function to an alkene double bond is thermodynamically unfavorable by at least 20 kcal/mol, as we have seen in this chapter. What is thermodynamic “driving force” which makes this reaction possible? (That is, which provides more than 20 kcal/mol of negative free energy change?)


q      It is, of course, the conversion of the Wittig reagent to triphenylphosphine. In particular, the Wittig reagent has carbanion character, but the phosphine oxide has oxyanion character, the latter being very much more favorable for stabilization of the negative charge.


Resonance Treatment of the Wittig Reagent.


q      The canonical structure written above for the Wittig reagent places negative charge on carbon and positive charge on the tetravalent phosphorous atom. Why not move the electron pair of the carbanion center in between the carbon and phosphorous atom to form a second bond, i.e., a pi bond?



q      Actually, for phosphorous this covalent structure is a valid (i.e., legal) canonical structure, because phosphorous is a third row element which has vacant 3d orbitals. Although it has already used all of its valence shell 3s and 3p AO’s to form four sp3 covalent bonds (three to phenyl groups and one to the carbanion carbon), it still has 3d AO’s, which are vacant and can overlap with the filled carbon 2p AO which  contains the electron pair. We thus can have a two electron pi bond structure.


q      Normally, since the covalent structure has one additional covalent bond, one would think that this structure would be the lower energy structure, but in this case, interestingly, the ionic structure is the one of lower energy. There are two reasons for this:


q      First, the pi bond which is formed is very weak indeed, because the 3d AO is quite high in energy, and overlap with that orbital yields relatively little stabilization.


q      Secondly, the first structure, although lacking a pi covalent bond, does have a relatively strong ionic (i.e., electrostatic) pi bond. Of course, ionic bonding can be quite strong.


q      Consequently, we usually draw the ionic structure to represent the Wittig reagent, but you should know that a covalent structure also contributes to the resonance hybrid, so that carbon doesn’t have a full unit of negative charge.


q      Suppose we had an analogous reagent in which the phosphorous was replaced by nitrogen? Would the covalent structure be valid? The answer is NO! Nitrogen is in the second row of the Periodic Table and the valence shell is the second main shell, which has no d type orbitals.


Preparation of the Wittig Reagent.


q      Some Wittig reagents are commercially available, but most are preferably synthesized prior to use in the laboratory. There are two steps in the generation of an appropriate Wittig reagent:


q      Step 1: Reaction of triphenylphosphine with an alkyl halide, in an SN2 reaction. Recall that P, as a heavier atom, is especially nucleophilic, although not very basic (the “polarization” effect), so these displacement reactions work very efficiently. This reaction results in the formation of a stable phosphonium salt (tetravalent phosphorous is analogous to tetravalent nitrogen of ammonium salts).  The simplest case, reaction with methyl iodide, is illustrated below, but analogous reactions with primary or secondary halides are feasible.




q      Step 2: Removal of a modestly acidic proton from the carbon atom alpha to the positively charged phosphorous atom. This proton is removable because the resulting “carbanion” will be substantially stabilized by the resulting electrostatic bond present in the Wittig reagent and to a modest extent by the weak pi bonding present in the Wittig reagent. Any of a variety of bases can be used, but relatively strong ones like butyl lithium are more commonly used.



Mechanism of the Wittig Reaction.






q      Sketch an efficient synthesis of the following alkene:



q      Answer:





q      You already know that alkene pi bonds can be readily hydrogenated to alkanes by reaction with dihydrogen in the presence of platinum, palladium, or nickel catalysts. These reactions occur under very mild conditions: room temperature with just 1 atmosphere of hydrogen.



q      You also may know that benzene and aromatic hydrocarbons require more forcing conditions for the hydrogenation of the highly resonance stabilized, aromatic ring. Forcing conditions refers to the use of much higher temperatures and pressures.


q      In much the same way as alkene pi bonds, carbonyl pi bonds can be hydrogenated to alcohols. The conditions required, however are quite different, since the carbonyl pi bond is much more stable thermodynamically than the alkene pi bond. For the hydrogenation of a carbonyl group, forcing conditions are required.



q      Consequently, a molecule which contains an alkene pi bond and a carbonyl pi bond can, if desired, be selectively hydrogenated at the alkene pi bond. In the example shown, the starting material has an aromatic ring, a ketone function, and an alkene function. Under mild conditions only the alkene function is reduced.







Example 1: Sketch an efficient synthesis of the target compound shown below using any organic compound having four carbon atoms or less, and any inorganic reagents, solvents and reaction conditions necessary.






q      Note: The use of the three  carbon Grignard reagent would require a five carbon aldehyde, which is not permitted under these ground rules.


Example 2.  Sketch an efficient synthesis of the following target alcohol using any organic halide and any carbonyl compound.






Example 3. Sketch an efficient synthesis of the following carboxylic acid ( Hint: Use carbon dioxide as a carbonyl component).