Chapter 20, Problem 31P

Write a structural formula for the principal organic product or products of each of the

following reactions:

Propanoyl chloride and sodium propanoate

Butanoyl chloride and benzyl alcohol

$\text{p-Chlorobenzoyl}$ chloride and ammonia

and water

and aqueous sodium hydroxide to give ${\text{C}}_{\text{4}}{\text{H}}_{\text{4}}{\text{Na}}_{\text{2}}{\text{O}}_{\text{4}}$

and aqueous ammonia to give ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}{\text{N}}_{\text{2}}{\text{O}}_{\text{3}}$

Methyl benzoate and excess phenylmagnesium bromide, then ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$

Acetic anhydride and $\text{3-pentanol}$

Ethyl phenylacetate and lithium aluminum hydride, then ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$

and aqueous sodium hydroxide to give ${\text{C}}_{\text{4}}{\text{H}}_{\text{7}}{\text{NaO}}_{\text{3}}$

and aqueous ammonia

and lithium aluminum hydride, then ${\text{H}}_{\text{2}}\text{O}$

and excess methylmagnesium bromide, then ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$

Ethyl phenylacetate and methylamine ${\text{(CH}}_{\text{3}}{\text{NH}}_{\text{2}}\text{)}$

and aqueous sodium hydroxide to give ${\text{C}}_{\text{5}}{\text{H}}_{\text{10}}{\text{NNaO}}_{\text{2}}$

and aqueous hydrochloric acid, heat to give ${\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{12}}{\text{NO}}_{\text{2}}\right]}^{\text{+}}$

and aqueous hydrochloric acid, heat to give ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}$+ ${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}\text{ClN}$

and aqueous sulfuric acid, heat to give ${\text{CH}}_{\text{7}}{\text{NO}}_{\text{4}}\text{S}$+ ${\text{C}}_{\text{7}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}$

and ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$

and aqueous hydrochloric acid, heat

$\text{p-Methoxybenzonitrile}$ and aqueous sodium hydroxide, heat to give ${\text{C}}_{\text{8}}{\text{H}}_{\text{7}}{\text{NaO}}_{\text{3}}$

Propanenitrile and methylmagnesium bromide, then ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$, heat

Interpretation Introduction

Interpretation:

Structural formulae of the principal organic products of the given reactions are to be written.

Concept Introduction:

As carboxylic acid derivatives contains a carbonyl group, they undergo reactions typical of carbonyl compounds.

Reactivity of carboxylic acid derivative toward nucleophilic acyl substitution depends on its structure. Acid chlorides are most reactive, followed by anhydrides, esters and amides.

The reactivity order can be explained on the basis of resonance stabilization of the carbonyl group by the acyl substituent.

Nucleophilic substitution reactions of carboxylic acid derivatives generally proceed by an addition-elimination mechanism involving a tetrahedral intermediate.

Answer to Problem 31P

Solution:

e)

f)

g)

h)

i)

j)

k)

l)

m)

n)

o)

p)

q)

r)

s)

t)

u)

v)

Explanation of Solution

a) Propanoyl chloride and sodium propanoate:

The propanoate anion acts as a nucleophile, adding to the carbonyl carbon from propanoyl chloride. Chloride is displaced in the next step, reforming the carbonyl to yield the mixed anhydride acetic propanoic anhydride.

b) Butanoyl chloride and benzyl alcohol:

The oxygen of benzyl alcohol adds to the electrophilic carbon of the carbonyl group in butanoyl chloride. Loss of the chloride and deprotonation yields benzyl butyrate.

c) $\text{p-Chlorobenzoyl chloride}$ and ammonia:

Ammonia is the nucleophile in this reaction and adds to the carbonyl carbon of $\text{p-chlorobenzoyl chloride}$. Loss of the chloride and deprotonation of ammonium group yields the product, $\text{p-chlorobenzamide}$

d) Succinic anhydride and water:

Succinic anhydride hydrolyzes to succinic acid

e) Succinic anhydride and aqueous sodium hydroxide to give ${\text{C}}_{\text{4}}{\text{H}}_{\text{4}}{\text{Na}}_{\text{2}}{\text{O}}_{\text{4}}$:

Succinic anhydride undergoes hydrolysis in basic aqueous conditions. The sodium salt of the acid is formed under these conditions.

f) Succinic anhydride and aqueous ammonia to give ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}{\text{N}}_{\text{2}}{\text{O}}_{\text{3}}$:

Ammonia attacks one of the carbonyl carbons. The bond between this carbonyl and the acyl group breaks. An amide group if formed with another molecule of ammonia extracting a proton. Cleavage of the bond between this carbonyl and the anhydride oxygen results in a carboxylate group at the other end of the carbon chain. This forms an ammonium salt.

g) Methyl benzoate and excess phenylmagnesium bromide, then ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$:

Grignard reagents are good nucleophiles and attack the electrophilic carbonyl carbon in esters. A carbon-carbon bond is formed between the carbonyl carbon and the phenyl group from the Grignard reagent, phenylmagnesium bromide. Loss of methoxy group from the ester leads to initial formation of diphenyl ketone. As an excess of phenylmagnesium bromide is used, another phenyl group is added to the initially formed ketone to produce triphenylmethoxy ion. Acidic work up yields triphenyl methanol.

h) Acetic anhydride and $\text{3-pentanol}$:

The alcohol acts as a nucleophile and adds to one of the carbonyl carbons from the anhydride. The loss of acyl group followed by deprotonation of alcohol oxygen yields the ester $\text{pentan-3-yl acetate}$

Ethyl phenylacetate and lithium aluminum hydride, then ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$:

Lithium aluminum hydride is a source of hydride ion. It is used for reduction of a variety of functional groups including esters. The hydride functions as a strong nucleophile, and initially adds to the carbonyl carbon of the ester. The loss of ethoxy group gives $\text{2-phenylacetaldehyde}$.

As the acetaldehyde also contains a carbonyl group, addition of a second hydride to it converts it to $\text{2-phenylethoxy}$ ion. Acidic work up converts this and the ethoxy ion formed in the first step to $\text{2-phenylethanol}$ and ethanol.

j) $\text{dihydrofuran-2(3H)-one}$($\text{γ-buterolactone}$) and aqueous sodium hydroxide to give ${\text{C}}_{\text{4}}{\text{H}}_{\text{7}}{\text{NaO}}_{\text{3}}$

Lactones are cyclic esters. They undergo hydrolysis when treated with aqueous solution of bases like $\text{NaOH}$. The hydroxide ion is a good nucleophile and adds to the carbonyl carbon. Rearrangement of the tetrahedral intermediate leads reformation of the carbonyl group and the breaking of the bond between the carbonyl and the ${\text{sp}}^{\text{3}}$ oxygen. Under the basic conditions, the carboxylic group formed loses a proton to form (sodium) carboxylate and an alcoholic $-\text{OH}$ at the other end of the carbon chain.

k) $\text{dihydrofuran-2(3H)-one}$($\text{γ-buterolactone}$) and aqueous ammonia:

Ammonia molecule adds to the carbonyl carbon, and forms an amide group on deprotonation. The ester bond breaks, forming an alkoxy group at the other end of the carbon chain. Protonation of this forms an alcoholic $-\text{OH}$ at the other end to yield the product $\text{4-hydroxybutanamide}$.

l) $\text{dihydrofuran-2(3H)-one}$($\text{γ-buterolactone}$) and lithium aluminum hydride, then ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$:

Lithium aluminum hydride reduces lactones to diols. A hydride ion adds to the carbonyl carbon of the lactone. Breaking of the ester $\text{C}-\text{O}$ bond results in formation of (metal salt) of $\text{butane-1, 4-bis(olate)}$.

Acidic work up converts this into $1\text{,4-butanediol}$.

m) $\text{dihydrofuran-2(3H)-one}$($\text{γ-buterolactone}$) and excess methylmagnesium bromide, then ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$

Like open chain esters, lactose also reacts with two moles of Grignard reagent per mole lactone. Because a lactone is in internal, cyclic ester, a diol is formed instead of two molecules of alcohol. The reaction starts with the methyl group from the Grignard reagent adding to the carbonyl carbon. This results in breaking of the lactone ring, with a ketone group at one end and an alkoxy group at the other end of the carbon chain. The ketone group reacts with another molecule of methylmagnesium bromide and is converted to a secondary alcohol on acidic work up. Therefore, this reaction gives the product $\text{4-methylpentane-1, 4-diol}$.

n) Ethyl phenylacetate and methylamine (${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$):

This reaction alo is addition-elimination, resulting in the formation of an amide and an alcohol. Methyl amine acts as a nucleophile and adds to the carbonyl carbon of the ester. Loss of the ethoxy group, and deprotonation of the protonated amide yields $\text{N-methyl-2-phenylacetamide}$ and ethanol

o) $\text{1-Methylpyrrolidin-2-one}$ and aqueous sodium hydroxide to give ${\text{C}}_{\text{5}}{\text{H}}_{\text{10}}{\text{NNaO}}_{\text{2}}$

$\text{1-Methylpyrrolidin-2-one}$ is a lactam, a cyclic amide. Treatment with aqueous NaOH results in hydrolysis, leading to ring opening. The hydroxide ion attack on carbonyl carbon and adds to the carbonyl carbon. The amide bond is broken, to form a carboxylate and an amine group at the end of the carbon chain. Under the basic conditions, the carboxylate forms a sodium salt.

p) $\text{1-Methylpyrrolidin-2-one}$ and aqueous hydrochloric acid, heat to give ${{\text{[C}}_{\text{5}}{\text{H}}_{\text{12}}{\text{NO}}_{\text{2}}\text{]}}^{\text{+}}$:

Lactams are poor electrophiles because of the planar, ${\text{sp}}^2$ character of the amide group. The first step is therefore activation of the amide group by protonation of the oxygen atom. This draws away electron density from the carbonyl carbon, making it a better electrophile. A molecule of water adds to this carbon in the next step. Deprotonation of the added water molecule converts it to a geminal diol. Subsequent rearrangement breaks the amide bond, creating two functional groups, a carboxylic acid and an amine, at the two ends of the carbon chain. Presence of the strong acid HCl converts the amine group to secondary ammonium chloride.

Therefore, the product of the reaction is the hydrochloride of $\text{4-(methylamino)butanoic acid}$.

q) $\text{N-phenylacetamide}$ and aqueous hydrochloric acid, heat to give ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}{\text{ + C}}_{\text{6}}{\text{H}}_{\text{8}}\text{ClN}$

Acid hydrolysis of an amide results in the formation of a carboxylic acid and an amide (as amide hydrochloride). The amide group is activated by protonation of the oxygen. Addition of water molecule to the carbonyl carbontakes place. Subsequent proton transfers and breaking of the amide bond yields acetic acid and aniline hydrochloride.

r) $\text{N-methylbenzamide}$ and aqueous sulfuric acid, heat to give ${\text{CH}}_{\text{7}}{\text{NO}}_{\text{4}}{\text{S + C}}_{\text{7}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}$:

Acidic hydrolysis of $\text{N-methylbenzamide}$ starts with protonation of the carbonyl oxygen in the amide. This activates the amide group, and a molecule of water adds to the carbonyl carbon. This is followed by deprotonation and breaking of the amide bond to yield benzoic acid and methyl ammonium hydrogen sulfate.

s) Cyclopentanecarboxamideand ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$:

${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ (shown here as ${\text{P}}_2{\text{O}}_5$ for convenience) is a strong dehydrating agent. It removes the two hydrogens from the primary amine along with the carbonyl carbon as ${\text{H}}_{\text{2}}\text{O}$ to yield a nitrile.

The reaction starts with the nucleophilic oxygen of amide attacking the phosphorus atom. Another oxygen bonded to this phosphorus deprotonates the amine group, with the bond pair moving to form the nitrile triple bond. The remaining proton attached to the nitrogen is extracted by another of the negatively charged oxygens in phosphorus pentoxide to give the product.

t) $\text{3-Methylbutane nitrile}$ and aqueous hydrochloric acid, heat.

Acid hydrolysis of nitriles is reverse reaction of the synthesis of nitrile from an amide. Under acidic conditions, the nitrile is initially hydrolyzed to an amide. The amide undergoes further hydrolysis to a carboxylic acid and ammonium salt.

u) $\text{p-Methoxybenzonitrile}$ and aqueous sodium hydroxide, heat to give ${\text{C}}_{\text{8}}{\text{H}}_{\text{7}}{\text{NaO}}_{\text{3}}$:

Hydrolysis of nitrile initially forms an amide, with the hydroxide attacking the nitrile carbon. The amide formed undergoes further hydrolysis to form the carboxylate salt and ammonia.

v) Propanenitrile and methylmagnesium bromide, then ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$:

The strongly basic methylmagnesium bromide attacks the nitrile carbon, forming an imine on protonation of the nitrogen. Imine then undergoes hydrolysis to ketone and ammonia.