Chapter 19, Problem 19.40AP

Interpretation Introduction

(a)

Interpretation:

The product when acetone reacts with ${\text{H}}_3{\text{O}}^{+}$ is to be predicted.

Concept introduction:

The replacement or substitution of one functional group with another different functional group in any chemical reaction is termed as a substitution reaction. In a nucleophilic substitution reaction, nucleophile takes the position of leaving group by attacking the electron-deficient carbon atom.

Answer to Problem 19.40AP

The gem-diol product is formed when acetone reacts with ${\text{H}}_3{\text{O}}^{+}$. The reaction is shown below.

Explanation of Solution

The hydrated product is formed known as gem-diol. The gem-diol is an unstable compound. Therefore, the reaction is always is in a backward direction. The overall reaction is shown below.

Figure 1

Conclusion

The product of hydration of acetone is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation:

The product when acetone reacts with ${\text{NaBH}}_{\text{4}}$ in ${\text{CH}}_{\text{3}}\text{OH}$ then ${\text{H}}_2\text{O}$ is to be predicted.

Concept introduction:

Sodium borohydride is a reducing agent which is selective in nature. It selectively reduces the carbonyl compounds into corresponding alcohols. It converts ketone into their respective alcohols. This reducing agent is used in the presence of methanol.

Answer to Problem 19.40AP

The $\text{propan-2-ol}$ product is formed when acetone reacts with ${\text{NaBH}}_{\text{4}}$ in ${\text{CH}}_{\text{3}}\text{OH}$, then ${\text{H}}_2\text{O}$. The overall reaction is shown below.

Explanation of Solution

Sodium borohydride is selective in nature which is used as a reducing agent. The conversion of acetone into $\text{propan-2-ol}$ takes place with the help of selective reducing agent. This reaction proceeds through nucleophilic substitution reaction. The sodium borohydride acts as a nucleophile. The overall reaction is given below.

Figure 2

Conclusion

The product of reduction of acetone is shown in Figure 2.

Interpretation Introduction

(c)

Interpretation:

The product when acetone reacts with ${\text{CrO}}_3$, pyridine is to be predicted.

Concept introduction:

Chromium trioxide is the acidic anhydride of chromic acid. It is used as an oxidizing agent which is an inorganic compound. It resists oxidizing the ketones in the presence of pyridine. It shows that chromium trioxide is selective in nature.

Answer to Problem 19.40AP

The product is not formed when acetone reacts with ${\text{CrO}}_3$. The overall reaction is shown below.

Explanation of Solution

Chromium trioxide is used as an oxidizing agent which is an inorganic compound. It resists oxidizing the ketones in the presence of pyridine. Therefore, acetone is not oxidized. It shows that chromium trioxide is selective in nature. The reactant is unreacted due to the inertness of the chromium trioxide. The overall reaction is shown below.

Figure 3

Conclusion

The reaction of acetone with the chromium trioxide is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

The product when acetone reacts with $\text{NaCN}$, $\text{pH}\text{10}$, ${\text{H}}_{\text{2}}\text{O}$ is to be predicted.

Concept introduction:

Sodium cyanide solution in water acts as a base. It is corrosive in nature. As well as reacts very rapidly with acid. It also reacts with water, moisture, carbon dioxide. The $\text{pH}$ is maintained at $10$. When it is reacted with ketone, then the product is formed is cyanohydrins.

Answer to Problem 19.40AP

The product formed is $\text{2-hydroxy-2-methylpropanenitrile}$ when acetone reacts with $\text{NaCN}$, $\text{pH}\text{10}$, ${\text{H}}_{\text{2}}\text{O}$. The overall reaction is shown below.

Explanation of Solution

The acetone reacts with sodium cyanide solution in water. The pH is maintained at $10$ It gives $\text{2-hydroxy-2-methylpropanenitrile}$ which is also known as cyanohydrin. The reaction between acetone and, NaCN $\text{pH}\text{10}$ ${\text{H}}_{\text{2}}\text{O}$ is a base-catalyzed hydration reaction.

There is a nucleophilic addition where cyanide is act as a nucleophile. The overall reaction is given below.

Figure 4

Conclusion

Acetone and sodium cyanide reacts to give the product as shown in Figure 4.

Interpretation Introduction

(e)

Interpretation:

The expected product when acetone reacts with ${\text{CH}}_{\text{3}}\text{OH}$ (excess), ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ (trace).

is to be predicted.

Concept introduction:

A dehydration reaction is defined as the removal of ${\text{H}}_{\text{2}}\text{O}$ in the reaction. Dehydration is the vice versa opposite concept of hydration. In the dehydration process, the alcohol is converted into an alkene.

There are several dehydrating agents in the chemistry some of them are sulfuric acid and ammonia.

Answer to Problem 19.40AP

The product formed is $\text{2, 2-dimethoxypropane}$ when acetone reacts with excess of ${\text{CH}}_{\text{3}}\text{OH}$ in the trace of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$.is shown below.

Explanation of Solution

When acetone reacts with excess of ${\text{CH}}_{\text{3}}\text{OH}$ in the trace of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ form acetal, the reaction is generally used as a way of protecting the carbonyl group in reactions. The carbonyl group can be generated back by hydrolysis of the acetal. The reaction is given below.

Figure 5

Conclusion

The acetone reacts with excess of ${\text{CH}}_{\text{3}}\text{OH}$ in the trace of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ gives $\text{2, 2-dimethoxypropane}$.which is shown in Figure 5.

Interpretation Introduction

(f)

Interpretation:

The expected product when acetone reacts with pyrrolidine and a trace of acid is to be predicted.

Concept introduction:

Pyrrolidine is a colorless water-soluble. It is a heterocyclic compound. Enamine is formed by the condensation process of an aldehyde or ketone. Enamine is also known as $\alpha \text{,}\beta \text{-}$ unsaturated tertiary amine.

Answer to Problem 19.40AP

The product formed is $\text{1-}\left(\text{prop-1-en-2-yl}\right)$ when acetone reacts with pyrrolidine with a trace of acid. The overall reaction is shown below.

Explanation of Solution

When acetone reacts with pyrrolidine in the presence of acid. This reaction is a nucleophilic addition reaction. Pyrrolidine acts as a nucleophile which attacks at the electron-deficient site that is carbonyl carbon. In the final step, protonation takes place. The overall reaction is shown below.

Figure 6

Conclusion

The reaction is shown in Figure 6.

Interpretation Introduction

(g)

Interpretation:

The expected product when acetone reacts with semicarbazide with diluted acid is to be predicted.

Concept introduction:

Semicarbazide is the derivative of urea. Semicarbazide is water-soluble and white in color. Protonation takes place by using acid in the reaction. This reaction is a nucleophilic addition reaction. Semicarbazide is used to convert ketone into amides.

Answer to Problem 19.40AP

The product formed is $\text{2-}\left(\text{propan-2-ylidene}\right)$ when acetone reacts with semicarbazide with diluted acid. The overall reaction is shown below.

Explanation of Solution

The acetone reacts with semicarbazide with diluted acid. This reaction is a nucleophilic addition reaction. Semicarbazide is used to convert ketone into amides. In the first step, the nitrogen of semicarbazide is act as a nucleophile which attacks on the electrophilic site of ketone that is carbonyl carbon. In the second step, rearrangements take place. In the final step, the protonation takes place. The overall reaction is shown below.

Figure 7

Conclusion

The rection is shown in Figure 7.

Interpretation Introduction

(h)

Interpretation:

The expected product when acetone reacts with ${\text{CH}}_{\text{3}}\text{MgI}$, ether, then ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is to be predicted.

Concept introduction:

Grignard reagents are organometallic compounds which are prepared using alkyl halides in the presence of magnesium metal in dry ether. These reagents act as strong nucleophiles and bases. The Grignard reagent is used to increase the chain length of the carbon-carbon bond.

Answer to Problem 19.40AP

The product formed is $\text{2-methylpropan-2-ol}$ when acetone reacts with ${\text{CH}}_{\text{3}}\text{MgI}$, ether then ${\text{H}}_{\text{3}}{\text{O}}^{+}$. The overall reaction is shown below.

Explanation of Solution

The acetone reacts with the Grignard reagent. The Grignard reagent acts as a nucleophile. This reaction is a nucleophilic addition reaction. In the first step, the nucleophile attacks on the electrophilic part of the ketone which is carbonyl carbon. In the final step, protonation takes place. The overall reaction is shown below.

Figure 8

Conclusion

The product formed is $\text{2-methylpropan-2-ol}$ when acetone reacts with ${\text{CH}}_{\text{3}}\text{MgI}$ is shown in Figure 8.

Interpretation Introduction

(i)

Interpretation:

The expected product obtained when $\text{propan-2-ol}$ reacts with ${\text{Na}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ in ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is to be predicted.

Concept introduction:

Oxidation is the process of addition of oxygen. Oxidation is also be defined as loosing of electrons. In the oxidation process, there is increasing in the oxidation state. The acid is added in the reaction for protonation.

The oxidizing agent is defined as the species which oxidizes others and itself gets reduced.

Answer to Problem 19.40AP

The product formed is acetone when $\text{propan-2-ol}$ reacts with ${\text{Na}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ in ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$. The overall reaction is shown below.

Explanation of Solution

The product $\text{propan-2-ol}$ is obtained by the reduction of acetone. It is known that the sodium dichromate is an oxidizing agent. It means that $\text{propan-2-ol}$ is loosing electrons. The sodium dichromate is getting self reduced. The overall reaction is shown below.

Figure 9

Conclusion

The product formed is acetone when $\text{propan-2-ol}$ reacts with ${\text{Na}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ is shown in Figure 9.

Interpretation Introduction

(j)

Interpretation:

The expected product obtained when $\text{2-methylpropan-2-ol}$ reacts with ${\text{H}}_{\text{2}}{\text{SO}}_4$ is to be predicted.

Concept introduction:

A dehydration reaction is defined as the removal of ${\text{H}}_{\text{2}}\text{O}$ in the reaction. Dehydration is the vice versa opposite concept of hydration. In the dehydration process, the alcohol is converted into an alkene.

There are several dehydrating agents in the chemistry some of are like sulfuric acid and ammonia.

Answer to Problem 19.40AP

The product formed is acetone when $\text{2-methylpropan-2-ol}$ reacts with ${\text{H}}_{\text{2}}{\text{SO}}_4$. The dehydration process takes place. The overall reaction is shown below.

Explanation of Solution

In the dehydration process, the alcohol is converted into an alkene. The double bond formation takes place in this reaction. In the first step, the hydroxyl group attracts one proton of an acid. In the second step, the proton is taken by ${\text{HSO}}_4{}^{-}$. In the final step, water is eliminated and alkene is formed as a product.

Figure 10

Conclusion

The dehydrated product of $\text{2-methylpropan-2-ol}$ is shown in Figure 10.

Interpretation Introduction

(k)

Interpretation:

The expected product obtained when acetone reacts with ${\text{H}}_{\text{2}}\cdot {\text{PtO}}_{\text{2}}$ is to be predicted.

Concept introduction:

Hydrogenation is defined as the addition of hydrogen. Platinum is used as a surface catalyst. Mostly, catalysts are insoluble in metals like ${\text{Pd-C, PdO}}_{\text{2}}$ and nickel in used as $\text{Ra-Ni}$. The alkene is absorbed in the surface for hydrogenation.

Answer to Problem 19.40AP

The product formed is $\text{propan-2-ol}$. Hydrogenation of acetone takes place. The overall reaction is shown below.

Explanation of Solution

Hydrogenation is defined as the addition of hydrogen. Platinum is used as a surface catalyst. ${\text{PdO}}_{\text{2}}$ catalyst isis insoluble in metals. The acetone is absorbed in the surface for hydrogenation.

The product formed is $\text{propan-2-ol}$. The overall reaction is shown below.

Figure 11

Conclusion

The reductive product of acetone is shown in Figure 11.

Interpretation Introduction

(l)

Interpretation:

The expected product obtained when acetone reacts with ${\text{H}}_{\text{2}}\text{C}={\text{PPh}}_{\text{3}}$ is to be predicted.

Concept introduction:

The Wittig reaction is the chemical reaction involves a change in an aldehyde or ketone converted into an alkene. The chemical name of Wittig reagent is triphenyl phosphonium ylide. In the Wittig reaction, aldehydes or a ketone reacts with triphenyl phosphonium ylide to generate an alkene and triphenylphosphine oxide. The Wittig reagent gives good yields of alkene even when other functional groups are present on the aldehydes or ketone. However, as the steric hindrance of the aldehydes or ketone increases, the yield of alkene decreases.

Answer to Problem 19.40AP

The product formed is $\text{2-methylprop-1-ene}$ using Wittig reagent. The overall reaction is shown below.

Explanation of Solution

In the Wittig reaction, acetone reacts with triphenyl phosphonium ylide to generate an alkene and triphenylphosphine oxide. The Wittig reagent gives good yields of alkene even when other functional groups are present on the aldehydes or ketone. However, as the steric hindrance of the aldehydes or ketone increases, the yield of alkene decreases. The overall reaction is shown below.

Figure 12

Conclusion

The treatment of Wittig reagent with acetone is shown in Figure 12.

Interpretation Introduction

(m)

Interpretation:

The expected product obtained when acetone reacts with $\text{Zn}$ amalgam in the presence of $\text{HCl}$ is to be predicted.

Concept introduction:

Clemmensen reduction is defined as the reduction in which aldehydes or ketones are converted into alkane or hydrocarbon. The catalyst is used in this reaction is zinc amalgam in the presence of concentrated hydrochloric acid. This is one of the easy methods to convert aldehydes or ketones into hydrocarbon.

Answer to Problem 19.40AP

The product formed is propane with the help of $\text{Zn}$ amalgam. The overall reaction is shown below.

Explanation of Solution

The acetone is converted into propane with the help of zinc-amalgam in the presence of concentrated hydrochloric acid. Zinc, $\text{Zn}$ is oxidized and the lost electrons are attracted by carbonyl carbon. Then protonation takes place and carbocation is formed. Carbocation takes two electrons and is neutralized by acid. The reaction is shown below.

Figure 13

Conclusion

The reduction of acetone with the help of $\text{Zn}$ amalgam is shown in Figure 13.