Chapter 23, Problem 23.50P

Draw the organic products formed in each reaction.

a. e.

b. f.

c. g.

d. h.

Interpretation Introduction

(a)

Interpretation: The organic products formed in given reaction are to be drawn.

Concept introduction: Decarboxylation occurs in carboxylic acids.[a1] It involves a cleavage of alpha carbon with loss of carbon dioxide along with heating.

Answer to Problem 23.50P

The organic product formed in given reaction is,

Figure 1

Explanation of Solution

Decarboxylation occurs when a carboxyl group is attached to alpha carbon of another carbonyl group. It removes the carboxylic group from the alpha carbon of the carbonyl group. The corresponding chemical reaction is shown below.

Figure 2

Hence, the organic product formed in given reaction is $\text{4-oxo-1, 2, 3, 4-tetrahydronaphthalene-1-carboxylic acid}$.

Conclusion

The organic product formed in given reaction is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation: The organic products formed in given reaction are to be drawn.

Concept introduction: Lithium diisopropylamide (LDA) is a sterically hindered strong base. In an unsymmetrical ketone, it abstracts hydrogen from least substituted carbon to form kinetic product.

Answer to Problem 23.50P

The organic product formed in given reaction is,

Figure 3

Explanation of Solution

LDA abstracts a proton from the alpha carbon of ethylbutanoate to form enolate. This enolate ion acts as a nucleophile and attacks on electrophilic centre of iodoethane and forms $\text{ethyl 2-ethylbutanoate}$. The corresponding chemical reaction is shown below.

Figure 4

Hence, the organic product formed in given reaction is $\text{ethyl 2-ethylbutanoate}$.

Conclusion

The organic product formed in given reaction is shown in Figure 3.

Interpretation Introduction

(c)

Interpretation: The organic products formed in given reaction are to be drawn.

Concept introduction: Alkyl amines are formed by the reaction of alkyl halides with amines. This reaction is known as alkylation of amines.

Answer to Problem 23.50P

The organic product formed in given reaction is,

Figure 5

Explanation of Solution

In the first step, $\text{2-bromooctahydronapthalen-1}\left(2H\right)\text{-one}$ reacts with $\text{propan-2-amine}$ to form a salt of secondary amine. Then, $\text{propan-2-amine}$ removes a proton from this secondary salt and forms $\text{2-}\left(\text{isopropylamino}\right)\text{octahydronapthalen-1}\left(2H\right)\text{-one}$. The corresponding chemical reaction is shown below.

Figure 6

Hence, the organic product formed in given reaction is $\text{2-}\left(\text{isopropylamino}\right)\text{octahydronapthalen-1}\left(2H\right)\text{-one}$.

Conclusion

The organic product formed in given reaction is shown in Figure 5.

Interpretation Introduction

(d)

Interpretation: The organic products formed in given reaction are to be drawn.

Concept introduction: Lithium diisopropylamide (LDA) is a sterically hindered strong base. In an unsymmetrical ketone, it abstracts hydrogen from least substituted carbon to form kinetic product.

Answer to Problem 23.50P

The organic product formed in given reaction is,

Figure 7

Explanation of Solution

LDA abstracts a proton from the less hindered alpha carbon of $\text{1-cyclohexylpropan-1-one}$ to form enolate ion. This enolate ion acts as a nucleophile and attacks on electrophilic centre of iodoethane and forms $\text{1-cyclohexyl-2-methylbutan-1-one}$. The corresponding chemical reaction is shown below.

Figure 8

Hence, the organic product formed in given reaction is $\text{1-cyclohexyl-2-methylbutan-1-one}$.

Conclusion

The organic product formed in given reaction is shown in Figure 7.

Interpretation Introduction

(e)

Interpretation: The organic products formed in given reaction are to be drawn.

Concept introduction: Monobromo product is formed by the reaction of carbonyl compounds with bromine in the presence of acetic acid and further treatment of monobromo with base yield alkenes.

Answer to Problem 23.50P

The organic product formed in given reaction is,

Figure 9

Explanation of Solution

Monobromo product is formed by the reaction of $\text{2, 2, 6-trimethylcyclohexanone}$ with bromine in the presence of acetic acid. On the further treatment of monobromo product with lithium carbonate (base) and forms an alkene. The corresponding chemical reaction is shown below.

Figure 10

Hence, the organic product formed in given reaction is $\text{2, 2, 6-trimethylcyclohex-2-enone}$.

Conclusion

The organic product formed in given reaction is shown in Figure 9.

Interpretation Introduction

(f)

Interpretation: The organic products formed in given reaction are to be drawn.

Concept introduction: Haloform is produced by the halogenation of a methyl ketone in the presence of a base. This reaction is also known as haloform reaction.

Answer to Problem 23.50P

The organic product formed in given reaction is,

Figure 11

Explanation of Solution

The enolate ion reacts with iodine and form $\text{1-cyclopropyl-2-iodoethanone}$. An enolate ion is formed by the abstraction of a proton by hydroxide ion forms $\text{1-cyclopropyl-2-iodoethanone}$.

$\text{1-cyclopropyl-2, 2-diiodoethanone}$ is formed by the reaction of enolate ion with iodine. Again, an enolate ion is formed by the abstraction of a proton by hydroxide ion from $\text{1-cyclopropyl-2, 2-diiodoethanone}$.The enolate ion reacts with iodine and form $\text{1-cyclopropyl-2, 2, 2-triiodoethanone}$. Tetrahedral intermediate is formed by the abstraction of a proton by hydroxide ion from $\text{1-cyclopropyl-2, 2, 2-triiodoethanone}$. In the final step, the decomposition of tetrahedral intermediate into sodium cyclopropanecarboxylate and iodoform takes place. The mechanism is shown below.

Figure 12

Conclusion

The organic product formed in given reaction is shown in Figure 11.

Interpretation Introduction

(g)

Interpretation: The organic products formed in given reaction are to be drawn.

Concept introduction: Sodium hydride is a strong base. It abstracts a proton and forms carbanion. This carbanion acts as a nucleophile and attacks on the electrophilic centre of alkyl halide.

Answer to Problem 23.50P

The organic product formed in given reaction is,

Figure 13

Explanation of Solution

Sodium hydride abstracts a proton from the carbon which is attached to cyanide and forms secondary carbanion. The next step is intra molecular cyclization. Now this carbanion acts as nucleophile and attack to the carbon (electrophile) which is directly attach to the electronegative atom i.e. chlorine. The corresponding chemical reaction of given organic compound with a strong base is shown below

Figure 14

Hence, the organic product formed in given reaction is cyclopentanecarbonitrile.

Conclusion

The organic product formed in given reaction is shown in Figure 13.

Interpretation Introduction

(h)

Interpretation: The organic products formed in given reaction are to be drawn.

Concept introduction: Bromination is a type of radical substitution reaction. It is highly specific reaction. In this reaction, an alkyl halide is formed by the replacement of hydrogen atom from highly substituted carbon by ${\text{Br}}_{}$.

Answer to Problem 23.50P

The organic product formed in given reaction is,

Figure 15

Explanation of Solution

An alkyl halide is formed by the replacement of hydrogen atom from highly substituted carbon by ${\text{Br}}_{}$. But in the given compound, there are two alpha protons; one is sterically hindered by the three methyl groups. Thus, bromine abstracts a proton from another alpha carbon and in excess of bromine; it will again abstracts a proton from the less hindered alpha carbon and forms a product $\text{4, 4-dibromo-2, 2-dimethylhexan-3-one}$. The corresponding chemical reaction is shown below.

Figure 16

Hence, the organic product formed in given reaction is $\text{4, 4-dibromo-2, 2-dimethylhexan-3-one}$.

Conclusion

The organic product formed in given reaction is shown in Figure 15.