chapter 3 personal practice questions Flashcards
everything on janice questions and solutions first to eight Edition
Problem 3.2:
Functional groups in shikimic acid:
Hydroxyl group (-OH): Present on the cyclic ring.
Carboxylic acid group (-COOH): Present on the side chain.
Problem 3.2:
Functional groups in oseltamivir:
Amine group (-NH2): Present as a primary amine.
Ester group (-COO-): Present in the side chain.
Ether group (-O-): Present in the cyclic structure.
Double bond: Part of the ring system.
Problem 3.3: Draw the structures of compounds fitting the description:
a) Aldehyde (C4H8O): Butanal.
b) Ketone (C4H8O): Butan-2-one.
c) Carboxylic acid (C4H8O2): Butanoic acid.
d) Ester (C4H8O2): Methyl propanoate.
Problem 3.4:
a) Two constitutional isomers (C5H10O) with different functional groups:
Pentanal (aldehyde).
Pentan-2-one (ketone).
b) Two constitutional isomers (C6H10O) with the same functional group:
Cyclohexanone (ketone).
Hex-2-en-1-one (ketone).
Problem 3.4: Classification of Alkyl Halides and Alcohols
(a) CH3CH2CH2CH2OH (1-butanol):
The alcohol is 1° (primary) because the -OH group is attached to a primary carbon.
(b) Cyclohexyl fluoride (C6H11F):
The alkyl halide is 2° (secondary) because the fluorine atom is attached to a secondary carbon.
(c) (CH3)3CBr (tert-butyl bromide):
The alkyl halide is 3° (tertiary) because the bromine atom is attached to a tertiary carbon.
(d) (CH3)3COH (tert-butyl alcohol):
The alcohol is 3° (tertiary) because the -OH group is attached to a tertiary carbon.
Problem 3.5
Intermolecular forces present in each compound:
a) Cyclohexane (C6H12):
London dispersion forces (weak, nonpolar molecule).
b) Tetrahydrofuran (THF):
London dispersion forces.
Dipole-dipole interactions (due to the polar C-O bond).
c) Triethylamine ((CH3CH2)3N):
London dispersion forces.
Dipole-dipole interactions (due to the polar C-N bond).
d) Vinyl chloride (CH2=CHCl):
London dispersion forces.
Dipole-dipole interactions (due to the polar C-Cl bond).
e) Propanoic acid (CH3CH2COOH):
London dispersion forces.
Dipole-dipole interactions (due to the polar C=O and O-H bonds).
Hydrogen bonding (between the O-H groups).
f) 2-Butyne (CH3-C≡C-CH3):
London dispersion forces (weak, nonpolar molecule).
Summary:
Hydrogen bonding: Present in e (Propanoic acid).
Dipole-dipole interactions: Present in b (THF), c (Triethylamine), d (Vinyl chloride), and e (Propanoic acid).
London dispersion forces: Present in all compounds.
Which compound in each pair has the higher boiling point?
a)
A (n-hexane) has a higher boiling point than B (2,2-dimethylbutane).
Reason: n-Hexane has a linear structure, allowing greater surface area for London dispersion forces. In contrast, 2,2-dimethylbutane is branched, reducing these forces and lowering the boiling point.
b)
D (2-propanol) has a higher boiling point than C (propane).
Reason: 2-propanol has an -OH group, enabling hydrogen bonding, which significantly increases its boiling point. Propane, on the other hand, relies only on weaker London dispersion forces.
Problem 3.6
Which compound in each pair has the higher boiling point?
a) (CH3)2C=CH2 (isobutene) vs. (CH3)2C=O (acetone):
(CH3)2C=O (acetone) has the higher boiling point.
Reason: Acetone has dipole-dipole interactions due to the polar C=O bond, while isobutene exhibits only London dispersion forces.
b) CH3CH2COOH (propanoic acid) vs. CH3COOCH3 (methyl acetate):
CH3CH2COOH (propanoic acid) has the higher boiling point.
Reason: Propanoic acid forms strong hydrogen bonds, while methyl acetate primarily exhibits dipole-dipole interactions.
c) CH3(CH2)4CH3 (hexane) vs. CH3(CH2)5CH3 (heptane):
CH3(CH2)5CH3 (heptane) has the higher boiling point.
Reason: Heptane has a larger molecular size and greater surface area, leading to stronger London dispersion forces.
d) CH2=CHCl (vinyl chloride) vs. CH2=CHI (vinyl iodide):
CH2=CHI (vinyl iodide) has the higher boiling point.
Reason: Iodine is more polarizable than chlorine, resulting in stronger London dispersion forces.
Problem 3.7
Why does propanamide (CH3CH2CONH2) have a significantly higher boiling point than N,N-dimethylformamide (HCON(CH3)2) despite being isomeric amides?
Reason: Propanamide can form intermolecular hydrogen bonds due to its -NH2 group, which has both a hydrogen donor and acceptor. In contrast, N,N-dimethylformamide lacks N-H bonds, so it cannot engage in hydrogen bonding, relying only on dipole-dipole interactions. Hydrogen bonding in propanamide significantly increases its boiling point (213°C vs. 153°C).
Problem 3.8
Predict which compound in each pair has the higher melting point:
a) Linear alkane vs. Amine:
The linear alkane has the higher melting point.
Reason: Linear alkanes pack more efficiently in the solid state, leading to stronger London dispersion forces. The amine has polar interactions but cannot pack as tightly due to the presence of the -NH2 group.
b) Branched alkane (neopentane) vs. Linear alkane (n-pentane):
The linear alkane (n-pentane) has the higher melting point.
Reason: Linear molecules pack more efficiently into a crystalline lattice than highly branched molecules, which reduces their melting points despite similar molecular weights.
Problem 3.9
a) Intermolecular forces in acetic acid (CH3CO2H) and sodium acetate (CH3CO2Na):
Acetic acid:
Hydrogen bonding (due to -COOH group).
Dipole-dipole interactions.
London dispersion forces.
Sodium acetate:
Ionic bonding (between Na+ and CH3CO2− ions).
Ion-dipole interactions (in aqueous environments).
London dispersion forces.
b) Why is the melting point of sodium acetate (324 °C) much higher than acetic acid (17 °C)?
Reason: Sodium acetate is an ionic compound, and ionic bonds are much stronger than hydrogen bonds present in acetic acid. Breaking the ionic lattice in sodium acetate requires significantly more energy, resulting in a much higher melting point. Acetic acid, being molecular, has weaker intermolecular forces and therefore melts at a lower temperature.
Problem 3.10
Which compounds are water soluble?
a) CH3CH2OCH2CH3 (Diethyl ether):
Not water soluble.
Reason: While diethyl ether contains a polar C-O bond, the molecule lacks the ability to form strong hydrogen bonds with water due to its relatively large hydrophobic region.
b) CH3CH2CH2CH2CH3 (Pentane):
Not water soluble.
Reason: Pentane is a completely nonpolar hydrocarbon and cannot form hydrogen bonds or significant dipole-dipole interactions with water.
c) (CH3CH2CH2CH2)3N (Tri-n-butylamine):
Not water soluble.
Reason: The molecule has a nitrogen atom that can form weak hydrogen bonds, but the large hydrophobic alkyl groups outweigh its polarity, making it insoluble in water.
Key Points for Water Solubility:
Compounds capable of hydrogen bonding (e.g., -OH, -NH, or -COOH groups) tend to be water soluble.
Large hydrophobic regions reduce water solubility, even if polar groups are present.
Small, polar molecules with functional groups like -OH or -NH are more likely to dissolve in water.
Problem 3.11
Label the hydrophobic and hydrophilic portions of each molecule:
a) Norethindrone (oral contraceptive component):
Hydrophobic portion: The large steroid ring structure is hydrophobic because it consists mainly of nonpolar C-C and C-H bonds.
Hydrophilic portion: The hydroxyl group (-OH) and the triple bond near the ketone (C=O) are hydrophilic due to their polarity and ability to participate in hydrogen bonding.
b) Arachidonic acid (fatty acid):
Hydrophobic portion: The long carbon chain (CH3-(CH2)14-) is hydrophobic because it is made up of nonpolar C-H and C-C bonds.
Hydrophilic portion: The carboxylic acid group (-COOH) at the end of the molecule is hydrophilic because it can form hydrogen bonds with water
Problem 3.12
Predict the water solubility of each vitamin:
a) Vitamin B3 (Niacin):
Water Soluble
Reason: Vitamin B3 has a polar hydroxyl group (-OH) and a carboxylic acid group (-COOH), which can both form hydrogen bonds with water. These hydrophilic groups make the molecule highly soluble in water despite the aromatic ring.
b) Vitamin K1 (Phylloquinone):
Not Water Soluble
Reason: Vitamin K1 has a large hydrophobic tail consisting of a long chain of nonpolar hydrocarbons. Although it has some polar groups (C=O), the hydrophobic tail dominates, making it insoluble in water. It is soluble in fats and oils (fat-soluble vitamin).
Key Concept: Solubility
Water-soluble vitamins (e.g., Vitamin B3) typically have polar functional groups that can hydrogen bond with water.
Fat-soluble vitamins (e.g., Vitamin K1) have large hydrophobic regions that prevent water solubility but make them soluble in lipids.
Problem 3.14
Which of the following structures represent soaps? Explain your answers
.
a) CH3CO2⁻ Na⁺ (Sodium acetate):
Not a soap.
Reason: Sodium acetate is a short-chain carboxylate salt. Soaps require long hydrocarbon chains (hydrophobic tails) for their cleaning action.
b) CH3(CH2)14CO2⁻ Na⁺ (Sodium palmitate):
Soap.
Reason: This molecule has a long hydrophobic hydrocarbon chain (CH3(CH2)14) and a hydrophilic carboxylate group (CO2⁻), meeting the structural requirements for a soap.
c) CH3(CH2)12COOH (Lauric acid):
Not a soap.
Reason: This is a fatty acid, not its salt form. Soaps are salts of fatty acids, not the acids themselves.
d) CH3(CH2)9CO2⁻ Na⁺ (Sodium decanoate):
Soap.
Reason: Similar to sodium palmitate, this molecule has a long hydrophobic hydrocarbon chain and a hydrophilic carboxylate group, making it a soap.
Problem 3.15
Explain how the synthetic detergent cleans away dirt.
Structure Analysis:
The detergent molecule has a hydrophobic tail (long hydrocarbon chain) and a hydrophilic head (SO3⁻ Na⁺ group).
Cleaning Action:
Hydrophobic Tail:
The nonpolar tail dissolves and interacts with grease and oils (nonpolar substances) on dirty surfaces.
Hydrophilic Head:
The polar head remains attracted to water molecules.
Micelle Formation:
Detergent molecules surround grease or dirt particles, with the hydrophobic tails embedded in the grease and the hydrophilic heads facing outward into the water, forming micelles.
Dirt Removal:
The micelles are soluble in water, allowing the grease/dirt to be washed away during rinsing.
Problem 3.16
What two functional groups are present in nonactin and valinomycin?
Nonactin:
Ester group (-COOR): Found in the cyclic ester structure.
Ether group (-O-): Present as part of the cyclic backbone.
Valinomycin:
Ester group (-COOR): Found in the cyclic ester linkages.
Amide group (-CONH-): Present in the peptide linkages within the molecule.
Problem 3.17
Why does aspirin cross a cell membrane as a neutral carboxylic acid rather than as an ionic conjugate base?
Reason:
Nonpolar environment of membranes: Cell membranes consist of a lipid bilayer, which is largely nonpolar. Neutral molecules, like the carboxylic acid form of aspirin, are more lipophilic (fat-soluble) and can diffuse through the hydrophobic interior of the membrane.
Polarity of ionic conjugate base: The ionic form of aspirin (its conjugate base, carboxylate ion -COO⁻) is highly polar and water-soluble. It cannot easily cross the nonpolar lipid bilayer due to its inability to interact with hydrophobic regions.
Equilibrium: At the slightly acidic pH of the stomach, aspirin exists predominantly in its protonated, neutral form, favoring its absorption across membranes.
Problem 3.19
Considering only electron density, state whether the following reactions will occur:
a) CH3CH2-Br + ⁻OH →
Reaction will occur.
Reason: The hydroxide ion (⁻OH) is a strong nucleophile that can attack the electrophilic carbon in CH3CH2-Br (which is partially positive due to the electronegativity of bromine). Bromine acts as a leaving group, allowing the substitution reaction to proceed.
b) CH3-C≡C-CH3 + Br⁻ →
Reaction will not occur.
Reason: Bromide (Br⁻) is a weak nucleophile and a poor base. Since the alkyne carbon atoms in CH3-C≡C-CH3 are not electrophilic enough, no reaction takes place.
c) CH3-C(=O)-Cl + ⁻OCH3 →
Reaction will occur.
Reason: The carbon in CH3-C(=O)-Cl is highly electrophilic due to the withdrawing effects of both the oxygen and the chlorine. The methoxide ion (⁻OCH3) is a strong nucleophile that readily attacks the electrophilic carbon, leading to nucleophilic acyl substitution.
d) CH3-C≡C-CH3 + Br⁺ →
Reaction will occur.
Reason: The positively charged bromine (Br⁺) is an electrophile. It will react with the π-electrons of the alkyne in CH3-C≡C-CH3, leading to an electrophilic addition reaction.
Problem 3.18
Label the electrophilic and nucleophilic sites in each molecule:
a) Cyclohexyl bromide (C6H11Br):
Electrophilic site: The carbon atom attached to the bromine.
Reason: The electronegative bromine atom withdraws electron density from the carbon, making it partially positive and susceptible to nucleophilic attack.
Nucleophilic site: The bromine atom.
Reason: Bromine can act as a leaving group, carrying a lone pair of electrons.
b) Water (H2O):
Nucleophilic site: The oxygen atom.
Reason: Oxygen has lone pairs of electrons that can be donated to an electrophile.
c) Benzene (C6H6):
Nucleophilic site: The π-electrons in the aromatic ring.
Reason: The delocalized π-electrons in benzene can be donated to an electrophile during electrophilic aromatic substitution.
d) Cyclohexyl imine (C6H11N=CH3):
Electrophilic site: The carbon atom in the C=N bond.
Reason: The nitrogen atom withdraws electron density from the carbon, making it partially positive.
Nucleophilic site: The nitrogen atom in the C=N bond.
Reason: Nitrogen has a lone pair of electrons that can act as a nucleophile.
Problem 3.19
Considering only electron density, state whether the following reactions will occur:
a) CH3CH2-Br + ⁻OH →
Reaction will occur.
Reason: The hydroxide ion (⁻OH) is a strong nucleophile that can attack the electrophilic carbon in CH3CH2-Br (which is partially positive due to the electronegativity of bromine). Bromine acts as a leaving group, allowing the substitution reaction to proceed.
b) CH3-C≡C-CH3 + Br⁻ →
Reaction will not occur.
Reason: Bromide (Br⁻) is a weak nucleophile and a poor base. Since the alkyne carbon atoms in CH3-C≡C-CH3 are not electrophilic enough, no reaction takes place.
c) CH3-C(=O)-Cl + ⁻OCH3 →
Reaction will occur.
Reason: The carbon in CH3-C(=O)-Cl is highly electrophilic due to the withdrawing effects of both the oxygen and the chlorine. The methoxide ion (⁻OCH3) is a strong nucleophile that readily attacks the electrophilic carbon, leading to nucleophilic acyl substitution.
d) CH3-C≡C-CH3 + Br⁺ →
Reaction will occur.
Reason: The positively charged bromine (Br⁺) is an electrophile. It will react with the π-electrons of the alkyne in CH3-C≡C-CH3, leading to an electrophilic addition reaction.
Problem 3.20: Analysis of Aspartame
(a) Identify the functional groups in aspartame:
Amine group (-NH2): Found on the left side of the molecule.
Carboxylic acid group (-COOH): Found near the center of the molecule.
Ester group (-COOCH3): Found on the right side of the molecule (methyl ester).
Amide group (-CONH): Found in the peptide linkage between the amine and the aromatic ring.
Aromatic ring (benzene): Found as part of the phenylalanine residue.
Problem 3.20: Analysis of Aspartame
(b) Label all of the sites that can hydrogen
bond to the oxygen atom of water:
The hydrogen atoms of the amide group (-CONH) can form hydrogen bonds with water’s oxygen.
The hydrogen atoms of the amine group (-NH2) can also form hydrogen bonds with water’s oxygen.
The hydrogen atom of the carboxylic acid (-COOH) can form hydrogen bonds with water’s oxygen.
(c) Label all of the sites that can hydrogen bond with the hydrogen atom of water:
The oxygen atom of the carboxylic acid (-COOH) can accept hydrogen bonds from water.
The oxygen atom of the ester group (-COOCH3) can accept hydrogen bonds from water.
The oxygen atom of the amide group (-CONH) can accept hydrogen bonds from water.
The oxygen atom of the amine group (-NH2) can also accept hydrogen bonds from water.
Problem 3.41: Predict the water solubility of each organic molecule
a) Caffeine
Prediction: Highly water-soluble.
Reason: Caffeine contains multiple polar groups, including carbonyl groups (C=O) and nitrogen atoms with lone pairs, which can form hydrogen bonds with water. Its structure makes it highly hydrophilic despite its aromatic rings.
b) Mestranol
Prediction: Poorly water-soluble.
Reason: Mestranol is mostly hydrophobic due to its large steroid structure with nonpolar rings. Although it has a hydroxyl (-OH) group and a methoxy (-OCH3) group, these polar regions are insufficient to overcome the molecule’s predominantly hydrophobic character.
c) Sucrose (table sugar)
Prediction: Highly water-soluble.
Reason: Sucrose has numerous hydroxyl (-OH) groups that can form extensive hydrogen bonds with water, making it extremely hydrophilic and readily soluble in water.
d) Carotatoxin
Prediction: Poorly water-soluble.
Reason: Carotatoxin has a long hydrophobic carbon chain and only one hydroxyl (-OH) group. The large nonpolar region dominates, making it insoluble in water.
Summary of Water Solubility:
Highly water-soluble: Caffeine, sucrose.
Poorly water-soluble: Mestranol, carotatoxin
Problem 3.1: Reactions of CH₃CH₂OH and CH₃CH₃ with the Given Reagents
(a) Reaction of CH₃CH₂OH (ethanol) with H₂SO₄:
Reaction: Dehydration of ethanol to form ethene (C₂H₄).
Mechanism:
Step 1: Protonation of the hydroxyl group by H₂SO₄, forming a good leaving group (H₂O).
Step 2: Loss of water generates a carbocation.
Step 3: A hydrogen atom is eliminated from the adjacent carbon, forming a double bond (alkene).
Product: Ethene (C₂H₄).
(b) Reaction of CH₃CH₂OH (ethanol) with NaH:
Reaction: Formation of sodium ethoxide (CH₃CH₂O⁻Na⁺).
Mechanism:
Sodium hydride (NaH) reacts with ethanol by removing a proton (H⁺) from the hydroxyl group.
This produces sodium ethoxide and hydrogen gas (H₂).
Product: Sodium ethoxide (CH₃CH₂O⁻Na⁺) and H₂ gas.
Reactions of CH₃CH₃ (ethane) with the Same Reagents
(a) With H₂SO₄:
No reaction occurs.
Reason: Ethane lacks a functional group (like -OH) to interact with sulfuric acid.
(b) With NaH:
No reaction occurs.
Reason: Ethane has no acidic hydrogen that can react with NaH.
Problem 3.2: Classification of Carbon and Hydrogen Atoms
(a) Classify the carbon atoms in each compound as 1°, 2°, 3°, or 4°:
Compound [1]:
Both terminal carbons are 1° (primary).
The two middle carbons are 2° (secondary).
Compound [2]:
Three methyl carbons attached to the central carbon are 1° (primary).
The central carbon is 4° (quaternary).
Compound [3]:
All four terminal carbons are 1° (primary).
The two carbons attached to the central carbon are 2° (secondary).
The central carbon is 3° (tertiary).
Compound [4]:
All carbons in the ring are 2° (secondary) except the two methyl group carbons, which are 1° (primary).
(b) Classify the hydrogen atoms in each compound as 1°, 2°, or 3°:
Compound [1]:
Hydrogens attached to terminal carbons are 1°.
Hydrogens attached to secondary carbons are 2°.
Compound [2]:
Hydrogens on methyl groups are 1°.
Compound [3]:
Hydrogens on the terminal carbons are 1°.
Hydrogens on secondary carbons are 2°.
Compound [4]:
Hydrogens on the ring carbons are 2°.
Hydrogens on methyl carbons are 1°.
Problem 3.3: Classification of
𝑠𝑝3
-Hybridized Carbons in Bilobalide
1° (Primary) Carbons:
Carbons directly bonded to only one other carbon. These are the methyl groups.
2° (Secondary) Carbons:
Carbons bonded to two other carbons. Most of the carbons in the rings fall under this category.
3° (Tertiary) Carbons:
Carbons bonded to three other carbons. The carbon bonded to the hydroxyl (-OH) group and two other carbons qualifies as tertiary.
4° (Quaternary) Carbons:
Carbons bonded to four other carbons. Any carbon at a branching point in the rings qualifies.
Problem 3.5: Classification of OH Groups and Halogens in Dexamethasone
OH Groups:
The -OH group attached to a secondary carbon is 2° (secondary).
The -OH group attached to a primary carbon is 1° (primary).
Halogens (F):
The fluorine atom attached to a secondary carbon is 2° (secondary).
problem 3.6: Classification of Amines
(a) Spermine:
Amines are classified based on the number of carbon atoms bonded to the nitrogen atom:
The terminal amines (-NH2) are 1° (primary).
The internal amine (-NH-) is 2° (secondary).
(b) Meperidine:
The nitrogen atom bonded to two carbon atoms is a 2° (secondary amine).
Problem 3.7: Drawing Structures for Given Descriptions
(a) Compound with molecular formula C4H11NO that contains a 1° amine and a 3° alcohol:
Example: (CH3)3COCH2NH2 (tert-butylaminomethanol).
The -NH2 group (1° amine) and the -OH group on the tertiary carbon (3° alcohol) meet the requirements.
(b) Compound with molecular formula C4H11NO that contains a 3° amine and a 1° alcohol:
Example: (CH3)3NCH2OH (trimethylaminomethanol).
The -N(CH3)3 group (3° amine) and the -OH group on a primary carbon (1° alcohol) meet the requirements.
Problem 3.8: Classification of Amides in Dolastatin
Classification of Amides:
1° Amides (Primary):
None are present in dolastatin, as all amide groups have at least one nitrogen bond to a carbon.
2° Amides (Secondary):
Amides with one hydrogen atom bonded to nitrogen and two other bonds to carbon atoms.
Found throughout the dolastatin structure where the nitrogen in the amide group (-CONH-) is bonded to one hydrogen and two carbon atoms.
3° Amides (Tertiary):
Amides with nitrogen bonded to three carbon atoms.
Found in dolastatin where the nitrogen in the amide group is substituted with two alkyl or aryl groups and no hydrogens.
Problem 3.9: Functional Groups in Oseltamivir and Shikimic Acid
Functional Groups in Shikimic Acid:
Alcohol (-OH): Found on the hydroxyl groups attached to the ring.
Carboxylic acid (-COOH): Present as the acidic functional group.
Functional Groups in Oseltamivir:
Ester (-COO-): Found in the side chain.
Amine (-NH2): Present as a functional group attached to the ring.
Ether (-O-): Found in the structure, connecting parts of the molecule.
Alkyl groups: Non-polar chains attached to the structure.
Problem 3.10: Draw a Compound Matching Each Description
An aldehyde with molecular formula C₄H₈O:
Example: Butanal (CH₃CH₂CH₂CHO).
A ketone with molecular formula C₄H₈O:
Example: Butanone (CH₃COCH₂CH₃).
A carboxylic acid with molecular formula C₄H₈O₂:
Example: Butanoic acid (CH₃CH₂CH₂COOH).
An ester with molecular formula C₄H₈O₂:
Example: Methyl propanoate (CH₃CH₂COOCH₃).
Problem 3.11: Draw Isomers and Name Functional Groups
(a) Two constitutional isomers with molecular formula C₅H₁₀O that contain different functional groups:
Pentanal (aldehyde): CH₃CH₂CH₂CH₂CHO.
2-Pentanone (ketone): CH₃COCH₂CH₂CH₃.
(b) Two constitutional isomers with molecular formula C₆H₁₂O that contain the same functional group:
Hexanal (aldehyde): CH₃CH₂CH₂CH₂CH₂CHO.
2-Hexanone (ketone): CH₃COCH₂CH₂CH₂CH₃.
