Apply Organic Functional Groups - IB Chemistry

1. A molecule has the molecular formula C₃H₆O. Which pair of functional groups could be present in two of its structural isomers?
   Answer: Aldehyde and ketone The molecular formula C₃H₆O corresponds to a degree of unsaturation of 1. Aldehydes (propanal) and ketones (propanone) are functional group isomers with the general formula CₙH₂ₙO, which for n=3 is C₃H₆O. A: Alcohols and ethers with 3 carbons have the formula C₃H₈O. C: Carboxylic acids and esters with 3 carbons have the formula C₃H₆O₂. D: Alkenes and alkynes are hydrocarbons and contain no oxygen.
2. Which compound is correctly classified as a secondary amine?
   Answer: N-methylethanamine Amines are classified by the number of alkyl groups directly attached to the nitrogen atom. A secondary amine has two alkyl groups on the nitrogen. N-methylethanamine has a methyl group and an ethyl group attached to the nitrogen. A: Propan-2-amine is a primary amine, as the nitrogen is only attached to one carbon atom. C: Propanamide is an amide, not an amine. D: N,N-dimethylethanamine is a tertiary amine, as the nitrogen has three alkyl groups attached.
3. During a multi-step synthesis, an intermediate compound must contain both an electrophilic carbon center and a nucleophilic heteroatom within the same molecule. The compound should also be capable of intramolecular cyclization under basic conditions. Which combination of functional groups would best satisfy these synthetic requirements?
   Answer: Ester group with pendant alcohol for transesterification-based cyclization reactions The correct answer is C. An ester provides an electrophilic carbonyl carbon, and a pendant alcohol provides a nucleophilic oxygen that can attack intramolecularly under basic conditions to form a cyclic ester (lactone). Option A involves SN2 displacement but not electrophilic carbon centers. Option B describes hemiacetal formation which typically requires acidic conditions and reversible equilibria. Option D involves addition to nitriles which requires harsh conditions and doesn't typically lead to cyclization under mild basic conditions.
4. A compound with molecular formula $C_8H_{10}O_2$ exhibits the following properties: it is soluble in aqueous sodium hydroxide, forms a white precipitate with iron(III) chloride solution, and produces a sweet-smelling ester when heated with ethanol in the presence of concentrated sulfuric acid. Which combination of functional groups is most consistent with these observations?
   Answer: Aromatic ring with both hydroxyl and carboxyl groups present The correct answer is A. The solubility in NaOH indicates an acidic functional group (carboxyl), the white precipitate with FeCl₃ indicates phenolic hydroxyl groups, and ester formation confirms the carboxyl group. This describes a hydroxybenzoic acid. B is incorrect because methoxy groups don't give positive FeCl₃ tests. C is incorrect because aldehydes don't dissolve in NaOH or form esters directly. D is incorrect because amino groups would give basic properties and different FeCl₃ behavior.
5. The molecule vanillin is responsible for the flavour of vanilla. Its structure contains a phenyl group, a hydroxyl group, an alkoxy group (-OCH₃), and an aldehyde group (-CHO), all attached to the benzene ring. Which of the following molecular formulas is consistent with vanillin?
   Answer: C₈H₈O₃ To determine the molecular formula, we sum the atoms from each group. The base is a benzene ring (C₆H₆). Attaching three groups means removing three H atoms, leaving a C₆H₃ core. The groups are: aldehyde (-CHO), hydroxyl (-OH), and methoxy (-OCH₃). Total carbons: 6 (ring) + 1 (aldehyde) + 1 (methoxy) = 8. Total hydrogens: 3 (ring) + 1 (aldehyde) + 1 (hydroxyl) + 3 (methoxy) = 8. Total oxygens: 1 (aldehyde) + 1 (hydroxyl) + 1 (methoxy) = 3. The molecular formula is C₈H₈O₃.
6. How many structural isomers with the molecular formula C₄H₉Cl are classified as primary halogenoalkanes?
   Answer: 2 A primary halogenoalkane has the chlorine atom attached to a carbon atom that is bonded to only one other carbon atom. There are two possible carbon skeletons for C₄: butane and 2-methylpropane. For the straight-chain butane skeleton, the chlorine can be on an end carbon, giving 1-chlorobutane (primary). For the branched 2-methylpropane skeleton, the chlorine can be on one of the three equivalent end carbons, giving 1-chloro-2-methylpropane (primary). Therefore, there are two primary isomers. The other isomers are 2-chlorobutane (secondary) and 2-chloro-2-methylpropane (tertiary). The total number of structural isomers is 4.
7. A pharmaceutical compound contains multiple functional groups that affect its bioactivity. The molecule exhibits basic properties ($pK_b ≈ 9$), shows strong UV absorption at 280 nm, and forms hydrogen bonds with biological targets. Analysis reveals it contains exactly two nitrogen atoms and one oxygen atom in addition to carbon and hydrogen. Which structural features most likely contribute to these observed properties?
   Answer: Heterocyclic aromatic system with amino substitution providing basicity, UV absorption, and hydrogen bonding respectively The correct answer is B. A heterocyclic aromatic system (like pyridine or pyrimidine) with amino substitution accounts for: two nitrogens, aromatic UV absorption at 280 nm, basic properties from the amino group, and hydrogen bonding capability. Option A only accounts for one nitrogen. Option C wouldn't show strong UV absorption at 280 nm (no aromatic character). Option D is incorrect because nitriles are not basic (pKb ≈ 9) and typically show different UV absorption patterns.
8. Amino acids contain both a primary amino group and a carboxylic acid group. Which property is expected for a simple amino acid like glycine (H₂NCH₂COOH)?
   Answer: It exists predominantly as a zwitterion in solid form and in neutral aqueous solution. Molecules with both an acidic carboxyl group and a basic amino group undergo an internal acid-base reaction to form a zwitterion, which has both a positive (-NH₃⁺) and a negative (-COO⁻) charge. This ionic nature results in strong intermolecular forces, making them crystalline solids with high melting points, not volatile liquids. It also makes them soluble in polar solvents like water and insoluble in non-polar solvents. Because it has both acidic and basic groups, it can act as both a Brønsted-Lowry acid and base (it is amphiprotic).
9. Which property is a defining characteristic of the members of a homologous series?
   Answer: They all exhibit similar chemical properties and can be represented by a general formula. A homologous series is a series of compounds with the same functional group and similar chemical properties in which the successive members differ by a CH₂ group. The same functional group dictates similar chemical reactions. A is incorrect; for example, in alcohols, methanol (CH₄O) has a different empirical formula than ethanol (C₂H₆O). C is incorrect; physical properties like boiling point show a gradual trend, increasing with chain length. D is incorrect; members of a homologous series have different molecular formulas, so they cannot be isomers of each other.
10. The artificial sweetener aspartame contains several functional groups, including an ester, an amide, a carboxylic acid, and an amino group. Which statement about the functional groups in aspartame must be correct?
    Answer: The molecule contains at least two carbonyl (C=O) groups in different chemical environments. Aspartame contains a carboxylic acid (-COOH), an ester (-COOR), and an amide (-CONH-). Each of these functional groups contains a carbonyl (C=O) group. Since these functional groups are different, the chemical environments of the carbonyl groups are also different. Therefore, the molecule contains at least two carbonyl groups in different environments. A: Aspartame contains a phenyl group, but not a phenol (hydroxyl on the ring). B: This describes an amide. While aspartame has an amide group, it also has a primary amino group where the nitrogen is not bonded to a carbonyl. C: This statement is correct. D: The -OH group in aspartame is part of a carboxylic acid, not an alcohol.
11. An unknown organic compound undergoes the following reaction sequence: Step 1 - treatment with $LiAlH_4$ produces compound X; Step 2 - compound X reacts with $PCC$ to give compound Y; Step 3 - compound Y gives a positive Tollens' test. If the original compound contains a six-carbon chain, which functional group was most likely present in the starting material?
    Answer: Ester group requiring reduction followed by controlled oxidation steps The correct answer is B. LiAlH₄ reduces esters to primary alcohols (X), PCC oxidizes primary alcohols to aldehydes (Y), and aldehydes give positive Tollens' tests. Option A is incorrect because a primary alcohol wouldn't need LiAlH₄ treatment. Option C is incorrect because LiAlH₄ reduces ketones to secondary alcohols, which PCC would oxidize back to ketones (negative Tollens' test). Option D is incorrect because carboxylic acids would also be reduced to primary alcohols, but the question context suggests a more complex starting functional group.
12. The compounds butan-1-ol, butanal, ethoxyethane, and pentane all have similar molar masses (74, 72, 74, and 72 g mol⁻¹ respectively). Which lists the compounds in order of increasing boiling point?
    Answer: pentane < ethoxyethane < butanal < butan-1-ol Boiling point is determined by the strength of intermolecular forces. Pentane only has weak London dispersion forces (LDF). Ethoxyethane (an ether) is slightly polar and has dipole-dipole interactions in addition to LDF. Butanal (an aldehyde) has a more polar carbonyl group, leading to stronger dipole-dipole interactions. Butan-1-ol (an alcohol) can form strong hydrogen bonds, giving it the highest boiling point. The correct order from weakest to strongest intermolecular forces is pentane < ethoxyethane < butanal < butan-1-ol.
13. An unlabeled bottle is known to contain either pentanoic acid or its isomer ethyl propanoate. Both compounds have the molecular formula C₅H₁₀O₂. Which simple test and observation would distinguish between them?
    Answer: Add sodium carbonate solution; only pentanoic acid will cause effervescence. Pentanoic acid is a carboxylic acid and will react with a carbonate (like Na₂CO₃) in an acid-base reaction to produce carbon dioxide gas, which is observed as effervescence (fizzing). Ethyl propanoate is an ester and is neutral, so it will not react. B is incorrect; the carboxylic acid has a much higher boiling point due to hydrogen bonding dimers. C is incorrect; esters are difficult to reduce, and this test would not be a simple distinguishing feature. D is incorrect; both molecules have significant nonpolar character and would be soluble in a nonpolar solvent like hexane.
14. Which compound contains both an alkene and a carboxylic acid functional group?
    Answer: Pent-4-enoic acid The IUPAC name indicates the functional groups present. In pent-4-enoic acid, 'pent-' indicates a 5-carbon chain, '-en-' indicates a C=C double bond (alkene), and '-oic acid' indicates a -COOH group (carboxylic acid). A contains an alkene and a ketone. C contains a hydroxyl (alcohol) and a carboxylic acid. D is an ester.
15. An organic compound undergoes hydrolysis under both acidic and basic conditions, but yields different products in each case. Under acidic conditions, it produces a carboxylic acid and an alcohol. Under basic conditions, it produces a carboxylate salt and the same alcohol. The compound also shows characteristic IR absorption at 1740 $cm^{-1}$. However, when treated with ammonia at elevated temperature, it forms an amide and releases the alcohol. Which functional group classification and mechanistic pathway best explains this behavior?
    Answer: Ester functionality undergoing nucleophilic acyl substitution with different nucleophiles under varying conditions The correct answer is A. The compound is an ester, evidenced by: IR at 1740 cm⁻¹ (ester C=O), hydrolysis to carboxylic acid + alcohol (acidic) or carboxylate + alcohol (basic), and aminolysis to form amide + alcohol. All reactions proceed via nucleophilic acyl substitution. Option B is incorrect because anhydrides typically show two C=O stretches. Option C is incorrect because acetals don't show 1740 cm⁻¹ absorption and don't react with ammonia to form amides. Option D is incorrect because amides don't typically react with ammonia to form different amides.
16. Which molecular formula could correspond to a saturated, acyclic molecule containing a primary amide functional group?
    Answer: C₄H₉NO A primary amide has the functional group -CONH₂. For a 4-carbon saturated acyclic amide, the structure would be based on a C₄ alkane chain. For example, butanamide is CH₃CH₂CH₂CONH₂. Counting the atoms gives: C=4, H=3+2+2+2=9, N=1, O=1. The molecular formula is C₄H₉NO. The general formula for a saturated acyclic primary amide is CₙH₂ₙ₊₁NO. For n=4, this is C₄H₉NO. A (C₄H₇NO) is unsaturated. C (C₄H₁₁N) is a saturated amine. D (C₄H₈O₂) could be a carboxylic acid or ester.
17. What is the correct IUPAC name for the compound with a four-carbon chain that has a hydroxyl group on the second carbon and a methyl group on the third carbon?
    Answer: 3-methylbutan-2-ol The principal functional group is the hydroxyl group (-OH), making it an alcohol. The longest carbon chain containing the -OH group is four carbons long (butan-). The -OH group should be given the lowest possible number. Numbering from the end closer to the -OH group gives it position 2 (butan-2-ol). This places the methyl group on carbon 3. Therefore, the name is 3-methylbutan-2-ol. A uses the wrong numbering direction. C and D incorrectly treat the alcohol as a substituent ('hydroxy').
18. Consider the structures of propan-1-ol and methoxyethane. Which statement best describes the relationship between these two molecules?
    Answer: They are functional group isomers because they have the same molecular formula but different functional groups. Both propan-1-ol (CH₃CH₂CH₂OH) and methoxyethane (CH₃OCH₂CH₃) have the molecular formula C₃H₈O. However, propan-1-ol is an alcohol and methoxyethane is an ether. Isomers that have the same molecular formula but different functional groups are called functional group isomers. A describes positional isomers (e.g., propan-1-ol and propan-2-ol). B describes chain isomers (e.g., butane and methylpropane). D is incorrect as they have the same molecular and empirical formula (C₃H₈O).
19. The skeletal formula of a molecule is represented by a zig-zag line of four connected segments, with a double line between the second and third vertices from the left end. What is the IUPAC name of this molecule?
    Answer: Pent-2-ene In a skeletal formula, each vertex and each end of a line represents a carbon atom. A zig-zag line of four connected segments has five carbon atoms in total (two ends and three vertices). The molecule is therefore a pentene. The double bond is between the second and third carbon atoms. According to IUPAC rules, we number the chain to give the double bond the lowest possible number, so numbering from the left makes it pent-2-ene. B miscounts the number of carbons. C uses incorrect numbering (the lowest number rule gives 2, not 3). D ignores the double bond.
20. Compound Y is a structural isomer of pentan-1-ol (Mᵣ = 88.15 g mol⁻¹). Compound Y has a significantly lower boiling point than pentan-1-ol and does not react with acidified potassium dichromate(VI). What is a possible identity of compound Y?
    Answer: Ethoxypropane Pentan-1-ol has the formula C₅H₁₂O. Its high boiling point is due to hydrogen bonding. An isomer with a significantly lower boiling point is likely an ether, which cannot hydrogen bond with itself. Ethoxypropane (CH₃CH₂OCH₂CH₂CH₃) has the formula C₅H₁₂O and is an ether. Ethers do not react with oxidizing agents like acidified dichromate. A: Pentan-2-ol is an alcohol, can H-bond, and would have a similar boiling point. B: Pentanal (C₅H₁₀O) is not an isomer. D: 2-methylbutan-2-ol is a tertiary alcohol. While it resists oxidation, it can still hydrogen bond and would have a boiling point closer to pentan-1-ol than an ether would.

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