Organic Chemistry Nitrogen Compounds, Spectroscopy, and Biomolecules Syllabus

Nitrogen Compounds, Spectroscopy, and Biomolecules syllabus — full chapter and topic list

Expand any chapter to see its topics and sub-topics. This is the whole examinable outline for Nitrogen Compounds, Spectroscopy, and Biomolecules in Organic Chemistry, not a summary of it.

1. 1

   Amines

   4 topics

   • Structure, Basicity, and Nomenclature
   • Synthesis of Amines
   • Reactions of Amines
   • Diazonium Salts and Aryl Substitution
2. 2

   Infrared Spectroscopy

   3 topics

   • Bond Vibrations and the IR Region
   • Characteristic Functional Group Frequencies
   • Interpreting IR Spectra
3. 3

   Nuclear Magnetic Resonance Spectroscopy

   4 topics

   • Proton NMR Chemical Shift
   • Integration and Signal Splitting
   • Carbon-13 NMR
   • Structure Determination from Spectra
4. 4

   Mass Spectrometry

   3 topics

   • Molecular Ion and Fragmentation
   • Isotope Patterns
   • Determining Molecular Formula
5. 5

   Biomolecules

   4 topics

   • Carbohydrates
   • Amino Acids, Peptides, and Proteins
   • Lipids and Fatty Acids
   • Nucleic Acids

Nitrogen Compounds, Spectroscopy, and Biomolecules flashcards for Organic Chemistry

25 of 55 cards from the Nitrogen Compounds, Spectroscopy, and Biomolecules deck — real questions with worked answers.

1. How are amines classified as primary, secondary, or tertiary?

   By the number of carbon groups bonded to nitrogen: primary ($1^\circ$) has one carbon on N ($\ce{RNH2}$), secondary ($2^\circ$) has two ($\ce{R2NH}$), and tertiary ($3^\circ$) has three ($\ce{R3N}$). Note this differs from alcohol classification, which counts carbons on the carbon bearing the OH.

2. Why are amines basic, and what species forms when an amine accepts a proton?

   The nitrogen lone pair can accept a proton (Bronsted base) or share electrons (Lewis base). Protonation gives an ammonium ion, e.g. $\ce{RNH2 + H+ -> RNH3+}$. Basicity is measured by $pK_b$ or by the $pK_a$ of the conjugate acid (higher $pK_{aH}$ = stronger base).

3. Rank the basicity of ammonia, a primary alkylamine, an arylamine (aniline), and an amide in water.

   Alkylamine $>$ ammonia $>$ arylamine $>$ amide. Alkyl groups donate electron density (raising basicity); in aniline the lone pair is delocalized into the ring, and in amides into the carbonyl, both lowering basicity.

4. Why is aniline a much weaker base than cyclohexylamine?

   In aniline the nitrogen lone pair is delocalized into the aromatic ring (resonance), making it less available to bond a proton. Cyclohexylamine's lone pair is fully localized on nitrogen, so it is far more basic ($pK_{aH} \approx 4.6$ for anilinium vs $\approx 10.6$ for cyclohexylammonium).

5. How do electron-donating and electron-withdrawing ring substituents affect the basicity of substituted anilines?

   Electron-donating groups (e.g. $\ce{-CH3}$, $\ce{-OCH3}$) increase basicity by pushing electron density toward nitrogen; electron-withdrawing groups (e.g. $\ce{-NO2}$, $\ce{-CN}$) decrease basicity by pulling density away. A $p$-nitroaniline is much weaker than aniline.

6. How is a simple primary amine named in IUPAC nomenclature?

   Name the parent chain and replace the terminal '-e' of the alkane with '-amine', using a locant for the nitrogen position, e.g. $\ce{CH3CH2CH2NH2}$ is propan-1-amine. As a substituent, $\ce{-NH2}$ is the 'amino' prefix.

7. What does the 'N-' locant designate when naming secondary and tertiary amines?

   'N-' indicates a substituent attached directly to the nitrogen atom rather than to the carbon chain, e.g. $\ce{CH3NHCH2CH3}$ is N-methylethanamine.

8. Give three general laboratory methods to synthesize primary amines from other functional groups.

   (1) Reduction of nitro compounds ($\ce{ArNO2 -> ArNH2}$, e.g. $\ce{Sn}$/HCl or $\ce{H2}$/catalyst); (2) reduction of nitriles or amides with $\ce{LiAlH4}$; (3) reductive amination of aldehydes/ketones, or the Gabriel synthesis via phthalimide.

9. What is the product and limitation of the direct alkylation of ammonia with an alkyl halide?

   $\ce{NH3}$ does an $S_N2$ on $\ce{RX}$ to give a primary amine, but the product amine is also nucleophilic, so it reacts further to give secondary, tertiary amines, and quaternary ammonium salts. This overalkylation gives a mixture, making it a poor method for a single pure amine.

10. Describe reductive amination and what carbonyl/amine combination it requires.

    A ketone or aldehyde condenses with ammonia or an amine to form an imine (or iminium ion), which is then reduced (e.g. $\ce{NaBH3CN}$ or $\ce{H2}$/catalyst) to an amine. It converts $\ce{C=O}$ into $\ce{C-N}$ and is a controlled way to make $1^\circ$, $2^\circ$, or $3^\circ$ amines.

11. How does the Gabriel synthesis produce a pure primary amine?

    Potassium phthalimide (a nitrogen nucleophile) does $S_N2$ on an alkyl halide; hydrolysis (or hydrazinolysis) then releases the primary amine. Because the nitrogen can only be alkylated once, no overalkylation occurs, giving a clean $1^\circ$ amine.

12. What product forms when a primary amine reacts with an acyl chloride or acid anhydride?

    An amide (acylation). For example $\ce{RNH2 + R'COCl -> R'CONHR + HCl}$. The amine's lone pair attacks the carbonyl carbon; a base is often added to neutralize the acid produced.

13. What happens when an amine reacts with excess alkyl halide, and what is the final product called?

    Exhaustive alkylation (methylation) occurs, adding alkyl groups until a quaternary ammonium salt ($\ce{R4N+ X-}$) forms. The nitrogen bears four carbon groups and a positive charge and has no remaining lone pair, so it can react no further.

14. Why are amines used to form water-soluble salts in drug formulation?

    Protonating a basic amine with acid gives an ammonium salt ($\ce{RNH3+ X-}$), which is ionic and far more water-soluble than the neutral amine. This is why many drugs are sold as hydrochloride salts.

15. How are aromatic (aryl) diazonium salts prepared, and under what conditions?

    By treating a primary aromatic amine with nitrous acid ($\ce{HNO2}$, generated in situ from $\ce{NaNO2}$ + HCl) at low temperature, typically $0$–$5\,^\circ\text{C}$: $\ce{ArNH2 + HNO2 + HCl -> ArN2+Cl- + 2H2O}$. Low temperature is required because aryl diazonium salts decompose when warmed.

16. Why can aryl diazonium salts be isolated at low temperature while alkyl diazonium salts cannot?

    Aryl diazonium ions are stabilized by resonance with the aromatic ring, so they survive at $0$–$5\,^\circ\text{C}$. Alkyl diazonium ions have no such stabilization and immediately lose $\ce{N2}$ (an excellent leaving group), giving carbocations and product mixtures.

17. List three substitution products obtainable by replacing the $\ce{-N2+}$ group of an aryl diazonium salt.

    The diazonium group can be replaced by: $\ce{-OH}$ (warm aqueous acid, phenol), $\ce{-Cl}$ or $\ce{-Br}$ (Sandmeyer, $\ce{CuCl}$/$\ce{CuBr}$), $\ce{-CN}$ ($\ce{CuCN}$), $\ce{-I}$ ($\ce{KI}$), or $\ce{-H}$ ($\ce{H3PO2}$). $\ce{N2}$ gas is lost in each case.

18. What is an azo coupling reaction and what type of compound results?

    A diazonium salt acts as an electrophile toward an activated aromatic ring (e.g. phenol or an arylamine), giving an azo compound with a $\ce{-N=N-}$ linkage bridging two rings. These conjugated azo dyes are intensely colored and used as dyes/indicators.

19. What molecular property must change during a vibration for it to absorb infrared radiation?

    The vibration must produce a change in the molecule's dipole moment. Vibrations that do not change the dipole (e.g. symmetric stretch of a symmetric bond like $\ce{N#N}$ or symmetric $\ce{O=C=O}$ stretch) are IR-inactive.

20. What does the wavenumber of an IR absorption represent, and what are its units?

    Wavenumber $\tilde{\nu}$ is the reciprocal of wavelength, $\tilde{\nu} = \frac{1}{\lambda}$, with units of $\text{cm}^{-1}$. It is proportional to the vibration frequency and thus to the energy of the transition; higher wavenumber means higher energy.

21. Over what wavenumber range does a typical IR spectrum run, and where is the fingerprint region?

    A typical IR spectrum spans roughly $4000$ to $400\,\text{cm}^{-1}$. The fingerprint region is below about $1500\,\text{cm}^{-1}$, containing complex bending/skeletal vibrations unique to each molecule.

22. How do bond strength and atomic mass affect the IR stretching frequency of a bond?

    Frequency follows $\tilde{\nu} \propto \sqrt{\frac{k}{\mu}}$, where $k$ is the bond force constant (stronger/stiffer bond = higher $\tilde{\nu}$) and $\mu$ is the reduced mass ($\mu = \frac{m_1 m_2}{m_1 + m_2}$; lighter atoms = higher $\tilde{\nu}$). Thus $\ce{C-H}$ absorbs high, and triple bonds absorb higher than double bonds.

23. What characteristic IR absorption identifies a carbonyl ($\ce{C=O}$) group, and roughly where does a ketone appear?

    A strong, sharp $\ce{C=O}$ stretch appears around $1650$–$1750\,\text{cm}^{-1}$; a typical ketone is near $1715\,\text{cm}^{-1}$. Its high intensity comes from the large dipole change.

24. How do the IR stretches of an O-H (alcohol) and N-H (amine) differ in shape and position?

    The alcohol $\ce{O-H}$ stretch is broad and strong near $3200$–$3550\,\text{cm}^{-1}$ (hydrogen bonding). The amine $\ce{N-H}$ stretch is around $3300$–$3500\,\text{cm}^{-1}$, sharper, and shows one band for a secondary amine but two bands for a primary amine ($\ce{-NH2}$).

25. How does a carboxylic acid O-H stretch appear in the IR, and how does it distinguish an acid from an alcohol?

    A carboxylic acid shows a very broad $\ce{O-H}$ stretch spanning roughly $2500$–$3300\,\text{cm}^{-1}$ (often obscuring $\ce{C-H}$ peaks), combined with a strong $\ce{C=O}$ near $1710\,\text{cm}^{-1}$. The extreme breadth (from strong H-bonded dimers) distinguishes it from an alcohol's narrower O-H.

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