CSIR NET Chemical Sciences Organic Chemistry Flashcards

24 sample cards from the Organic Chemistry deck

Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.

1. State the Woodward-Hoffmann rules for electrocyclic reactions.

   For 4n pi electrons: thermal = conrotatory, photochemical = disrotatory. For 4n+2 pi electrons: thermal = disrotatory, photochemical = conrotatory. (e.g., thermal ring closure of a triene, 6 e, is disrotatory.)

2. Define conrotatory and disrotatory ring closure.

   In an electrocyclic reaction, the two terminal p-orbitals rotate to form the new sigma bond. Conrotatory = both ends rotate in the same direction; disrotatory = they rotate in opposite directions. The mode determines the stereochemistry of substituents.

3. Rank pyrrole, furan, and thiophene by aromaticity and explain the trend.

   Aromaticity order: thiophene > pyrrole > furan. Greater aromaticity correlates with the heteroatom's lower electronegativity and better lone-pair donation into the ring; oxygen holds its lone pair most tightly, making furan least aromatic and most diene-like.

4. Where does electrophilic aromatic substitution occur on pyrrole/furan/thiophene, and on pyridine?

   For five-membered heterocycles (pyrrole, furan, thiophene), EAS occurs preferentially at the alpha (C-2) position. Pyridine is deactivated toward EAS and reacts at C-3 (beta); it instead favors nucleophilic substitution at C-2/C-4.

5. Why is pyridine a stronger base than pyrrole?

   In pyridine the nitrogen lone pair is in an sp2 orbital in the ring plane, not part of the aromatic sextet, so it is available for protonation. In pyrrole the lone pair is delocalized into the aromatic ring; protonating it would destroy aromaticity, so pyrrole is a very weak base.

6. Name the heterocycle synthesis: the Paal-Knorr and the Fischer indole syntheses.

   Paal-Knorr: 1,4-dicarbonyl compounds cyclize to furans (acid), pyrroles (with amines/NH3), or thiophenes (with P2S5/Lawesson's). Fischer indole synthesis: an arylhydrazine plus an aldehyde/ketone under acid gives an indole via a [3,3]-sigmatropic step.

7. Classify carbohydrates as aldoses/ketoses and by ring size (furanose/pyranose).

   Aldoses contain an aldehyde group; ketoses contain a ketone group. A five-membered cyclic hemiacetal ring is a furanose; a six-membered ring is a pyranose (e.g., glucose exists mainly as glucopyranose).

8. What is the anomeric effect and what are alpha/beta anomers?

   Anomers are cyclic sugar stereoisomers differing in configuration at the new anomeric carbon (C-1 in aldoses): alpha has the OH trans to the CH2OH reference (axial in glucopyranose), beta is cis (equatorial). The anomeric effect is the tendency of an electronegative C-1 substituent to prefer the axial position.

9. Distinguish reducing and non-reducing sugars; classify glucose, maltose, and sucrose.

   Reducing sugars have a free anomeric (hemiacetal) carbon that can open to an aldehyde and reduce Tollens'/Fehling's reagent. Glucose and maltose are reducing; sucrose is non-reducing because both anomeric carbons are joined in the glycosidic bond.

10. Describe the four levels of protein structure.

    Primary: the amino acid sequence (peptide bonds). Secondary: local folding into alpha-helices and beta-sheets stabilized by H-bonds. Tertiary: overall 3D fold of one chain (disulfides, hydrophobic, ionic, H-bonds). Quaternary: assembly of multiple polypeptide subunits.

11. What is the zwitterionic form and isoelectric point (pI) of an amino acid?

    A zwitterion has both a protonated amino group (-NH3+) and a deprotonated carboxyl group (-COO-), so it is dipolar but net neutral. The isoelectric point (pI) is the pH at which the amino acid has zero net charge; for a neutral amino acid pI = (pKa1 + pKa2)/2.

12. What forces define the geometry of the peptide bond?

    The peptide (amide) bond has partial double-bond character due to resonance, making it planar and restricting rotation; it is usually in the trans configuration. This planarity defines the phi/psi backbone torsion angles described by the Ramachandran plot.

13. Describe the Watson-Crick base pairing in DNA and the helix dimensions.

    Adenine pairs with thymine via 2 hydrogen bonds; guanine pairs with cytosine via 3 hydrogen bonds. DNA is an antiparallel right-handed double helix (B-form) with ~10 base pairs per turn and the two strands running 5'->3' opposite each other.

14. List the structural differences between DNA and RNA.

    DNA: deoxyribose sugar, thymine base, usually double-stranded. RNA: ribose sugar (2'-OH), uracil instead of thymine, usually single-stranded. The 2'-OH makes RNA more reactive/less stable than DNA.

15. Classify lipids and define a triglyceride.

    Lipids include fatty acids, triglycerides (fats/oils), phospholipids, glycolipids, steroids, and waxes. A triglyceride is a triester of glycerol with three fatty acids; saturated fats (no C=C) are solids, unsaturated (with C=C) are oils.

16. What is saponification and what is the iodine value of a fat?

    Saponification is base-catalyzed hydrolysis of a triglyceride into glycerol and fatty acid salts (soaps). The iodine value is the grams of iodine absorbed per 100 g of fat, measuring the degree of unsaturation (higher value = more C=C double bonds).

17. State the Beer-Lambert law used in UV-visible spectroscopy.

    A = epsilon x c x l, where A is absorbance, epsilon is the molar absorptivity (L mol-1 cm-1), c is concentration (mol/L), and l is path length (cm). Absorbance is linearly proportional to concentration.

18. Define chromophore, auxochrome, and bathochromic shift in UV-Vis.

    A chromophore is the unsaturated group responsible for absorption (e.g., C=C, C=O). An auxochrome is a saturated group with lone pairs (-OH, -NH2) that intensifies/shifts absorption when attached to a chromophore. A bathochromic (red) shift moves lambda_max to longer wavelength; hypsochromic (blue) shifts it shorter.

19. Give characteristic IR stretching frequencies for O-H, C=O, C≡N, and C≡C.

    Broad O-H: ~3200-3600 cm-1; C=O (carbonyl): ~1700-1750 cm-1; nitrile C≡N: ~2200-2260 cm-1; alkyne C≡C: ~2100-2260 cm-1. (N-H ~3300-3500; sp3 C-H ~2850-2960 cm-1.)

20. What molecular property must change for a vibration to be IR active?

    A vibrational mode is IR active only if it produces a change in the molecule's dipole moment. Symmetric vibrations in symmetric molecules (e.g., N2, the symmetric stretch of CO2) give no dipole change and are IR inactive.

21. In 1H NMR, what do chemical shift, integration, and the n+1 rule tell you?

    Chemical shift (ppm, relative to TMS) reflects the electronic environment/deshielding. Integration gives the relative number of protons. The n+1 multiplicity rule: a proton with n equivalent neighboring protons is split into n+1 peaks (e.g., 3 neighbors give a quartet).

22. Give approximate 1H NMR chemical shifts for TMS, aliphatic CH, vinyl/aromatic H, and aldehyde/COOH protons.

    TMS = 0 ppm (reference); aliphatic C-H ~0.8-2 ppm; allylic/adjacent to C=O ~2-3 ppm; vinylic ~4.5-6.5 ppm; aromatic ~6.5-8 ppm; aldehyde ~9-10 ppm; carboxylic acid ~10-12 ppm.

23. In mass spectrometry, what are the molecular ion peak and the base peak?

    The molecular ion (M+) peak corresponds to the intact molecule minus one electron and gives the molecular mass. The base peak is the most intense peak in the spectrum (assigned 100% relative abundance), representing the most stable/abundant fragment ion.

24. What does the nitrogen rule and the M+2 isotope pattern indicate in mass spectrometry?

    Nitrogen rule: an odd molecular ion mass implies an odd number of nitrogen atoms. M+2 peaks indicate halogens: chlorine gives an M:M+2 ratio of ~3:1, bromine gives ~1:1, due to the natural abundances of their isotopes.

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