Organic Chemistry II

1. The hydrogen deficit

Learning goal: Use the formula to count rings and pi bonds before drawing anything.

The degrees of unsaturation (DBE) count how many pairs of hydrogens a formula lacks compared with a saturated acyclic hydrocarbon. For a formula with c carbons, h hydrogens, n nitrogens and x halogens, DBE = (2c + 2 + n - h - x) / 2. Oxygen does not change the count. Each ring or pi bond adds one.

A DBE of 0 means no rings and no pi bonds. A DBE of 4 or more with six or more carbons suggests a benzene ring, but it is only a hint. Use DBE to narrow the candidate list, then let other data choose.

A worked habit: always compute DBE before drawing. For C7H8O, DBE = (14 + 2 - 8) / 2 = 4, which matches a benzene ring and nothing else, so the carbon skeleton outside the ring is saturated. A student who draws a ring plus a C=O has spent five and has made an arithmetic error. Also check that the answer is a whole number: a half-integer DBE means a miscounted atom or a radical, not a real neutral molecule.

Worked example

A saturated acyclic ketone is C5H10O. Check the DBE and say which feature uses it.

1. DBE = (2 x 5 + 2 - 10) / 2 = 1.
2. Saturated and acyclic means no ring and no C=C.
3. The only unsaturation left is the carbonyl, C=O.
4. So the single DBE is already spent on the C=O.

Mental model: DBE says how many rings and pi bonds to place; it does not say which.

Common trap: Treating oxygen as if it changes the count.

2. Neighbors talk: the n+1 rule

Learning goal: Predict first-order splitting from the number of neighboring hydrogens.

In ideal first-order proton NMR, a set of equivalent hydrogens with n equivalent neighbors on adjacent atoms appears as n + 1 lines. One neighbor gives a doublet, two a triplet, three a quartet. The area (integration) of a signal is proportional to the number of hydrogens that cause it.

Real spectra can overlap or show extra coupling, so the rule is a prediction for clean cases. Hydrogens on the same atom as each other, or equivalent hydrogens, do not split each other in the simple picture.

Integration needs a scale. Pick the signal whose hydrogen count you are sure of, such as an isolated CH3 singlet equal to 3 H, and scale every other area to it. Total the scaled areas and compare with the formula's hydrogen count. If they differ, an exchangeable O-H or N-H may be broad or missing, which is common. Splitting is also mutual: if signal A has a quartet from signal B, B must be a triplet, so check the pair together as a consistency test.

Worked example

An ideal spectrum shows a triplet and a quartet. State which group each is and what fragment they point to.

1. The triplet has two neighbors, so its group sits next to a CH2.
2. The quartet has three neighbors, so its group sits next to a CH3.
3. Splitting is mutual: a CH3 next to a CH2 gives the triplet, and the CH2 gives the quartet.
4. Together they suggest an ethyl group, CH3CH2-.

Mental model: Lines = neighbors + 1, and splitting is mutual.

Common trap: Counting methyl groups instead of hydrogens.

3. Two clues and a whole molecule

Learning goal: Combine formula, DBE and NMR into one structure.

Each data type removes candidates. The formula gives the atoms and the DBE. NMR gives groups, their neighbors and their sizes. A good answer must satisfy all clues at once, including the total hydrogen count.

Work in a fixed order: formula, DBE, signal count, integration, splitting, then draw. If a drawn structure explains the data but needs a special assumption, check whether a simpler candidate does not.

A useful discipline is to write one line per clue. Formula C4H8O2 gives DBE 1. A triplet 3 H, a singlet 3 H and a quartet 2 H give 8 H in total. A CH3 singlet next to oxygen and an ethyl group point to ethyl ethanoate or methyl propanoate, and the shifts of the CH2 and CH3 groups decide which. Ask which clue eliminates which candidate, and write the elimination down so you can defend the final structure.

Worked example

Ideal 1H NMR of a saturated acyclic ketone C5H10O shows only a 6H triplet and a 4H quartet. Identify it.

1. DBE is 1, used by C=O.
2. A 6H triplet means two equivalent CH3 groups, each next to a CH2.
3. A 4H quartet means two equivalent CH2 groups, each next to a CH3.
4. CH3CH2-CO-CH2CH3, pentan-3-one, and 6 + 4 = 10 H matches the formula.

Mental model: Combine every clue; the DBE decided between a ketone and an ether here.

Common trap: Fitting one clue and ignoring the formula.

4. Arrows are bookkeeping

Learning goal: Use curved arrows to track electrons and check that charge is conserved.

A curved arrow starts at a pair of electrons (a lone pair or a bond) and points to where the pair ends up. It never shows atoms moving. Every step must conserve atoms and total charge, so the sum of formal charges before an arrow-pushing step equals the sum after.

In nucleophilic addition to a carbonyl, a nucleophile donates a pair to the carbonyl carbon and the C=O pi bond moves onto oxygen as a lone pair. Oxygen becomes an alkoxide anion. A separate acid step protonates it.

Practice with three checks after each arrow. Atoms: every atom you started with is still there. Charge: the sum of formal charges did not change. Electrons: every atom still has a sensible octet or valid hypervalent count. Resonance arrows are different from reaction arrows: they show alternative drawings of one structure, not two species. Mixing them up is the commonest arrow-pushing error, so mark which kind you are drawing.

Worked example

Show charge conservation when R- (a carbanion nucleophile) adds to a neutral ketone and gives an alkoxide.

1. Before: ketone charge 0 + nucleophile -1 = -1.
2. The nucleophile pair forms a new C-R bond at the carbonyl carbon.
3. The C=O pi pair moves onto oxygen, giving O-.
4. After: the alkoxide has charge -1, equal to the total before.

Mental model: Arrows move electrons; total charge and atoms stay constant.

Common trap: Drawing an arrow from an atom instead of from electrons.

5. Counting the carbon nucleophile

Learning goal: Do Grignard stoichiometry, limiting reagent and percent yield.

A Grignard reagent RMgX acts as a source of a carbon nucleophile. With a ketone it adds once, and after acid workup gives a tertiary alcohol. The ratio is one reagent per ketone, so the smaller number of moles is limiting if the reaction is clean.

Theoretical yield is limiting moles times the product molar mass. Percent yield is isolated mass divided by theoretical mass times 100. The reaction needs dry, aprotic conditions because water or alcohols destroy the reagent.

Percent yield can mislead when the starting amounts are not equal. Compute moles of each reagent first, find the limiting reagent, then multiply its moles by the product molar mass. If you add 1.2 equivalents of Grignard to 1.0 of ketone, the ketone limits and 0.2 of the reagent remains unused after workup. Remember that real isolated yields drop through side reactions, enolization and loss in workup, so the theoretical mass is a ceiling and not a prediction.

Worked example

Cyclohexanone (20.0 mmol) reacts with EtMgBr (25.0 mmol) in dry ether, then acid workup. The product 1-ethylcyclohexan-1-ol has molar mass 128.21 g/mol. What mass is the theoretical yield, and what is the mass at 70.0% isolated yield?

1. One equivalent per ketone, so the ketone (20.0 mmol) is limiting.
2. Theoretical mass = 0.0200 mol x 128.21 g/mol = 2.5642 g.
3. Isolated mass = 2.5642 g x 0.700 = 1.7949 g.
4. Report 1.79 g to three significant figures.

Mental model: Limiting reagent, then molar mass, then yield.

Common trap: Basing yield on the excess reagent.

6. Add once, or replace and add again

Learning goal: Tell carbonyl addition from acyl substitution and count Grignard equivalents.

A ketone has no good leaving group on the carbonyl carbon, so a Grignard adds once and the alkoxide stays until workup. An ester has an alkoxy group that can leave. The first addition gives a tetrahedral intermediate that collapses to a ketone, which reacts again with a second equivalent.

So an ester with excess Grignard gives a tertiary alcohol in which two of the three groups on the carbinol carbon come from the Grignard. Stopping at the ketone is hard because the ketone is more reactive than the ester toward the nucleophile in this setting.

Count the Grignard carefully. One equivalent of reagent with an ester gives mixed products because the intermediate ketone competes for the remaining reagent. For a clean tertiary alcohol, use at least two equivalents per ester and watch for a third equivalent if the reagent is partly quenched. A formate ester is a special case because the first product is an aldehyde, which on further addition gives a secondary alcohol with two identical groups from the Grignard.

Worked example

Ethyl acetate reacts with excess methylmagnesium bromide, then acid workup. Give the product and the number of Grignard equivalents consumed per ester.

1. First CH3 adds to C=O, giving a tetrahedral intermediate.
2. Ethoxide leaves, forming acetone.
3. Acetone reacts with a second CH3-, giving a tertiary alkoxide.
4. Workup gives (CH3)3C-OH, 2-methylpropan-2-ol, after two equivalents.

Mental model: Add, eject the leaving group, add again.

Common trap: Expecting a ketone to stop the reaction.

7. Reagents that cannot be in the same flask

Learning goal: Spot incompatible groups before choosing a synthesis route.

A Grignard reagent is a strong base. Any group with an acidic hydrogen (alcohol O-H, carboxylic acid, N-H, water) protonates it quickly and wastes it. The reagent is also incompatible with its own carbonyl partner on the same molecule.

Plan the route so that incompatible groups are absent, or protected, when the Grignard is made or used. Protection means converting a group to a form that survives the step and then restoring it. The order of steps is part of the answer.

Protecting-group logic is an ordering problem. If a molecule has both a ketone and an alcohol, a Grignard would deprotonate the alcohol first. A silyl ether or acetal masks the alcohol or the ketone, the Grignard step is performed, and a mild acid or fluoride removes the group at the end. Count the steps: protect, react, deprotect adds two steps, so a route that avoids protection by changing the order is shorter and has a better overall yield.

Worked example

4-Hydroxybutan-2-one (HOCH2CH2COCH3) is treated with ethylmagnesium bromide. How many equivalents are consumed per molecule, and why?

1. The O-H is acidic and protonates one equivalent, giving an alkoxide and ethane.
2. The ketone C=O then adds a second equivalent.
3. So two equivalents are consumed per molecule.
4. Protecting the alcohol first would allow only the ketone to react.

Mental model: Acid-base first, addition second.

Common trap: Counting only the carbonyl and forgetting the O-H.

8. Counting electrons in a ring

Learning goal: Apply the 4n+2 rule to ask whether a ring is aromatic.

A ring is aromatic when it is cyclic, planar and fully conjugated, with 4n + 2 pi electrons for a whole number n. Benzene has six pi electrons, which fits n = 1. Naphthalene has ten, which fits n = 2.

Counting needs care. A lone pair in a ring heteroatom counts only if it sits in the p orbital that joins the pi system, as in pyrrole. A carbon that is sp3 breaks conjugation. A ring with 4n pi electrons that is conjugated and planar is antiaromatic, which is unstable, so some rings pucker to avoid it.

Heterocycles reward careful counting. Pyrrole has one nitrogen lone pair in the pi system and two C=C bonds, so six pi electrons and it is aromatic. Pyridine's nitrogen lone pair sits in an sp2 orbital in the plane, so it is not part of the pi count: pyridine has six pi electrons from three C=C-like bonds. Furan counts one oxygen lone pair. Cyclooctatetraene has eight pi electrons and is not planar, so it is not antiaromatic in practice.

Worked example

Apply 4n + 2 to naphthalene (ten pi electrons). What is n, and is the pi count consistent with aromaticity?

1. 4n + 2 = 10.
2. 4n = 8, so n = 2.
3. n is a whole number, so the count fits.
4. The ring must still be planar and cyclic, which naphthalene is.

Mental model: Aromatic needs structure plus a 4n + 2 count.

Common trap: Using the count without checking planarity.

9. Where the next group lands

Learning goal: Predict the product sites for electrophilic substitution on a substituted benzene.

An existing substituent directs the next electrophile. Groups that donate electron density into the ring, such as methyl and methoxy, direct to the ortho and para positions. Groups that withdraw, such as nitro, direct meta and make the ring less reactive.

Counting products means counting distinct positions after symmetry. In toluene the two ortho positions are equivalent, as are the two meta positions. That gives three distinct mononitration products: ortho, meta and para, though meta is expected to be minor. Exact ratios depend on conditions and are not predicted here.

Ring reactivity can be ranked qualitatively. Activating groups such as OH, OCH3 and alkyl groups speed up electrophilic substitution; deactivating groups such as NO2 and CF3 slow it. Halogens are an exception: they deactivate but still direct ortho and para. When two groups disagree, the stronger activator usually decides the position, and steric crowding between two adjacent groups disfavors the position between them. Treat these as guides, because real mixtures depend on conditions.

Worked example

How many distinct mononitration products does toluene give if ortho, meta and para are all counted?

1. Toluene has a methyl on one carbon; the two ortho carbons are equivalent by symmetry.
2. The two meta carbons are equivalent.
3. The para carbon is unique.
4. So there are three distinct products: ortho, meta, para.

Mental model: Directing effect chooses the type of position; symmetry sets the number of products.

Common trap: Counting all four C-H positions as different.

10. Counting mirror images

Learning goal: Count stereocenters and the stereoisomers they allow.

A carbon with four different groups is a stereocenter. A molecule with n stereocenters has at most 2^n stereoisomers. Many molecules have fewer because internal symmetry makes some pairs identical.

A meso compound has stereocenters but an internal mirror plane, so it is not chiral and equals its own mirror image. 2,3-Dibromobutane has two stereocenters, so the maximum is four, but one pair collapses into one meso form, leaving three distinct stereoisomers.

A sweep for stereocenters: mark each carbon with four different groups, then count 2^n and look for symmetry. Tartaric acid has two stereocenters, so up to four stereoisomers, but one pair is the meso form, leaving three: the two enantiomers and the meso compound. Enantiomers rotate plane-polarized light in opposite directions by equal amounts, while a meso compound does not rotate it. A 50:50 mixture of enantiomers is a racemate and also shows no net rotation, but it is a mixture, not a single compound.

Worked example

How many stereoisomers does 2,3-dibromobutane have?

1. It has two stereocenters, C2 and C3, so the upper limit is 4.
2. (R,R) and (S,S) are enantiomers.
3. (R,S) has an internal mirror plane and is meso, identical to (S,R).
4. So there are three distinct stereoisomers.

Mental model: 2^n is the maximum; symmetry reduces it.

Common trap: Assuming 2^n is always exact.

11. Mass and IR clues: what the formula parity says

Learning goal: Combine the nitrogen rule, isotope patterns and IR bands with DBE to screen structures.

The nitrogen rule says that a neutral molecule with an odd number of nitrogen atoms has an odd nominal molecular mass, and one with zero or an even number of nitrogens has an even mass. If the parent peak is at m/z 73 for a compound with only C, H, N and O, there is at least one nitrogen.

Chlorine appears as two peaks two mass units apart in about a 3:1 ratio because of the natural isotopes 35Cl and 37Cl. Bromine gives a nearly 1:1 pair. These patterns are quick to read and tell you which halogen is present before you look at any NMR.

In infrared spectra, typical textbook positions are a broad O-H band around 3200 to 3550 cm-1, a strong carbonyl C=O band near 1700 to 1750 cm-1 and a nitrile C-N triple bond band near 2250 cm-1. Exact positions vary with the environment, so use ranges as screens, not proofs.

Order the clues. Mass gives the formula parity and halogens, DBE gives the unsaturation count, IR names functional groups, and NMR assigns the skeleton. A candidate that fails any one clue is out.

Worked example

A compound shows a parent peak at m/z 87 and contains only C, H, N and O. What does the nitrogen rule say about the number of nitrogens?

1. Odd mass means an odd number of nitrogens.
2. Count: 87 is odd.
3. So at least one N.
4. Not zero.

Mental model: Parity, isotope pattern, DBE, IR and NMR each remove candidates.

Common trap: Trusting one clue and ignoring the rest.

12. Enolates: let the pKa pick the base

Learning goal: Use pKa differences to judge whether a base will deprotonate the alpha carbon.

A hydrogen on the carbon next to a carbonyl is acidic because the conjugate base, an enolate, spreads its charge onto oxygen. Typical textbook pKa values are about 17 for aldehydes, 19 to 20 for ketones and about 25 for esters. These are approximate and depend on the solvent.

For a proton transfer, log10 K equals the pKa of the conjugate acid of the base minus the pKa of the acid. Ethoxide, whose conjugate acid ethanol has pKa about 16, deprotonates a ketone only slightly, with K near 10 to the power of minus 3 or minus 4. A strong base such as LDA, whose conjugate acid diisopropylamine has pKa near 36, deprotonates it almost completely.

That matters for planning. With a weak base, enolate and carbonyl coexist, so the enolate can attack another carbonyl in an aldol reaction. With a strong base, the ketone is fully converted first, and you add the electrophile afterward, which gives more control.

Pick the base by comparing pKa values. A difference of three or more in the favorable direction gives mostly product at equilibrium. Always check the solvent and conditions before trusting a table value.

Worked example

A ketone alpha-H has pKa 20 and the base's conjugate acid has pKa 16. What is log K?

1. pKa of BH minus pKa of HA.
2. 16 - 20.
3. = -4.
4. K is about 10 to the minus 4.

Mental model: log K = pKa(BH) - pKa(HA). Large positive means nearly complete deprotonation.

Common trap: Assuming ethoxide fully converts a ketone.

13. Diels-Alder: counting a ring-forming reaction

Learning goal: Count electrons, bonds and moles in a Diels-Alder reaction.

The Diels-Alder reaction joins a diene with four pi electrons and a dienophile with two pi electrons to make a cyclohexene. Six electrons move in a cyclic transition state, which fits the 4n + 2 rule you used for aromaticity. The reaction is thermally allowed.

The diene must be able to adopt the s-cis shape, with both double bonds on the same side of the single bond between them. Cyclopentadiene is locked in that shape, which is why it reacts fast. Electron-poor dienophiles such as maleic anhydride react well with electron-rich dienes.

Count the bonds. Two new sigma bonds form and one new pi bond remains in the ring, while two pi bonds are lost. The overall change is exchanging two pi bonds for two sigma bonds, which is why the reaction is usually favorable. The product ring has six atoms.

For yields, do the usual accounting. Convert masses to moles, find the limiting reagent, multiply by the adduct molar mass. For cyclopentadiene (66.10 g/mol) and maleic anhydride (98.06 g/mol) the adduct is 164.16 g/mol, because the reaction is an addition with no atoms lost.

Worked example

Maleic anhydride 9.81 g (0.100 mol) reacts with excess cyclopentadiene at 100% yield. Mass of adduct?

1. Moles 0.100.
2. Molar mass 164.16.
3. 0.100 x 164.16.
4. 16.4 g.

Mental model: 4 + 2 pi electrons, two sigma bonds formed, adduct mass is the sum.

Common trap: Forgetting the diene must be s-cis.