Organic Chemistry Functional Group Chemistry: Alcohols, Ethers, and Carbonyls Syllabus

What drives the ring-opening reactivity of epoxides compared to ordinary ethers?

The strained three-membered ring ($\approx 60^\circ$ bond angles) has significant ring strain, which is relieved on opening, making epoxides far more reactive than acyclic ethers.

How do you name a carbonyl as an aldehyde vs a ketone using IUPAC suffixes?

Aldehydes use the suffix -al (e.g. propanal, $\ce{CH3CH2CHO}$); ketones use -one (e.g. propan-2-one). The aldehyde carbon is always C1.

Compare the electrophilicity and steric accessibility of aldehydes versus ketones toward nucleophiles.

Aldehydes are more reactive than ketones: they are less sterically hindered and have less electron donation to the carbonyl carbon (only one alkyl group vs two), making the carbonyl more electrophilic.

Describe the general mechanism of nucleophilic addition to a carbonyl.

The nucleophile attacks the electrophilic carbonyl carbon, breaking the $\pi$ bond and forming a tetrahedral alkoxide intermediate; protonation gives the addition product (e.g. an alcohol).

What is a cyanohydrin and how is it formed?

Addition of $\ce{HCN}$ (from $\ce{NaCN}$/$\ce{H+}$) to an aldehyde or ketone gives a cyanohydrin: a carbon bearing both an $\ce{-OH}$ and a $\ce{-CN}$ group, $\ce{R2C(OH)(CN)}$.

What product forms when an aldehyde/ketone reacts with a primary amine, and what is required?

An imine (Schiff base), $\ce{R2C=NR'}$, plus water. The reaction is acid-catalyzed and works best around $\ce{pH} \approx 4\text{-}5$.

What is formed when an aldehyde or ketone reacts with a secondary amine?

An enamine, $\ce{C=C-NR2}$, plus water (a $\beta$-carbon C=C forms because there is no N–H proton to lose to form an imine).

What acetal forms when an aldehyde reacts with two equivalents of alcohol under acid, and why is this useful?

An acetal, $\ce{R-CH(OR')2}$, plus water. Acetals are stable to base and nucleophiles, so they serve as protecting groups for carbonyls; they are removed by aqueous acid.

Write the Wittig reaction and identify the two products.

A phosphorus ylide (Wittig reagent, $\ce{R2C=PPh3}$) reacts with an aldehyde/ketone to give an alkene plus triphenylphosphine oxide ($\ce{Ph3P=O}$): $\ce{R2C=O + R'2C=PPh3 -> R2C=CR'2 + Ph3P=O}$.

How is a Wittig ylide prepared?

An alkyl halide reacts with $\ce{PPh3}$ ($S_N2$) to give a phosphonium salt, which is deprotonated at the $\alpha$-carbon by a strong base (e.g. $\ce{BuLi}$, NaH) to form the ylide.

What is the key regiochemical advantage of the Wittig reaction over other alkene syntheses?

The new C=C double bond forms at a defined position (where the carbonyl carbon was), giving unambiguous placement of the double bond with no isomer scrambling.

What products result from adding a Grignard reagent to formaldehyde, to another aldehyde, and to a ketone?

Grignard ($\ce{RMgX}$) + formaldehyde $\to 1^\circ$ alcohol; + other aldehyde $\to 2^\circ$ alcohol; + ketone $\to 3^\circ$ alcohol (after aqueous workup).

Why must Grignard reagents be prepared and used under strictly anhydrous conditions?

The carbanion-like $\ce{R-MgX}$ is a very strong base; any protic source (water, alcohols, terminal alkynes, N–H, O–H, COOH) protonates it to the alkane $\ce{R-H}$, destroying the reagent.

What happens when two equivalents of a Grignard reagent react with an ester?

Two additions occur: the first gives a ketone intermediate, the second adds to give (after workup) a $3^\circ$ alcohol with two identical R groups from the Grignard.

What mild reagent oxidizes an aldehyde to a carboxylic acid but leaves ketones untouched (Tollens' test)?

Tollens' reagent, $\ce{Ag(NH3)2+}$ (ammoniacal silver nitrate). Aldehyde is oxidized to carboxylate and $\ce{Ag+}$ is reduced to a silver mirror ($\ce{Ag^0}$). Ketones do not react.

State the acidity trend and give the approximate $\mathrm{p}K_a$ of a typical carboxylic acid versus an alcohol.

Carboxylic acids ($\mathrm{p}K_a \approx 4\text{-}5$) are far more acidic than alcohols ($\mathrm{p}K_a \approx 16\text{-}18$) because the carboxylate anion is resonance-stabilized over two equivalent oxygens.

How do electron-withdrawing substituents affect carboxylic acid acidity? Give an example comparison.

Electron-withdrawing groups stabilize the carboxylate and increase acidity. E.g. trichloroacetic acid ($\mathrm{p}K_a \approx 0.7$) is much stronger than acetic acid ($\mathrm{p}K_a \approx 4.76$).

Rank the relative reactivity of carboxylic acid derivatives toward nucleophilic acyl substitution.

Acid chloride > anhydride > ester $\approx$ carboxylic acid > amide (most reactive to least reactive). This tracks with leaving-group ability and resonance donation from the attached heteroatom.

Describe the general mechanism of nucleophilic acyl substitution.

Nucleophile adds to the carbonyl to form a tetrahedral intermediate; then the leaving group is expelled as the carbonyl re-forms. Net result: replacement of the leaving group $\ce{X}$ by the nucleophile (addition-elimination).

How is an acid chloride made from a carboxylic acid?

Treat the carboxylic acid with $\ce{SOCl2}$ (or $\ce{PCl3}$/$\ce{PCl5}$/oxalyl chloride): $\ce{RCOOH ->[SOCl2] RCOCl}$, releasing $\ce{SO2}$ and $\ce{HCl}$.

What product forms when an acid chloride reacts with (a) water, (b) an alcohol, (c) ammonia/amine?

(a) carboxylic acid, (b) ester, (c) amide. Each is a nucleophilic acyl substitution releasing $\ce{HCl}$ (usually with a base to neutralize).

State the Fischer esterification reaction and note its key features.

$\ce{RCOOH + R'OH <=>[H+] RCOOR' + H2O}$. It is acid-catalyzed and reversible; an excess of one reactant or removal of water (Le Chatelier) drives ester formation.

How is an ester hydrolyzed under basic conditions (saponification), and is it reversible?

$\ce{RCOOR' + NaOH -> RCOO^- Na+ + R'OH}$. Saponification is irreversible because the carboxylate product is deprotonated and unreactive toward the alcohol.

What is the product of reacting a carboxylic acid with an amine directly upon heating, versus using DCC?

Direct heating of $\ce{RCOOH}$ + amine first gives an ammonium carboxylate salt; strong heating drives off water to form an amide. A coupling reagent like DCC forms the amide under mild conditions by activating the acid.

Why are amides the least reactive carboxylic acid derivatives and the most stable?

The nitrogen lone pair donates strongly into the carbonyl (strong resonance), reducing carbonyl electrophilicity, and $\ce{-NH2}$/$\ce{-NR2}$ is a poor leaving group. This stability underlies the peptide bond.

How is a nitrile ($\ce{R-C#N}$) synthesized from an alkyl halide?

$S_N2$ reaction of $\ce{NaCN}$ (or KCN) with a $1^\circ$/$2^\circ$ alkyl halide: $\ce{R-X + CN^- -> R-C#N + X^-}$, adding one carbon to the chain.

What products result from (a) complete hydrolysis and (b) $\ce{LiAlH4}$ reduction of a nitrile?

(a) Acidic or basic hydrolysis gives a carboxylic acid (via amide): $\ce{R-C#N -> RCOOH}$. (b) $\ce{LiAlH4}$ reduces the nitrile to a primary amine: $\ce{R-C#N -> R-CH2-NH2}$.

What does a Grignard reagent form when added to a nitrile?

Addition gives an imine salt that, on aqueous acidic workup, hydrolyzes to a ketone: $\ce{R-C#N + R'MgX -> R-C(=O)-R'}$.

Define keto-enol tautomerism and state which tautomer usually predominates.

Tautomers are constitutional isomers interconverting by shifting an $\alpha$-H and the double bond: keto form ($\ce{-CH2-C=O}$) $\rightleftharpoons$ enol form ($\ce{-CH=C-OH}$). The keto form is normally far more stable and predominates.

Why are $\alpha$-hydrogens (adjacent to a carbonyl) acidic, and roughly what is their $\mathrm{p}K_a$?

Removing an $\alpha$-H gives an enolate whose negative charge is resonance-delocalized onto the electronegative carbonyl oxygen. Typical $\alpha$-H $\mathrm{p}K_a \approx 20$ (much more acidic than ordinary C–H, $\mathrm{p}K_a \approx 50$).