Aldehydes and Ketones

Question 1

What is the suffix when an aldehyde is attached to a ring?

Answer 1

-carbaldehyde

Question 2

What is the prefix when an aldehyde is not the priority group?

Answer 2

formyl-

Question 3

What’s the prefix for ketone substituent?

Answer 3

oxo-

Question 4

Three most common ways to prepare aldehydes and ketones?

Answer 4

Oxidation of alcohols, Ozonolysis of alkenes, friedel-crafts reaction

Question 5

How to make an aldehyde/ketone from an alcohol?

Answer 5

If a terminal alcohol, use mild oxidizing reagent PCC (stronger will oxidize all the way to COOH). For secondary alcohols, use dichromium salt, Jone’s oxidation (CrO3 in dilute H2SO4 in acetone), or PCC.

Question 6

How to make a ketone/aldehyde from alkene?

Answer 6

Ozonolysis (O3 with CH2Cl2 in Zn/H2O or S(CH3)2) will cleave the double bond and oxidize terminal carbons to aldehydes and non-terminal to ketones.

Question 7

How to make ketone from aromatic hydrocarbon?

Answer 7

An acyl halide is turned into a carbocation by a Lewis acid, and the carbocation is incorporated into the ring while booting out a H+.

Question 8

Describe what happens to ketones in solution?

Answer 8

They tautomerize between enols and ketones, with ketones being favored. Owed to the acidity of the alpha carbon’s H, with pKa of about 20.

Question 9

Explain a Michael addition

Answer 9

An enolate carbanion is formed by deprotonation by a strong base (remember the hydrogen on the alpha carbon of a carbonyl is weakly acidic). This then attacks an alpha or beta unsaturated carbonyl compound. The enolate carbanion is often formed from 1,3-dicarbonyl since the 2-H is extra acidic because there are two carbonyls to delocalize negative charge.

Question 10

What the key to carbonyl’s reactivity?

Answer 10

The polarization of the bond makes the carbonyl capable of nucleophilic and electrophilic attack.

Question 11

Describe the general process of nucleophilic addition to a carbonyl

Answer 11

A nucleophile attacks the partially positive carbon. The pi electrons from the C=O bond are displaced to the oxygen, which is happy to have those extra electrons for a time. Then, if there’s a good leaving group, the carbonyl will reform and the leaving group will leave. Otherwise the oxygen will keep the negative charge, but will likely pick up a H+ from solution to form a OH.

Question 12

Explain the formation of geminal diols, hemiacetals, hemiketals, acetals, and ketals from aldehydes/ketones

Answer 12

In all cases, the O in an OH (either from an alcohol or water) makes a nucleophilic attack on the carbonyl carbon and dumps its excess H+ to the now negative carbonyl O. Hemiketals and hemiacetals are formed by reacting 1 equivalent of alcohol/water with ketone/aldehyde. Ketals/acetals formed by reacting two equivalents. In basic conditions, only hemi reaction. Proceeds all the way under anhydrous acidic conditions.

Question 13

How is a cyanohydrin formed?

Answer 13

Acidic H on HCN pops off and CN- attacks carbonyl carbon. Then negative carbonyl O grabs the H+ from solution. Cyanohydrins are pretty stable because of their C-C bond.

Question 14

Describe carbonyl condensation with ammonia derivatives

Answer 14

Through nucleophilic substitution on the carbonyl. The nucleophilic free electrons on the N of NH3 attack the carbonyl carbon. The O- gains a couple protons and leaves as water. N loses another H and donates the electrons to form a double bond with the C.

Question 15

Describe aldol formation and aldol condensation

Answer 15

And aldehyde acts as both nucleophile and electrophile. A strong base produces an enolate ion from the aldehyde, which attacks the positive carbon of another aldehyde, forming an aldol. With strong base and high temps, the hydroxyl is lost as water and an alpha beta double bond is formed.

Question 16

What’s the simplest thing to remember about the Wittig reaction?

Answer 16

It swaps a C=O for a C=C bond

Question 17

Explain the Wittig reaction

Answer 17

1) Formation of phosphonium salt from SN2 rxn of alkyl halide with nucleophile triphenylphosphine.
2) Deprotonation of salt with strong base to yield zwitterion ylide.
3) Negative ylide carbon attacks carbonyl carbon, forming intermediate betaine (oxaphosphetane).
4) Oxaphosphetane decomposes to yield alkene and triphenylphosphine.

Question 18

Describe the Wolf-Kishner reduction

Answer 18

Reduces ketone all the way to alkane. Carbonyl converted to hydrazone (=NNH2). Heated with base to release N2, forming alkane. Useful only when product is stable under basic conditions.

Question 19

Explain the Clemmensen reduction

Answer 19

Reduces ketone/aldehyde all the way to alkane. Aldehyde or ketone heated with amalgamated Zn [Zn(Hg)] in HCl.