Aldehydes + Ketones (no mechanisms)

How to know whoch ketone is more reactive

Steric effects (want to be able to easily attack it and not be super crowded(

Electronic effects: (Prefer EWG as it makes the C=O more positive)

How to draw a cyclic hemiacetal from a intramolecular molecule

1. Count carbones starting at the double bonded O and end at the OH

2. Make a ring structure with an O embedded with the amount of carbons you just added

3. Add the other elements to the molecule

4. since it is a hemiacetal the double bonded O will turn into a hydroxyl group (on carbon 1)

What is the nucleophile during a ketal reaction

The O from the OH

• it will directly attach

What is a dead giveaway that it will be a intramolecular ketal/acetal

embedded O in cyclic ring

Ketones vs. esters

Ketones will be attacked first so sometimes you must protect it to prevent unwanted reactions, while esters are generally less reactive due to resonance stabilization.

How to name aldehydes

-al ending

higher priority than other molecules

-CHO is carbon #1

How to names ketones

just ends in -one

Important note about carbonyls

vulnerable to attack but also very stable as a C=O

• acts as nucleophile (the O) and electrophile (the C)

Quick notes about acadic contondtions for carbonyls

• protonate carboyl before attacking

• no strong bases can leave

• RO- should never be a Nu: or LG

Quick notes about basic contondtions for carbonyls

• attack carbonyl first

• H2O and ROH can be present

• NO -OH₂⁺ can be present

relative reactivties

Ketone → Aldehyde → Carbonyl

LAH vs NaBH₄ when it comes to esters and ketones

LAH is too strong and will attack both

NaBH₄ is weak and will only attack the ketone

(A negative triple bond will also attack both a classic LG and a ketone)

N nucleophiles with a primary amine

Primary amine with have 2 H so they leave with the O to form water

Imine is formed

replace the double bonded O with a double bonded NR

N nucleophiles with a secondary amine

Secondary amine only has 1H

the double bond will end up moving as an H is taken from the orginal moelcule to form H2O

Hydroxylamine

HONH₂

replaces double bonded O with double bonded NOH

forms oxime

Hydrazine

H₂NNH₂

replace double bonded O with double bonded NNH₂

forms hydrazone

Wolff-Kishner

H₂NNH₂ / KOH + heat

replaces double bonded O with two hydrogens and also N₂ gas is formed

Important note about acetal/ketal, imine and enamine

Can be hydrolyzed back to the ketone/aldhyde with excess H₂O and catalytic H+

Reductive Animation

making amines via reduction reactions

Using R-Br with a N-Nucleophile yeilds a mixture of products

• R groups are EDG and make N more nucleophilic/basic

• Trick is to make imine or enamine then reduce it

Aldehyde + N nucleophile

the N group (with two H) replaces the double bonded O with a double bonded N still conntected to whatever R group (no H)

An H₂O leaves

• usually gets reduced (double bond dissapears)

Ketone + N nucleophile

The N group replaces double bonded O, and makes double bonded N with whatever R group

An H₂O leaves

• usally gets reduced (minus double bond)

What is NaBH₃CN

a reducing agent that is selective and reduced Imine or enamines

• takes away double bonds from double bonded N’s

C nucleophiles

• Includes Gringard

• Inlcudes lithium

• Cn-

Gringard and Lithium regeants with aldehydes and ketones

ONLY ATTACK ONCE

• Make double bonded O into an OH

• R group slides right on in there with everyon else

Cyanohydrins

CN- ion is a very good nucleophile

First step: turns double bonded O into HO and CN slides right in

second step: Gets reduced and CN is replaced with NH₂

Secon step option two: H₃O+ amd CN is replaced with a carboxylic acid

Wittig Reactoin

Ketones and aldehydes can make alkenes in the presence of phosphonium ylide (Ph₃P)

Ph₃P and the double vonded O basically disappears the two double bonds merge into one double bond

Ph₃P is a good nucleophile

Important note about Wittig Reaction

Want the PPh₃ on a primary alkyl halide to ensure high yields.

Wittig reagents are made via SN2

regeants should be made from the less substituded side of the double bond

Steps for cyclic hemiacetals

1. Count carbons starting at double bonded O and end at the OH

2. draw O in ring with however many carbons counted

3. Add other groups based on that

• hemiacetals will have OH still present from the double bond

• Double bonds turns into OH and OH turns into O in ring