Chem Mid2

• 3H triplet

• 2H quartet

ethyl (CH3-CH2)

• 6H doublet

• 1H septet

isopropyl (HC(CH3)2)

• 9H singlet

tert-butyl

• 3H singlet (2-2.7 ppm)

acetyl methyl (O=C-CH3)

• 3H singlet (3.3-3.8 ppm)

methoxy (OCH3)

• 1H (9-10 ppm)

aldehyde proton (CHO)

• broad 1H (10-12 ppm)

carboxylic acid proton (COOH)

5H with 6.5-8.5 ppm

monounsubstituted benzene (Ph-X)

• characteristic: messy multiplet

4H with 6.5-8.5 ppm

disubstituted benzene

3H or 2H with a 6.5-8.5 ppm

tri or tetrasubstituted benzene

2 2H doublets with high ppm

same integration (2+2) —> para-disubstituted benzene (1,4)

4 signals with high ppm

no symmetry with messy multiplets, no equal integration pairs —> ortho-disubstituted benzene (1,2)

3-4 signals, uneven splitting, partial symmetry but no clean pairing

meta-disubstituted benzene (1,3)

6.5-7.2 ppm

electron donating substituents like OCH3

7.2-8.2 ppm

normal benzene

8.0-8.8 ppm

electron withdrawing like a carbonyl or nitro

3H singlet 3.7-3.9 ppm

methoxy on benzene (OCH3)

3H singlet 2.0-2.6 ppm

acyl methyl (COCH3)

1H 9-10 ppm

CHO

2.3-3.0

benzylic CH2

nomenclature of aldehydes and ketones

aldehyde: -al

ketone: -one

given aldehyde, alcohol, alkene, alkyne, ketone, or halide, what takes president?

aldehyde or ketone, given both, aldehyde —> ketone gets the prefix “oxo-”

preparing carbonyls

oxidation of alcohol —> alcohol —> carbonyl

ozonolysis of alkenes —> alkene —> carbonyl

hydration of alkynes —> alkyne —> carbonyl

friedel-crafts acylation —> benzene

primary alcohol oxidation with H2CrO4

Jones reagent triggers formation of a carboxylic acid

primary alcohol oxidation with PCC

stops at aldehyde

secondary alcohol with H2CrO4

ketone

secondary alcohol with PCC

ketone

tertiary alcohol with PCC

violates octet rule, tertiary alcohols can’t react

alcohol oxidation mechanism

OH is partial negative, Cr is partial positive, especially after removing the first Hydrogen directly form the carbon —> activates it

O lone pairs form a bond with Cr causing the Cr on H2CrO4 to jump back to O within the molecule

O on the original chain is assigned a positive charge, keeping its lone pair

lone pair on OH on H2CrO4 bind to H, causing the electrons of the OH bond on the original chain to go back to O on the original chain

Left with H2O+ bound to H2CrO4, neutral oxygen on the original chain with a lone pair —> chromate ester state

Weak base like water grabs a H off the C attached to O on the original chain, leaving the electrons, which are attracted to O so they form a pi bond between C and O

Since O now has too many bonds, bond between O and the H2CrO4 breaks (breaking the chromate ester) and leaving the original chain with a double bond attaching the O instead of it being connected to an H —> FINAL PRODUCT ketone if starting with a SECONDARY ALCOHOL

There is still an alpha hydrogen if starting with a PRIMARY ALCOHOL, meaning the mechanism repeats:

• H+ from the protonated environment binds to O to charge/make it more reactive, assigning it a lone pair and positive charge, C it is attached to is partial positive as a result

• Partially negative O on H2O can attack, giving a protonated diol (double bond to O breaks, H2O+ bound, H still bound)

• Water comes in and attacks the charged H2O+ (attacking a hydrogen on the positively charged O), causing one H to leave

• given a gemdiol intermediate (2 OH bound to 1 C —> reactive) will react with H2CrO4, lone pair on one of the alcohols attacks the Cr, breaking the bond between Cr and the activated (+ charged) O

• original chain now bound to H2CrO4 through the positively charged OH, allowing a lone pair on the OH on the H2CrO4 to attack the H on the OH attached to the original chain, allowing electrons from the bond to jump back to the O

• the O on water then attacks the remaining H bound to the C, allowing the electrons in the bond to jump to O on the original chain, forming a double bond and leaving a COOH

ozone

zwitterion —> O=O+-O-

what happens in ozonolysis and how can it be used

double bond in alkene is cleaved, forming 2 carbonyls instead (essentially imagine the double bond splitting down the middle and getting an O stuck to either end of the break)

terminal alkenes can be made into a carbonyl

performing on a cyclic alkene pulls it apart and gives a linear product

hydration of alkynes reagents

H2O, H2SO4, HgSO4

what does hydration of alkynes do?

Adding water across a triple bond —> OH goes to more substituted C, H goes to less substituted —> produces an enol intermediate (en = double bond, ol = OH) —> rearranges for stability into a ketone

OH on markovnikov —> more substituted

friedel craft’s acylation reagents

R-C(=O)-Cl in AlCl3

friedel-crafts acylation

carbonyl Cl attacks Al (lewis acid, acceptor) —> covalent bond between Al- and Cl+ (higher energy) —> carbon chlorine bond is lost to stabilize the + on Cl —> carbocation intermediate (acylium, more reactive) —> benzene reacts, causing electrons to move up a bond and up to the O to neutralize