PROTON NMR SPECTROSCOPY

Equivalent Protons

Check for symmetry and stereochem, must see other atoms/groups constitutionally and stereochemically the same. cis and trans in ring

Saturated Protons, SP3 C - SP3 CH3

0-1.5

Alkyne, Allylic vinyl c allyl H3, Benzyllic, Alpha protons (carbonyl - ch3)

1.5-2.5

E-neg Region

Proton set connected to electronegative atom, o-ch2 this is 2.5-4.5, ether protons, ester protons amine protons N is less en so lower ester highest ether 3.0-3.5

Vinyl region

4.5-6.5, H directly to PI bond, diamagentic anisotropy (circulating pi electrons induce a local magnetic fied that can shielf or deshield protons)

Aryl Ring (benzene H)

6.5-8.0

Phenolic OH

9

regular hexane oh

2-6

Aldehyde

10

Carboxylic Acid

12

Amide (sp2 h-C =o-Nh2) or Ester OCH3 is not as strong as

aldehyde in 10, usually 8.0 ppm

because of RESONANCE AND ATOM SPREADS E DENSITY

Methine Methylene Methyl WHEN SAME TYPE OF PROTONS (LIKE ALPHA OR BENZYLIC)

CH←—CH2—-CH3

MORE DOWNFIELD

Deuterium Proton Swap

1. write out the compound twice

2. switch D IN ONE OF EACH COMPOUNDS

3. Then see what replacement does to both

Deuterium swap for equivalent protons

Homotopic Protons

D makes Chrial carbons w RR or SS config

Does not make chiral center

Enantiotopic Protobs

-D replacement makes chiral carbons with R and S configurations

making enatiomers

DO deutrium swaps for equivalent but only rly need R R OR RS

Non equivalent protons

Diastereotopic protons results in a pair of diastereomers

Simply non eqiv, just constiutionally diff

d is more than H btw

short cut

check if equivalent through stereochem and constiutionally, just replace either H with D, look at R and S config

Homotopic would do RR OR SS, no need to look at other chiral centers

Diastereotopic H

1. Restriction in rotation

2. exisiting cirality

3. diff cis/trans relationships

4. DIFF STEREOCHEM

Free rotation

if H’s are connected to C with single bonds or free rotation no cis/trans diasteroeosmerisn

with double bond, the groups may see 2 H as either cis or trans , write lin through double bond and look for symmetry, if yes symmetry than no stereeoisomerism

so 1. constitional relationship 2. stereochem/chirality or cis/trans 3. free rotation for diassteroeotopic

what splits

adjacent c with nonequivalent H!! not equivalent H

Complex Splitting w J

1. Assume singlet

2. Then look at two different H’s and possible n+1 splitting

3. if J’s are queivalent then regular n+1 rule

4. If one J is more, then split first according to the more J n+1 then split again according to weaker n+1

A proton set that has two neighboring c will be split by both adjacent proton sets. equivalent to each other, then just regular n + M+ 1 applied

if non equivalent n+1 x m+1

ok

Ortho (1,2 Di sub benzene ring)

4 signals, no singlet 2 D, 2 DD

Meta (1,3)

4 Signals, singlet present. 1 DD, 2 D, 1 S

Para (1,4)

2 signals, 2 Hd, 2 Hd., ALL D

oh/NH SIGNAL WILL APPEAR AS

SINGLET due to avg of spin states, and will not split other prootons

Degree of Saturation

2C + 2 +N -H - X /2

IHD OF 1

1 RING OR 1 DOUBLE BOND

IHD OF 2

1 CYCLOALKANE RING AND 1 PI BOND OR TWO PI BONDS OR TWO RINGS

4 DEGREES

1 RING

, 3 PI BONDS (BENZENE)

6 DEGREES

1 RING, 5 PI BONDS BENZENE AND 2 ADDITONAL PI BONDS

STRATEGIC APPROACH TO NMR ELUCIDATION

1. CALCULATE IHD

2. USING CHEMICAL SHIFT DATA TO DISCERN THE TYPES OF PROTONS PRESENT

   1. READ DOWNFIELD TO UPFIELD

   2. DOWNFIELD INDUCTION

3. Splitting and integration data to build alkyl chains

   1. integration for signals often provided, this is the RELATIVE number of protons, symmetric

   2. Splitting discerns neighboring C or carbons

   3. then check w constiutional structure and ihd

4. with your final structure, check how many signals it would do, check chemical shifts, splitting oaterns and consistent with those shown in spectrum

C-13 NMR

EACH UNIQUE TYP OF C GIVES ONE C13 SIGNAL

C 13 C LACKING H USUALLY SHOW

LOWER INTENSITIES

Ethyl group (–CH₂–CH₃)

Triplet (3H, ~1 ppm) + Quartet (2H, ~2–4 ppm)

Isopropyl group (–CH(CH₃)₂)

Doublet (6H, ~1 ppm) + Septet (1H, ~2–4 ppm)

tert-Butyl group (–C(CH₃)₃)

Singlet (9H, ~1 ppm)

Benzyl group (–CH₂–Ph)

Singlet (2H, ~2.3–3.0 ppm)

Allyl group (–CH₂–CH=CH₂)

• Doublet of doublets / multiplets in alkene region (4.5–6.5 ppm).

• CH₂ next to double bond = ~2 ppm, often complex.

Monosub benzene

multiplet (5H, ~7.2 ppm).

Para-disubstituted

OPPOSITE 2 clean doublets.

Aldehyde (–CHO)

Singlet (1H, ~9–10 ppm).

Carboxylic acid (–COOH)

Broad singlet (1H, ~10–12 ppm).

Alcohol (–OH) and Amine

Broad singlet (1H, ~1–5 ppm)

Methyl next to electronegative group

Singlet (3H, ~3–4 ppm).

Methyl next to carbonyl (–COCH₃)

Singlet (3H, ~2 ppm).

Long alkyl chains (–CH₂–CH₂–CH₃)

• Terminal CH₃ = triplet (3H, ~0.9 ppm).

• Middle CH₂’s = multiplets (~1.2–1.5 ppm).

• CH₂ near heteroatom = shifted downfield (3–4 ppm).