Give the formula and structure of the compound with this IR and a molecular ion peak at 116.

How many peaks in the 13C NMR spectra?

OHOH

F

OH

F

F

2 3 4 3

Cl

FHO

Cl

FCl

Cl

ClCl

6 4 2

Which compound matches this 13C NMR spectrum?

Br

BrBr

Br

Br

Br

Br

Br Br

Br

Br

Br

O

OH

AB

C

OA

BC

Br

B

CA

Rank the labeled C’s of each compound in order of increasing chemical shift?

B ~ 33ppm

C ~ 35ppm

A ~ 115ppm

A ~ 8ppm

B ~ 37ppm

C ~ 210ppm

C ~ 9ppm

B ~ 30ppm

A ~ 177ppm

Information gained from a 1H NMR Spectrum

• Number of Signals

• Position of Signals

• Intensity of Signals

• Spin-Spin Splitting of Signals

How many different types of protons are in each compound?

H3C

O

CH3

HAHA

1 type

H3C

H2C

ClHA

HB

2 types

H3C

H2C

O

CH3

HA

HB

HC

3 types

H3C

CH

H3C

CH2

H2C CH3HA

HA

HB

HC

HD

HE

5 types

Determining the number of different protons in compounds with bonds.

Cl

Cl

H

H

cis to Cl

cis to Cl

1 type

Cl

Br

H

H

cis to Cl

trans to Cl

2 types

Cl

H

H

H

cis to Cl

cis to H

3 types

Determining the number of different protons in compounds with rings..

H

H

H

H

H

H

All H’s are equivalent, 1 type

H

H

H

H

Cl

H

cis to Cl

cis to H3 types

H

H

H

H

CH3

CH3

Protons on methyls are equivalent

Each of these is cis to a methyl 2 types

O

OO

OH

O

O HO

OH

Spin-Spin Coupling

When determining the spin-spin coupling, look at the number of protons on the adjacent carbon. For the methyl group, look at the methylene group. There are 2 protons, so using the N+1 rule tells us that the peak should be a triplet in a 1:2:1 ratio.

For the methylene group, look at the methyl group. There are 3 protons, so using the N+1 rule tells us that the peak should be a quartet in a 1:3:3:1 ratio.

Protons attached via a double bond show a unique splitting pattern., a doublet of doublets.

O

HO

Hc

Ha

Hb

OOH

O

OH

An unknown molecule A has 4 signals in the 1H NMR spectrum. Which of the following corresponds to molecule A

How many nonequivalent protons does the following structure have?

4

Reading from left to right, what multiplicity would be found for the three nonequivalent sets of protons in the 1H NMR spectrum of the following compound?

d, d, s

Introduction

Homolytic bond cleavage leads to the formation of radicals(also called free radicals)

Radicals are highly reactive,, short lived species

Single headed arrows are used to show the movement fo single electrons.

Production of Radicals

Homolysis of relatively weak bonds such as O-O and X-X bonds can occur with the addition of energy in the form of heat or light.

RCH2 R2CH R3CH

Carbon radicals are categorized as primary (1°), secondary (2°) and tertiary (3°) based on the number of attached R groups.

1° 2° 3°A carbon radical is sp2 hybridized with a trigonal planar geometry with the unpaired electron in the unhybridized p orbital.

Bond dissociation energy is used as a measure of radical stability.

Two different radicals can be formed with the cleavage of a C-H bond.

Basically, the more alkyl groups attached to the radical carbon the more stable it is. Also the more stable the radical, the less energy it takes to break the C-H bond.

What type of radical are each of the following?

H3C

H3C

CH2

CH3

•

H3C

CH

H3C

CH

CH3

•H3C

CH

H2C

CH2

CH3•

1°2°3°

Of these three radicals, which is the most stable?

H3C

H3C

CH2

CH3

•

•H + X + H X•

Radical Reactions of Alkanes

Abstraction of a H from a C-H bond in which one electron is sued to form H-X while the other is left on the new alkyl radical.

X

X

•

•

A radical can also add to a alkene by adding onto a double bond and leaving the other carbon that was part fo the double bond as a radical.

X + X X X

Radicals are highly reactive and unstable and usually react quickly with a sigma or pi bond. However sometimes they can react with another radical.

O O + X O O X

When oxygen, a diradical, is present it acts as a radical inhibitor or scavenger. Meaning it prevents the radical from attacking any alkanes or alkenes.

• •

• • • •

In the presence of heat and light, alkanes and halogens will react to form alkyl halides.

H3C H + Cl2 H3C Cl + HCl

H+ Br2

Br

+ HBr

H3CH2C CH3 + 2Cl2 H3C

H2C CH2Cl

+

H3CHC CH3

Cl

Predict the products from the monobromination of the foloowing compound?

CH CH

H3C

H3C

CH3

CH3Br2

hv/ heat CH CH

H3C

H3C

CH2Cl

CH3CH C

Cl

H3C

H3C

CH3

CH3

CH CH

H3C

H3C

CH3

CH2Cl CH CH

ClH2C

H3C

CH3

CH3

Step 1 - Initiation

Cl Cl hv/heat2Cl •

Step 2 – Propagation

H3CH2C H + Cl H3C CH2 + HCl

H3C CH2 +Cl Cl H3CH2C Cl + Cl

• •

• •

Step 3 - Termination

Cl Cl+ Cl Cl

H3C CH2 CH3H2C+ H3CH2C

H2C CH3

H3C CH2+ Cl H3C CH2Cl

In each step of the propagation a bond is broken and formed. And because the overall step has a -H it is exothermic. Step 1 is called the rate determini g step because it is higher in energy.

Transition States

Cl----H----CH2CH3 Cl---Cl---- CH2CH3

H3CH2C CH3 +Cl Cl H3C

H2C CH2Cl + H3C

HClC CH3

There are 6 Methyl H’s and 2 Methylene H s. Based on this, the ratio fo the two products should be 3:1(primary to secondary).

However, the ratio is 1:1.

The more stable the radical being formed is, the easier it is to cleave the C-H bond.

Which C-H bond in each compound is most reactive?

H H

H

H

Chlorination Vs. Bromination

1:1

H3CH2C CH3 + Cl2 H3C

H2C CH2Cl + H3C

HClC CH3

H3CH2C CH3 + Br2 H3C

H2C CH2Br + H3C

HBrC CH3

99%

Chlorination is faster and nonselective. This is due to it’s rate determining step being exothermic.

Bromination is slower and chooses the most stablew radical. This due to it’s rate determining step being endothermic.

H3C

H2C CH

CH3

CH3

+ BR2

H3C

H2C CCl

CH3

CH3

+ Br2Br

Halogenation is useful in the formation of alkenes.

+ Cl2

Cl

+ K+-OH

An elimination in the presence of a strong base is responsible for the formation of the alkene.

Cl -OH

H

+ H2O + KBr

H+

+

+ -HSO4

HOHO

H

H

-HSO4

OH

aq. H2SO4

OH+

Conversion of the alkene to an alcohol via nucleophillic substitution is an extension of the utility of radical halogenation.

+HgOAc

+HgOAc

MeOH

H

CH3

HgOAc

OCH3

HgOAc

NaBH4

OCH3

Hg(OAc)2

NaBH4

OCH3

Oxymercuration-demercuration of an alkene results in the formation of an ether.

Radical halogenations give a racemic mixture of pRodcuts when possible. This halogenation of an achiral compound results in 3 products. A primary and secondary alkyl halide. The secondary halide exist as a pair of enantiomers due to the creation of a stereogenic center upon halogenation.

H3CH2C

H2C

Cl2H3C

H2C

H2C H3C

HClC

H2C+CH3 CH2Cl CH3

H3CCH2CH3

BrH

Cl

H3C

Br

CH2CH3

Cl2

Cl2

H3C CH2CH3

Br Cl

H3C CH2CH3

Cl Br

H3C

CH2

CH3

Br H

Cl

H3C CH

CH3

Br H

Cl2

Cl2

H3CCH3

Br

Cl

H

H

H3CCH3

Br

H

H

Cl

enantiomers

diastereomers

H3C CH2

CH3

Br HCl2

ClH2C CH2

CH3

Br H

+

H3C CH2

CH2Cl

Br H

Only achieve enantiomers or diastereomers if the halogenation takes place at a stereogenic center.