Short Answer:

1) What type of electromagnetic radiation is used in
nuclear magnetic resonance?

 

 

radio

2) What is the most abundant peak in a mass
spectrum called?

 

Base
peak

3) What would the proton NMR peak look like
for the indicated hydrogen?

 

 

CH

CH

3

CH

3

O

H

3

C

Because the two sets of
adjacent protons are
equivalent this peak would
follow the n+1 rule and be a
septet.

Nuclear

Magnetic

Resonance

Spectroscopy

1

H NMR—Spin-Spin Splitting

When two sets of adjacent protons are different from
each other (n protons on one adjacent carbon and m
protons on the other), the number of peaks in an NMR
signal = (n + 1)(m + 1).

Figure 14.8

A splitting diagram for the H

b

protons in 1-bromopropane

4) To which end of an alkene does the hydrogen add
in hydrohalogenation without a radical initiator?

The hydrogen adds to the least substituted end of
the double bond so that the most stable carbocation
is formed.

5) What is another name for the 1,4-addition
product?

Conjugate addition
product

Predict the products.

1)

This cation is 2° and 3°. Other cation is 2° and 2°.
Thermodynamic product is most substituted alkene.

H

Br

+

+

Br-

Br

2)

Br

Br

H

Br

Br

+ Br

3
)

Br

H

Br

Br

4)

H

3

CO

O

O

H

3

CO

H

H

5)

H

3

C

CH

3

CH

3

CH

3

Cl

2

hv

H

2

C

H

Cl

CH

2

Cl

CH

2

Cl

6)

NBS

hv

H

Br

Br

Br

Br

+

Br

Mechanism: Draw out the Mechanism for the
following reaction.

1)

H

3

C

C

H

2

CH

H

3

C

CH

3

Br

2

ROOR

H

3

C

C

H

2

C

H

3

C

CH

3

Br

Step 1 -
Initiation

Br

Br

Br

RO

Step 2-
Propagation

H

Br

+ HBr

Br

Br

Br

+ Br

Step 3 - Termination

Br

Br

Br

2

Br

bR

2)

+

H

3

CO

2

C

CO

2

CH

3

heat

CO

2

CH

3

CO

2

CH

3

CO

2

CH

3

H

3

CO

2

C

CO

2

CH

3

CO

2

CH

3

+

CO

2

CH

3

CO

2

CH

3

Spectroscopy

1)      Using the MS and IR spectra attached (1A and
1B) propose the formula and structure of this
compound.

 

MS shows a molecular ion peak at 106 and a M+2
peak at 108..

So 106-35=71 so 71/12=5 carbons so 71-60=11
hydrogens so

C

5

H

11

Cl 2(5)+2-11-1=0

However, there is a carbonyl peak in the IR

So need to add an oxygen.

-CH

4

gives C

4

H

7

ClO so 2(4)+2-7-1=2/2=1

So this is taken by the C=O bond.

One more thing, there is no peak at 2750 so no
aldehyde, our carbonyl is a ketone

O

Cl

O

Cl

O

Cl

The first one can be eliminated because of the
base peak at 43 in the MS, a loss of 63 accounts
for the loss of a –C

2

H

4

Cl group.

2) Using the MS, IR and proton NMR (2A, 2B, 2C)
propose a possible formula and structure.

So the molecular ion peak is 165. Odd number
means a nitrogen. 165-14=151/12=12 carbons so
151-144=7 hydrogens

So C

12

H

7

N 2(12)+2-7+1=20/2=10 way to high

So lets take off a C and add 12 Hs C

11

H

19

N 2(11)+2-

19+1=6/3

Now lets look at the IR. There is a small peak at
3243 which tell sus thee is some sort of amine
present. Also there is a peak at 1649 and with a
nitrogen present this tells us there is an amide.Also
there is a C=C double bond at 1622.

So we need to add an oxygen

-CH4 and Get C

10

H

15

NO so 2(10)+2-15+1=8/2=4

Alright now looking at the proton NMR we see a
doublet of doublets in the aromatic region (6 to
7ppm) which tells that we have a para substituted
benzene ring.

So the last structure we came up with, , had a DOUS
of 4. This only covers the benzene ring so we need
to ad one more degree for the C=O in the amide.

We know adding an oxygen well take care of this and
give us

C

9

H

11

NO

2

so 2(9)+2-11+1=10/2=5

O

NH

2

So there are 5 peaks in the proton NMR, two of those
are the doublet of doublets in the aromatic region and
another one is the small peak the furthest downfield
which is the amine proton.

So that leaves 2 carbons and 1 oxygen. And we
know that there is only the para position on the
ring to add anything to.

So lets look at the remaining two peaks in the
NMR they are both singlets which tells us there
are no adjacent protons that have an effect.

This means that they are either attached to an
oxygen or nitrogen.

O

NH

2

O

So that only leaves one place to add the other
carbon.

O

HN

O