Chemistry 242
Clark College
Homework 1: Alkynes and NMR SOLUTION
All spectra obtained from
SDBS Spectral Database for Organic Compounds, http://www.aist.go.jp/RIODB/SDBS/cgibin/cre_index.cgi, (various dates).
1. Nomenclature. Name the following substituted alkynes.
H
Br
Cl
OH
4-bromo-4-methyl-2-pentyne
3-methylhex-5-yn-2-ol OR
R-2-chloro-3-hexyne
3-methyl-5-hexyn-2-ol
**Remember – Non-carbon functional groups take priority!**
2. Predict the product or give the missing reagents of the following reactions. If more than one
product can be formed, give both products.
1) BH3
2) CH3COOH
-ORH2
Lindlar's cat.
Either reagent can be used
to generate a Z alkene.
O
HgSO4
H2SO4/H2O
Two ketone products
are generated from
an internal alkyne.
O
Excess H-Br
Br Br
H
H
1) HB(sia)2
2) NaOH, H2O2
H2O
O
1) NaNH2
2) CH3CH2Br
Like alkene additions, this
reaction is Markovnikov!
A sterically-hindered boron
reagent is used to obtain the
H anti-Markovnikov product.
Chain extension
Br
Br2 (1 eq)
LiBr, CH3COOH
Stoichiometry is used to
help control the reaction.
Br
Homework 1
Winter 2009
Page 1 of 4
Chemistry 242
Clark College
3. Perform the following synthesis. Create the product given from the starting material shown,
using any reagent needed to affect the transformation.
1) BH3 2) H2O2, NaOH, H2O
-orHgSO4
H2SO4
1) NaNH2
H
2) CH3CH2Br
O
H
Br
H
Br2, CH2Cl2
1) 3 NaNH2
Br 2) H2O
1) HB(sia)2
2) NaOH, H2O2,
H2O
O
4. Using the NMR and IR data provided, give the identities of the starting material, product and the
reagents used for the transformation
Compound A
NMR: 2.08 ppm, 2H, d
1.96, 1H, s
1.825, 1H, m
0.990, 6H, d

What reagents were used?
IR: 3310 cm-1, sharp
2120 cm-1, w Alkyne
peaks! (C-H and CC)
H
Compound B
9.6 ppm. 1H, s Aldehyde!
2.4, 2H, t
1.90, 2H, m
1.74, 1H, m
0.97, 6H, d
IR: 1725 cm-1, s C=O
O
1) HB(sia)2
2) NaOH, H2O2, H2O
H
Starting with compound B, the NMR peak at 9.6 ppm is definitely an aldehyde, and the IR verifies
this. The only way we know how to make an aldehyde at this point is to start with an alkyne. This
allows us to put two carbons in the structure of the reactant and in the product. Also, it is a good
idea to cross off the peaks in the NMR that you have identified – so you do not keep looking at
them. At this point, we have identified 1.96 ppm peak in A (alkyne H), and the 9.6 and 2.4 (alphaCH) peaks in B. Looking at the remaining NMR peaks in either compound shows an isopropyl
group, with the characteristic 6H, d and 1H, m pattern. The only remaining NMR peaks belong to a
CH2 group, and we can use that to link the alkyne to the isopropyl group!
Homework 1
Winter 2009
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Chemistry 242
Clark College
5. Road Map. Give the structures of products A - D along the following synthetic pathway.
Br
1) NaNH2
2) CH3CH2CH2Br
H
1) 3 NaNH2
2) H2O
Br2
CH2Cl2
Br
A
B
H
Na0,
NH3
-78°C
H
C
D
6. Describe and/or sketch the 1H-NMR spectrum for the following molecules. Be sure to indicate
approximate chemical shifts, splitting patterns, and integrations.
The spectrum sketches leave a little to the imagination!
b
c
This ether has three unique hydrogens. The methyl next to the oxygen would be a singlet
located around 3.3 ppm. The -CH2- on the other side of the oxygen would be a quartet around
3.5 ppm. The methyl group at the end of the molecule will be a triplet around 1.2 ppm.
a
O
a
b
3.5
O
c
b
a
O
c
c
3.3
1.2
This ester also has three unique hydrogens, for 3 peaks. The ester methyl group (attached to
the oxygen) should be a singlet and is shifted to approximately 3.7 ppm. The C-H (!-carbon)
next to the carbonyl would be a septet (mulitplet) around 2.2 ppm, and the two methyls would
be a doublet around 1.0 ppm.
a
c
b
3.7
Homework 1
2.2
Winter 2009
1.0
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Chemistry 242
O
c
b
a
O
c
Clark College
This ester also has three unique hydrogens, for 3 peaks. The ester methyl group (attached to
the oxygen) should be a singlet and is shifted to approximately 3.7 ppm. The C-H (!-carbon)
next to the carbonyl would be a septet (mulitplet) around 2.2 ppm, and the two methyls would
be a doublet around 1.0 ppm.
a
c
b
3.7
Homework 1
2.2
Winter 2009
1.0
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