Chemistry 121

Assignment #1

Answers

Marked answers are given in red.

Q1

Wrong name IUPAC name
2-ethylpentane 3-methylhexane
3-isopropylpentane 3-ethyl-2-methylpentane
2-methyl-2-hydroxybutane 2-methylbutan-2-ol
hepta-3,6-diyne hepta-1,4-diyne
4-bromo-5-isopropylnona-1-yne-trans-7-ene (E)-6-bromo-5-isopropylnon-8-yne-2-ene
3-oxopentan-1-ol 1-hydroxypentan-3-one
5-oxopentan-1-ol 5-hydroxypentanal
5-chloropentane 1-chloropentane
3-methyl-5-ethyloctane 5-ethyl-3-methyloctane
1-chloropropan-3-oic acid 3-chloroprpanoic acid
trans-1-chloro-1-bromocyclopentane 1-bromo-1-chlorocyclopentane
5-hydroxy-5-methyl-8-oxooctan-3-one 4-hydroxy-4-methyl-6-oxooctanal
heptan-4-yne heptan-3-yne
trans-2-methylhexan-5-ene 5-methylhex-1-ene
cis-4,6-dimethylcyclohexanone cis-2,4-dimethylcyclohexanone
1-iodo-2-chlor-3-fluoro-4-bromohex-5-ene 3-bromo-5-chloro-4-fluoro-6-iodohex-1-ene
hexa-1,6-dioic acid hexanedioic acid
sodium 3-bromo-2-iodo-3-ethyl-2-butyldecan-6,8-dienoate sodium 3-bromo-2-butyl-3-ethyl-2-iododecan-6,8-dienoate


Q2

From top left, across the page:

2,4-dimethylhexane
(Z)-6,6-dimethyloct-2-ene
10-hydroxy-4-methyldecanal
propanedioic acid
2-brombutanoic acid

From bottom left, across the page:
1,1,1-trifluoropropanone
trans-1,2-diethylcyclohexane
1-isopropylcyclopentene
4-cycloheptyl-7-cyclopropyl-6-methyl-5-propyloct-1-ene
(Z)-5-ethyl-2,4-dimethylhepta-4,6-dien-3-one


Q3.1

a. A major driving force introduced in this class is the reaction of an electron rich centre with an electron poor centre.

b. Haloalkanes have an electron deficient carbon centre due to the polarization of the C - Halogen bond. Any group reacting with a haloalkane needs to be electron rich. CN- is electron rich and so will react. H+ is electron poor and so will not react.

c. The p bond of an alkene is electron rich and will react readily with electron poor centres.NO+ is electron poor and so will react. NO2- is electron rich and so will not react.


Q3.2

a. haloalkane to alkanol: use hydroxide ion.

b. a secondary alkanol can be oxidized with KMnO4 to give an alkanone.

c. An alkanone can be reduced to an alkanol using LAH followed by a water work up.

d. An alkanal can be oxidized to an alkanoic acid with KMnO4.

e. An alkene will add HCl to give a chloroalkane.

f. An alkyne can be reduced using Hydrogen gas with a Platinum catalyst to gave an alkane.

g. A primary haloalkane can be converted into an alkanoic acid in two steps: first convert the haloalkane into an alkanol using hydroxide ion (as in part (a)); second oxidize the primary alkanol to the alkanoic acid with KMnO4.

h. A secondary alkanol can be converted into an oxime in two steps: first oxidize the secondary alkanol to an alkanone with KMnO4; second react the alkanone with hydroxylamine to give the oxime.


Q3.3

reactant 1 reactant 2 product
a hex-1-yne anion 1-bromobutane dec-5-yne
b hex-5-yn-1-ene anion bromethane oct-1-en-5-yne
c hex-1-yne anion propanal non-4-yn-3-ol
d propyne anion (2 equivs) propanedial non-2,7-diyne-4,6-diol
e (step 1) but-1-yne anion 1-bromopropane hept-3-yne
e (step 2) hept-3-yne hydrogen/platinum heptane
f (step 1) but-1-yne anion butanal oct-5-yn-4-ol
f (step 2) oct-5-yn-4-ol hydrogen/platinum octan-4-ol


Q3.4

a. Primary alkanol and alkanal: use 2,4-DNP. Alkanal reacts.

b. Alkene and alkanone: use KMnO4. Alkene reacts

c. Alkanone and alkanal: use KMnO4. Alkanal reacts.


Q4.

Oxidation by KMnO4:

reactant product
cyclohexene (cold KMnO4 cis-cyclohexane-1,2-diol
3-methylpentan-3-ol No reaction (tertiary alkanol)
4-oxopentanal 4-oxopentanoic acid
chlorocyclohexane No reaction
cyclohexanol cyclohexanone


Q5.

reactant reagent(s) product
a (E)-hex-3-ene H2 / Pt hexane
b butanone 1. LAH; 2. H2O Butan-2-ol
c (E)-hex-3-ene HCl 3-chlorohexane
d butan-1-ol ZnCl2 / HCl 1-chlorobutane
e pent-1-yne Br2 (2 equivs) 1,1,2,2-tetrabromopentane
f
g 1-chlorohexane LAH hexane
h 1-bromo-3-methylbutane hydroxide ion 3-methylbutan-1-ol
i NH2OH
j 1-bromobutane hex-1-yne anion dec-5-yne