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Revision Work Organic (Nov 09)D
Name :
70
Group : ________
1
(a) (a) Aldehydes and ketones can serve as starting points for a number of chemical transformations. Use your
knowledge of the carbonyl group chemistry to show how the following changes can be achieved :
(Please specify the reagents and conditions necessary for these reactions)
Aldehyde/ketone +
(i)
NaBH4 in aq. methanol + heat
(ii)
H2NNH2, Hydrazine (or a derivative)
(iii) HCN* + heat
(iv)
(v)
at r.t.
* or H2SO4 added to KCN(aq) + heat
Alcohol
→
Hydrazone
gem
→
RNH2,i.e.an Amine at r.t.
.
→ Hydroxynitrile .
→
PCl5 at r.t./gentle heat
.
→
dihalolkane .
Imine
.
(5 marks)
2
What tests would you use to distinguish between the following pairs of isomeric compounds. (Give tests and the
expected results) :
(i)
But-1-yne and buta-1,3-diene (formulae : C4H6)
CuCl dissolved in ammonia will give a rusty brown precipitate with terminal alkynes like but-1yne but shows no visible reaction with butadiene. (A similar reaction is that with aq.Ag(NH3)2+
but this gives a white precipitate with but-1-yne and nothing with buta-1,3-diene.)
(ii)
Chloroethanal and ethanoyl chloride (formulae : C2H3ClO)
Ethanoyl chloride reacts exothermically with water to give acidic solutions while the
chloroethanal readily produces the neutral gem diol in water. So a pH check of a few drops
of the substances in water should easily tell them apart – the acidic solution being that
obtained from the acid chloride. An alternative method is to add a drop of each of the
materials to AgNO3(aq) + dil HNO3 and the one that gives the white (AgCl) precipitate would
be the acid chloride again. The Chloroethanal would not release its chlorine that easily.
(iii)
3-methylpentan-2-one and 2-methylpentan-3-one (formulae : C6H12O)
3-Methylpentan-2-one gives a positive iodoform test i.e. would produce the yellow
precipitate having an antiseptic smell (CHI3) when a drop or two of this material is added to
I2(aq) + dilute NaOH(aq) (or KI(aq) + NaOCl(aq)). 2-Methylpentan-3-one gives no visible reaction
under the same conditions (The α hydrogens are still substituted but the carbon skeleton
remains intact).
(iv)
Butanenitrile and 4-aminobut-1-yne (formulae : C4H7N)
Heating a nitrile with dilute aqueous NaOH gives ammonia vapour after some time but the 4aminobut-1-yne remains unchanged in a similar edium under the same conditions.
Alternatively, addition of a small amount to a solution of Cu2+ gives a dark blue water soluble
comples with amino compounds i.e. with the 4-aminobut-1-yne but not the nitrile.
The tests for a terminal alkyne as seen in 2(i) above can also be used here because the 4aminobut-1-yne is a terminal alkyne whilt the butane nitrile is not and will not give any
precipitate with either Cu(NH3)4+ or the Ag(NH3)2+
(8 marks)
3
(a) Using named reagents, give the mechanism that operates in a Unimolecular Nucleophilic Substitution :
1
A Unimolecular Nucleophilic Substitution mechanism i.e the SN1 mechanism is seen when a
tertiary haloalkane, such as 2-bromo-2-methylpropane for example, reacts with nucleophiles,
such as the hydroxide ions from KOH. The Bromine atom leaves as a Bromide ion Br- and
leaves behind a partially stabilised carbonium ion (CH3)3C +. The nucleophiles present such as
the OH- from the KOH (but could equally be water molecules) attack the carbocations to give
the alcohol
CH3
C
CH3
CH3
fast
slow
(r.d.s.)
Br
CH3
+
C
CH3
+ Br
C
CH3
CH3
CH3
_
HO
_
+
Br
CH3
HO
(6 marks)
(b) Fill in the missing details/structures in the table below :
A
CH3
CH2
C
CH
CH3
O
CH2
CH2
Reagents : Cr2O7 2 - / H+
Conds.:
warm / gentle reflux
OH
HO
C
C
C
CH3
CN
CH2
O
CN
C
O
CH2
Reagents : dil. NaOH(aq)
Conds.:
heat under reflux
then cool and acidify
Cl
CH3 C
COOH
HO
C
C
CH3
O
CH2
HO
C
CH3
H
dil. NaOH(aq)
warm
O
C
CH
OH
O
H
OH
CH3
CH3 C
H
OH
B
O
Reagents : Cr2O7 2 - / H+
Conds.:
distil immediately
OH
HO
C
CH2
2
O
HCN(aq)
heat under reflux
OH
CH3 C
COOH
PCl5 (or SOCl2)
heat
then add H2O
HO
C
CH2
O
HO
Cl
CH3 C
D
COOH
C
HO
NH2
Reagents : conc. aq. NH3
Conds.:
stir at r.t. for several hours
O
CH2
C
CH3 C
COOH
NH2
CH3 C
C2H5OH /H+
heat under reflux
COOC2H5
CH2
O
CH2
COOC2H5
(18 marks)
4
(a) Examine the scheme below. (In this scheme, ethanol is converted to malonic acid)
O
CH3 CH2Br
I
CH3
II
A
C
H
C
B
III
F then H2O
F
O
IV
C
CH3
C2H5
E
D
(i) For each of the steps I to IV indicated above, give (i) the reagents and (ii) the conditions used
Step
Reagents
Conditions
I:
dilute NaOH(aq)
heat under reflux
.
2+
II :
Cr2O7 /H
distil immediately
.
Mg metal
heat under reflux in dry ether .
III :
IV :
Cr2O7 2- /H+
heat (can reflux gently)
(8 marks)
)
(ii)
Give structures for compounds B and D
Br
Mg
CH3 CH2 OH
CH3 CH2
B
D
(2 marks)
(iii) What is the organic substance that forms when the compound E in the above scheme is dissolved in dry
methanol and dry HCl gas is passed through this solution until no further reaction occurs?
The ketal, 2,2-dimethoxybutane, is obtained eventually, going initially through the hemi ketal
2-methoxybutan-2-ol.
Give the structure for it :
3
O
CH3
CH3
O CH3
C
C2H 5
(3 marks)
(b) Give one structure of a carbonyl compound that gives a resin with strong NaOH and another one which gives a
Cannizzaro. Briefly explain what determines what type of reaction will be given with strong alkali.
An aldehyde that lacks alpha hydrogens in its structure will give a Cannizzaro reaction (a
disproportionation reaction) in strong alkaline solutions while Aldehydes that have alpha
hydrogens will give resinous products in (hot) strong alkalis. Examples for these two reactions
are : ethanal gives a resin while methanal gives the Cannizzaro.
O
O
CH3 C
H
H
C
H
gives a resin
gives the Cannizzro
(4 marks)
4