Please see the attached file for full problem description.
1) Ethanal is the common name for acetaldehyde. And it has the formula C(-CH3, =O, -H) (I will show
carbon compounds this way, first the carbon and then its constituents, "-" is for single bond and "=" for
double bond) and benzaldehyde has the formula C(-Ar=O, -H) (-Ar is used for the aromatic ring).
(a) With KMnO4 these aldehydes will oxidate to acids. Ethanol will oxidate to acetic acid C(-CH3, =O, -OH), and benzaldehyde will oxidate to benzoic acid C(-Ar, =O, -OH).
(b) With KCN the CN- ion will attack the slightly positive carbon and the result will be cyanohydrins.
For ethanal the result is C(-CH3, -H, -OH, -CN). And for benzaldehyde C(-Ar, -H, -OH, -CN).
(c) With NaOH the result will be gem-diols. For ethanol the result is C(-CH3, -H, -OH, -OH). And for
benzaldehyde C(-Ar, -H, -OH, -OH).
(d) 2,4 DNP is a reagent for distinguishing aldehydes and ketones. Again they will react and the results
will be for ethanal, C(-CH3, -H, =NNHC6H5). And for benzaldehyde, C(-Ar, -H, =NNHC6H5).
2) Homologous series: A series of compounds in which each member differs from the next member by a
constant unit. For example CH3-CH2-CH2-CH2-CH2-CH3 Isomerism: Different molecules that have the same molecular formula are isomers.
(a) The three molecules with the formula C3H8O will be
1) Propyl alcohol HO-CH2-CH2-CH3
2) 2-propanol or iso-propyl alcohol CH3-CH(-OH)-CH3
3) Ethyl methyl ether CH3-O-CH2-CH3
For distinguishing them first we can make the Iodine test, the ether(3) will react. And then to
distinguish between a primary and secondary alcohol (1 and 2) we use Lucas Test (ZnCl2/HCl) the primary alcohol(1) will give no reaction while the secondary (2)will give a slow reaction. To synthesize the Ethyl methyl ether from ethanol, we take the ethanol and react it with NaH, so we get a very reactive CH3CH2O:(-) Na(+) and then react it with CH3I, so the ethoxide attacks the iodoalkane with an SN2 reaction and the Ethyl methyl ether is the result.
(b) If 25 cm^3 alkene reacts with oxygen to give 100 cm^3 CO2, then the ratio is 25/100 = 1/4 so there
must be 4 moles of CO2 thus 4 Carbons in the alkene. Ozonolysis reaction is for alkenes, and it creates =O oxygen in the place of the alkene and divides it from there. So the ozonolysis creates methanal C(-H, =O, -H) and Y(C3H6O) can have 2 possible structures one is proponal an aldehyde and the other is acetone a ketone. But since it does not give ppt from Tollens rxn it cannot be an aldehyde and it is a ketone thus Y is acetone. And then we can solve for X by combining methanal and acetone by removing the oxygens (going back for ozonolysis) and putting a double bond combining them, we get the alkene X as 2 -methyl-1-propene.
The expert analyzes the homologous series of compounds. The reagents for distinguishing aldehydes and ketones are determined.