November 06
21
37
39
Nov 06 p121. This one, you need to do it step by step. Find the number of chloroethanes when n=1, 2, 3 and 4 in the formula C
2H
6-nCl
nWhen n=1 ----> C
2H
5Cl : You can have only
one type.
When n=2 ----> C
2H
4Cl
2 : You can have
2 types, one when both chlorine atoms are found on the same carbon and two, when each carbon atoms in the compound carries one chlorine.
When n=3 ----> C
2H
3Cl
3 : You can have
2 types, one when all three chlorine atoms are found on the same carbon and two when one carbon carries 2 chlorine while the other carbon carries one chlorine.
When n=4 -----> C
2H
2Cl
4 : Again
2 types. One when one carbon carries 3 Cl while the other carries one Cl. and two when both carbon atoms carries 2 Cl.
So in all that makes
7 types.
Answer is
C37. To have two organic products upon oxidation, you need to have two double bonds so that when both breaks two different compounds are obtained.
So
number two is rejected since only one product will be formed.
Now, to have a ketone, you need to have a secondary alcohol formed upon oxidation of the alkenes.
The answer is
DThis is because the double bonds are arranged in such a way that upon oxidation you'll be getting two different compounds each carrying a ketone.
The
third option however will produce a compound carrying two ketone groups while the other compound will form a carboxylic acid. So only
compound 1 is appropriate.
39. Decolourise KMnO
4 ----> he's talking about alkenes.
So it's going to be a dehydration process. Therefore number 3 being a carboxylic acid will not be dehydrated.
Both number 1 and 2 are alcohols and will be dehydrated by sulfuric acid to form alkenes.
Hence answer is
BNOTE : I've not had time to confirm my answers with the marking scheme. Do check it out and let me know if there are mistakes.