Questions from aqa jun 10 Chem 4- Please help!
Chemistry discussion, revision, exam and homework help.
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Re: Questions from aqa jun 10 Chem 4- Please help!
8a) Friedal Crafts acylation - electrophilic substituion reaction.
so you know to get electrophile reaction is:
RCOCL + ALCL3 --) RCO3+ + ALCL4-
And so one of the answers is alcl3
and now just look at product given in question to work out what the other is..
as it is benzene--c=0--ch2ch3 thats product given so from that you know ch3ch2c=0cl must have been added.
so overalll reaction to get electrophile:
ch3ch2c=0cl + alcl3 -) ch3ch2c0+ + alcl4-
ch3ch2c0+ is electrophile its used in electrophile substitution which is next q..Last edited by wenger16; 08-06-2012 at 07:40. -
Re: Questions from aqa jun 10 Chem 4- Please help!
4b im not sure need to revise that again.
3e)
Its primary amine so its gonna be
R-N-H2
And because its carbon nmr focus on carbons.
There are 3 peaks so 3 different carbons in 3 different environments.
So theres ch2 - thats 1 environment
c conected to 2 ch3's is another environment as its all same environment as you cant tell which is which so same enviroment.
and other ch3 is 3rd carbon different enviroment so thats 3rd peak...
so:
NH2--CH2--C(CH3)2--CH3
3bii)
Tertiary amines are like this:
N-R3 connected to 3 r groups.
Theres a doublet so yy=ou know because of n+1 rule there has to be a carbon in there connected to only 1 H.
Also N has to be connected to 3 r groups but if its connected to only 1 carbon then no adjacent hydrogen for that particular carbon so no peak.
So its:
CH3-N-CH3 NO PEAKS GIVEN FROM THIS
AND THEN OTHER PART CONNECTED TO N IS :
CH3-CH-CH3
CH GIVES THE DOUBLET, 3 SETS OF HYDROGENS GIVE 3 PEAKS.