argh. question 2 screwed me up too. zzz. no use crying over spilt milk guys! good luck for chem on tues since math is over!
i have econs on tues also. zzzzzzzz worse thing that can happen to a timetable.
Another P3 question from the H2 'A' level 2008 paper :
(Or something close to it, according to my students)
List and explain the following alkyl halides (halogenoalkanes) according to their boiling points :
Alkyl bromide, alkyl iodide, alkyl chloride.
Solution :
There are two opposing factors, which you have to describe and explain in the exam to score full marks.
Electronegativity and hence strength of permanent dipole-dipole interactions would point to (highest boiling point) alkyl chloride, alkyl bromide, alkyl iodide (lowest boiling point).
However, no. of electrons and hence strength of induced dipole-dipole van der Waals interactions would point to (highest boiling point) alkyl iodide, alkyl bromide, alkyl chloride (lowest boiling point).
As it turns out, based on experimental evidence, van der Waals interactions trumps/owns/pwnz/outweighs permanent dipole-dipole interactions (specifically in THIS context of alkyl iodide versus alkyl bromide versus alkyl chloride; because we're talking about a lot of electrons here).
Hence, the correct answer is (highest boiling point) alkyl iodide, alkyl bromide, alkyl chloride (lowest boiling point).
A similar question, this time involving hydrogen halides, may be found on my "Collection of 'A' Level Chem Qns" here :
http://www.sgforums.com/forums/2297/topics/320107?page=3
Do a "find" (control-F) for "hydrogen iodide".
nvm...it is not like 8 marks anyway. I knew they are trying to be funny. I also know the yield will be low. I couldnt think of anything else, since I know in H2 syllabus, the only way to get phenylamine is to reduce nitrobenzene using conc.HCl/Sn.
So I guess, there are more than one possible answer, but strictly speaking it has to be sodium sulphide, since it makes perfect common sense, even if question did not mention that, we will select the reagents/conditions whereby you get a good yield of products.
I wonder what is their 'standard' answer key for this question? Hmm....
How do Cambridge really mark anyway?
'A' level 2008 Qn
The molecules of compond P, C7H15Br, are chiral. On treatment with NaOH(aq), P produces alcohol Q, C7H15OH, which does not react with hot, acidified Na2Cr2O7(aq). The elimination of HBr from compound P produces a mixture of 4 different isomeric alkenes with the formula C7H14, only two of which are geometrical isomers of each other. Suggest the structural formulae of compound P and the 4 alkanes.
Solution :
P is 3-bromo-2,3-dimethyl-pentane.
Originally posted by UltimaOnline:'A' level 2008 Qn
The molecules of compond P, C7H15Br, are chiral. On treatment with NaOH(aq), P produces alcohol Q, C7H15OH, which does not react with hot, acidified Na2Cr2O7(aq). The elimination of HBr from compound P produces a mixture of 4 different isomeric alkenes with the formula C7H14, only two of which are geometrical isomers of each other. Suggest the structural formulae of compound P and the 4 alkanes.
Solution :
P is 3-bromo-2,3-dimethyl-pentane.
i got P = 3-bromo-3-methyl-hexane.
Originally posted by deathmaster:
i got P = 3-bromo-3-methyl-hexane.
Simple test to see if your answer is correct :
Are you able to draw out 4 different isomers (at the end of the reaction pathway), of which 2 of them are geometrical isomers?
ya.
cos bromine is on 3rd carbon, you will get cis-trans isomerism.
Originally posted by deathmaster:ya.
cos bromine is on 3rd carbon, you will get cis-trans isomerism.
The question specifies you must only get
1) Cis isomer
2) Trans isomer
3) Another separate isomer (no cis-trans)
4) Yet another separate isomer (no cis-trans)
deathmaster, can you draw out your 4 isomers above, and post (their IUPAC names) here on this thread?
you will get 4 isomers, 2 of them cis-trans.
1) when water removed from the 3 carbon side, you get 3-methyl-hex-3-ene
2) when water removed from the 2 carbon side, you get cis-trans isomers of
3-methyl-hex-2-ene. (2 isomers)
3) when water is removed from methyl side, you get 2-ethyl-hex-1-ene
total 4 isomers.
Originally posted by deathmaster:you will get 4 isomers, 2 of them cis-trans.
1) when water removed from the 3 carbon side, you get 3-methyl-hex-3-ene
2) when water removed from the 2 carbon side, you get cis-trans isomers of
3-methyl-hex-2-ene. (2 isomers)
3) when water is removed from methyl side, you get 2-ethyl-hex-1-ene
total 4 isomers.
>>> 1) when water removed from the 3 carbon side, you get 3-methyl-hex-3-ene <<<
That C has no H bonded to it. (From the name of your compound P, that C is bonded to 3 other Cs and to 1 Br.)
Originally posted by UltimaOnline:
>>> 1) when water removed from the 3 carbon side, you get 3-methyl-hex-3-ene <<<
That C has no H bonded to it. (From the name of your compound P, that C is bonded to 3 other Cs and to 1 Br.)
don't get you.
3-bromo-3-methyl-hexane,
carbon 3 is attached to a 1-carbon methyl, 2 carbon ethyl, 3-carbon propyl, and all these adjacent carbon do have hydrogen attached wad.
my compound is just an isomer of wad u suggested wad, except that the 3-carbon propyl on mine is a methyl-ethyl on urs. not much diff actually.