Hi UltimaOnline, Sorry to trouble you again, but here are some
questions which I have 
NJC/2012/P1/Q22
Why can’t B be an answer, given that acyl chlorides can be made
by reacting thionyl chlorides with carboxylic acid?
NJC/2012/P1/Q25
I understand that C is the answer
(http://www.chemguide.co.uk/organicprops/haloalkanes/making.html),
but why can’t the other options be prepared with distillation?
NJC/2012/P1/Q27
I think there are 3 ways which the elimination can take place
given that there are 3 B hydrogens. Applying Saytzeff’s Rule, I can
only eliminate the option where the hydrogen is abstracted from the
methyl substituent. The other options (where the hydrogen is
abstracted from the CH2, and where the hydrogen is abstracted from
the cyclohexane substituent) seem equally substituted to me. So
what would be the major product of the reaction? Further, wouldn’t
either of the products contain 2 double bonds, thereby making the
total number of stereoisomers 4?
Once again, thank you for your help. :)
Hi Gohby, no prob :)
NJC/2012/P1/Q22 : B is incorrect because the amine group present
is even more nucleophilic than the OH of COOH, and hence would
react even more readily with the thionyl halide electrophile, in
this case the primary amine would undergo 2 consecutive
nucleophilic substitutions with the thionyl bromide to generate the
N=S=O group, eliminating 2 molecules of H-Br. This is beyond the A
level H2 syllabus, and so isn't completely fair, but nonetheless
Cambridge *could* ask this question as a challenging A grade
question, to determine if the candidate understands the concepts of
relative nucleophilicities and electrophilicities, and choose the
best possible option among the 4 presented.
Cambridge has done indeed so many times in the past, especially
in recent years, deliberately presenting 2 likely options : 1
option going beyond the A level H2 syllabus so candidates cannot be
sure if it's correct, the other option within the A level H2
syllabus so competent candidates should be sure if it's correct or
wrong, and therefore choose accordingly.
NJC/2012/P1/Q25 : Options A & D can be carried out at room
temperature, and hence doesn't require this setup (and therefore
should not be used). B requires heating under reflux. The setup
shown is heating with distillation to separate out the desired
product which has a lower boiling point compared to the
reactants.
NJC/2012/P1/Q27 : This reaction involves competition between the
Zaitsev aka Saytzeff elimination, and the Hofmann elimination.
While the Bronsted-Lowry base employed is the sterically unhindered
OH- ion, nonetheless the cyclohexane ring does pose some degree of
steric hindrance, but in this case the Zaitsev product should
ultimately still dominate (and conveniently also because the
Hofmann elimination isn't taught in the H2 syllabus).
The major product which is the Zaitsev product, is therefore the
product that has the double bond with the cyclohexane ring, as this
product would be a more substituted internal alkene (with no H
atoms bonded to its sp2 C atoms), compared to the alternative
product that has the double bond with the alkyl side chain (with 1
H atom bonded to its sp2 C atoms).
The number of stereoisomers of the major product is therefore 2
(ie. as a result of the pre-existing alkene double bond in the
longer side chain), instead of 4 (cis-cis, cis-trans, trans-cis,
trans-trans).