CN is an electron withdrawing group, there are significant

E2 solution
1 (a) Brief explanation with reference to appropriate resonance structures
C
C
Even though OEt is an electron donating
group, and –CN is an electron withdrawing
group, there are significant resonance
contributors that impart single bond
character to the C=C
N
N
O
O
1 (b) Brief explanation with reference to appropriate resonance structures
O
O
OCH3
OCH3
2(a)
3-ethyl-3-hexene
c
2(b)
isopropyloxybenzene,
C6H5-OCH(CH3)2
g
2(c)
4’-methylacetophenone,
para CH3-C6H4-COCH3
h
3. compound A
1-octene
In the p-isomer, electron density from the
OMe group is delocalized through to the
C=O. In the m-isomer, although there are a
number of reasonable resonance
contributors, the electron density cannot be
delocalized through to the C=O.
trans-2-octene
b
a
4-isopropylphenol,
para (CH3)2CH-C6H4-OH
d
2-(1-propyl)phenol,
ortho CH3(CH)2-C6H4-OH
f
4-(1-propyl)phenol,
para CH3(CH)2-C6H4-OH
e
4-methoxystyrene,
para CH3O-C6H4-CH=CH2
k
3. compound B
propiophenone,
2-phenylpropanal,
PhCOCH2CH3
C6H5CH(CH3)CHO
j
i
3. compound C
O
Ph
COOH
O
SODAR = 5, IR⇒ o-disubstituted
benzene, no C=C oop
SODAR = 2, IR⇒non-congugated
COOH, no C=C
SODAR = 5; IR⇒benzene ring,
monosubstituted; high C=O freq
⇒phenyl on O of ester.
4. compound D
4. compound E
5. compound F
See below
5. compound F
It is difficult to reconcile all the data…so, credit for the following clues:
1. IR ⇒ C≡C? (or C≡N?)
2. IR ⇒ conjugated C=O (i.e., rel. low freq.);
3. IR ⇒ Monosubstituted phenyl ring
4. MS ⇒ M+ suggests even number of N (or zero)
5. M+1 suggests C9
6. NMR⇒benzene ring, and one other type of proton (CH3?), but what is the integral ratio? 6:4?
7. 5:3 would be better since you know that there is a monosubstituted benzene ring. In which case, the
3H signal is likely to be a CH3.
Now, if you insist on a real structure….
C6H5, C≡N, C=O and CH3 adds up to mass of 146, but C9H8NO cannot be a molecular formula!
What about C6H5, C≡C, C=O and CH3? C10H8O (M = 144)? Possibly an error in measuring height of tiny
(2.2%) M+1 peak in MS?
…, so…?
… you got lots of credit for clues