Chemistry_Today_April_2017_vk_com_stopthepress

(Z) ⇒
+
OH
(W) has chiral carbon;
Two optically active isomers are possible.
(X) doesn't have any chiral carbon.
(Y) has chiral carbon;
Br
+ C
– ..
O
NGP
O
HO O
NaHCO 3
Br 2
O
Br
+
HO
O
Br
Two optically active isomers are possible.
(Z) doesn't have any chiral carbon.
2. (b) : Idea is, after dehydrohalogenation who gives
the most stable product.
Cl
after elimination
(I) (Highly strained ;
not possible)
(II)
Cl
H
after elimination
(Lovely ; it is aromatic
and highly stable)
O
C O
∴ Correct option is (d).
4. (c) : H +
Attacks this double
bond as it will give
more stable carbocation
+
Br
∴ Correct option is (c).
5. (a) : H 2
+ Pd will reduce double bond as well as
–NO 2
group.
∴ Correct option is (a).
6. (d) :
Br –..
OH oxymercuration
+ Hg ( OAc)
2 –
–OAc
+
(III)
(IV)
Cl
Cl
after elimination
tough to remove H
from sp
2 hybridised Impossible !!
carbon
after elimination
tough as –Cl is with sp
2
hybridised carbon
So, correct option is (b), (II).
Terrible !!
3. (d) : There will be the formation of a six membered
ring through Diels–Alder reaction.
CHEMISTRY TODAY | APRIL ‘17 67