XIII(XYZ) ORGANIC CHEMISTRY REVIEWTEST-6
PAPER-1
PART-A
Select the correct alternative(s). (One or More than one is/are correct) [15 × 5 = 75]
There is NEGATIVE marking. 1 mark will be deducted for eachwrong answer.
Q.31 (A)


 
NaOH
Reactant (A) is
(A) 3
5
2
3 CH
C
)
CH
(
C
CH
|
|
|
|
O
O



 (B*) H
C
)
CH
(
C
CH
|
|
|
|
O
O
4
2
3 



(C) H
C
)
CH
(
C
H
|
|
|
|
O
O
5
2 


 (D) OH
CH
)
CH
(
C
CH
|
|
O
2
4
2
3 



[Sol.

 
 H
O
  
 

O
H2
O
H2



 ]
Q.33 3
2
3 CH
CH
C
CH
|
|
O


 

 
 2
2N
CH
A (Major) 


 

H
CO
CH 3
3
B (Major)
Product B is:
(A) 3
2
2
3 CH
CH
CH
C
O
CH
|
|
O




 (B*) 3
2
2
3 CH
CH
CH
O
C
CH
|
|
O





(C) 3
2
2
3 CH
CH
O
C
CH
CH
|
|
O




 (D) 3
2
3 CH
CH
O
C
CH
|
|
O




[Sol.  
)
A
(
N
)
Major
(
CH
CH
CH
C
CH
|
|
O
2
3
2
2
3




H
O
O
C
CH
|
|
O
3 


 l




 H
O
O
C
CH
|
|
O
3


+ H  
)
B
(
OH
C
CH
pr
O
C
CH
|
|
|
|
O
O
3
3 




  pr
O
C
CH
|
|
H
O
3 




+

O
C
CH
|
|
O
3 
 ]
Q.35


 
 5
PCl
(A)
Identifyproduct (A). Name of the reaction.
(A) C
NH
|
|
O
 Cl, Beckmann rearrangement
(B) NH
C
|
|
O
 Cl,Hoffmannbromamidereaction
(C) C
NH
|
|
O
 Cl, Hoffmannbromidereaction
(D*) NH
C
|
|
O
 Cl, Beckmann rearrangement
[Sol. 
 
 5
PCl 
O
|
|
PhCl
NH
C
Ph 

 l
Beckmann'sRearrangement ]

Q.37 Compound (X) C4
H8
O, which gives 2,4-Dinitrophenyl hydrazine derivative (orange orred or yellow
colour) andnegativehaloform test.
(A) 3
2
3 CH
CH
C
CH
|
|
O


 (B*)
3
3
CH
|
CHO
CH
CH 

(C) (D*) CH3
– CH2
– CH2
– CHO
[Sol. CompoundB&Dhave
O
C
|
H


groupso,givetestwith2,4-Dinitrophenylhydrazinebutdonotcontain

 C
CH
|
|
O
3 & 
 CH
CH
|
OH
3 group so give –ive iodoform test. ]
Q.39 Whichoffollowingreaction productisDiastereomer.
(A*)
4
2
CCl
Br


 (B*)



 

H
)
ii
(
NaCN
)
i
(
(C)
4
CCl
HBr

 
 (D*)
Et
|
Ph
CH
CH
CH
CH3 



peroxide
HCl

 

[Sol. (A) + Br2 
 
 4
CCl +
(B)
O
H
NaCN
2

 
 +
(D) HCl does not show peroxide effect
Et
|
HCl
Ph
CH
CH
CH
CH3 



 
H
Et
Cl
Et
|
|
|
|
Ph
C
CH
C
CH
Ph
C
CH
C
CH
|
|
|
|
Cl
H
H
H
2
3
2
3 







 ]

Q.41 Whichofthefollowingreactionis notrepresentingmajorproduct.
(A*)




H
CH3 3
CH
NH
C
|
|
O


(B*) 
 
 Li
Ph C
14
C CH3
(C) 2
NH
C
Ph
|
|
O


KOH
Br2


 Ph–NH2
(D*)
4
2
3
SO
H
HN

 

[Sol. (B) 
 
PhLi  H3
C C
14
C
Major
(D)
4
2
3
SO
H
H
N

 
 ]

Q.43 Inthegivenreaction:
[X]
[X]is:
(A) 2 moles (B*)
(C) one mole (D)All ofthese
[Sol. BuLi
)
ii
(
P
Ph
)
i
( 3


 

+ 2Ph3
P = O ]
PAPER-2
PART-B
MATCH THE COLUMN [3 × 10 = 30]
There is NEGATIVE marking. 0.5 Marks will be deducted for each wrongmatch.
INSTRUCTIONS:
Column-Iand column-IIcontainsfour entries each. Entries ofcolumn-Iareto bematchedwith some
entriesofcolumn-II.Oneormorethanoneentriesofcolumn-Imayhavethematchingwiththesameentries
ofcolumn-IIandoneentryofcolumn-Imayhaveoneormorethanonematchingwithentriesofcolumn-II.
Q.9 ColumnI ColumnII
(Reaction) (Type of intermediate formed)
(A) Ph–CHCl2 

 


HO
(A) (P) Carbocation
(B) R–Br ether
dry
Na

 (Q) Carbanion
(C) 3
3 CH
C
CH
|
|
O






 

,
SO
H
)
ii
(
O
H
/
Mg
)
i
(
4
2
2 (R) Free – radical
(D)




H
(S) Carbene
[Ans. (A) Q,S, (B) Q, R, (C) P,R (D) P]

[Sol. (A) Ph–CHCl2

 
 H
O


(B) Carbanion & freeradical
(C) Carbocation & freeradical (Pinacol formation)
(D) , Carbocation ]
PART-C
SUBJECTIVE:
ThereisNO NEGATIVE marking.
Q.19





 

O
H
)
ii
(
MgBr
CH
x
)
i
(
3
3
(reaction-1)





 

O
H
)
ii
(
MgBr
CH
y
)
i
(
3
3 (reaction-2)
Et
O
C
O
Et
|
|
O








 

H
)
ii
(
PhMgBr
z
)
i
(
Ph
|
Ph
C
Ph
|
OH


(reaction-3)
What is the value of x + y+ z = ? [6]
[Ans. 12]
[Sol. In reaction Ist 6 mole are used
2 mole are used for acid-base reaction
so, x = 6
In 2nd reaction 3 moles are used
y =3
In 3rd reaction 3 moles are used
z = 3
x + y + z = 6 + 3+ 3
= 12 Ans. ]