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Byju's Answer
Standard XII
Chemistry
Bromine Water Test
CH3-CH2-Cl|CH...
Question
C
H
3
−
C
H
2
−
C
|
H
C
l
−
C
H
3
A
Meso form
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B
Racemic mixture
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C
d-form
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D
l-form
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Solution
The correct option is
B
Racemic mixture
2-chlorobutane is optically active racemic mixture containing (R) and (S)- enantiomers, as the chlorine substituted carbon is the Chiral center.
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Similar questions
Q.
Pick out the correct reactions:
A
.
C
H
3
−
C
H
=
C
H
2
+
H
C
l
⟶
C
H
3
−
C
|
C
l
H
−
C
H
3
B
.
C
H
3
−
C
H
=
C
H
2
+
H
B
r
⟶
C
H
3
−
C
H
2
−
C
H
2
B
r
C
.
C
H
3
−
C
H
=
C
H
2
+
H
B
r
p
e
r
o
x
i
d
e
−
−−−−
→
C
H
3
−
C
H
2
−
C
H
2
−
B
r
D
.
C
H
3
−
C
H
=
C
H
2
+
H
I
p
e
r
o
x
i
d
e
−
−−−−
→
C
H
3
−
C
|
I
H
−
C
H
3
Q.
Out of
C
H
3
−
C
H
−
C
H
2
−
C
l
|
C
H
3
and
C
H
3
−
C
H
2
−
C
H
−
C
l
|
C
H
3
,
which is more reactive towards
S
N
1
reaction and why?
Q.
Select the member of each pair that shows faster rate of
S
N
2
reaction with
K
I
in acetone.
(a)
C
H
3
−
C
H
2
−
C
H
2
−
C
H
2
−
C
l
a
n
d
C
H
3
−
C
H
∣
C
H
3
−
C
H
2
−
C
l
(
I
)
(
I
I
)
(b)
C
H
3
−
C
H
2
−
C
H
2
−
C
l
a
n
d
C
H
3
−
C
H
2
−
C
H
2
−
B
r
(
I
)
(
I
I
)
(c)
C
H
3
−
C
H
∣
C
H
3
−
C
H
2
−
C
H
2
−
C
l
a
n
d
C
H
3
−
C
H
3
∣
C
∣
C
H
3
−
C
H
2
C
l
(
I
)
(
I
I
)
(d)
C
H
3
−
C
H
2
−
C
H
2
−
B
r
∣
C
H
−
C
H
3
a
n
d
C
H
3
−
C
H
3
∣
C
H
−
C
H
2
−
B
r
∣
C
H
−
C
H
3
(
I
)
(
I
I
)
Q.
Which of the following reaction is example of
S
N
2
reaction using
c
o
n
c
.
H
C
l
:
(
a
)
C
H
3
−
C
H
2
−
O
H
A
n
h
y
Z
n
C
l
2
/
△
−
−−−−−−−−
→
C
H
3
−
C
H
2
−
C
l
(
b
)
C
H
3
−
C
H
3
|
C
|
C
H
3
−
C
H
2
O
H
A
n
h
y
Z
n
C
l
2
/
△
−
−−−−−−−−
→
C
H
3
−
C
H
1
|
C
|
C
H
3
−
C
H
2
−
C
H
3
(
c
)
C
H
3
−
O
H
|
C
|
C
H
3
−
C
H
3
A
n
h
y
Z
n
C
l
2
/
△
−
−−−−−−−−
→
C
H
3
−
C
l
|
C
|
C
H
3
−
C
H
3
(
d
)
C
H
2
=
C
H
−
C
H
2
−
O
H
A
n
h
y
Z
n
C
l
2
/
△
−
−−−−−−−−
→
C
H
2
=
C
H
−
C
H
2
−
C
l
Q.
Identify the type of the following reaction of carbon compounds.
a. CH
3
-CH
2
-CH
2
-OH → CH
3
-CH
2
-COOH
b. CH
3
-CH
2
-CH
3
→ 3CO
2
+ 4H
2
O
c. CH
3
-CH = CH -CH
3
+ Br
2
→ CH
3
-CHBr - CHBr -CH
3
d. CH
3
-CH
3
+ Cl
2
→ CH
3
-CH
2
-Cl + HCl
e. CH
3
-CH
2
-CH
2
-CH
2
-OH → CH
3
-CH
2
-CH=CH
2
+ H
2
O
f. CH
3
-CH
2
-COOH + NaOH → CH
3
-CH
2
-COO - Na
+
+ H
2
O
g. CH
3
-COOH + CH
3
-OH → CH
3
-COO- CH
3
+ H
2
O
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