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CBSE Questions for Class 12 Medical Chemistry Alcohols, Phenols And Ethers Quiz 11 - MCQExams.com
CBSE
Class 12 Medical Chemistry
Alcohols, Phenols And Ethers
Quiz 11
Which of the following statements is/are true about phenol?
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0%
It is a stronger acid than water
0%
It forms salt with
N
a
H
C
O
3
0%
Passing
C
O
2
through an aqueous solution of sodium phenoxide gives phenol
0%
On exposure to air, it turns pink in colour
Explanation
Phenol is a weak acid. However, its acidity is higher than that of water.
It doesn't form a salt with
N
a
H
C
O
3
.
When carbon dioxide gas is dissolved in water, it gives carbonic acid which reacts with sodium phenoxide to form phenol.
C
O
2
+
H
2
O
→
H
2
C
O
3
H
2
C
O
3
+
C
6
H
5
O
N
a
→
C
6
H
5
O
H
+
N
a
H
C
O
3
On exposure to air, phenol turns pink in colour due to slow oxidation.
Hence, options
A
,
C
and
D
are correct.
Select schemes A, B, C out of the given below.
I. Acid catalysed hydration
II. Hydroboration oxidation
III. Oxymecruation-demercuration
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0%
I in all cases
0%
I, II, III
0%
II, III, I
0%
III, I, II
Primary alcohols can be prepared from alkenes by:
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mercuration and demercuration of alkenes
0%
direct hydration of alkenes
0%
hydroboration of alkenes
0%
all of the above
Explanation
Primary alcohols can be prepared from alkene by hydroboration oxidation.
The hydroboration of propene followed by oxidation gives propanol.
The reagent used for hydroboration is diborane.
The reagent used for oxidation is alkaline hydrogen peroxide solution.
6
C
H
3
C
H
=
C
H
2
+
B
2
H
6
→
2
(
C
H
3
C
H
2
C
H
2
)
3
B
(
C
H
3
C
H
2
C
H
2
)
3
B
+
3
H
2
O
2
O
H
−
→
3
C
H
3
C
H
2
C
H
2
O
H
+
B
(
O
H
)
3
What is
A
?
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0%
T
s
O
−
C
H
2
−
C
≡
C
H
0%
T
s
O
2
−
C
H
2
−
C
≡
C
H
0%
T
s
O
−
C
H
2
−
C
−
C
H
3
0%
none of the above
Explanation
Catalytic p-toluenesulfonic acid
(
T
s
O
H
)
reacts as acid and gives elimination products with the formation of
T
s
O
−
C
H
2
−
C
≡
C
H
.
How many pi(
π
) electron are there in the following molecule?
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0%
4
0%
6
0%
8
0%
10
Reactions I and II are limited mainly to naphthalene compounds. These reactions are called, respectively:
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bucherer reaction and its reversal
0%
bischler-Napieralski reaction and its reversal
0%
birnbaum-Simonini reaction and its reversal
0%
borodine-Hunsdiecker reaction and its reversal
Explanation
The Bucherer reaction is the reversible conversion of a naphthylamine to a naphthol in the presence of an aqueous sulfite or bisulfite.
The reaction is given as shown above.
Which of the following statements is/are correct?
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0%
The compounds (B) is (I).
0%
The compounds (B) is (II)
0%
The compounds (C) is (I)
0%
The compounds (C) is (II)
Explanation
Resonance structure of 2-naphthoxide ion is an ambident nucleophile and a better nucleophile than given structure of A because negative charge is on more electronegative oxygen atom. In the presence of polar non-protic solvent (DMF, dimethyl formamide), O-alkylation is favoured to give product II (C). In the presence of protic solvent, hydrogen bonding with the O atom occurs, thus reducing the nucleophilicity of oxygen. Therefore, C-alkylation is favoured to give product I (B).
The compound (A) is:
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0%
0%
0%
0%
Explanation
Red
P
and
H
I
is a traditional reducing agent to convert alcohols into alkanes.
H
3
P
O
3
form in this reduction reaction. The product
A
of reaction is cyclobutandicarboxylic acid.
Which statement is correct for the conversion of
R
O
H
to
R
X
by reagents
(
A
)
(
S
O
C
l
2
,
P
B
r
3
P
C
l
3
P
l
3
,
T
s
C
l
)
compared to using
H
X
?
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0%
No rearrangement occurs with predictable stereochemistry by reagents
(
A
)
0%
Reasgents
(
A
)
are not useful for
3
∘
alcohols, while
H
X
reacts easily with
3
∘
alcohols involving no rearrangement.
0%
Reagents
(
A
)
must be used in anhydrous conditions because all react vigorously with
H
2
O
and they produce harmful gases
(
S
O
2
,
H
C
l
,
H
B
r
,
a
n
d
H
I
)
0%
All of the above
Explanation
All the three statements (A), (B) and (C) are correct. Hence, (D) is the correct option.
(A) No rearrangement occurs with predictable stereochemistry by reagents
(
A
)
. However if
H
X
is used, the rearrangement is possible.
(B) Reagents
(
A
)
are not useful for tertiary alcohols, while
H
X
reacts easily with tertiary alcohols involving no rearrangement.
(C) Reagents
(
A
)
must be used in anhydrous conditions because all react vigorously with
H
2
O
and they produce harmful gases (
S
O
2
,
H
C
l
,
H
B
r
and
H
I
).
For example, phosphorous trichloride reacts with water to form orthophosphorous acid and
H
C
l
.
P
C
l
3
+
3
H
2
O
→
P
(
O
H
)
3
+
3
H
C
l
In given reaction A is :
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0%
0%
0%
0%
Which is the best method for the conversion of (A)
p
e
n
t
a
n
−
3
−
o
l
t
o
3
−
b
r
o
m
o
p
e
n
t
a
n
e
(B)?
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0%
0%
0%
(
A
)
H
B
r
→
(
B
)
0%
None of these
Consider the following reaction sequence.
C
H
2
=
C
H
2
c
o
l
d
→
d
i
l
.
K
M
n
O
4
A
e
x
c
e
s
s
→
P
C
l
5
B
The products
(
A
)
and
(
B
)
are, respectively:
Report Question
0%
C
H
3
C
H
2
O
H
and
C
H
3
C
H
2
C
l
0%
C
H
3
C
H
O
and
C
H
3
C
H
C
l
2
0%
C
H
2
O
H
C
H
2
O
H
and
C
H
2
C
l
C
H
2
C
l
0%
C
H
2
O
H
C
H
2
O
H
and
C
H
2
O
H
C
H
2
C
l
Explanation
The products (A) and (B) are
C
H
2
O
H
C
H
2
O
H
and
C
H
2
C
l
C
H
2
C
l
respectively.
C
H
2
=
C
H
2
c
o
l
d
→
d
i
l
.
K
M
n
O
4
C
H
2
O
H
C
H
2
O
H
A
e
x
c
e
s
s
→
P
C
l
5
C
H
2
C
l
C
H
2
C
l
B
When ethylene is treated with cold dilute potassium permanganate solution, two
−
O
H
groups are added to
C
atoms joined by double bond to form a vicinal diol.
Thus, it is an example of dihydroxylation.
In the next step, on treatment with phosphorous pentachloride, the
−
O
H
groups are replaced with
C
l
atoms to form vicinal dichloride.
Option C is correct.
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0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect
Explanation
When phenol is treated with chlorine in presence of iron, chlorination takes place at ortho and/or para positions. In aqueous medium, tri-substituted product is obtained and in organic medium, mono-substituted product is obtained.
Hence, both Assertion and Reason are incorrect.
The product (B) is :
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0%
0%
0%
0%
All
Explanation
The cleavage of (C- OH) bond is feasible rather than cleavage of (Ar-OH) bond because of resonance stabilistion in ArOH.
For the reaction given, what is product
(
A
)
?
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0%
0%
0%
0%
Explanation
Product (A) is benzene. During reaction, 3 molecules of water are lost. The product is obtained by rearrangement of carbon skeleton.
Which statement is correct for the conversion of ROH to RX by reagents
(
S
O
C
l
2
,
P
B
r
3
,
P
C
l
3
,
P
I
3
)
and TsCl compared to using HX?
Report Question
0%
No rearrangement occurs with predictable stereo-chemistry by reagents
0%
Reagents are not useful for
3
∘
alcohols, while HX reacts easily with
3
∘
alcohols involving no rearrangement
0%
Reagents must be used in anhydrous conditions because all react vigrously with
H
2
O
and they produce harmful gases (
S
O
2
, HCl, HBr, and HI)
0%
All
Explanation
All the statements given in options A, B and C are correct for the conversion. Hence option D is correct.
When benzene sulphonic acid and p-nitrophenol are treated with
N
a
H
C
O
3
, the gases released respectively :
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0%
S
O
2
,
N
O
2
0%
S
O
2
,
N
O
0%
S
O
2
,
C
O
2
0%
C
O
2
,
C
O
2
Explanation
P
h
S
O
3
H
+
N
a
H
C
O
3
→
P
h
S
O
3
N
a
+
C
O
2
+
H
2
O
.
p
−
O
2
N
−
C
6
H
4
−
O
H
+
N
a
H
C
O
3
→
p
−
N
O
2
−
C
6
H
4
−
O
N
a
+
C
O
2
+
H
2
O
Option D is correct.
The product (B) is:
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0%
0%
0%
0%
all
Consider the reaction:
C
3
H
7
−
O
H
+
E
t
O
⨁
B
F
4
⊝
.
Which of the following statements is wrong?
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0%
The nucleophile in the reaction is
\displaystyle C_{3}H_{7}O^-
0%
The nucleophile in the reaction is
\displaystyle BF_{4}^{{\circleddash }}
0%
The leaving group is
\displaystyle Et_{2}O
0%
\displaystyle SN^{2}
reaction occurs
Phenol is an organic compound even though it is soluble in water due to the presence of:
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0%
ionic bonding
0%
covalent bonding
0%
hydrogen bonding
0%
coordinate bonding
Explanation
Phenol is an organic compound even though it is soluble in water due to the presence of hydrogen bonding. Phenol has hydroxyl group which is involved in the formation of hydrogen bonds with water. It increases the solubility. Hydrogen bonding is possible when
H
atom is attached to electronegative
N,\ O
or
F
atom.
Correct statement(s) in case of n-butanol and t-butanol is (are):
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0%
both are having equal solubility in water.
0%
t-butanol is more soluble in water than n-butanol.
0%
boiling point of t-butanol is lower than n-butanol.
0%
boiling point of n-butanol is lower than t-butanol.
Explanation
More branching means less boiling point and high solubility.
Give the decreasing order of reactivity of the following compounds with
HBr
.
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0%
III > IV > II > I
0%
III > II > IV > I
0%
III> II > I > IV
0%
II > III > IV > I
Explanation
The decreasing order of reactivity of the following compounds with HBr is
\displaystyle III> II > I > IV
.
The
\displaystyle -OH
group of alcohol is protonated with
\displaystyle HBr
followed by loss of water molecule to form carbocation.
\displaystyle Ar-CH_2-OH + H^+ \rightarrow Ar-CH_2-OH_2^+
\displaystyle Ar-CH_2-OH_2^+ \rightarrow Ar-CH_2^+ + H_2O
Electron donating substituents (such as methyl and methoxy group in para position) on benzene ring stabilizes the carbocation and decreases the reactivity.
Electron withdrawing substituent (such as methoxy group in meta position) on benzene ring destabilizes the carbocation and decreases the reactivity.
A compound (X) reacts with thionyl chloride to give a compound (Y). (Y) reacts with
\displaystyle Mg
to form a Grignard reagent, which is treated with acetone and the product is hydrolysed to give 2-methyl-2- butanol. What are structural formulae of (X) and (Y) ?
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0%
X = Acetic acid Y = Ethyl Chloride
0%
X = Acetaldehyde Y = Ethyl Chloride
0%
X = Ethyl Alcohol Y = Ethyl Chloride
0%
X = Propyl alcohol Y = Propyl Chloride
Explanation
The reaction can be written as, shown in the above image.
Hence,
X
is Ethyl alcohol.
and
Y
is Ethyl chloride.
Select the incorrect synthesis.
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0%
0%
0%
0%
Explanation
The synthesis of option D is incorrect as it is nucleophilic aromatic substitution reaction which requires drastic conditions. Under these conditions, the sodium alkoxide undergoes elimination reaction to form an alkene.
In which of the following solvents, KI has highest solubility? The dielectric constant
(\epsilon)
of each liquid is given in parentheses.
Report Question
0%
C_{6}H_{6}(\epsilon =0)
0%
(CH_{3})_{2}CO(\epsilon =21)
0%
CH_{3}OH(\epsilon =32)
0%
CCl_{4}(\epsilon =0)
Explanation
KI is an ionic compound. Hence, it will be most soluble in apolar solvent, which has a high value of dielectric constant. Hence,
CH_3OH
with the highest value of dielectric constant among the given options will provide high solubility.
For the reaction the major product is:
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0%
0%
0%
0%
Explanation
A ring expansion creates the six membered carbocation intermediate.
The boiling point of phenols are higher than the hydrocarbons of comparable masses due to:
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0%
more polarising power
0%
presence of hydrogen bonding
0%
resonance stabilisation
0%
acidic character
Explanation
The boiling point of compound is effected by forces of attraction existing between molecules. Hydrogen bonding is a stronger force of attraction existing between molecules of phenols as compared to vander waals forces of attraction existing between alkane molecules. (they are weaker forces)
\therefore
Boiling point of phenol is more than hydrocarbon due to presence of hydrogen bonding.
PhCH=C{ H }_{ 2 }\overset { ArC{ O }_{ 3 }H }{ \underset { C{ H }_{ 2 }{ Cl }_{ 2 } }{ \longrightarrow } } A\overset { 1.LiAl{ H }_{ 4 }\cdot { Et }_{ 2 }O }{ \underset { 2.{ H }_{ 2 }O }{ \longrightarrow } } B
The end product (B) is
:
Report Question
0%
0%
PhCH(OH)C{ H }_{ 3 }
0%
PhC{ H }_{ 2 }C{ H }_{ 2 }COAr
0%
PhC{ H }_{ 2 }C{ H }_{ 2 }OH
Explanation
In the first step, the O atom is added across C=C bond to form epoxide.
In the second step, the epoxide ring is reduced to form secondary alcohol.
\displaystyle PhCH=C{ H }_{ 2 }\overset { ArC{ O }_{ 3 }H }{ \underset { C{ H }_{ 2 }{ Cl }_{ 2 } }{ \longrightarrow } } \underset {\text {A, 2-phenyloxirane}}{C_8H_8O} \overset { 1.LiAl{ H }_{ 4 }\cdot { Et }_{ 2 }O }{ \underset { 2.{ H }_{ 2 }O }{ \longrightarrow } } \underset {\text {B}}{PhCH(OH)C{ H }_{ 3 }}
Option B is correct.
Propene is allowed to react with m-chloroperoxybenzoic acid. The product (A) is then reduced with
\displaystyle { LiAlH }_{ 4 }
in dry ether to give (B).
\displaystyle { CH }_{ 3 }CH={ CH }_{ 2 }\xrightarrow { m-CPBA }\ A\xrightarrow [ 2.{ H }_{ 3 }{ O }^{ + } ]{ 1.{ LiAlH }_{ 4 } } B
The structure of the product (B) is:
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0%
\displaystyle { CH }_{ 3 }{ CH(OH)CH }_{ 2 }OH
0%
\displaystyle { CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }OH
0%
\displaystyle { CH }_{ 3 }{ CH }(OH)CH_{ 3 }
0%
Explanation
The product (A) is 2-methyloxirane and the product (B) is propan-2-ol. Alkenes reacts with peracids to form oxiranes. The oxirane on reduction forms alcohols.
Option C is correct.
Tertiary alcohols are more acidic than phenol.
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0%
True
0%
False
Complete the blanks by identifying
X
and
Y
.
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0%
Ethanal, ethene
0%
Ethanol, ethene
0%
Ethanal, ethyne
0%
Ethanol, ethyne
Explanation
Since,
Y
undergo hydrogenation and give alkane with two carbon.
Hence,
Y
must contain double bond and it is Ethene.
Since,
KMnO_4
is an oxidizing agent, it oxidizes alcohol to aldehyde and further oxidises into acetic acid.
Option B is correct.
Identify the correct sequence of given steps for the conversion of calcium carbide to methyl alcohol.
(a) Reaction with aqueous
KOH
(b) Hydrolysis
(c) Reaction with soda lime
(d) Reaction with
Ha{SO}_{4}/{H}_{2}{SO}_{4}
(e) Reaction with
Hg{SO}_{4}
(f) Reaction with
P{Cl}_{5}
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0%
b\ d\ e\ c\ f\ a
0%
b\ d\ c\ a\ f\ e
0%
b\ f\ d\ e\ a\ c
0%
d\ b\ c\ e\ f \ a
Explanation
Steps for the conversion of calcium carbide to methyl alcohol:
(b) Hydrolysis
(d) Reaction with
Ha{SO}_{4}/{H}_{2}{SO}_{4}
(e) Reaction with
Hg{SO}_{4}
(c) Reaction with soda lime
(f) Reaction with
P{Cl}_{5}
(a) Reaction with aqueous
KOH
CaC_2 + 2 H_2O\rightarrow C_2H_2 + Ca(OH)_2 \Rightarrow C_2H_2 + H_2SO_4/ HgSO_4 \rightarrow CH_3CHO \Rightarrow CH_3CHO + H_2SO_4\rightarrow CH_3COOH \Rightarrow CH_3COOH + NaOH + CaO \rightarrow CH_3COONa \Rightarrow CH_3COONa + PCl_5 \rightarrow CH_3Cl\Rightarrow CH_3Cl + KOH \rightarrow CH_3OH
Ethanol is produced by/from :
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0%
ethene
0%
fermentation of sugars
0%
both A and B
0%
none of the above
Explanation
Formation of Ethanol:
\mathrm{\underset{Ethene}{H_2C=CH_2} \xrightarrow[H_2O]{H_2SO_4} \underset{Ethanol}{H_3C-CH_2OH}}
\mathrm{\underset{Glucose(Sugar)}{C_6H_12O_6} \xrightarrow{Zymase \ enzyme} \underset{Ethanol}{C_2H_5OH}}
So, Ethanol can be formed from Ethene and Fermentation of sugar.
Hence, Option "C" is the correct answer.
What will be product in given reaction?
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0%
0%
0%
0%
Which has the smallest size ?
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0%
Na^+
0%
Mg^{2+}
0%
Al^{3+}
0%
P^{5+}
R-OH\xrightarrow{PBr_3}A\xrightarrow[]{Mg/Dry \ ether}B\xrightarrow[]{HCHO}C\xrightarrow[]{H^+/H_2O}D
What are
A,B,C
&
D
respectively?
Report Question
0%
R-Br,R-Br-Mg,R-CH_2Mg-Br,R-CH_2CHO
0%
R-Br,R-Mg-Br,R-CH_2-O-Mg-Br,R-CH_2CHO
0%
R-Br,R-Mg-Br,R-CH_2-O-Mg-Br,R-CH_2OH
0%
R-Br,R-Mg-Br,R-CH_2-Br,R-CH_2OH
Explanation
R-OH\overset{PBr_3}{\rightarrow}R-Br\xrightarrow[]{Mg/Dry ether}R-Mg-Br\xrightarrow[]{HCHO}R-CH_2-O-Mg-Br\xrightarrow[]{H^+/H_2O}RCH_2OH
So the correct order is
R-Br,R-Mg-Br,R-CH_2-O-Mg-Br,R-CH_2OH
Hence option C is correct.
Identify E:
Report Question
0%
ethene
0%
ethane
0%
ethyne
0%
propyne
Which one of the following is produced when acetone is saturated with
HCl
gas?
Report Question
0%
Acetone alcohol
0%
Phorone
0%
Mesityl oxide
0%
Benzene
CH_3CH_2-OH\xrightarrow{A}CH_3-CH_2-Cl\xrightarrow{B}CH_2=CH_2.
A and B in this sequence of reactions are
:
Report Question
0%
KOH(aq) and
PCl_5
0%
PCl_5
and KOH(aq)
0%
Cl_2
and KOH(alc)
0%
PCl_5
and KOH(alc)
Explanation
A and B in this sequence of reactions are
\displaystyle PCl_5
and KOH(alc).
\displaystyle CH_3CH_2OH\xrightarrow{PCl_5(A)}CH_3CH_2Cl\xrightarrow{alc.KOH(B)}CH_2=CH_2
Ethanol reacts with
PCl_5
to give ethyl chloride. Ethyl chloride on heating with alcoholic potash undergoes dehydrohalogenation to give ethene.
Hence the correct option is D.
R-OH\xrightarrow {P+I_2}A\xrightarrow[alc. \ KCN]{Heating}B\xrightarrow[Na/alcohol]{LiAlH_4}C\xrightarrow[NaNO_2+dil\ HCl]{HNO_2}D
What are
A,B,C
&
D
respectively?
Report Question
0%
R-I
,
R-CN
,
R-NH_2
,
R-CH_2OH
0%
R-I
,
R-CN
,
R-CH_2-NH_2
,
R-CH_2OH
0%
R-I
,
R-NC
,
R-CH_2-NH_2
,
R-CH_2OH
0%
R-I
,
R-CN
,
R-CH_2-NO_2
,
R-CH_2OH
Explanation
R-OH\xrightarrow{P+I_2}R-I\xrightarrow[alc KCN]{Heating}R-CN\xrightarrow[Na/alchohol]{LiAlH_4}R-CH_2-NH_2\xrightarrow[NaNO_2+dil\ HCl]{HNO_2}R-CH_2OH
Hence option B is correct.
R-OH\xrightarrow {PBr_3}A\xrightarrow[]{Mg/Dry \ ether}B\xrightarrow[CH_2-CH_2-O]{Grignard\ reaction}C\xrightarrow[]{H^+/H_2O}D
What are
A,B,C
and
D
are respectively?
Report Question
0%
R-Br,R-Mg-Br,R-CH_2-O-Mg-Br,R-CH_2CH_2-OH
0%
R-Br,R-Mg-Br,R-CH_2-O-CH_2-Mg-Br,R-CH_2CH_2-OH
0%
R-Br,R-Mg-Br,R-CH_2-CH_2-O-Mg-Br,R-CH_2CH_2-OH
0%
R-Br,R-Mg-Br,R-CH_2-CH_2-O-Mg-Br,R-CH_2OH
Explanation
R-OH\overset{PBr_3}{\rightarrow}R-Br\xrightarrow[]{Mg/Dry\ ether}R-Mg-Br\xrightarrow[CH_2-CH_2-O]{Grignard\ reaction}R-CH_2-CH_2-O-Mg-Br\\ \quad \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \xrightarrow[]{H^+/H_2O}RCH_2-CH_2-OH
PBr_3
converts primary or secondary alcohol to alkyl halide. Mg and dryether forms Grignard reagent R-Mg-X , which on reaction with epoxide like ethylene oxide (
CH_2CH_2O
) forms
R-CH_2CH_2-O-Mg-Br
and then in presence of acid gives alcohol.
So the order is
R-Br, R-Mg-Br, \ R-CH_2-CH_2-O-Mg-Br,\ R-CH_2CH_2-OH
.
Hence option C is correct,
CH_3-OH\xrightarrow{P+I_2}A\xrightarrow[alc\ KCN]{Heating}B\xrightarrow[Na/alcohol]{LiAlH_4}C\xrightarrow[NaNO_2\ + \ dil \ HCl] {HNO_2}D
What is
D
?
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CH_3CHO
0%
CH_3COOH
0%
CH_3CH_2OH
0%
CH_3-O-CH_3
Explanation
CH_3-OH\xrightarrow{P+I_2}CH_3-I\xrightarrow[alc \ KCN]{Heating}CH_3-CN\xrightarrow[Na/ethyl \ alchohol]{LiAlH_4}CH_3-CH_2NH_2\xrightarrow[NaNO_2+dil \ HCl]{HNO_2}CH_3-CH_2OH
Hence option C is correct.
Product formed in the following reaction is/are :
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0%
0%
0%
both (A) and (B)
0%
none is correct
Explanation
Y
is less soluble than
\left(X\right)
due to lack of symmetry of chiral carbon. This is reduced to
-C{H}_{2}OH
.
Under suitable condition temperature,pressure and catalyst,methanol can be obtain by:
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0%
partial oxidation of hydrocarbons
0%
reacting
H_2
with
CO
0%
reacting
H_2
with
CO_2
0%
All the above
Explanation
Partial oxidation of hydrocarbons produces methanol.
CO + 2H_2O \longrightarrow CH_3OH
CO_2 + H_2O \longrightarrow CH_3OH + H_2O
All the three processes produces mathanol.
In the preparation of alkene from alcohol using
Al_{2}O_{3}
which is effective factor?
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Porosity of
Al_{2}O_{3}
0%
Temperature
0%
Concentration
0%
Surface area of
Al_{2}O_{3}
Explanation
Temperature is the effective factor for dehydration of alcohols by
{ Al }_{ 2 }{ O }_{ 3 }
.
R-C{ H }_{ 2 }-C{ H }_{ 2 }-OH\xrightarrow [ { 350 }^{ o }-{ 380 }^{ o }C ]{ { Al }_{ 2 }{ O }_{ 3 } }
R-CH=C{ H }_{ 2 }+{ H }_{ 2 }O
At
{ 220 }^{ o }-{ 250 }^{ o }
it forms ether.
Choose the correct statement.
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0%
A
is more volatile than
B
0%
B
is more volatile than
A
0%
A
is formed more rapidly at a higher temperature
0%
A
is formed in higher yield at a low temperature
Explanation
Boiling point of of
A
is lower than that of
B
. In
A
, there is intramolecular
H-bonding
and in
B
, there is intermolecular
H-bonding
). Thus,
A
is more volatile than
B
.
In the given reaction, number of possible structure is:
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0%
1
0%
2
0%
5
0%
6
Explanation
Therefore, in
X
order of stability of alkene is as follows:
II> I> III
All of these alkenes, with
{Br}_{2}/C{Cl}_{4}
, produce additive product having molecular formula
{C}_{4}{H}_{6}{Br}_{2}
. The possible products are
(i)
{ CH }_{ 3 }-\underset { |\\ Br }{ CH } -\underset { |\\ Br }{ CH } -{ CH }_{ 3 }
(ii)
{ CH }_{ 3 }-\underset { |\\ Br }{CH}-\underset { |\\ Br }{ CH } -{ CH }_{ 3 }
(iii)
Br{ H }_{ 2 }C-{ CH }_{ 2 }-CHBr-{ CH }_{ 3 }\quad
(iv)
{ CH }_{ 2 }Br-{ CH }_{ 2 }-{ CH }_{ 2 }-{ CH }_{ 2 }Br
(v)
{ CH }_{ 3 }-{ CH }_{ 2 }-{ CH }Br-{ CH }_{ 2 }Br
The major product of the reaction is:
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0%
I
0%
II
0%
III
0%
IV
The major product of the following reaction is:
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0%
0%
0%
0%
Explanation
The major product of the following reaction is represented by the option (C).
\displaystyle Phenolic
\displaystyle -OH
group is acidic in nature and on deprotonation gives
\displaystyle phenoxide
ion which then attacks
\displaystyle methyl \: iodide
. Thus net reaction is methylation of phenolic
\displaystyle -OH
group.
Complete the equation:-
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0%
0%
0%
0%
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Practice Class 12 Medical Chemistry Quiz Questions and Answers
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