An acid weaker than carbonic acid

An acid stronger than carbonic acid

MCQ Questions for Class 12 Medical Chemistry Alcohols, Phenols And Ethers Quiz 9

But-

$2$ -ene,

$1, 4$ -diol is converted in two steps through an intermediate X into oxobutanedioic acid.

$\underset{but-2-ene-1, 4-diol}{HOCH_2CH=CHCH_2OH}\overset{step 1}{\rightarrow} x\underset{hot, acidified KMnO_4}{\overset{step 2}{\rightarrow}}\underset{oxobutanedioic acid}{HO_2CCOCH_2CO_2H}$
What could be the reagent for step

$1$ and what is the intermediate X?

	reagent for step $1$	X
A	cold, acidified $KMnO_4$	$HOCH_2CH_2CH(OH)CH_3OH$
B	hot, acidified $K_2Cr_2O_7$	$HO_2CCH=CHCO_2H$
C	steam and concentrated $H_2SO_4$	$HOCH_2CH(OH)CH_2CH_2OH$
D	warm, acidified $K_2Cr_2O_7$	$OHCCH(OH)CH_2CHO$

0%
A
0%
B
0%
C
0%
D

Explanation

Alkenes react with concentrated sulphuric acid to produce alkyl hydrogen sulphates which on hydrolysis gives alcohol.

Thus but-

$2$ -ene-

$1, 4$ -diol will react with steam and concentrated

${H}_{2}S{O}_{4}$ and will produce butan-

$1, 2, 4$ -triol (X} which on further reaction with hot and acidified

$KMn{O}_{4}$ gives oxobutandioic acid.

Hence (C) is correct.

Menthol is a naturally occurring alcohol. When menthol is heated with concentrated sulfuric acid it reacts, and the products formed is:

What is the major product formed when compound

$Q$ is warmed with excess

$HBr$ ?

Explanation

The

$-OH$ group will react with

$HBr$ to undergo nucleophilic substitution reaction and will replace

$OH$ with

$Br$ and the double bond in the compound will also react with

$HBr$ and undergo addition rection.

Hence (C) will be the major product formed.

Which reagent is suitable for this conversion?

0%
$Zn-Hg/HCl$
0%
$LiAlH_4$
0%
$NH_2-NH_2/OH^-$
0%
Red $P+HI$

Explanation

Suitable Reagent for this conversion is

$LiAlH_4$ .

Ether which is liquid at room temperature is:

0%
$C_2H_5 - O - CH_3$
0%
$CH_3 - O - CH_3$
0%
$C_2H_5 - O - C_2H_5$
0%
None of these

Heating together sodium ethoxide and ethyl chloride will give:

0%
ether
0%
ethyl alcohol
0%
acetaldehyde
0%
acetic acid

The reaction,

$(CH_{3})_{3}C - OH \underset {Heat}{\xrightarrow {H_{2}SO_{4}}} (CH_{3})_{2}C = CH_{2}$ , is an example of

0%
Sulphonation
0%
Decomposition
0%
Dehydration
0%
Alkylation

An olefin may be converted into alcohol by:

0%
$B_{2}H_{6}$ and $H_{2}O_{2}$
0%
$LiAlH_{4}$
0%
$Acidic \ KMnO_{4}$
0%
$HOCl$

Which of the following will not give anisole?

0%
Phenol is reacted with dimethyl sulphate in presence of a base
0%
Sodium phenoxide is treated with methyl iodide
0%
Diazomethane is reacted with phenol
0%
Methyl magnesium iodide is treated with phenol

An equimolar quantities of ethanol and propanol are heated with conc.

$H_2SO_4$ . The products formed is/are:

0%
$C_2H_5OC_2H_5$
0%
$C_3H_7OC_3H_7$
0%
$C_2H_5OC_3H_7$
0%
all of these

The compound(s) which will give propylene on dehydration with

$conc.H_{2}SO_{4}$ at

$440\ K$ is/ are

0%
Isobytyl alcohol
0%
$n-propyl$ alcohol
0%
Isopropyl alcohol
0%
$n-butyl$ alcohol

Anisole on treatment with

$Br_2/CS_2$ gives:

0%
bromobenzene
0%
methyl-2-bromophenyl ether
0%
O- and p-bromoanisole
0%
phenol

State whether the following statement is True or False.
Ethylene can be used for the synthesis of ethyl alcohol.

0%
True
0%
False

Methyl phenyl ether can be obtained by reacting:

0%
phenolate ions and methyl iodide
0%
methoxide ion and bromobenzene
0%
methanol and phenol
0%
bromobenzene and methyl bromide

State, whether the following statements are True or False:
Phenol evolves carbon dioxide from sodium bicarbonate.

0%
True
0%
False

Explanation

Phenol does not decompose sodium carbonate or sodium bicarbonate, i. e.,

$CO_2$ is not evolved because phenol is weaker than carbonic acid.

$False$

State, whether the following statements are True or False:
Ethers are not affected by atmosphere.

0%
True
0%
False

State, whether the following statements are True or False:
Phenol forms

$2,4,6-Tribromophenol$ with bromine water but form only monobromophenol with

$Br_{2}/CS_{2}$ .

0%
True
0%
False

Explanation

Water is a polar substance and so in bromine water, the phenol gives a strongly stable phenoxide ion. This ion has negative charge on oxygen and that transfers electrons to the benzene ring and that becomes much activated that substitution reaction occurs at three positions and tri-substituted product is obtained.

But,

$CS_2$ is less polar solvent so the solvent effect is minimum and the formation of phenoxide ion is not much stable so the activation of ring also does not occur and thus only monosubstituted product is obtained.

img, on reaction with

$CH_3OH$ in presence of

$H^+$ gives :

$CO+H_2\overset{X(catalyst)}{\underset{300^oC/300 atm}{\rightarrow}}CH_3OH$ , the catalyst X is?

0%
$Fe$
0%
$Cr_2O_3/ZnO$
0%
$V_2O_5$
0%
$Al_2O_3$

Explanation

A mixture of zinc oxide and chromium oxide

$(ZnO−Cr_2O_3)$ converts water gas in methanol

$(CH_3OH)$ .

The temperature of the reaction is 573 K or 300

$^\circ C$ and 300 atm.

$CO+2H_2 \overset{ZnO−Cr_2O_3}{\longrightarrow}CH_3OH$

Phenol on standing in air develops a red colour due to formation of:

0%
Cyclohexane
0%
Resorcinol
0%
Phenoquinone
0%
Quinol

Wood spirit if known as?

0%
Ethanol
0%
Methanol
0%
Acetone
0%
Benzene

Phenol is

0%
A base weaker than ammonia
0%
An acid stronger than carbonic acid
0%
An acid weaker than carbonic acid
0%
A neutral compounds

Reaction,

$\underset{Water gas}{\underbrace{CO+H_2}}+H_2\overset{673K, 300 atm.}{\underset{Cr_2O_3-ZnO}{\rightarrow}}?$ may be used for manufacture of:

0%
$HCHO$
0%
$CH_3COOH$
0%
$HCOOH$
0%
$CH_3OH$

Explanation

Preparation of methanol includes catalytic reduction of carbon monoxide. For this purpose, water gas is used.

Water gas is an equimolar mixture of

$CO and H2.$

A mixture of zinc oxide and chromium oxide

$(ZnO−Cr_2O_3)$ converts water gas in methanol

$(CH_3OH)$ .

The temperature of the reaction is 673 K and pressure is 300 atm.

$CO+2H_2\overset{ZnO−Cr_2O_3}{\longrightarrow} CH_3OH$

The only alcohol that can be prepared by the indirect hydration of alkene is:

0%
Methyl alcohol
0%
Ethyl alcohol
0%
Propyl alcohol
0%
Isobutyl alcohol

Explanation

Except for ethyl alcohol, no other primary alcohol can be prepared by this method as the addition of

$H_2SO_4$ follows Markownikoff's rule. Generally, secondary and tertiary alcohols are obtained.

So option B is correct.

1666271_1724787_ans_5761aa1e4f8f40eca31ece3d1636f8c8.png

Acid catalysed hydration of alkenes except ethene leads to the formation of.

0%
Primary alcohol
0%
Secondary or tertiary alcohol
0%
Mixture of primary and secondary alcohols
0%
Mixture of secondary and tertiary alcohols

Explanation

Alkenes react with water in the presence of acid as a catalyst to form alcohol. In case of unsymmetrical alkenes, OH is added to the carbon having fewer hydrogen atoms according to Markovnikov's rule.

Hence, acid catalyzed hydration of alkenes except ethene leads to the formation of secondary or tertiary alcohol. To obtain primary alcohol, hydroboration oxidation is used.

Hence, option

$B$ is correct.

1626816_1724783_ans_d392aecba5334321a05c6a9d99862352.JPG

Among the following compounds which can be dehydrated very easily?

0%
$CH_3CH_2CH_2CH_2CH_2OH$
0%
0%
0%

Phenol on treatment with

$dil. HNO_{3}$ gives:

0%
Picric acid
0%
$o-$ and $p-nitrophenols$
0%
$o-$ and $m-nitrophenols$
0%
$p-$ and $m-nitrophenols$

When phenol reacts with benzene diazonium chloride, the product obtained is:

0%
Phenylhydrazine
0%
p-amino azobenzene
0%
Phenol hydroxylamine
0%
p-hydroxy azobenzene

During dehydration of alcohols to alkenes by heating with conc.

$H_2SO_4$ , the initiation step is:

0%
Elimination of water
0%
Formation of an ester
0%
Formation of carbocation
0%
Protonation of alcohol molecule

Explanation

Dehydration of alcohol to alkene in presence of concentrated

$H_2SO_4$ involves following steps :

Thus, the initiation step is protonation of alcohol.

1666276_1724764_ans_ce88c5cb3f9d4f9b9a72e3409348c0ba.png

Which of the following is soluble in water?

0%
$CS_2$
0%
$C_2H_5OH$
0%
$CCl_4$
0%
$CHCl_3$

Explanation

Ethanol

$(C_2H_5OH)$ forms hydrogen bonding with water.
Hence, it is soluble in water. Option (B) is correct.

The compound which gives the most stable carbonium ion on dehydration is?

0%
$(CH_3)_2CH-CH_2OH$
0%
$CH_3CH_2CH_2OH$
0%
$CH_3-CH(OH)-CH_2CH_3$
0%
$(CH_3)_3C-OH$

Explanation

Hint: The Stability order of Carbonium ion or Carbocation is $3^{\circ} > 2^{\circ} > 1^{\circ} > methyl$ .

Correct Answer: (D)

$(CH_3)_3C-OH$

Explanation of Correct Option:

Dehydration of primary, secondary, tertiary alcohol gives primary, secondary and tertiary carbocations respectively.
Dehydration of $(CH_3)_2CH-CH_2OH$ gives primary carbocation which is $(CH_3)_2CH-CH_2^+$

Dehydration of $CH_3CH_2CH_2OH$ gives primary carbocation which is $CH_3CH_2CH_2^+$

Dehydration of $CH_3-CH(OH)-CH_2CH_3$ gives secondary carbocation which is $CH_3-CH^+-CH_2CH_3$

Dehydration of $(CH_3)_3C-OH$ gives tertiary carbocation which is $(CH_3)_3C^+$

As in the case of option D, there is the formation of tertiary carbonium ion which has 9 $\alpha$ -hydrogens. So the carbonium ion is most stable.

Isopropyl benzene on air oxidation in the presence of dilute acid gives

0%
$C_{6}H_{5}COOH$
0%
$C_{6}H_{5}COCH_{3}$
0%
$C_{6}H_{5}CHO$
0%
$C_{6}H_{5}OH$

Mark the correct statement:

0%
Water gas is used in the manufacture of methyl alcohol.
0%
Water gas has the highest calorific value.
0%
Water gas burns with luminuous flame.
0%
The production of water gas is exothermic process

The alcohols which gives the most stable carbonium ion on dehydration is?

0%
$(CH_3)_2CHCH_2OH$
0%
$(CH_3)_3C-OH$
0%
$CH_3CH_2CH_2CH_2OH$
0%

Which of the following alcohols will dehydrate most rapidly when treated with conc.

$H_2SO_4$ ?

0%
0%
0%
0%
$CH_3-CH_2-CH_2-CH_2-OH$

Phenol is more acidic than alcohol because:

0%
Phenol is more soluble in polar solvents
0%
Alcohol does not lose hydrogen atom
0%
Phenoxide ion is stabilised by resonance
0%
Phenoxide ion does not exhibit resonance

Butane-2-ol is:

0%
Primary alcohol
0%
Secondary alcohol
0%
Tertiary alcohol
0%
Aldehyde

Explanation

Butan-2-ol is an example of secondary alcohol, as -OH group connected to secondary carbon (connected to two carbon atom).

Hence, option

$B$ is correct.

1626899_1762347_ans_c9443404031d406d858926de633f412e.JPG

What is obtained when chlorine is passed in boiling toluene and product is hydrolysed

0%
o-Cresol
0%
p-Cresol
0%
2, 4-Dihydroxytoluene
0%
Benzyl alcohol

During dehydration of alcohols to alkenes by heating with conc.

$H_2SO_4$ the initiation step is

0%
Protonation of alcohol molecule
0%
Formation of carbocation
0%
Elimination of water
0%
Formation of an ester

Explanation

$H_2SO_4 \rightarrow H^+ + HSO_4^-$

In the first step protonation of alcohol molecule takes place.

Option A is correct.
1658824_1763262_ans_211af3f4c49e4b15b24827c796fbe525.png

1658824_1763262_ans_211af3f4c49e4b15b24827c796fbe525.png

Dehydration of ethanol gives

0%
Acetic acid
0%
Ethane
0%
Ethylene
0%
Acetylene

Explanation

$CH_3-CH_2-OH \xrightarrow [170^0]{Conc.H_2SO_4} CH_2 =CH_2 +H_2O$

Option C is correct.

In the following series of chemical reactions, identify Z

$C_3H_7OH \quad \xrightarrow [ 160-180^{ 0 }C ]{ Conc.H_{ 2 }SO_{ 4 } } X\xrightarrow [ ]{ Br_2 }Y\xrightarrow [ Excess\ of ]{ Alc. KOH } Z$

0%
$CH_3 - \underset{NH_2}{\underset{\!\!\!\!|}{CH}} - \underset{NH_2}{\underset{\!\!\!\!|}{CH_2}}$
0%
$CH_3 - \underset{OH}{\underset{\!\!\!\!|}{CH}} - \underset{\!\!\!\!OH}{\underset{\!\!\!\!\!\!\!|}{CH_2}}$
0%
$CH_3 - \underset{OH}{\underset{|}{C}} = CH_2$
0%
$CH_3C \quad \equiv \quad CH$

Explanation

$CH_3CH_2CH_2OH \xrightarrow [160-18060 C]{conc.H_2SO_4} CH_3CH =CH_2 \xrightarrow []{Br_2} CH_3 - \underset{Br}{\underset{\!\!\!\!|}{CH}} - \underset{Br}{\underset{\!\!\!\!|}{CH_2}}\xrightarrow {Alc.KOH} \underset{Propyne}{CH_3-C\equiv CH}$

The reaction of

$C_2H_5OH$ with

$H_2SO_4$ does not give

0%
Ethylene
0%
Diethyl ether
0%
Acetylene
0%
Ethyl hydrogen sulphate

Wine (alcoholic beverages) contains:

0%
$CH_3OH$
0%
Glycerol
0%
$C_2H_5OH$
0%
2-propanol

Explanation

Alcoholic beverage, any fermented liquor, such as wine, beer, or distilled spirit, that contains ethyl alcohol, or ethanol

$(CH_3CH_2OH)$ , as an intoxicating agent.

Hence, option

$C$ is the answer.

Propan-1-ol can be prepared from propene by alcohol

0%
$H_2O/H_2SO_4$
0%
$Hg(OAc)_2/H_2 O$ followed by $NaBH_4$
0%
$B_2H_6$ followed by $H_2O_2$
0%
$CH_3CO_2H/H_2SO_4$

Explanation

$6CH_3-CH=CH_2+B_2H_6 \xrightarrow []{H_2O_2} CH_3-CH_2-CH_2OH$
Preparation of Propan-1-ol is shown in the above reaction.

Option C is correct.

In the following sequence of reactions

$CH_3CH_2CH_2Br \overset{KOH(alc)}{\longrightarrow} (A) \overset{HBr}{\longrightarrow} (B) \overset{KOH (aq.)}{\longrightarrow} (C).$
The product (C) is

0%
Propan - 2 - ol
0%
Propan - l - ol
0%
Propyne
0%
Propene

One of the following that cannot undergo dehydro - halogenation is

0%
Iso - propyl bromide
0%
Ethanol
0%
Ethyl bromide
0%
None of these

Primary alcohols on dehydration give

0%
Alkenes
0%
Alkanes
0%
Both (a) and (b)
0%
None of these

Explanation

$\begin{matrix} R-CH_{ 2 }-CH_{ 2 }-OH \\ 1^{ 0 }\quad alcohol \end{matrix}\xrightarrow [ 170^{ 0 }C ]{ Conc.H_{ 2 }SO_{ 4 } } \begin{matrix} R-CH=CH_{ 2 }+H_{ 2 }O \\ Alkene \end{matrix}$

Which of the following is most soluble in water?

0%
n-butyl alcohol
0%
Isobutyl alcohol
0%
Tertiary butyl alcohol
0%
Secondary butyl alcohol

Explanation

The solubility of alcohols and phenols in water due to their ability to form hydrogen bonds with water molecules.

The solubility decreases with an increase in size of alkyl group which is a hydrophobic group, which makes the alcohol less hydrophilic.

In the case of isomers, the order of solubility in water is

$1^0>2^0>3^0$ due to decrease in polar character.

Hence, among all, n-butyl alcohol is the most soluble in water.

The answer is option

$A$ .

1648139_1763756_ans_cc3993ff2c014088893068c4d0846ab5.JPG

Which of the following is not characteristic of alcohols

0%
They are lighter than water
0%
Their boiling points rise fairly uniformly with increasing molecular weight
0%
Lower members are insoluble in water and organic solvents but solubility regularly increases with molecular weight
0%
Lower members have pleasant smell and burning taste, while higher members are odourless and tasteless

Which of the following is most acidic

0%
Phenol
0%
Benzyl alcohol
0%
m-chlorophenol
0%
Cyclohexanol

Explanation

Benzyl alcohol and cyclohexanol are not acidic while phenol and m-chlorophenol are acidic due to the presence of electron withdrawing groups like

$-NO_2,-Cl,-CN$ increases the acidic character of phenols. hence m-chlorophenol is more acidic than phenol.
Option C is correct.

CBSE Questions for Class 12 Medical Chemistry Alcohols, Phenols And Ethers Quiz 9 - MCQExams.com

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Practice Class 12 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Medical Chemistry Alcohols, Phenols And Ethers Quiz 9 - MCQExams.com

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Practice Class 12 Medical Chemistry Quiz Questions and Answers

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