MCQ Questions for Class 12 Engineering Chemistry Alcohols,Phenols And Ethers Quiz 3

In the following dehydration of diol with

$\displaystyle { H }_{ 3 }{ PO }_{ 4 }$ , following product is:

Explanation

Tertiary alcohol at

$\displaystyle { C }_{ 2 }$ is more basic than secondary alcohol at

$\displaystyle { C }_{ 5 }$ . Hence, intramolecular dehydration takes place such that H of secondary alcoholic group at

$\displaystyle { C }_{ 5 }$ eliminates

$\displaystyle OH$ of tertiary alcoholic group at

$\displaystyle { C }_{ 2 }$ (with isotopic

$\displaystyle \overset { 18 }{ O }$ ) to form a cyclic ether.

Lower alcohols are more soluble in water than higher alcohols.

0%
True
0%
False

Explanation

Lower alcohols are more soluble in water than higher alcohols. With the increase in the molar mass, solubility decreases. In alcohols,

$-OH$ group is polar and forms hydrogen bonds with water. This makes alcohol water soluble. The alkyl chain (hydrocarbon chain) is non-polar and decreases the solubility of alcohol in water.

Consider the reaction between

$\displaystyle C_{2}H_{5}O^{\circleddash }$ and dimethyl sulphate. The leaving group in this reaction is

0%
$\displaystyle MeOSO_{2}O^{\circleddash }$
0%
$\displaystyle MeO^{\circleddash }$
0%
$\displaystyle Me^{\circleddash }$
0%
$\displaystyle MeOSO_{2}^{\circleddash }$

The correct reactivity order of alcohols towards

$H-X$ will be:

0%
$\;II>I>III>IV$
0%
$\;IV>III>II>I$
0%
$\;II>IV>I>III$
0%
$\;II>IV>III>I$

Explanation

The correct reactivity order of alcohols towards H−X will be

$\displaystyle II>IV>III>I$

Alcohol II has maximum reactivity as the carbocation formed will be stabilized by resonance with adjacent C=C double Bond.

Alcohol I has minimum reactivity as the carbocation formed will have positive charge on

$\displaystyle sp^2$ carbon atom.

Alcohol IV is more reactive than alcohol III because alcohol IV gives secondary carbocation whereas alcohol III gives primary carbocation.

Secondary carbocation is more stable than primary carbocation.

Phenol on reaction with

$\displaystyle { NH }_{ 3 }$ gives :-

0%
Aniline
0%
Nitrobenzene
0%
Benzaldehyde
0%
Anilinium chloride

Explanation

Phenol on reaction with ammonia gives aniline. In this reaction, -OH group is replaced with $-NH_2$ group.

Which of the following is a primary alcohol ?

0%
Butan - 2 - ol
0%
2 - Methylpropan -1- ol
0%
2 - Methylpropan - 2- ol
0%
None of the above

Explanation

$CH_{3}-CH(CH_{3})-CH_{2}-OH$ is a primary alcohol as hydroxal group; -OH is attached to the primary carbon.

The major product of the above reaction is:

Carbinol is a trivial name of :-

0%
$\displaystyle { C }_{ 2 }{ H }_{ 5 }OH$
0%
$\displaystyle { CH }_{ 3 }OH$
0%
$\displaystyle HCOOH$
0%
$\displaystyle { CH }_{ 3 }COOH$

Explanation

Carbinol is a trivial name of methanol $CH_3OH$ .

The high boiling point of ethanol

$(78.2^o C)$ compared to dimethyl ether

$(-23.6^o C)$ , though both having the same molecular formula

$\displaystyle { C }_{ 2 }{ H }_{ 6 }O$ , is due to:

0%
hydrogen bonding
0%
ionic bonding
0%
co-ordinate covalent bonding
0%
resonance

Explanation

The high boiling point of ethanol

$(78.2 ^oC)$ compared to dimethyl ether

$(-23.6^oC)$ , though both having the same molecular formula

$C_2H_6O$ , is due to Hydrogen bonding.
Hydrogen bonding is observed when H atom is attached to more electronegative N, F or O atom.
Ethanol molecules are associated due to hydrogen bonding. This results in higher boiling point.
However, hydrogen bonding is not possible in dimethyl ether.

Heating together of sodium ethoxide and ethyl iodide will give:

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

For phenol, which of the following statements is correct?

0%
It is insoluble in water
0%
It has higher boiling point than toluene
0%
It does not shows acidic property
0%
It has lower melting point compared to aromatic hydrocarbons of comparable molecular weight

Mark the correct order of decreasing acid strength of the following compounds.

0%
e > d > b > a > c
0%
b > d > a > c > e
0%
d > e > c > b > a
0%
e > d > c > b > a

Explanation

The correct order of decreasing acid strength is $b > d > a > c > e$ p-nitrophenol is most acidic and p-methoxy phenol is least acidic. When an electron withdrawing group is para to OH group, the acidity is maximum. When an electron releasing group is para to OH group, the acidity is minimum.

A vicinal diol has two alcoholic groups present :

0%
on the same carbon
0%
on adjacent carbon atoms
0%
anywhere along the carbon chain
0%
none of the above

Electrophilic substitution reaction in phenol takes place at:

0%
para position
0%
meta position
0%
ortho position
0%
ortho and para position

Explanation

Electrophilic substitution reaction in phenol takes place at ortho and para position.

$-OH$ is an ortho para directing group.

The electron density at ortho and para positions will be higher than the electron density at meta position.

The IUPAC name of

$CH_3\, -\, C(CH_3) (OH) CH_2\, -\, CH(CH_3)\, CH_3$ is :

0%
2, 4 - Dimethylpentan - 2 - ol
0%
2, 4 - Dimethylpentan - 4 - ol
0%
2, 2 - Dimethylbutane
0%
Isopentanol

The major product obtained on acid - catalysed hydration of 2-phenylpropene is:

0%
2-phenylpropan-2-ol
0%
2-phenylpropan-1-01
0%
3-phenylpropan-2-ol
0%
1-phenylpropan-1-ol

Explanation

The major product obtained on acid - catalysed hydration of 2-phenylpropene is 2-Phenylpropan-2-ol. A molecule of water is added to

$\displaystyle C=C$ double bond.

$\displaystyle -OH$ group is added to more substituted C atom.

Among ethanol, dimethyl ether, methanol and propanal, the isomers are

0%
ethanol, dimethyl ether, methanol and propanal
0%
ethanol and methanol
0%
ethanol, dimethyl ether, and methanol
0%
ethanol and dimethyl ether

Explanation

Ethanol

$(C_2H_5OH)$ and dimethyl ether

$(CH_3-O-CH_3)$ have same molecular formula but different functional groups, so they are isomers.

What are the types of reactions occurring in the formation of X, Y and Z respectively?

0%
Oxidation, Reduction, Substitution
0%
Substitution, Esterification, Dehydration
0%
Reduction, Substitution, Oxidation
0%
Dehydration, Oxidation, Esterification

Explanation

Reactions:

$CH_3CH_2OH \xrightarrow {conc. H_2SO_4} CH_2 = CH_2$ dehydration

$CH_3CH_2OH \xrightarrow {oxidation} CH_3COOH$ oxidation
esterification

Refer the given reactions and answer the following questions:

Identify X, Y and Z respectively:

0%
$\displaystyle CH_{3}COOH,CH_{2}= CH_{2},CH_{3}COOCH_{3}$
0%
$\displaystyle CH_{2}= CH_{2},CH_{3}COOH,CH_{3}COOCH_{2}CH_{3}$
0%
$\displaystyle HCHO,CH_{3}CH_{3},CH_{3}CH_{2}COOH$
0%
$\displaystyle CH_{3}CH_{3},HCHO,CH_{3}COOH$

Explanation

Reactions:

$CH_3CH_2OH \xrightarrow {conc. H_2SO_4} CH_2 = CH_2$

$CH_3CH_2OH \xrightarrow {oxidation} CH_3COOH$

Methanol is now prepared from:

0%
water gas
0%
sugar through fermentation
0%
wood
0%
both $A$ and $B$

Identify the stereoisomeric pair from the following choices

0%
$CH_3 CH_2 CH_2 OH$ and $CH_3 CH_2 OCH_3$
0%
$CH_3 CH_2 CH_2 Cl$ and $CH_3 CHClCH_3$
0%
0%

What is the common name of methyl alcohol?

0%
Wine
0%
Wood spirit
0%
Dilute acid
0%
None of the above

Among the four compounds (i) acetone, (ii) propanol, (iiii) methyl acetate and (iv) propionic acid, the two that are isomeric are

0%
methyl acetate and acetone
0%
methyl acetate and propanol
0%
propionic acid and methyl acetate
0%
propionic acid and acetone

Explanation

Propionic acid and methyl acetate both have same molecular formula

$(C_3H_6O_2)$ but different functional groups, so they are isomers.

How many structures of F is possible?

0%
2
0%
5
0%
6
0%
3

The product formed by decolourization of

$KMn{O}_{4}$ by ethylene is _________ which is used as ______.

0%
ethane, oxidant
0%
ethyl alcohol, coolant
0%
ethylene glycol, antifreeze
0%
ethanol, reductant

Explanation

When ethylene react with

$KMnO_4$ , than pink colour of

$KMnO_4$ gets disappear and ethylene gets converted with ethylene glycol.
1175537_399041_ans_bc430365b5784ec09f0afad669a43e29.PNG

Which among the following reactions does not give methyl alcohol?

0%
The reaction of ethylene with ${H}_{2}{SO}_{4}$ at ${80}^{o}C$
0%
The reaction of water gas with hydrogen at high temperature
0%
Alkaline hydrolysis of methyl bromide
0%
Both $A$ and $B$

Explanation

The reaction of ethylene with

${H}_{2}{SO}_{4}$ at

${80}^{o}C$ gives ethyl hydrogensulphate.

$CH_2=CH_2+H_2SO_4 \rightarrow CH_3CH_2OSO_2OH$
Water gas (carbon monoxide and hydrogen) at high temperature gives methanol.

$CO + 2H_2 \rightarrow CH_3OH$ .
Alkaline hydrolysis of methyl bromide gives methanol

$CH_3Br + H_2O \rightarrow CH_3OH + HBr$

The molecular formula of ethanol is :

0%
$(CH_3)_2 CHOH$
0%
$C_2 H_5 OH$
0%
$CH_3 OH$
0%
$CH_3 CH_2 CH_2 OH$

Explanation

$(CH_3)_2CHOH$ is 2-propanol or isopropyl alcohol,

$CH_3OH$ is methanol,

$CH_3CH_2CH_2OH$ is 1- propanol or isopropyl alcohol.

The compound in option B is ethanol. It is hydroxyl derivative of ethane.

Glycol is a:

0%
primary alcohol
0%
dihydric alcohol
0%
secondary alcohol
0%
trihydric alcohol

Identify the major product X obtained in the following reaction.

$2, 3-\text{dimethyl butan}-2-ol\xrightarrow []{Conc.H_{ 2 }SO_{ 4 }} X$

Explanation

Dehydration of alcohol occurs when the reaction proceeds in the presence of sulphuric acid.

$({ CH_{ 3 }) }_{ 2 }HC-C(OH)({ CH_{ 3 }) }_{ 2 }\longrightarrow { (CH_{ 3 }) }_{ 2 }C=C{ (CH_{ 3 }) }_{ 2 }+{(CH_3)}_2CHC(CH_3)=CH_2$

2,3-dimethylbut-2-ene and 2,3-dimethylbut-1-ene are formed in which 2,3-dimethylbut-2-ene is major product.

In the following reation,

$\underset{(in\ excess)}{Ethyl \ alcohol}+\underset{(from\ H_2SO_4)}{H^+}\overset{413K}{\rightarrow}$ product, the product is:

0%
ethene
0%
ethyl hydrogen sulphate
0%
diethyl ether
0%
acetylene

Explanation

Ethyl alcohol when reacts with conc.

$H_2SO_4$ , it forms diethyl ether as a product.

$C_2H_5-OH \xrightarrow[conc.H_2SO_4]{413K} \underset{Diethyl\ ether}{C_2H_5-O-C_2H_5}$

At higher temperature, the ethyl undergoes elimination reaction to from ethene as a product.

The suffix used in the nomenclature of alcohol is:

0%
'al'
0%
'ol'
0%
'yl'
0%
'one'

Propene on hydration gives

0%
propan-1-ol
0%
propan-2-ol
0%
propanal
0%
methoxy ethane

The compound which reacts fastest with Lucas reagent at room temperature, is

0%
butan-1-ol
0%
butan-2-ol
0%
2-methyl propan-1-ol
0%
2-methyl propan-2-ol

The correct order of boiling point for primary

$\left( { 1 }^{ } \right)$ , secondary

$\left( { 2 }^{ } \right)$ and tertiary

$\left( { 3 }^{ } \right)$ alcohols is:

0%
${ 1 }^{ } > { 2 }^{ } > { 3 }^{ }$
0%
${ 3 }^{ } > { 2 }^{ } > { 1 }^{ }$
0%
${ 2 }^{ } > { 1 }^{ } > { 3 }^{ }$
0%
${ 2 }^{ } > { 3 }^{ } > { 1 }^{ }$

Explanation

Alcohols with same molecular weight are expected to have almost same boiling point however two more factors other than molecular weight are important, they are namely H-bonding and surface area of molecule. Both these factors are least in

$3$ alcohols and maximum in

$1$ alcohols. Hence,

$3$ alcohols have least boiling point while

$1$ alcohols have maximum boiling point.

What amount of bromine will be required to convert

$2\ g$ of phenol into

$2, 4, 6-tribromophenol$ ?

0%
$4.00$
0%
$6.00$
0%
$10.22$
0%
$20.44$

Explanation

Write equation for chemical change, find molecular masses of reactants and products and solve it.
Molecular weight of phenol

$= 12\times 6 + 1 \times 6 + 16 = 94$
Molecular wt. of

$Br_{2} = 3\times 160 = 480$

$\because 94$ of phenol requires

$= 480\ g$ of

$Br_{2}$

$\therefore 2\ g$ of phenol requires

$= \dfrac {480}{94} \times 2 = 10.22\ g$

The correct answer is 10.22 g

1056479_461928_ans_f36e0ab2bf844e8e9faab61125e74f3b.png

The bond angle of the bond

$C-O-C$ in methoxy methane is:

0%
$111.7$
0%
$109$
0%
$108.9$
0%
$180$

Explanation

bond angle =

$111.7^{o}$
Ethers feature C–O–C linkage defined by a bond angle of about 110° and C–O bond length of about 140 pm.

In methoxymethane, the bond angle is

$111.7^{o}$ . Methoxymethane show positive deviation in the bond angle. This is basically due to the steric crowding on carbon atoms of methyl group.

Due to the crowding, the repulsion between bond pair of carbons and lone pair of oxygen is low. Thus, the bond angle of methoxymethane is

$111.7^0$ .

Sodium phenoxide

$\xrightarrow [ pressure ]{ { CO }_{ 2 } } A \xrightarrow [ ]{ { { H }_{ 3 }O }^{ + } }$ o-hydroxy benzoic acid
This product can also be obtained by the reaction of which of the following?

0%
Phenol + chloroform +alkali
0%
Phenol + pyrene + alkali
0%
Phenol + acetyl chloride + $Al{Cl}_{3}$
0%
Phenol + methyl chloride + $Al{Cl}_{3}$

The miscibility of ether with _______ resembles those of alcohol of the same molecular mass.

0%
methanol
0%
water
0%
DCM
0%
benzene

Explanation

The miscibility of ether with water resembles those of alcohol of the same moleculer mass. It is generally due to the hydrogen bonding with water in alcohol. So the correct option is

$B$ .

With ethoxyethane which compound is miscible to almost the same extent?

0%
Formaldehyde
0%
Bhutanone
0%
Butan-1-ol
0%
Propan-1-ol

Explanation

The correct option is

$C$ . Ethoxyethane is an ether group which is miscible to the alcohol that is Butan

$-1-$ ol. This is because they have the same molecular mass.

Among the following which has the highest boiling point?

0%
$CH_3CH_2OH$
0%
$CH_3CH_2CH_3$
0%
$CH_2=CHCH_3$
0%
$CH_3OCH_3$

Explanation

Among the given molecules, ethanol

$CH_3CH_2OH$ has the highest boiling point.

This is due to formation of intermolecular hydrogen bonds which leads to molecular association.

Option A is correct.

Which of the following alcohols forms a ketone when oxidized?

0%
1-Propanol
0%
Methanol
0%
2-Methyl-2-propanol
0%
2-Propanol

Explanation

2-propanol (a secondary alcohol) forms a ketone when oxidized.

$\displaystyle CH_3 - \underset { \underset {\displaystyle OH}{|} }{C} H- CH_3 \xrightarrow { \text {oxidation}} CH_3 - \underset { \underset {\displaystyle O}{||} }{C} - CH_3$

1-propanol and methanol (primary alcohols) forms aldehydes when oxidized.

$\displaystyle CH_3 -CH_2- \underset { \underset {\displaystyle OH}{|} }{C} H_2 \xrightarrow { \text {oxidation}} CH_3-CH_2 - \underset { \underset {\displaystyle O}{||} }{C} - H$

$\displaystyle CH_3 -OH \xrightarrow { \text {oxidation}} H- \underset { \underset {\displaystyle O}{||} }{C} - H$

2-methyl-2-propanol (

$\displaystyle (CH_3)_3 C -OH$ ) is a tertiary alcohol whose oxidation requires drastic conditions. The number of

$C$ atoms in the oxidation product are less than the number of

$C$ atoms in the starting material.

In iodine charge transfer test the color of ether compound formed is:

0%
blue
0%
pale yello
0%
brown
0%
dark red

Explanation

The correct option is

$C$ . Ether forms charge transfer complexes with Iodine to give a brown solution.

$R-O-R'\longrightarrow R'I-+RI+H_2O$

Which of the following is the functional group of an ether?

0%
$R - OH$
0%
$R - O - R$
0%
$R-CO-R'$
0%
$R-COOH$
0%
$RCHO$

Explanation

The correct option is B.

$R-O-R$ is generally the functional group of the ether which is non-polar and does not form hydrogen bonding.

The_____ ethers are highly volatile and flammable.

0%
Higher
0%
Lower
0%
Both A and B
0%
None of these

Methanol cannot be prepared by the hydration of an alkene.

0%
True
0%
False

Ethers show dipolar nature due to the presence of ______.

0%
central C atom
0%
central O atom
0%
central N atom
0%
none of these

$CH_3OH\xrightarrow{P+I_2}A \xrightarrow{Mg/dry\ ether}B\xrightarrow{HCHO}C\xrightarrow{{H^+/H_2O}}D$

What are

$B$ &

$D$ respectively?

0%
$CH_3MgI$ , $CH_3CHO$
0%
$CH_3MgI$ , $CH_3CH_2OH$
0%
$CH_3MgI$ , $CH_3COOH$
0%
$CH_3MgI$ , $CH_3COCH_3$

Explanation

$CH_3OH\xrightarrow{P+I_2}CH_3I \xrightarrow{Mg/dry\ ether}CH_3MgI\xrightarrow{HCHO}CH_3CH_2OMgI\xrightarrow{{H^+/H_2O}}CH_3CH_2-OH$

Hence

$B$ is the correct answer.

_____________ is prepared on a large seals from water gas.

0%
Methanol
0%
Ethanal
0%
Methane
0%
Ethanol

Explanation

Methanol is generally prepared on a large scale using water gas. Water gas is also called syn gas is the mixture of

$(CO+{ H }_{ 2 }).$

On treatment with dilute

$H_2SO_4$ , ethene forms ethanol.

0%
True
0%
False

Explanation

On the reaction of ethane with dilute

${ H }_{ 2 }SO_{ 4 }$ , it terms ethanol.
1157157_522560_ans_2febd731de324143aa3998e60ea3bb53.PNG

Metallic sodium reacts with methanol and ethanol to give

$O_2$ gas.

0%
True
0%
False

Explanation

A metallic solution which reacts with methanol and ethanol does not give

${ O }_{ 2 }$ but give the solution which is basic in solution and give

${ O }_{ 2 }$ gas.

CBSE Questions for Class 12 Engineering Chemistry Alcohols,Phenols And Ethers Quiz 3 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Alcohols,Phenols And Ethers Quiz 3 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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