MCQ Questions for Class 12 Engineering Chemistry Alcohols,Phenols And Ethers Quiz 8

Ethene reacts with

$Y$ to produce ethanol.

$ethene + Y \rightarrow ethanol$
What is

$Y$ ?

0%
Hydrogen
0%
Oxygen
0%
Steam
0%
Yeast

Explanation

Solution:- (C) Steam

Ethanol can be manufactured by hydration of ethene.

Ethene is heated with steam in the presence of a catalyst of phosphoric acid to produce ethanol.

${C}_{2}{H}_{4} + {H}_{2}O \xrightarrow{\text{high temperature}} \underset{\left( \text{Ethanol} \right)}{{C}_{2}{H}_{5}-OH}$

Hence

$Y$ is steam.

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.

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.

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%
$HOCl$

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

State whether the following statement is True or False.

$C_2H_5OH$ can accept a proton to give protonated,

$C_2H_5\overset{+}{O}H_2$ , i.e., it can act as an acid.

0%
True
0%
False

Explanation

It can act as a base, because it abstracts

$H^+$

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

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

0%
True
0%
False

The boiling point of thioether are ________ than those of ethers.

0%
higher
0%
lower
0%
equal
0%
nearly same

Explanation

An organic sulfide (British English sulphide) or thioether is a functional group in organosulfur chemistry with the connectivity C–S–C as shown on right. A sulfide is similar to an ether except that it contains a sulfur atom in place of the oxygen.

The electronegativity of sulfur is only

$2.6$ whereas hydrogen is

$2.2.$ That means there isn't much of a dipole in the S-H bond, and hydrogen bonding is not significant. Contrast

$H_2O$ , with a boiling point of

$100^{\circ}C,$ with H2S and its boiling point of

$–60^{\circ}C$

The weak polarity of ethers do not appreciably affect their boiling points which are comparable to those of the alkenes of comparable molecular mass.hence have less oiling of

$34.6^{\circ}C$

$\therefore$ The boiling point of thioether higher than those of ethers.

$\mathbf{Hence\ the\ correct\ answer\ is\ option\ (A)}$

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

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

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.

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

The boiling point of ethanol is higher as compared to the boiling point of diethyl ether though both have the same molecular formula. This is due to:

0%
Resonance
0%
H-bonding
0%
R-group
0%
Covalent bonding

Wood spirit if known as?

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

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

0%
True
0%
False

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.

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

State, whether the following statements are True or False:

$C_2HSOC_2H5$ is a linear molecule having zero dipole moment.

0%
True
0%
False

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%

An organic compound 'X' on treatment with acidified

$K_2Cr_2O_7$ gives a compound 'Y' which reacts with

$I_2$ and

$Na_2CO_3$ to form tri-iodomethane. The compound 'X' is?

0%
$CH_3OH$
0%
$CH_3COCH_3$
0%
$CH_3CHO$
0%
$CH_3CHOHCH_3$

Explanation

The compound Y should be a ketone as it is responding to iodoform test. Since it is formed by treating X with potassium dichromate(acidic) X should be a secondary alcohol. X is

$CH_3CH(OH)CH_3$

Boiling point of

$C_2H_5OH$ is higher as compared to

$C_2H_5SH$ on account of:

0%
Association
0%
Dissociation
0%
Low molecular mass
0%
None of these

Explanation

Due to intermolecular hydrogen bonding, ethyl alcohol remains in the associated form and therefore, boils at a higher temperature compared to

$C_2H_5SH$ .

The correct option is (A).

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.

The correct order of boiling points for primary

$(1^o)$ , secondary

$(2^o)$ and tertiary alcohol

$(3^o)$ is:

0%
$1^o > 2^o > 3^o$
0%
$3^o > 2^o > 1^o$
0%
$2^o > 1^o > 3^o$
0%
$2^o > 3^o > 1^o$

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 the molecule. Both these factors are least in

$3^\circ$ alcohols and maximum in

$1^\circ$ alcohols. Hence,

$3^\circ$ alcohols have a least boiling point while

$1^o$ alcohols have a maximum boiling point.

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

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

Cresols are

0%
Hydroxy toluenes
0%
Dihydric phenols
0%
Trihydric phenols
0%
Trihydric alcohols

IUPAC name of tertiary butyl alcohol is

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

Wood spirit is known as

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

The IUPAC name of picric acid is

0%
$2,4,6-$ trinitrophenol
0%
$2,4,6-$ trinitrobenzoic acid
0%
$4-$ nitrophenol
0%
None of these

Explanation

The IUPAC name of picric acid is

$2,4,6-$ trinitrophenol.
1680239_1761556_ans_5a4332f492334ffabdddf104948f51b1.png

The

$IUPAC$ name of compound

$\begin{matrix} \quad \quad O{ CH }_{ 3 } \\ | \\ { CH }_{ 3 }-{ CH }_{ 2 }-CH-{ CH }_{ 2 }-OH \end{matrix}$ is

0%
$2-$ methoxy $-1-$ butanol
0%
$3-$ methoxy $-1-$ butanol
0%
$2-$ methoxy $-1-$ butanol
0%
$1,2-$ methoxy $-$ butanol

The structural formula of cyclohexanol is

3-pentanol is a:

0%
Primary alcohol
0%
Secondary alcohol
0%
Tertiary alcohol
0%
None of these

Explanation

3-pentanol is a secondary alcohol as it is connected to secondary carbon (conected to two carbon atom).

Hence, option

$B$ is correct.

1626909_1762350_ans_ea06d5cf9360427683030ef16eeaa6a6.JPG

General formula of primary alcohol is

0%
$>CHOH$
0%
0%
$-CH_2OH$
0%

Explanation

$-OH$ group is attached to primary carbon.

Which of the following is phenolic

0%
Phthalic acid
0%
Phosphoric acid
0%
Picric acid
0%
Phenylacetic acid

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

Boiling point of alcohol is comparatively higher than that corresponding alkane due to:

0%
Intermolecular hydrogen bonding
0%
Intramolecular hydrogen bonding
0%
Volatile nature
0%
None of these

Explanation

In alkanes, the only intermolecular forces are Van der Waals dispersion forces.

Due to high electro-negativity of oxygen in alcohols, inter molecular hydrogen bond is possible. Hydrogen bonds are much stronger than Van der Waals and therefore it takes more energy to separate alcohol molecules than it does to separate alkane molecules. So due to strong inter molecular H-bonds in alcohols, they have higher boiling points than ethers.

Hence, option

$A$ is correct.

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.

The reaction of

$C_2H_5OH$ with

$H_2SO_4$ does not give

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

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.

Acetyl bromide reacts with excess of

$CH_3MgI$ following by treatment with a saturated of

$NH_4Cl$ gives

0%
2-methyl-2-propanol
0%
Acetamide
0%
Acetone
0%
Acetyl iodide

Explanation

$\underset{Acetyl \ bromide}{CH_3 - \overset{O}{\overset{||}{C}} - Br }\xrightarrow[(ii) Saturated \ NH_4 Cl]{(i) Excess - CH_3 MgI} \underset{2 - methyl \ 2-propanol}{CH_3 - \underset{CH_3}{\underset{|}{\overset{CH_3}{\overset{|}{C}}}} - OH}$

Option A is correct.

CBSE Questions for Class 12 Engineering Chemistry Alcohols,Phenols And Ethers Quiz 8 - MCQExams.com

0:0:2

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Alcohols,Phenols And Ethers Quiz 8 - MCQExams.com

0:0:2

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.