MCQ Questions for Class 12 Engineering Chemistry Alcohols,Phenols And Ethers Quiz 1

The major product of the above reaction is :

Which of the following compounds will most readily be dehydrated to give alkene under acidic condition?

0%
2-Hydroxycyclopentanone
0%
1-Pentanol
0%
3-Hydroxypentan-2-one
0%
4-Hydroxypentan-2-one

Explanation

3-Hydroxypentan-2-one and 4-Hydroxypentan-2-one will most readily be dehydrated to give alkene under acidic condition. This will lead to formation of

$\alpha - \beta$ unsaturated ketone. But out of both 4-Hydroxypentan-2-one will most readily be dehydrate to give alkene under acidic condition.
Note: Dehydration of 2-Hydroxycyclopentanone will lead to

$C=C$ double bond in 5 member ring that will cause strain and product is less likely to be formed.

Which of the following is not the product of dehydration of :

Compound

$A, C_{8}H_{10}O$ , is found to react with

$NaOI$ (produced by reacting

$Y$ with

$NaOH$ ) and yields a yellow precipitate with characteristic smell.

$A$ and

$Y$ are respectively are:

Explanation

A and Y are 1-phenylethanol and iodine respectively.
NaOI oxidizes 1-phenylethanol to methyl ketone (acetophenone) which gives positive iodoform test. The yellow precipitate is of iodoform

$(CHI_3)$

The compound

$A$ on treatment with

$Na$ gives

$B$ , and with

$PCl_{5}$ gives

$C. B$ and

$C$ react together to give diethyl ether.

$A, B$ and

$C$ are in the order.

0%
$C_{2}H_{5}Cl, C_{2}H_{6}, C_{2}H_{5}OH$
0%
$C_{2}H_{5}OH, C_{2}H_{6}, C_{2}H_{5}Cl$
0%
$C_{2}H_{5}OH, C_{2}H_{5}ONa, C_{2}H_{5}Cl$
0%
$C_{2}H_{5}OH, C_{2}H_{5}Cl, C_{2}H_{5}ONa$

Explanation

The compound A is ethanol

$\displaystyle C_{2}H_{5}OH$ . The compound A on treatment with

$Na$ gives sodium ethoxide

$\displaystyle (C_{2}H_{5}ONa)$ which is compound B.

$\displaystyle C_{2}H_{5}OH \xrightarrow [-H_2]{Na} C_{2}H_{5}ONa$

The compound A on treatment with

$PCl_5$ gives ethyl chloride

$\displaystyle ( C_{2}H_{5}-Cl)$ which is compound C.

$\displaystyle C_{2}H_{5}OH \xrightarrow {PCl_5} C_{2}H_{5}-Cl+HCl+POCl_3$

B and C react together to give diethyl ether.

$\displaystyle C_{2}H_{5}ONa+C_{2}H_{5}Cl \rightarrow C_{2}H_{5}OC_{2}H_{5}+NaCl$

A,B and C are in the order

$\displaystyle C_{2}H_{5}OH, C_{2}H_{5}ONa, C_{2}H_{5}Cl$ .

Option C is correct.

The reaction can be classified as:

0%
williamson ether synthesis reaction
0%
alcohol formation reaction
0%
dehydration reaction
0%
williamson alcohol synthesis reaction

Among the following sets of reaction which one produces anisole?

0%
$CH_3CHO; RMgX$
0%
$C_6H_5OH; NaOH, CH_3I$
0%
$C_6H_5OH; neutral FeCl_3$
0%
$C_6H_5-CH_3; CH_3COCl; AlCl_3$

Explanation

Among the following sets of reaction, the reaction B produces anisole:

(A) Acetaldehyde reacts with grignards reagent followed by acidic hydrolysis to form secondary alcohol.

(B) Phenol reacts with NaOH to form sodium phenoxide. It reacts with methyl iodide in

$SN^2$ reaction to form
anisole (methoxy benzene).

$C_6H_5- OH + NaOH \rightarrow C_6H_5 - ONa\underset {S_N}{\xrightarrow {CH_3I}}\underset {(Anisole)}{C_6H_5-OCH_3}$ (Williamson's synthesis)

$3C_6H_5OH + neutral FeCl_3\rightarrow \underset {\text {Violet colour complex} } {(C_6H_5O)_3Fe}+3HCl.$

(D) Toluene reacts with acetyl chloride in presence of anhydrous

$AlCl_3$ to form p-methyl acetophenone. This is friedel craft's reaction.

$C_6H_5-CH_3+CH_3COCl+AlCl_3\rightarrow \underset { \text {Para- methyl aceto phenone} } { p-{CH_3-C_6H_4- COCH_3}}$

Degree of unsaturation of

$A=2$ and it contains no double or triple bonds. Identify the structure of A.

0%
0%
0%
0%
None of these

Explanation

The degree of unsaturation of

${C}_{10}{H}_{18}O=2$ but it contains no double or triple bond. Hence, there are two rings-one six-membered as indicated by product and the other three-membered which is cleaved by

$HCl$ due to strain. Hence

$A$ has given structure.

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%
Assertion and Reason both are incorrect.

Explanation

Dehydration of alcohols can be carried out either with protonic acids such as

$conc.H_{2}SO_{4}, H_{3}PO_{4}$ or catalysts such as anhydrous zinc chloride or alumina.

$H_2O^+$ is a good leaving group than

$OH^-$

The strongest acid among the following is:

0%
o-methoxyphenol
0%
m-methoxyphenol
0%
p-methoxyphenol
0%
phenol

Explanation

m-Methoxy phenol is the strongest acid.

Methoxy group at meta position exerts an only-I effect.

Due to this, it withdraws electrons from all the nuclear position of the benzene ring including the one holding the

$O-H$ group consequently, electron density in the

$O-H$ bond decreases.

Hence, m-methoxy phenol becomes more acidic than o and p-methoxy phenols.

Option B is correct.

How many structures of X is possible ?

0%
$4$
0%
$5$
0%
$6$
0%
$3$

$\left[ X \right]$ is

The compound which gets dissolved in water is :

0%
$C_{2}H_{6}$
0%
$C_{2}H_{4}$
0%
$C_{2}H_{2}$
0%
$C_{2}H_{5}OH$
0%
$C_{3}H_{8}$

Methyl amine with nitrous acid gives

0%
methyl alcohol
0%
methyl nitrite
0%
dimethyl ether
0%
methyl nitrite and dimethyl ether

An ester can be prepared by the reaction of:

0%
two alcohols
0%
an alcohol and an aldehyde
0%
an alcohol and an organic acid
0%
an organic acid and an aldehyde
0%
an acid and a ketone

A and B are -

Ethyl ether when reacted with chlorine in dark yields

0%
${CH}_{3}CHClO{CH}_{2}{CH}_{3}$
0%
${CH}_{3}{CH}_{2}O{CH}_{2}{CH}_{2}Cl$
0%
$Cl{CH}_{2}{CH}_{2}O{CH}_{2}{CH}_{2}Cl$
0%
${CH}_{3}CHClOCHCl{CH}_{3}$

Explanation

Ethyl ether, when reacted with

${Cl}_{2}$ in dark, substitution, occurs at

$\alpha$ -carbon atoms.

1295183_461871_ans_7de5dfb16ed04c79ad0fb02297c53ec4.png

Molecular formula of ethanol is ?

0%
$C{H}_{3}OH$
0%
${C}_{2}{H}_{4}{\left(OH\right)}_{2}$
0%
${C}_{2}{H}_{5}OH$
0%
$HCHO$

Explanation

The correct option is C. Ethanol is alcohol and is a primary

$1$ degree monohydric alcohol with moleculer formula

$C_2H_5OH.$

The IUPAC name of the given acid is:

0%
1, 2, 3-propane tricarboxylic acid
0%
3-hydroxy-1, 5-pentanendioic acid
0%
3-hydroxy-3-keto-1, 5-pentanedioic acid
0%
1, 3, 5-trihydroxypentane-1, 5-dione

Identify the compound

$'Y"$ in the following sequence of reaction:

$HC \equiv CH \xrightarrow[ii) H_2O/Zn]{i) O_3} X \xrightarrow{Zn/CH_3COOH} Y$

0%
0%
0%
0%
$CH_{3}COOH$

Which of the following cannot be dissolved in alcohol?

0%
Dyes and drugs
0%
Soaps and varnishes
0%
Resins and varnishes
0%
Rubber and plastics

The IUPAC name of the compound is

0%
1, 1-Dimethyl-3-cyclohexanol
0%
1, 1-Dimethyl-3-hydroxycyclohexane
0%
3,3-Dimethyl-1-cyclohexanol
0%
3, 3-Dimethyl-1-hydroxycyclohexane

.How many number of compounds, which have same IUPAC name?

0%
$0$
0%
$1$
0%
$2$
0%
$3$

The IUPAC name is:

0%
2-ethyl-2-butanol
0%
3-Methyl-3-pentanol
0%
3-ethyl-3-methyl-3-pentanol
0%
1,1-diethyl ethanol

IUPAC name of the tertiary butyl alcohol is:

0%
1-butanol
0%
2-butanol
0%
2-methyl-1-propanol
0%
2-methyl-2-propanol

The correct IUPAC name of the compound is:

0%
Isopropyl butanol
0%
2-methylbutane-3,4-diol
0%
3-methylbutane-1,2-dial
0%
3-methylbutane-1,2-diol

Explanation

The correct IUPAC name of the compound is 3-methyl butane-1,2-diol.

A methyl substituent is present on third C atom and

$-OH$ substituents are present on first and second C atoms.

Note: The longest chain of C atoms containing the

$-OH$ group is selected as the parent alkane and numbered from the end closer to

$-OH$ group.

The IUPAC name of the compound is:

0%
1 - dimethylbutane - 1, 3 - diol
0%
1, 3, 3 -trimethylpropane - 1, 2 -diol
0%
2 - methylpentane - 2, 4 - diol
0%
1, 3, 3 -trimethylpropane - 1, 3 -dial

2-methyl pentane-1-ol is a:

0%
$1^{0}$ alcohol
0%
$2^{0}$ alcohol
0%
$3^{0}$ alcohol
0%
none of the above

Explanation

As shown in the structure the

$-OH$ functional group is attached to 1st degree carbon, hence it is

$1^0$ alcohol.

$CH_3-CH_2-\overset{OH}{\overset{|}{C}H}-CH_2-CH_2-\underset{CH_3}{\underset{|}{C}H}-CH_2-CH_3$

The IUPAC name of the compound is:

0%
6-methyl-3-hexanol
0%
6-Methyl-3-octanol
0%
1-methyl-4-hexanol
0%
4-methyl-1-octanol

Explanation

The parent chain will be of 8 carbons. The funtional group is

$-OH$ , so the name of the parent chain will be octanol.

The numbering will start from left side,

$-OH$ will get number 3 and

$-CH_3$ will get number 6.

Therefore, IUPAC name is 6-Methyl-3-octanol.

Option B is correct.

Which of the following is a tertiary alcohol?

0%
$(CH_{3})_{2}CHCH_{2}OH$
0%
$CH_{3}CH_{2}CH_{2}CH_{2}OH$
0%
$CH_{3}CH_{2}CH_{2}OH$
0%
$(CH_{3})_{3}COH$

Explanation

Hint: The degree of alcohol is determined by the number of carbon atoms attached to the carbon having the alcohol functional group.

Explanation

Step 1: Identify the degree of alcohol in each option.

$Isobutyl alcohol$ - primary

It is primary alcohol as the $OH$ is attached to the $C$ which is attached to one more carbon.

$n-butyl alcohol$ - primary

It is primary alcohol as the $OH$ is attached to the $C$ which is attached to one more carbon.

$n-propyl alcohol$ - primary

It is primary alcohol as the $OH$ is attached to the $C$ which is attached to one more carbon.

$tert-butyl alcohol$ - tertiary

tert-butyl alcohol is the tertiary carbon as $OH$ attached to that $C$ which is attached to 3 different carbons.
2-methyl-2-propanol is the IUPAC name for tert-butyl alcohol.

Correct Option:

$D$

The IUPAC name is:

0%
2, 4 -dimethyl pentan-2-ol
0%
2, 4 - dimethyl pentan-4-ol
0%
2, 2 - dimethyl butan-2-ol
0%
butan-2-ol

The IUPAC name of

$(CH_{3})_{2}CHCH_{2}CH_{2}OH$ is:

0%
3-Methyl butan-1-ol
0%
2-methyl pentan-1-ol
0%
3-ethyl pentan-1-ol
0%
2-ethyl butan-1-ol

Explanation

$H-\overset{\overset{4}{CH_3}}{\overset{|}{\underset{CH_3}{\underset{|}{C_3}}}}-\underset{2}{CH}-CH_2-OH$

3-Methyl butan-1-ol

Option A is correct.

IUPAC name of the following compound is:

$CH_{3}-CH(C_{2}H_{5})-CH_{2}-CH(OH)-CH_{3}$

0%
4-Methyl -2- hexanol
0%
4-ethyl -2- pentanol
0%
2-ethyl -2- pentanol
0%
3-methylhexanol

Explanation

IUPAC name of

$CH_{3}-CH(C_{2}H_{5})-CH_{2}-CH(OH)-CH_{3}$ is 4-methyl -2- hexanol.
The parent compound contains a 6 C atom chain and a hydroxyl group. Hence, it is a derivative of hexanol. The hydroxyl group is present on the second C atom and methyl substituent is present on the fourth C atom.

Note: The longest chain of C atoms containing the

$-OH$ group is selected as the parent alkane and numbered from the end closer to

$-OH$ group.

Hence, the correct option is

$\text{A}$

$CH_2OH$

$|$

$CH_2OH$ is a :

0%
$1^{0}$ alcohol
0%
$2^{0}$ alcohol
0%
$3^{0}$ alcohol
0%
carbinol

Explanation

$-OH$ group is attached to

$1^{o}$ carbon atom.

$1^o CH_2-OH$

$|$

$1^o CH_2-OH$

Option A is correct.

The gas liberated during fermentation is:

0%
alcoholic vapours
0%
$H_2$
0%
$CO$
0%
$CO_{2}$

Explanation

The gas liberated is

$CO_2$ .

Option D is correct.

Which of the following soluble in water?

0%
$CH_{3}OH$
0%
$CHCl_{3}$
0%
$CCl_{4}$
0%
$CS_{2}$

Explanation

Only

$CH_3OH$ has hydrogen connected to an electronegative element, this enables it for hydrogen bonding with water. Others have

$Cl$ connected to carbon, which makes the bond less polar, disqualifies

$Cl$ from showing any H bonding, (moreover, only a few cases are observed where

$Cl$ shows H Bonding).

Thus, restricting them to a weak polar or Van der Waal's interactions.

Enzymes involved used in the fermentation of cane sugar to alcohol?

0%
Diastase, invertase
0%
Invertase, zymase
0%
Diastase, zymase
0%
Maltase, zymase

Explanation

Invertase and zymase enzymes are used in the fermentation of cane sugar to alcohol.

Invertase converts sucrose present in the sample to glucose and fructose, while zymase converts it finally to ethanol and

$CO_2$ .

Option B is correct.

$C_{2}H_{5}OH\rightarrow C_{2}H_{5}Br$

In this conversion, the reagent is:

0%
$HBr$
0%
$Br_{2} + CCl_{4}$
0%
$Br_{2}$ water
0%
$KBr$

Explanation

To convert

$C_2H_5OH$ to

$C_2H_5Br$ ,

$HBr$ is used. Nucleophilic substitution reaction occurs in this reaction.

The correct option is 'A'.

Which group can be completely displaced by a halogen group?

0%
Hydroxyl group(OH)
0%
Aldehyde $(-CHO)$ group
0%
Nitro $(-NO_{2})$ group
0%
Keto $(C=O)$ group

Explanation

Hydroxyl group (-OH) present in an alcohol or an acid can be completely displaced by a halogen atom.

$2C_2H_5OH+PCl_3 \rightarrow 2C_2H_5Cl + H_3PCl_3$

The IUPAC name of sec. butyl alcohol is:

0%
1-butanol
0%
2-butanol
0%
2-methyl-l-butanol
0%
2-methyl-2-butanol

Explanation

The IUPAC name of the sec. butyl acohol is 2-butanol because the longest continue chain contain 4-carbon as well as the

$-OH$ group is attached to 2nd carbon , hence the name is 2-butanol.

Which alcohol is prepared from water gas?

0%
$C_{2}H_{5}OH$
0%
$CH_{3}CH_{2}CH_{2}OH$
0%
$CH_{3}OH$
0%
$C_{6}H_{5}CH_{2}OH$

Explanation

Water gas is an equimolar mixture of

$CO$ and

$H_2$ .

A mixture of zinc oxide and chromium oxide

$(ZnO-Cr_{2}O_{3})$ converts water gas in methanol

$(CH_3OH)$ .

The temperature of the reaction is 573K.

$CO +2 H_2 \xrightarrow {ZnO-Cr_{2}O_{3}} CH_3OH$

Option C is correct.

Which one of the following reagents can not cause

$C- O$ bond fission in alcohols?

0%
$HCl /ZnCl_{2}$
0%
$PCl_{5}$
0%
$SOCl_{2}$
0%
Sodium

Explanation

Unlike other reagents which promote a substitution reaction, Na only abstracts the proton. Hence, no

$C-O$ bond fission takes place.

An example for a polyhydric alcohol is:

0%
methyl alcohol
0%
neopentyl alcohol
0%
sec butyl alcohol
0%
mannitol

The structure of compound P is :

The correct acidic order of following is:

0%
I > II > III
0%
III > I > II
0%
II > III > I
0%
I > III > II

Explanation

Acidic strength depends on the stability of anion when acidic

$H$ atom is removed because if anion is stable it will readily lose

$H^+$ .

$-NO_2$ is an electron withdrawing group which stabilizes the anion.

$-CH_3$ is an electron donating group which destabilizes the anion.

Therefore the correct option will be B.

The most suitable method of separation of equal (

$1:1$ ) mixture of ortho-nitrophenol and para-nitrophenol is:

0%
Crystallisation
0%
Distillation
0%
Sublimation
0%
Chromatography

On reacting with grignard reagent acetone gives?

0%
$1^0 alcohol$
0%
$2^0 alcohol$
0%
$3^0 alcohol$
0%
None of these

Phenol reacts with bromine water in carbon disulphate at low temperature to give

0%
o-bromophenol
0%
o- and p-bromophenols
0%
p-bromophenol
0%
2, 4, 6-tribromophenol

Explanation

$-OH$ is ring activating group because it is electron-donating so it is ortho and para director with compound

$o-$ and

$p-bromophenols$

The compound in which hydrogen bonding is not possible is:

0%
$CH_{3}OCH_{3}$
0%
$CH_{3}CH_{2}OH$
0%
$H_{2}O$
0%
$CH_{3}COOH$

Hydrolysis of an ester gives acid A and alcohol B. The acid reduces Fehlings solution. Oxidation of alcohol B gives acid A. The ester is:

0%
Mehthyl formate
0%
Ethyl formate
0%
Methyl acetate
0%
Ethyl acetate

Explanation

A
A:- HCOOH
B:-

$CH_{3}OH$
Therefore the ester is

$HCOOCH_{3}$

CBSE Questions for Class 12 Engineering Chemistry Alcohols,Phenols And Ethers Quiz 1 - 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 1 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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