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

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

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:

$D$

An example for a polyhydric alcohol is:

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

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

$\text{A}$

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 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%
$CH_{3}COOH$

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.

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

$HCOOCH_{3}$

Which of the following is used as freezing mixture?

0%
Mixture of ether and liquid $CO_{2}$
0%
Mixture of ether and ethyl alcohol
0%
Mixture of ether and dry ice
0%
Mixture of ethyl alcohol and dry ice

The IUPAC name of

$CH_{3} - C(CH_{3}) (OH) CH_{2} - CH (CH_{3})_2$ is:

0%
2,4-Dimethylpentan-2-ol
0%
2,4-dimethylpentan-4-ol
0%
2,2-dimethylbutane
0%
2,4,4-trimethylbutan-2-ol

Explanation

In the longest carbon chain,

$-OH$ group is closest to the left-hand side, therefore we number from the left-hand side. There is 5 carbon in the longest chain, in second and fourth positions methyl group is attached and in second position there is an alcoholic group. Thus IUPAC name is 2,4-Dimethyl pentan-2-ol.

Identify the final product in the following sequence of reactions.

0%
$CH_3CH(CH_{(c)})COOH$
0%
$CH_3CH_2CH_2CH_2OH$
0%
$CH_3CH_2CH_2COOH$
0%
$CH_3CH_2COOH$

Trivial name of methanol is _________ .

0%
wood spirit
0%
methyl alcohol
0%
wood alcohol
0%
all of the above

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

$p-bromophenols$

What is the IUPAC Name of t-butyl alcohol?

0%
Butan-2-ol
0%
2-Methyl-propan-2-ol
0%
Butan-1-ol
0%
Propan-2-ol

The major product obtained in the reaction $C{H_3}C{H_2}C \equiv CH\xrightarrow[{{H_2}O}]{{H{g^{ 2 + }},{H_3}{O^ + }}}$ is:

0%
$CH_3CH_2CH_2CH(OH)_2$
0%
$CH_3CH_2CH = CHOH$
0%
$CH_3CH_2CH_2CHO$
0%
$CH_3CH_2COCH_3$

Identify E :

IUPAC name of the compound

$CH_3-CH(CH_2CH_3)-CH_2-CH(OH)-CH_3$ is:

0%
4-Ethyl-2-pentanol
0%
4-Methyl-2-hexanol
0%
2-Ethyl-2-pentanol
0%
3-Ethyl-2-hexanol

The name of the compound :

0%
Pentan-1-ol
0%
Pentan-2-ol
0%
Pentan-4-ol
0%
None of these

On treatment with dilute

$H_{2}SO_{4}$ , ethene forms ethanol.

Is the above statement true?

0%
True
0%
False

Which of the following is used in the manufacture of ethanol?

0%
$C_{2} H_{6}$
0%
$C_{2}H_{2}$
0%
$CH_{4}$
0%
$C_{2}H_{4}$

Explanation

Alcohols can be prepared by the hydration of alkenes. Therefore, ethanol(

$C_{2}H_{5}OH$ ) can be prepared from the corresponding alkene;

$C_{2}H_{4}$ as:

$CH_{2}=CH_{2} + H_{2}O \overset{H^{+}}{\rightarrow} CH_{3}CH_{2}OH$

Decolorization of

$KMnO_4$ solution takes place when it reacts with ethene. The final product of this reaction which is used as anti-freeze is:

0%
ethylene glycol
0%
propylene glycol
0%
ethanol
0%
propanol

Explanation

Hydroxylation of ethene takes place in the presence of cold

$KMnO_{4}$ to form Ethylene glycol. The reaction can be written as:

$CH_{2}= CH_{2} \xrightarrow[25^{\circ}C]{MnO_{4}^{2-}} HOCH_{2}-CH_{2}OH + MnO_{2}$

0%
Both assertion and reason are CORRECT and reason is the CORRECT explanation of the assertion.
0%
Both assertion and reason are CORRECT, but reason is NOT THE CORRECT explanation of the assertion.
0%
Assertion is CORRECT, but reason is INCORRECT.
0%
Assertion is INCORRECT, but reason is CORRECT.

Alcohols containing only up to ___________ carbon atoms are completely miscible with water.

0%
three
0%
four
0%
five
0%
two

By treating alkylchloride with aq.

$KOH$ produce:

0%
alcohol
0%
$KCl$
0%
alkane
0%
Both $A$ and $B$

Explanation

In the presence of aqueous

$KOH$ , alkyl chloride gives the substitution product as alcohol. The reaction can be written as:

$RCl + KOH(aq) \rightarrow ROH + KCl$

__________ is prepared on a large scale from water gas.

0%
Methanol
0%
Ethanol
0%
Methane
0%
Ethane

Explanation

Methanol is prepared on a large scale from water gas.
Water gas is an equimolar mixture of

$\displaystyle CO$ and

$\displaystyle H_2$ . Water gas is heated with hydrogen (2:1 ratio) in presence of catalyst to form methanol.

$\displaystyle \underset { \displaystyle \text { synthetic gas } }{ \underset { \displaystyle \text { Water gas } }{ CO+H_2 } }+H_2 \xrightarrow [ 300^oC \text { under pressure}]{Cr_2O_3 -ZnO} CH_3-OH$

In the production of alcohol _____ are used.

0%
alkene
0%
alkynes
0%
alkanes
0%
both (a) and (b) above

Explanation

Hydration of alkene:

In this process, concentrated

$H_2SO_4$ is treated with alkene at room temperature to produce alkyl hydrogen sulphide. After addition of

$H_2O$ and heated it, alkyl hydrogen sulphate hydrolysis to alcohol.

IUPAC name is 3-phenylpropan-1-ol.

0%
True
0%
False

$CH_3CH = CHCH_3 \xrightarrow [(i)A]{(ii) B} CH_3-\underset{OH}{\underset{|}{C}H} - \overset{OH}{\overset{|}{C}H} - CH_3$ ,

$A\& B$ are

How would you get racemic mixture of 1, 2-butane diol from cis 2- butene?

0%
$\displaystyle A=KMnO_{4}/OH^{-}, B=H_{2}O$
0%
$\displaystyle A=CF_{3}/CO_{3}H, B=H_{2}O$
0%
$\displaystyle A=OsO_{4}/OH^{-}, B=H_{2}O$
0%
$\displaystyle A=O_{3}/H_{2}O, B=Ph_{3}P$

Dehydration of alcohols take place more rapidly with

$\displaystyle POCl_{3}$ than with

$\displaystyle H_{2}SO_{4}.$

Select the correct statements(s) about the above dehydration reaction :

0%
It does not involve carbocation.
0%
It involves $\displaystyle R-OPOCl_{2}$ with $\displaystyle -OPOCl_{2}$ as a better leaving group.
0%
It involves $E2$ mechanism as pyridine base abstracts proton from the adjacent carbon as the same time at which $\displaystyle -OPOCl_{2}$ is leaving.
0%
It is $E1$ reaction without formation of carbocation.

Explanation

1. Nucleophilic attack and deprotonation.

$E-2\quad \beta -$ Hydrogen elimination.

Hence the correct options are A, B and C.

1157025_137125_ans_b7f6c59cb8c0473b91051abea3197990.png

Ethanol on reaction with acetic anhydride gives:

0%
acetic ester
0%
formic ester
0%
ethanoic acid
0%
ethyl acetate and ethanoic acid

Explanation

The reaction of acetic anhydride with ethanol yields ethyl acetate and ethanoic acid.

$(CH_3CO)_2O + CH_3CH_2OH \rightarrow CH_3COOCH_2CH_3 + CH_3COOH$

IUPAC name is 1-phenyl propan-3-ol.

Is the above statement correct?

0%
True
0%
False

Compound which gives alcohol on reduction is/are:

Explanation

Acid chlorides, Epoxides and Acid anhydrides on reduction with

$LiAlH_4$ forms alcohol as end product. Only Acid amides when reduced form amines.

$CH_3CONH_2\overset{LiAlH_4}\rightarrow CH_3CH_2NH_2$
Therefore, Options A, C and D are correct.

What is IUPAC name of the following compound?
C

$H_3$ - C

$H_2$ - C

$H_2$ - OH

0%
Propane-1-ol
0%
Propane-2-ol
0%
Ethane-1-ol
0%
Ethane-2-ol

Explanation

${ CH }_{ 3 }^{ 3 }{ CH }_{ 2 }^{ 2 }{ CH }_{ 2 }^{ 1 }{ OH }$
Carbon with functional group gets position 1. Therefore, IUPAC name is Propan-1-ol.

$\displaystyle EtO-CH_2-C_6H_5+HI\rightarrow D$

Name the product D in the given reaction.

0%
Diethyl ether
0%
Benzyl iodide
0%
ethane
0%
none of these

Which of the following has a higher boiling point?

0%
Butanol
0%
Butanal
0%
Both of them have similar boiling point
0%
Cannot say anything

The following structures represents cyclopentyl methyl carbinol.

State whether the given statement is true or false:

0%
True
0%
False

Explanation

The IUPAC name of the above compound is 1-Cyclopentylethanol but the group (

$CH_3OH$ ) is commonly called as carbinol therefore the the above compound can also be named as cyclopentyl methyl carbinol.

Number of 1,2-hydride shifts occur while converting

$1-$ methylcyclohexanol to

$1-$ bromo

$-1-$ methylcyclohexane is:

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

The decreasing basic character of the following is:
(I)

$Ph{O}^{\ominus}$ (II)

$C{H}_{3}{S}^{\ominus}$ (III)

$\overset { \ominus }{ O }H$ (IV)

$Me{O}^{\ominus}$

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

Explanation

The increasing acidic character is

$PhOH > C{H}_{3}SH > MeOH > {H}_{2}O$ .
Phenols are most acidic (due to resonance stabilization in phenoxide ion) and water is least acidic.
The decreasing basic character is

$Ph{O}^{\ominus} < C{H}_{3}{S}^{\ominus} < Me{O}^{\ominus} < \overset { \ominus }{ O }H$ . As the acidic character increases, the basic character of conjugate base decreases. Since phenols are most acidic, phenoxide ions are least basic.

Among the three isomers of the nitrophenol, the one that is least soluble in water is:

0%
m-nitrophenol
0%
o-nitrophenol
0%
p-nitrophenol
0%
none of these

The IUPAC name of the following compound is:

0%
3-cyclohexylpentan-3-ol
0%
3-cyclohexyl,3-hydroxypentane
0%
3-hydroxypentyl cyclohexane
0%
none of the above

The IUPAC name of the following compound is:

0%
1-Phenylpropan-2-ol
0%
3-Phenylpropan-2-ol
0%
2-Hydroxypropylbenzene
0%
None of these

Explanation

The IUPAC name of the given compound is 1-Phenylpropan-2-ol.

The parent hydrocarbon contains 3 C atoms and is named as propane.

A hydroxyl group (

$-OH$ ) is present at second C atom.

Hence, the last alphabet e of propane is replaced with the suffix -2-ol.

A phenyl group is present as a substituent at first carbon atom.

The

$-OH$ group in alcohols can be replaced by:

0%
$\displaystyle SOCl_{2}$
0%
$\displaystyle Cl_{2}$
0%
$\displaystyle PCl_{5}$
0%
$\displaystyle P+Cl_{2}$

Explanation

$2P + 3Cl_2 \rightarrow 2PCl_3$ and

$PCl_3$ can also replace the

$-OH$ group in alcohol.

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

0:0:1

Practice Class 12 Medical Chemistry Quiz Questions and Answers

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