MCQ Questions for Class 12 Engineering Chemistry Alcohols,Phenols And Ethers Quiz 5

Among the following compounds, the one which can be dehydrated most easily is:

0%
$CH_{ 3 }-CH_{ 2 }-\overset { \overset { CH_{ 3 } }{ | } }{ \underset { \underset { OH }{ | } }{ C } }-CH_{ 2 }CH_{ 3 }$
0%
$CH_3-CH_2-CH_2-\underset { \underset { OH }{ | } }{ C } H-CH_3$
0%
$CH_3-CH_2-CH_2-CH_2-CH_2-OH$
0%
$(CH_3)_2CH-CH_2-CH(OH)-CH_3$

Explanation

The compound which can be easily dehydrated will form most stable carbocation after removal of

$H_{ 2 }O$ molecule.

Since stability of carbocation follows 3 degree> 2 degree> 1 degree,

so, compound

$A$ is most easily dehydrated.

Option A is correct.

1036615_771013_ans_ed4cd4dc24ec4078b3376c059ee2b08e.png

product A is:

0%
0%
0%
0%
all

Explanation

The given reaction is an example of dehydration reaction. In this reaction removal of

$H^+$ and

$OH^-$ takes place to form a double bond between the carbon atom. Thus we get a double bond in the ring of the given ketone.
1016027_791690_ans_93ac11e930e44af89a52b22c8df71c1c.png

1016027_791690_ans_93ac11e930e44af89a52b22c8df71c1c.png

Which of the following phenols has lowest solubility in water?

0%
Phenol
0%
o-Nitrophenol
0%
m-Nitrophenol
0%
p-Nitrophenol

Explanation

$O-$ nitrophenol undergoes creation because of intra-moleculer hydrogen bonding. So, it is least soluble.
1174666_776124_ans_702a802666e44d3c81599b2a6935a1ea.PNG

1174666_776124_ans_702a802666e44d3c81599b2a6935a1ea.PNG

An Alcohol X pass over Cu at 300C, then produced compound decolourised Bromine liquid, X is:

Explanation

Since the compound formed finally decolourises Bromine water, so it must be alkene. Now

$3$ degree alcohol were dehydrated with heated Cu forms alkene. So,

$X$ is a

$3$ degree alcohol.
1037654_856007_ans_91c86a5b2a0243d58256facf1362549b.PNG

(A)

$CH_3-\underset{\underset{\,\,\,\,CH_3}{|}}CH-CH_2-CH_2-OH$

(B)

$CH_3-\overset { \underset { | }{ \, \, \, \, \, OH } }{ \underset { \overset { | }{ \, \, \, \, \, CH_{ 3 } } }{ C } }-CH_2-CH_3$

(C)

$CH_3\overset{\overset{CH_3}{|\,\,\,\,\,}}{CH}-\overset{\overset{OH}{|\,\,\,}}{CH}-CH_3$

The order of given alkanols

$A,B,C$ in their reactivity towards acid catalysed dehydration is:

0%
$A> B> C$
0%
$B> A> C$
0%
$B> C> A$
0%
$C> B> A$

Explanation

Carbocation formed during reaction are referred to the image above:

Therefore stability of carbocations have the order

$3 degree> 2degree> 1 degree$ and

$CH_{ 3 }$ is

$+$ group so, the order of stability is

$: B>C>A$

(B)

$CH_3-\overset { \underset { | }{ \, \, \, \, OH } }{ \underset { \overset { | }{ \, \, \, \, CH_{ 3 } } }{ C } }-CH_2-CH_3 \rightarrow CH_3-\overset { \underset { | }{ \, \oplus} }{ \underset { \overset { | }{ \, \, \, \, \, CH_{ 3 } } }{ C } }-CH_2-CH_3$

$(2^o)$

(C)

$CH_3-\overset{\overset{\,\,\,CH_3}{|}}{C}H-\overset{\overset{OH}{|}}{C}H-CH_3 \rightarrow CH_3-\underset{\underset{\,\,CH_3}{|}}{C}-\underset{\oplus}{Ch}-CH_3\underset{(2^o)}{CH_3}$

1036711_781330_ans_dcdbb0a50bf34d399e6df9fecfb47ce9.png

Dehydration of the following in increasing order is:

0%
$I < II < III < IV$
0%
$II < III < IV < I$
0%
$I < III < II < IV$
0%
$I < IV < II < III$

Explanation

Phenol

$(I)$ is stable aromatic compound. So it is most stable. The diene

$(IV)$ undergoes quick dehydration because it will form stable phenol. The alkene

$(III)$ undergoes dehydration. less readily because it gives a more stable diene. Alcohol

$(II)$ undergoes much lesser dehydration to form alkene. . So, order is:

$4> 3> 2> 1.$

Diols (I - IV) which react with

$CrO_{3}$ in aqueous

$H_{2}SO_{4}$ and yield products that readily undergo decarboxylation on heating are:

0%
I and II
0%
II and III
0%
II and IV
0%
I and IV

Which of the following statements is incorrect regarding preparation of alcohols?

0%
Aldehydes and Ketones on catalytic reduction by [H] give primary alcohol and secondary alcohol respectively
0%
Ketones on reaction with Grignard’s reagent gives tertiary alcohol
0%
Aldehydes on reaction with Grignard’s reagent gives primary or secondary alcohol
0%
Alkyl halides on reaction alcoholic potash gives alcohol

Rate of dehydration when given compound is treated with conc,

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

0%
P > Q > R > S
0%
Q > P > R > S
0%
R > Q > P > S
0%
R > Q > S > P

The correct decreasing order of dehydration of following alcohols with conc.

$H_2SO_4$ is:

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

Methylphenyl ether can be obtained by reacting :

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

Correct representation of 3- methylpent-3-en-2-ol is :

Which of the following compounds is stable?

0%
$CH_3CH(OH)_2$
0%
$(CH_3)_2CH(OH)_2$
0%
$CCl_3CH(OH)_2$
0%
None of these

Explanation

$\bf{Hint} -$ Electron withdrawing groups stabilize the vicinal diols.

$\bf{Explanation-}$

$\bullet$ Structures of compound are as follow:

$CH_3 - CH-(OH)_2 \\ (CH_3)_2- CH-(OH)_2 \\ CCl_3 - CH-(OH)_2$

$\bullet$

$2$

$O-H$ groups on the same

$C$ is unstable and they can easily undergo dehydration but when

$C$ having

$2 \ O-H$ groups attached to the electron-withdrawing groups then the

$O-H$ group will be more stabilized.

$\bullet$ As

$CH_3$ has

$+I$ effect so it will increase the

$\delta^-$ charge at the next carbon bonded with it and there is the formation of a weak bond. so this compound will be unstable.

$\bullet$

$Cl$ has

$-I$ effect. So,

$3 \ Cl$ withdraws electrons from the

$C$ containing

$2 \ O-H$ group and that makes vicinal diols stable

$\bf{Conclusion-}$ Hence, option

$C$ is correct.

Which of the following is best method to prepare phenyl-t-butyl ether?

0%
$(CH_{3})_{3}C - O^{-}Na^{+} + C_{6}H_{5}Br$
0%
$C_{6}H_{5}ONa + (CH_{3})_{3}CCl$
0%
$(CH_{3})_{2}C = CH_{2}\xrightarrow[C_6H_5OH]{Hg(OCOCH_3)_{2}} \overset{NaBH_4}{\rightarrow}$
0%
None of these

Product of the above reaction is:

A hydrocarbon

$'A' (C_{3}H_{8})$ on treatment with chlorine in presence of sunlight yielded compound

$'B'$ as the major product. Reaction of

$'B'$ with aqueous

$KOH$ gave

$'C'$ which on treatment with concentrated

$H_{2}SO_{4}$ yielded

$'D'$ . Hydrogenation of

$'D'$ gave back

$'A'$ . The sequence of reactions involved in the above conversion is:

0%
Substitution, Substitution, Addition, Dehydration
0%
Substitution, Substitution, Dehydration, Addition
0%
Substitution, Dehydration, Addition, Addition
0%
Addition, Substitution, Dehydration, Substitution

Explanation

(i)

$C_{3}H_{8} + Cl_{2}\xrightarrow {hv} C_{3}H_{7}Cl + HCl [Substitution]$
(ii)

$C_{3}H_{7}Cl + Aq\ KOH \xrightarrow {hv} C_{3}H_{7}OH + KCl [Substitution]$
(iii)

$C_{3}H_{7}OH \xrightarrow {dil\ H_{2}SO_{4}} C_{3}H_{6} + H_{2}O [Dehydration]$
(iv)

$C_{3}H_{6} + H_{2} \rightarrow C_{3}H_{8} [Addition]$ .

Ethyl alcohol obtained by fermentation of starch or molasses is called wash, and what is it’s purity?

0%
15%
0%
45%
0%
55%
0%
35%

The correct IUPAC name of the compound is ?

0%
2-ethyl-1-chlorocyclohexanol
0%
4-chloro-5-ethylcyclhexanol
0%
4-hydroxy-2-ethyl-1-chlorocycloheane
0%
4chloro-3-ethylcyclohexanol

Which of the following alcohol will have the fastest rate of dehydration ?

Explanation

will have fastest rate of dehydration as the

$\alpha -H$ is more acidic and the product formed is resonance stabilized.
1174103_880696_ans_bfeb9855b44d4b60b41077f78515e16c.png

An alkene

$CH_3CH = CH_2$ is treated with

$B_2H_6$ in presence of

$H_2O_2$ . The final product formed is:

0%
$CH_3CH_2CHO$
0%
$CH_3CH(OH)CH_3$
0%
$CH_3CH_2CH_2OH$
0%
$(CH_3CH_2CH_2)_3B$

Explanation

Hydroboration-oxidation reaction follows anti-Markovnikov's addition of

$H-OH$ across

$C=C$ to give alcohol.

Thus an alkene

$CH_3CH=CH_2$ when treated with

$B_2H_6$ in presence of

$H_2O_2$ will yield the final product as

$CH_3CH_2CH_2OH$

Which of the following statements is not correctly showing the trend of the properties mentioned ?

0%
$CH_3CH_2OH > CH_3CH_2CH_2OH > \underset {(Solubility)}{CH_3CH_2CH_2CH_2OH}$
0%
$CH_3CH_2OH > CH_3CH_2CH_2OH > \underset {(Boiling point)}{CH_3CH_2CH_2CH_2OH}$
0%
$CH_3CH_2CH_2OH > CH_3{\underset{CH_3}{\underset{|}{CH}}}-CH_2OH >\underset{Boiling \, point}{CH_3-\underset{CH_3}{\underset{|}{\overset{CH_3}{\overset{|}C}}}-OH}$
0%
$CH_3-{\underset{CH_3}{\underset{|}{\overset{CH_3}{\overset{|}{C}}}}}-OH$ < $CH_3{\underset{CH_3}{\underset{|}{CH}}}-CH_2OH$ < $CH_3CH_2CH_CH_2OH$

Explanation

Boiling point and solubility of alcohols is determined by the ease and extent of H-bonding among the alcohol molecules. More is the hydrogen bonding higher is the boiling point and solubility in water. Thus, tertiary alcohols being highly hindered will have least hydrogen bonding and thus low boiling point/solubility as compared to less hindered secondary alcohol.

Therefore the order of boiling point is primary alcohol>secondary alcohol>tertiary alcohol as:

$CH_3CH_2CH_2OH>CH_3CH(CH_3)CH_2OH>CH_3C(CH_3)_2OH$

In case of linear versus branched alcohol, linear alcohol has higher melting/boiling points due to better stacking and surface area contact.

But in case of highly branched vs. branched molecules more sphere-like structure of the molecule better is the stacking thus higher will be the melting point. Thus order of melting point will be

$CH_3CH(CH_3)CH_2OH<CH_3C(CH_3)_2OH<CH_3CH_2CH_2CH_2OH$

A and B. Solubility and boiling point of alcohols depends on the chain length. Longer is the chain length, lesser is the solubility/boiling point as the extent of H-bonding decreases. Also, the hydrophobic nature of the hydrocarbon chain reduces the solubility.

thus order of solubility and boiling point will be:

$CH_3CH_2OH>CH_3CH_2CH_2OH>CH_3CH_2CH_2CH_2OH$

A compound

$C_6H_{14}O_2$ has two tertiary alcoholic groups. The IUPAC name of compound is:

0%
2,3 - dimethyl - 1,2 - butanediol
0%
3,3 - dimethyl - 1,2 - butanediol
0%
2,3 - dimethyl - 2,3 - butanediol
0%
2 - methyl - 2,3 - pentanediol

Explanation

$C_6H_{14}O_2$ with two tertiaryy alcoholic groups possible has the following structure and using the IUPAC nomenclature its name is:

Steps to name an organic molecule:

1. Identify the fuctional group: OH in the given molecule. two Alcoholic group i.e suffix "diol" will come.

2. Identify parent chain: The longest carbon chain that contains the functional group. Here its 4 C chain i.e butane

3. Assign locant number: lowest number to the main functional group and side chains i.e. 2,3-diol and 2,3-dimethyl

4. Identify side chains/groups: two methyl groups.

IUPAC name is: 2,3-dimethyl-2,3-butanediol or 2,3-dimethylbutane-2,3-diol.

Strength of acidity is in order :

0%
$II > I$
0%
$I > II$
0%
$I=II$
0%
cannot be determined

Explanation

Solution:-

As we know that the strength of acidity inversely depends upon the

$+I$ effect.

Thus, due to

$+I$ effect of methyl group

$\left( -C{H}_{3} \right)$ ,

$I$ is more acidic than

$II$ .

Hence the correct order of strength of acidity is-

$I > II$

The best method to prepare 3-methylbutan-2-ol from 3-methylbut-1-ene is:

0%
addition of water in presence of dil. $H_2SO_4$
0%
addition of $HCI$ followed by reaction with dil. $NaOH$
0%
hydroboration - oxidation reaction
0%
Reimer - Tiemann reaction.

Explanation

To prepare 3-methylbutan-2-ol from 3-methylbut-1-ene shouls involve Markovnikov's addition of

$H-OH$ across double bond

A. Addition of water in presence of dil.

$H_2SO_4$ follows Markovnikov's addition of

$H-OH$ across the alkene

B. Addition of

$HCI$ followed by reaction with dil.

$NaOH$ to an alkene can give substitution and elimination product after alkyl halide formation

C. Hydroboration-oxidation reaction follows Anti-Markovnikov's addition of

$H-OH$

D.Reimer - Tiemann reaction.is a chemical reaction used for the ortho-formylation of phenols

Thus the best method to prepare 3-methylbutan-2-ol from 3-methylbut-1-ene is the addition of water in presence of dil.

$H_2SO_4$ .

Primary, secondary and tertiary alcohols can be distinguish by:

0%
Baeyer's reagent
0%
Fehling's solution
0%
Sulphuric acid
0%
Lucas reagent

The decreasing order of boiling points of the following alcohols is?

0%
3-methylbutan-2-ol > 2-methylbutan-2-ol > pentan-1-ol
0%
pentan-1-ol > 3-methylbutan-2-ol > 2-methylbutan-2-ol
0%
2-methylbutan-2-ol > 3-methylbutan-2-ol > pentan-1-ol
0%
2-methylbutan-2-ol > pentan-1-ol >3-methylbutan-2-ol

Explanation

3-methylbutan-2-ol is a secondary alcohol

$CH_3CH(CH_3)CH(OH)CH_3$

2-methylbutan-2-ol is a tertiary alcohol

$CH_3CH_2C(CH_3)(OH)CH_3$

pentan-1-ol is a primary alcohol

$CH_3CH_2CH_2CH_2CH_2OH$

Boiling point of alcohols is determined by the ease and extent of H-bonding among the alcohol molecules. Tertiary alcohols being highly hindered will have least hydrogen bonding and thus low boiling point as compared to less hindered secondary alcohol.

Therefore the order of boiling point is primary alcohol>secondary alcohol>tertiary alcohol as:

pentan-1-ol > 3-methylbutan-2-ol > 2-methylbutan-2-ol.

Alcohols of low molecular weight are:

0%
soluble in water
0%
soluble in water on heating
0%
insoluble in water
0%
insoluble in all solvents.

Which of the following synthesis given

$3$ -methyl-

$1$ -hexanol?

0%
$2-Bromohexane \xrightarrow[dry \, ether]{Mg} \,\, \xrightarrow[(ii) H_3 O^+]{(i) HCHO}$
0%
0%
$3-Bromopentane \xrightarrow[dry \, ether]{Mg} \,\, \xrightarrow[(ii) H_3O^+]{(i) CH_3 CHO}$
0%
$1 - Bromobutane \xrightarrow[dry \, ether]{Mg} \,\, \xrightarrow[(ii) H_3 O^+]{(i) CH_3 COCH_3}$

IUPAC name of

$CH_3CH_2CH_2{\underset {CH_2OH}{\underset{|}{CH}}}-CH_2CH_3$ is:

0%
3 -propylbutan-1-ol
0%
2-Ethylpentan - 1 - ol
0%
3- methyl hydroxyhexane
0%
2- ethyl -2 - propyl ethanol

Vapours of an alcohol X when passed over hot reduced copper, produce an alkene, the alcohol is:

0%
primary alcohol
0%
secondary alcohol
0%
tertiary alcohol
0%
dihydric alcohol

Explanation

Given:

$X(ROH) +Cu(hot) \rightarrow alkene$

primary alcohol passed over Cu at

$300^{\circ}C$ is dehydrogenated to aldehydes and hydrogen gas liberates

secondary alcohols are dehydrogenated to ketones with liberation of hydrogen gas

tertiary alcohol is dehydrated to alkene.

This method can therefore be used to distinguish between primary, secondary and tertiary alcohol

thus X should be a tertiary alcohol

Arrange the following alcohols in order of increasing reactivity towards sodium metal.
(i)

$(CH_3)_3C-OH$ (ii)

$(CH_3)2CH - OH$ (iii)

$CH_3CH_2OH$

0%
(iii) < (ii) < (i)
0%
(ii) < (i) < (iii)
0%
(i) < (ii) < (iii)
0%
(iii) < (i) < (ii)

Arrange the following compounds in increasing order of boiling points . Propan-1-ol , butan-1-ol , butan-2-ol, pentan-1-ol.

0%
Propan-1-ol, butan-2-ol,butan-1-ol ,pentan-1-ol
0%
Propan-1-ol, butan-1-ol,butan-2-ol ,pentan-1-ol
0%
Pentan-1-ol , butan-2-ol , butan-1-ol , propan -1-ol
0%
Pentan-1-ol , butan-1-ol , butan-2-ol , propan -1-ol

Explanation

The boiling point increases as the moleculer mass of the alcohol increases. Further, among isomeric alcohols,

$1^0$ alcohols have higher boiling points than

$2^0$ alcohols.

Therefore the b.p.t increases in the order: propan-1-ol < butan-2-ol < butan-1-ol < pentan-1-ol.

Which of the following compounds does not react with

$NaOH$ ?

0%
$CH_3COOH$
0%
$CH_3CONH_2$
0%
$C_6H_5OH$
0%
$CH_3CH_2OH$

Explanation

$CH_3COOH+NaOH \rightarrow CH_3COO^-Na^++H_2O$

It is an acid-base neutralization reaction where salt forms with water.

$CH_3CONH_2+NaOH \rightarrow CH_3COO^-Na^++NH_3$

it is called hydrolysis of amide in an alkaline medium which gives carboxylic acid as the product

$C_6H_5OH+NaOH \rightarrow C_6H_5O^-Na^++H_2O$

It is an acid-base neutralization reaction where salt forms with water.

$CH_3CH_2OH+NaOH \rightarrow No\ reaction$

Alcohols are very weak acids and do not react with

$NaOH$ .

Phenol is more acidic than alcohol and therefore can react with

$NaOH$ to form sodium phenoxide.

Which of the following alcohols will give the most stable carbocation during dehydration?

0%
2-Methyl -1- propanol
0%
2-Methyl -2- propanol
0%
1-Butanol
0%
2-Butanol

Explanation

Alcohol

$+ H_2SO_4 \rightarrow$ Alkene

The dehydration reaction is an elimination reaction that goes via carbocation formation as shown in the figure.

Tertiary carbocations are most stable due to + Inductive effect of alkyl groups, then is secondary and least stable is primary. Therefore the reactivity and stability of carbocation follow the order as

Tertiary alcohol>secondary alcohol>primary alcohol

2-Methyl-1-propanol forms secondary carbocation as:

$(CH_3)_2CHCH_{2}^+$

2-Methyl-2-propanol forms tertiary carbocation as:

$(CH_3)_3C^+$

1-Butanol forms primary carbocation as:

$CH_3CH_2CH_2CH_2^+$

and 2-Butanol forms secondary carbocation as:

$CH_3CH^+CH_2CH_3$

Thus most stable carbocation is formed by 2-Methyl -2- propanol during its dehydration.

$(CH_3)_3C - CH_2OH\xrightarrow[170^0C]{Conc.H_2SO_4}$ X
In the reaction , X is:

0%
$(CH_3)_2C = CHCH_3$
0%
$(CH_3)_{2}C = CH_{2}$
0%
$(CH_3)_2CHCH_2CH_3$
0%
$CH_3-CH_2-\underset{CH_3}{\underset{|}C}=CH_2$

Explanation

$(CH_3)_3C - CH_2OH\xrightarrow[170^0C]{Conc.H_2SO_4}$ X
The reagent used in the above reaction is a dehydrating agent thus will form an alkene via carbocation formation as:

$(CH_3)_3C - CH_2OH \xrightarrow{+H^+} (CH_3)_3C - CH_2\overset {+}OH_2 \xrightarrow{-H_2O} (CH_3)_3C - \overset {+}CH_2 \xrightarrow{1,2-methyl\ shift} (CH_3)_2 \overset {+}C - CH_2CH_3 \xrightarrow{-H^+} (CH_3)_2C=CHCH_3$

thus overall reaction is:

$(CH_3)_3C - CH_2OH\xrightarrow[170^0C] {Conc.H_2SO_4} (CH_3)_2C=CHCH_3$ (X)

Unlike phenol, 2,4-dinitrophenol is soluble in sodium carbonate solution in water because ?

0%
presence of two - $NO_2$ groups in the ring makes 2 , 4 - dinitrophenol a stronger acid than phenol.
0%
presence of two - $NO_2$ groups in the ring makes 2 , 4 - dinitrophenol a weaker acid than phenol.
0%
presence of two $-NO_2$ groups make the hydrogen bonding easier, making 2 , 4-dinitrophenol soluble.
0%
nitro group reacts with $Na_2CO_3$ while - $OH$ group does not.

Explanation

Presence of two electron withdrawing

$-No_2$ groups in the ring makes 2,4-dinitrophenol a stronger acid than phenol. Hence it react with aqueous

$Na_2CO_3$ solution to form sodium salt thus making it soluble in

$Na_2CO_3$ .

Which of the following alcohols is dehydrated most easily with conc.

$H_2SO_4$ ?

0%
$p-O_2NC_6H_4CH(OH)CH_3$
0%
$p-ClC_6H_4CH(OH)CH_3$
0%
$p-CH_3OC_6H_4CH(OH)CH_3$
0%
$C_3H_5CH(OH)CH_3$

Explanation

The dehydration reaction is as:

$Alcohol + H_2SO_4 \rightarrow alkene$

is an elimination reaction that goes via carbocation formation.

Higher is the stability of carbocation more easily it can be dehydrated.

The corresponding carbocations formed by given molecules are shown in the figure.

Due to the --Inductive effect (-I)and --Mesomeric effect (-M) of

$NO_2$ group, it will decrease the electron density on benzene and will, therefore, destabilise the carbocation the most and makes it difficult to dehydrate.

In case of

$p−ClC_6H_4CH(OH)CH_3$ , the Cl has stronger -I effect (due to high electronegativity) than its +M effect (due to a lone pair of electrons) thus will destabilise the carbocation and unfavour dehydration.

In case of

$p−CH_3OC_6H_4CH(OH)CH_3$ , the

$CH_3O$ has the stronger +M effect due to a lone pair of electrons on O than its -I effect and will stabilise the carbocation by increasing the electron density in the benzene ring and will favour dehydration.

$C_3H_5CH(OH)CH_3$ has +I effect that stabilises the carbocation but its impact is lower than the +-mesomeric effect of the methoxy group.

The alcohol that is dehydrated most easily with conc.

$H_2SO_4$ is

$p−CH_3OC_6H_4CH(OH)CH_3$ .

An alcohol X on heating with concentrated

$H_2SO_4$ gives an alkene Y which can show geometrical isomerism.The alcohol X is?

0%
$CH_3CH_2OCH_3$
0%
$CH_3CH_2CHO$
0%
$CH_3CH_2CH_2OH$
0%
$CH_3CH_2CH(OH)CH_3$

Explanation

$CH_3CH_2CH(OH)CH_3 \xrightarrow {H_2SO_4} CH_3-CH=CH-CH_3$

$X$

$Y$

Y show cis and trans isomerism. Hence, option D is correct.

Methyl alcohol is industrially prepared by the action of ?

0%
$CH_3COCH_3$
0%
$CO + H_2$
0%
$CH_3COOH$
0%
$C_2H_5OH$

Explanation

The industrial method of preparation of methanol or methyl alcohol is by the action of

$CO+H_2$ as:

$CO+H_2 \overset {230-400 ^{\circ}C}{\rightarrow} CH_3OH$

Carbon monoxide reacts with hydrogen to produce methanol/methyl alcohol.

This reaction takes place at a temperature of

$230-400 ^{\circ}C$ and at 50-100 atm. In this reaction, the catalyst is can be copper oxide on oxide chromium(III).

Give IUPAC name of the compound given below :

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

0%
2- Chloro - 5 - hydroxyhexane
0%
2 - Hydroxy - 5 - chlorohexane
0%
5 - Chlorohexan - 2 - ol
0%
2 - Chlorohexan - 5 - ol

Explanation

Given molecule:

${ C }H_{ 3 }-\underset { Cl }{ \underset { | }{ { C } } } H-{ C }H_{ 2 }-{ C }H_{ 2 }-\underset { OH }{ \underset { | }{ { C } } } H-{ C }H_{ 3 }$

For IUPAC nomenclature following steps are followed:

1. Identify the fuctional group and their priority: -OH and -Cl are the side groups in the given molecule. OH has higher priority thus the name will have suffix "ol".

2. Identify parent chain: The longest carbon chain that contains the functional group. Here its 6 C chain i.e hexyl

3. Assign locant number: lowest number to the main functional group and side chains i.e. 2-ol

4. Identify side chains/groups: its chloro, and 5-chloro in the given molecule.

$\overset { 6 }{ C } H_{ 3 }-\underset { Cl }{ \underset { | }{ \overset { 5 }{ C } } } H-\overset { 4 }{ C } H_{ 2 }-\overset { 3 }{ C } H_{ 2 }-\underset { OH }{ \underset { | }{ \overset { 2 }{ C } } } H-\overset { 1 }{ C } H$

IUPAC name is: 5-chlorohexan-2-ol

Which compound is the most soluble in water?

0%
Chloroethane
0%
Ethanol
0%
Ethanal
0%
Ethylene glycol

Explanation

Among given compounds, ethylene glycol (

$HO-CH_2-CH_2-OH$ ) is the most soluble in water. Ethylene glycol has two hydroxy groups both of which form hydrogen bonds with water. Greater is the number of hydrogen bonds, greater is the extent of hydrogen bonding and greater is the solubility in water.

Which of the following alcohols is the least soluble in water?

0%
Ethanol
0%
1-Propanol
0%
1-Butanol
0%
1-Pentanol

Explanation

Alcohols are soluble in water due to hydrogen bonding. However, as the length of carbon chain increases, the extent of hydrogen bonding decreases due to bulky chain.
According to this,

$1-pentanol$ will be the least soluble in water, among the other given alcohols.

The order of reactivity of following alcohols with halogen acids is ______
(I)

$CH_3CH_2 - CH_2-OH$

(II)

$CH_3CH_2-(CH_3)CH-OH$

(III)

$CH_3CH_2 - {\underset{CH_3}{\underset{|}{\overset{CH_3}{\overset{|}{C}}}}}-OH$

0%
(I) > (II) > (III)
0%
(III) > (II) > (I)
0%
(II) > (I) > (III)
0%
(I) > (III) > (II)

Explanation

Order of reactivity of alcohol with halogen is in accordance with the stability of carbocation formed. More stable the carbocation, more will be reactivity of alcohol with halogen acid.

The order of stability of carbocation is:

$3^o> 2^o > 1^o$

Option B is correct.

1032434_939302_ans_9efe9692c5564a179d2f831272046dec.png

The general formula for alcohol series is

$C_nH_{2n+1}OH$ . Which is the formula for alcohol contains four carbon?

0%
$C_5H_{11}OH$
0%
$C_4H_{9}OH$
0%
$C_4H_{8}OH$
0%
$C_3H_{8}OH$

Explanation

The general formula for alcohol series is

$C_nH_{2n+1}OH$ . The formula for alcohol contains four carbon atoms is

$C_nH_{2n+1}OH$

$=C_4H_{2(4)+1}OH$

$=C_4H_{8+1}OH$

$=C_4H_{9}OH$ .

Compound

$C_2H_6O$ has two isomers X and Y. On reaction with

$HI$ , X gives alkyl iodide and water while Y gives alkyl iodide and alcohol.Compounds X and Y are respectively :

0%
$C_2H_5OC_2H_5\, and \,CH_3OC_2H_5$
0%
$CH_3OCH_3 \,and \,C_2H_5OCH_3$
0%
$C_2H_5OH \,and \, CH_3OCH_3$
0%
$CH_3OH \,and \, CH_3OCH_3$

Explanation

$\underset{(X)}{C_2H_5OH} + HI \rightarrow C_2H_5I +H_2O$

$CH_3O\underset{(Y)}CH_3+HI \rightarrow CH_3I+CH_3OH$

Ether is obtained from ethyl alcohol in presence of

$H_2SO_4$ at?

0%
113 K
0%
443 K
0%
413 K
0%
213 K

Explanation

Acid-catalyzed method of preparing symmetrical ethers from primary alcohols is temperature dependence. At

$110^{\circ}C$ or 383 K to

$130^{\circ}$ C or 403K, a

$S_N2$ reaction of the alcohol conjugate acid leads to an ether product. At higher temperatures (over

$150^{\circ}C$ or 423 K) an E2 elimination takes place and instead of ether, an alkene is obtained.

Thus, Ether is obtained from ethyl alcohol in presence of

$H_2SO_4$ at 413 K.

Diethyl ether when refluxed with excess of HI gives two molecules of (i). Ethers can be most commonly prepared by reaction of (ii) and (iii) . The method is called (iv).
(i), (ii),(iii) and (iv) respectively are?

0%
ethyl iodide, sodium alkoxide , alkyl halide, Williamson's synthesis
0%
ethanol, alcohol , alkyl halide , substitution
0%
methyl iodide, Grignard's reagent , alkyl halide , Williamson's synthesis
0%
ethyl iodide, phenol, ethyl iodide, esterification

Explanation

When ethers are treated with strong acid in the presence of a nucleophile, they can be cleaved to give alcohols and alkyl halides. If the ether is on a primary carbon this may occur through an

$S_N2$ pathway.

Thus Diethyl ether (

$C_2H_5OC_2H_5$ ) when refluxed with Hi gives two molecules as

$C_2H_5I+NaOC_2H_5$ . And in excess of HI, the alkoxide undergoes subtitution reaction and gives alkyl halide as

$C_2H_5I$

Thus (i)=

$C_2H_5I$ .

Ethers can be most commonly prepared by Williamson's synthesis where a reaction of alkyl halide and sodium alkoxide takes place. Therefore (ii) and (iii) are alkyl halide and sodium alkoxide. The method is called Williamson's synthesis=(iv)

(i), (ii),(iii) and (iv) respectively are

$C_2H_5I$ , alkyl halide, sodium alkoxide and Williamson's synthesis, respectively.

The major product formed during the reaction of 1- methyl cyclopentene with

$CH_3CO_3H$ is:

Explanation

As refer to the above image since,

$\overset {\bullet\bullet}{\underset {\bullet\bullet}{O}}$ is a good nucleophilic, the peroxygen attaches to the electron deficient double bond forming a

$3-$ membered ring.

1095184_989501_ans_a1e108c2fdca4a7a8ccf691acc13fdc9.png

Cis-3-hexene

$\xrightarrow{(a)}$ meso 3,4-hexanediol
trans-3-hexene

$\xrightarrow{(b)}$ meso 3,4-hexanediol.
Choose pair of reagent (a, b) for above conversions.

0%
Cold $KMnO_4, OsO_4$
0%
Cold $KMnO_4, RCO_3H/H_3O^\bigoplus$
0%
$RCO_3H/H_3O^\bigoplus, \, cold \, KMnO_4$
0%
None of these

Explanation

$Cis-3-Hexane \quad\underrightarrow{Cold\quad KMnO_4}\quad Meso-3-4-hexaanediol$

This is because syn addition to a cis molecule results in a meso compound and cold ${ KMnO }_{ 4 }$ is a syn reagent.

$Trans-3-hexane\quad\underrightarrow {RCO_3H/H_3O^+}\quad Meso-3-4-hexanediol$

Similarly anti addition to trans molecule forms meso compound

Reaction of

$R-2-$ butanol with

$p-$ toluenesulphonyl chloride in pyridine followed by reaction with

$LiBr$ gives:

0%
$R-1-$ butyl bromide
0%
$R-2-$ butyl tosylare
0%
$R-1-$ butyl tosylate
0%
$R-2-$ butyl bromide

Explanation

$R-2$

$- butylbromide$

The above reaction mechanism follows

$S_N1$ mechanism.

$-O-pTs$ is a good leaving group thus a carbocation formed, which gives the product in same stereochemistry as of Reactant.

1119615_989664_ans_d1ddb26ab81a481795fba481441f301e.JPG

CBSE Questions for Class 12 Engineering Chemistry Alcohols,Phenols And Ethers Quiz 5 - 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 5 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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