MCQ Questions for Class 12 Medical Chemistry Alcohols, Phenols And Ethers Quiz 6

(A)

C H_{3} - C | C H_{3} H - C H_{2} - C H_{2} - O H

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

(B)

C H_{3} - O H | C | C H_{3} - C H_{2} - C H_{3}

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

(C)

C H_{3} C H_{3} | C H - O H | C H - C H_{3}

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

The order of given alkanols

A, B, C

$A,B,C$ 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 d e g r e e > 2 d e g r e e > 1 d e g r e e

$3 degree> 2degree> 1 degree$ and

C H_{3}

$CH_{ 3 }$ is

+

$+$ group so, the order of stability is

: B > C > A

$: B>C>A$

(B)

C H_{3} - O H | C | C H_{3} - C H_{2} - C H_{3} \to C H_{3} - \oplus | C | C H_{3} - C H_{2} - C H_{3}

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

(2^{o})

$(2^o)$

(C)

C H_{3} - C H_{3} | C H - O H | C H - C H_{3} \to C H_{3} - C | C H_{3} - C h \oplus - C H_{3} C H_{3} (2^{o})

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

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)

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

(I V)

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

(I I I)

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

(I I)

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

4 > 3 > 2 > 1.

$4> 3> 2> 1.$

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

H_{2} S O_{4}

$H_2SO_4$ 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

The correct statements about squaric acid?

0%
It is stronger acid than phenol
0%
It can form two conjugate base
0%
It can liberate $CO_2$ from $NaHCO_3$
0%
The diconjugate base of squaric acid has all $C-C$ bond lengths are equal

Explanation

It is stronger than phenol. Phenol has

p K a

$pKa$ value

10.0

$10.0$ whereas squaric acid has

p K a

$pKa$ value

1.5

$1.5$ . So, lesser the

p K a

$pKa$ value more will be the acidity.
1119902_877196_ans_55072f8bb6f3433db1b46382c2351ab6.png

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

In the series of reaction,the product

D

$D$ is a/an :

0%
Alcohol
0%
Ketone
0%
Aldehyde
0%
Carboxylic acid

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

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 - O H

$H-OH$ across double bond

A. Addition of water in presence of dil.

H_{2} S O_{4}

$H_2SO_4$ follows Markovnikov's addition of

H - O H

$H-OH$ across the alkene

B. Addition of

H C I

$HCI$ followed by reaction with dil.

N a O H

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

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

H - O H

$H-OH$

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_{2} S O_{4}

$H_2SO_4$ .

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)

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

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

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.

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

Consider the following reaction,

$Phenol\overset{Zn\,dust}{\rightarrow}X \xrightarrow[anhyd.Alcl_3]{CH_3Cl} Y \overset{alkalineKMnO_4}{\longrightarrow}Z$
The product Z is:

0%
benzaldehyde
0%
benzoic acid
0%
benzene
0%
toluene

Explanation

$Phenol\overset{Zn\,dust}{\rightarrow}X \xrightarrow[anhyd.Alcl_3]{CH_3Cl} Y \overset{alkalineKMnO_4}{\longrightarrow}Z$

Phenol undergoes reduction in the presence of

$Zn$ dust and gives benzene i.e X.

Benzene in the presence of Alkyl halide and anhy.

$AlCl_3$ undergoes Friedel-craft alkylation and gives methyl substituted benzene i.e. toluene as Y.

Toluene in the presence of alkaline

$KMnO_4$ undergoes oxidation to COOH group resulting in the formation of Benzoic acid.

Overall reaction is represented as:

$\underset {Phenol}{C_6H_5OH}\overset{Zn\,dust}{\rightarrow} \underset {benzene}{C_6H_6} \xrightarrow[anhyd.AlCl_3]{CH_3Cl} \underset {toluene}{C_6H_5CH_3} \overset{alkalineKMnO_4}{\longrightarrow} \underset {Benzoic\ Acid}{C_6H_5COOH}$

the product is Benzoic acid.

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.

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:

$CH_3CH_2OH>CH_3CH_2CH_2OH>CH_3CH_2CH_2CH_2OH$

Which of the following has highest boiling point?

0%
Benzene
0%
Phenol
0%
Toluene
0%
Ethylbenzene

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

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.

$(CH_3)_3C - CH_2OH\xrightarrow[170^0C]{Conc.H_2SO_4}$ 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 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%
$1 - Bromobutane \xrightarrow[dry \, ether]{Mg} \,\, \xrightarrow[(ii) H_3 O^+]{(i) CH_3 COCH_3}$

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).

Which of the following compounds will react with sodium hydroxide solution in water ?

0%
$C_6H_5OH$
0%
$C_6H_5CH_2OH$
0%
$(CH_3)_3COH$
0%
$C_2H_5OH$

Explanation

Phenol

$(C_6H_5OH)$ will react with sodium hydroxide solution in water , as phenols are more acidic than alcohols.

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

$\xrightarrow [ ]{ mix\quad kBr+kBrO_{ 3 }\quad }$ ( major)

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

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

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

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.

Based on above isomers answer the following.

Which isomers on dehydration with conc.

$H_2SO_4$ give alkene which is capable to show geometrical isomerism?

0%
g, h, f
0%
b, d, a
0%
a, b, c.
0%
None of these

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

Based on above isomers answer the following.

Which isomer on dehydration with conc.

$H_3PO_4$ undergo maximum rearrangement?

0%
a
0%
c
0%
e
0%
b

Explanation

Rearrangement keeps occurring until the carbocation formed is stable.

Stability order

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

Hence, option

$C$ is correct.

1120096_991902_ans_62f963bf0abc433295318b69d5da864d.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

Propene

$CH_{3} - CH = CH_{2}$ can be converted into

$1$ -propanol by oxidation. Indicate which sets of reagents amongst the following is ideal to effect the above conversion?

0%
$KMnO_{4}$ (alkaline)
0%
Osmimum tetroxide $(OsO_{4}/CH_{2}Cl_{2})$
0%
$B_{2} H_{6}$ and alk $H_{2}O_{2}$
0%
$O_{3}$ and $Zn/H_{2}O$

Explanation

$CH_3-CH=CH_2\xrightarrow[]{B_2H_6}CH_3-\underset{H}{\underset{|}{C}}H-\underset{BH_2}{\underset{|}{C}H_2}\xrightarrow[]{H_2O_2} CH_3-CH_2-CH_2OH$

This reaction is called as hydroboration-oxidation.

Product of the reaction is :

Explanation

As refer to the above image addition reaction resulting in formation of diols. Here,

$OsO_4/NaHSO_3$ is the reagent used which occurs in a syn manner.

1095692_989548_ans_706b878f052043589f8dcd31e6045c7f.png

$CH_3-CH_2-CH_2-CH_2- \overset{CH_3}{\overset{|}{\underset{CH_3}{\underset{|}{C}}}}-CH_2-CH_2-OH\xrightarrow[\Delta]{H^+}\underset{(major)}{(A)}$
Product (A) is:

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$CH_3-CH_2-CH_2-CH_2-\underset{CH_3}{\underset{|}{C}}=CH-CH_3$
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$CH_3-(CH_2)_3-C(CH_3)=C(CH_3)_2$
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$CH_3-CH_2-CH_2-CH_2-\underset{CH_3}{\underset{|}{C}}=CH_2$
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$CH_3-CH_2-\underset{CH_3}{\underset{|}{CH}}-CH_2-\overset{CH_3}{\overset{|}{C}}=CH_2$

Trans - cyclohexane -1,2- diol can be obtained by the reaction of cyclohexene with :

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$KMnO_4$
0%
$OsO_4$
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peroxy formic acid / $H_3 O^+$
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$SeO_2$

Energy profile diagram for dehydration of 2-butanol using conc.

$H_2SO_4$ is given below.
Product (b) of above reaction is :

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1-butene
0%
cis-2-butene
0%
trans-2-butene
0%
iso-butene

Explanation

From energy profile

$Me-CH_2-\overset{+}{CH}-CH_3$ has the least energy while means it is most stable molecule. Hence, dehydration of 2-Butanol forms

$Me-CH_2-\overset{CH}-CH_3$ which on dehydrogenation gives

$Me-CH=Ch-CH_3$ (A is molecule)

Product (Q) in this reaction is?

Explanation

This is an example of friedal craft's reaction. it involves acylation of the aromatic ring. Acylating agents are acyl chlorides the viability of the friedal craft acylation depends on the stability of the acyl chloride regent these reaction proceeds through generation of an acylium enter followed by deprotonation of the areium ion

$AlCl^-_4$ regenarting the

$AlCl_3$ catalyst.
1100482_995536_ans_be3cd2fd44e1471a8e7e1c3cd0dffa5e.png

Which of the following procedures would be best for the preparation of phenyl benzyl ether?

$C_6H_5OCH_2C_6H_5$ .

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$C_6H_5Cl+C_6H_5H_2O^{(-)}Na^{(+)}$
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$C_6H_5O^{(-)}Na^{(+)}+C_6H_5CH_2Cl$
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$2C_6H_5Cl+Na_2O$
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$2C_6H_5MgBr+CH_2O$

An organic compound having the molecular formula

$C_8H_{10}O$ on being heated with

$I_2$ and dilute NaOH gives a yellow precipitate. The expected compound is?

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$C_6H_5CH_2CH_2OH$
0%
0%
0%

Explanation

organic compound with the molecular formula of

$C_8H_{10}O$ which on being heated with

$I_2$ and dilute

$NaOH$ gives a yelow precipitate is in the diagram.
basically acetyle group

$CH_3-C=O$ containing compounds and

$CH_3-CH-OH$ group containing compounds will give yellow ppt.(

$CHI_3$ iodoform) with

$I_2$ and

$OH$ . this is iodoform test.

Final product

$A$ is:

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0%
$CH_{2} = C = CH_{2}$
0%
0%

Explanation

Refer to Image

This is pinacol reaction. This reaction occurs in presence of Conc.

$H_2SO_4$ (acidic medium). Adjacent diols under goes this reaction.

1074799_992354_ans_14bbe8168e49475685c2f321a10e8ac7.png

Select the best method for the preparation of the following compounds :

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reaction of cyclohexanone with $CH_{3}Li$
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reaction of $1-$ methylcyclohexene with $Hg(OAc)_{2}$ followed by $NaBH_{4}$
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reaction of cyclohexene with $BH_{4} ; NaOH/H_{2}O_{2}$ , followed by $CH_{3}Br$
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reaction of cyclohexene with $MCPBA$ , followed by $CH_{ 3 }MgBr$

Explanation

$MCPBA$ (Meta-chloro perbenzoic acid) is used as an oxidizing agent to produce epoxides. Grignard reagents are used to produce alcohols from epoxides.

What is the product obtained by heating the following allylic ether of phenol?

The structure of the product formed in the reaction given below is :

CBSE Questions for Class 12 Medical Chemistry Alcohols, Phenols And Ethers Quiz 6 - 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 6 - MCQExams.com

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Practice Class 12 Medical Chemistry Quiz Questions and Answers

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