an acid weaker than carbonic acid

an acid stronger than carbonic acid

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

Ether is more volatile than alcohol having same molecular formula because of:

0%
dipolar character of ether
0%
alcohols having resonance structure
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intermolecular hydrogen bonding in ethers
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intermolecular hydrogen bonding in alcohols

Explanation

An ether is more volatile than an alcohol having same molecular formula because intermolecular hydrogen bonding in alcohols.
The

$\displaystyle -OH$ group of alcohol can form intermolecular hydrogen bonds wheres

$\displaystyle -O-$ group of ethers cannot form hydrogen bonds. Hydrogen bond formation is possible when H atom is attached to electronegative N, O or F atom. Intermolecular hydrogen bonds leads to molecular association.

$\underset{(major)}{Y}\overset{CH_3CH_2ONa}{\underset{(CH_3CH_2OH)}{\leftarrow}}H_3C-\overset{\overset{\displaystyle CH_3}{\displaystyle |}}{\underset{\displaystyle \underset{\displaystyle CH_3}{|}}{C}}-Br\overset{CH_3CH_2OH}{\rightarrow}\underset{(Major)}{X}$
X and Y are.

0%
$\displaystyle H_3C-\overset{\overset{\displaystyle CH_3}{\displaystyle |}}{\underset{\displaystyle\underset{\displaystyle CH_3}{|}}{C}}-OCH_2CH_3$ in both cases
0%
$\displaystyle CH_3\overset{\overset{\displaystyle CH_3}{|}}{C}=CH_2$ in both cases
0%
$\displaystyle H_3C-\overset{\overset{\displaystyle CH_3}{|}}{\underset{\underset{\displaystyle CH_3}{|}}{C}}-OCH_2CH_3$ and $H_3C-\overset{\overset{\displaystyle CH_3}{\displaystyle |}}{\underset{\displaystyle \underset{CH_3}{\displaystyle ||}}{C}}$
0%
$\displaystyle H_3C-\overset{\overset{\displaystyle CH_3}{|}}{\underset{\underset{\displaystyle CH_2}{||}}{C}}$ and $H_3C-\overset{\overset{\displaystyle CH_3}{|}}{\underset{\underset{\displaystyle CH_3}{||}}{C}}-OCH_2CH_3$

Isopropylbenzene is oxidized in the presence of air to compound

$'A'$ . When compound

$'A'$ is treated with dilute mineral acid, the aromatic product formed is:

0%
phenol
0%
benzene
0%
benzaldehyde
0%
acetophenone
0%
toluene

Write the IUPAC name of

$\begin{matrix} & OH & \\ & | & \\ C{ H }_{ 3 }-C{ H }_{ 2 }- & C & -C{ H }_{ 2 }-C{ H }_{ 3 } \\ & | & \\ & C{ H }_{ 3 } & \end{matrix}$

0%
3-Methylpentan-3-ol
0%
3-Hydroxyhexane
0%
3-Hydroxy-3-methyl pentane
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All of the above

Explanation

The parent chain has 5 C atoms and an -OH group. So it is named pentanol. It has methyl substituent. The numbering of chain is shown below.

$\underset { 3-methyl-\quad pentan-\quad 3-ol }{ \begin{matrix} & OH & \\ & | & \\ \overset { 5 }{ C{ H }_{ 3 } } -\overset { 4 }{ C{ H }_{ 2 } } - & { C }^{ 3 } & -\overset { 2 }{ C{ H }_{ 2 } } -\overset { 1 }{ C{ H }_{ 3 } } \\ & | & \\ & C{ H }_{ 3 } & \end{matrix} }$

Hydroxy is used when

$-OH$ group is written in prefix. So, choice (b) and (c) are wrong.

Which of the following statements is/(are) correct?

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The viscosity decreases along the series of liquids: $H_3PO_4 > H_2SO_4 > HClO_4$
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The boiling point decreases along with the series of liquid: p-dihydroxy benzene > m - dihydroxybenzene > o- dihydroxybenzene
0%
The solubility in water decreases along the series of liquids: $HOCH_2CH_2OH> HOCH_2 CH(OH) CH_2OH > HOCH_2(CHOH)_4CH_2OH$
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The vapors pressure decreases along the series of liquids: $H - O - H > CH_3 - CH_2-OH > CH_3 - O - CH_3$

Synthesis of cyclohexane-1, 2-diol from cyclohexene may be accomplished in two way:

${MnO_4}^-$ dilute,

$OH^-$ at

$273 K$ .
peracid epoxidation followed by

$NaOH$ opening of the epoxide ring.
Which of the following statements/are correct :

0%
Method 1 and 2 give same product.
0%
Method 1 gives resolvable racemic mixture while 2 will give non resolvable achiral product.
0%
Method 1 gives non resolvable optivally inactive compound while method 2 gives resolvable racemic mixture.
0%
Product obtain in 1 and 2 will have diasteriomeric relationship.

Acid catalysed dehydration of t-butanol is faster than that of n-butanol because:

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tertiary carbocation is more stable than primary carbocation
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primary carbocation is more stable than tertiary carbocation
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t-butanol has a higher boiling point
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- rearrangement takes place during dehydration of t-butanol.

Explanation

Alcohol

$+ H_2SO_4 \rightarrow$ Alkene

The dehydration reaction is an elimination reaction that goes via carbocation formation. Since tertiary carbocations are most stable due to + Inductive effect of alkyl groups, then is the secondary and least stable is primary.

Therefore the reactivity of alcohol is governed by the stability of carbocation that follows the order as

Tertiary > secondary > primary

Hence, acid catalysed dehydration of t-butanol is faster than that of n-butanol because tertiary carbocation formed by t-butanol is more stable than primary carbocation formed by n-butanol.

Propan

$-2-ol\xrightarrow[250^0C]{Ai_2O_3}A\xrightarrow[HBr]{H_2O_2}B\xrightarrow{aq.KOH}$

What is the final product in the following reaction sequence?

0%
Propan-2-ol
0%
Propan-1-ol
0%
Ethane 1, 2 diol
0%
Propane 1, 2-diol

The relationship between the reactant and the final product is:

0%
position isomers
0%
identical compounds
0%
enantiomers
0%
none of the above

Which of the following is not a characteristic of alcohol ?

0%
They are lighter than water.
0%
Their boiling points rise fairly uniformly with rising molecular weight.
0%
Lower members are insoluble in water and organic solvents but the solubility regularly increases with molecular mass.
0%
Lower members have a pleasant smell and burning taste, higher members are colourless and tasteless.
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Higher members have a pleasant smell and burning taste, lower members are colourless and tasteless.

Consider the following reaction sequence,

$CH_3CH(OH)CH_3 \xrightarrow[Heat]{Conc.H_2SO_4} X \xrightarrow[H_2O]{dil.H_2SO_4}Y$

X and Y in the reaction respectively are :

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$CH_3CH = CH_2, CH_3CH(OH)CH_3$
0%
$CH_3CH = CHCH_3, CH_3CH_2CH_2OH$
0%
$CH_3CH_2CH = CH_2,CH_3CH_2CH_2OH$
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$CH_3CH_2CH_2CH_3, CH_3CH(OH)CH_2CH_3$

Explanation

$CH_3 \underset {\underset {\displaystyle OH}{|}}{C}H-CH_3 \xrightarrow {Conc. H_2SO_4/Heat} \underset {(X)}{CH_3CH=CH_2} \xrightarrow {dil. H_2SO_4/H_2O} \underset {(Y)}{CH_3 \underset {\underset {\displaystyle OH}{|}}{C}H-CH_3}$

Which of the following is not possible ?

0%
0%
0%
Both A and B
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None of these

Which of the following reactions will not yield phenol?

The ease of dehydration in the following compounds is:

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

Major product obtained in the reaction is :

Compare rate of dehydration of (i), (ii), and (iii) by conc.

$H_2SO_4$ .

0%
(i) > (iii) > (ii)
0%
(i) > (ii) > (iii)
0%
(ii) > (i) > (iii)
0%
(ii) > (iii) > (i)

2.2 g of an alcohol (A) when treated with

$CH_3$ -MgI liberates 560 mL of

$CH_4$ at STP. Alcohol (A) on dehydration followed by ozonolysis gives ketone (B) along with (C). Oxime of ketone (B) contains 19.17% N. (A) on oxidation gives ketone (D) having same number of carbon atom .
Structure of (A) is ?

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$CH_3-\underset{CH_3}{\underset{|}C}H-CH_2-\underset{OH}{\underset{|}C}H-CH_3$
0%
$CH_3-\underset{CH_3}{\underset{|}C}H-CH-\underset{OH}{\underset{|}C}H-CH_3$
0%
$CH_3-CH_2-\underset{OH}{\underset{|}C}H-CH_2-CH_3$
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$CH_3-\overset{CH_3}{\overset{|}{\underset{CH_3}{\underset{|}C}}}-CH_2-OH$

Explanation

Given that:

$\underset {2.2g}{A(ROH)}+CH_3MgI \rightarrow \underset {560mL\ at\ STP}{CH_4}$

$A(ROH) \overset {(i)H^+(ii)Ozonolysis}{\rightarrow} B(ketone)+C$

$B+NH_2OH \rightarrow \underset {19.17N}{O(oxime)}$

$A(ROH) \overset {[O]}{\rightarrow} D(ketone$

560 mL of

$CH_4$ at STP:

number of moles of

$CH_4$ , n=560/22400=0.025 mol

1 mol of

$CH_4$ will be obtained by reaction of 1 mol of

$CH_3MgI$ with 1 mol of A

thus 0.025 mol of

$CH_4$ is obtained from 0.025 mol of A

if 0.025 mol of A=2.2g

mass of 1 mol of A=2.2/0.025=88g/mol

Molecular formula of an alcohol=

$C_nH_{2n+1}OH$

molar mass of alcohol=12n+2n+1+16+1

For A, molar mass=88,

thus, 12n+2n+1+16+1=88

and n=5

Thus A=

$C_5H_{11}OH$

Since A on dehydration will form an alkene which on ozonolysis gives a ketone. Therefore the alkene formed should have two alkyl groups attached on one of the doubly bonded carbon so as to get a ketone. Thus possible structure of alkenes are:

$CH_2={ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } -CH_2-C{ H }_{ 3 }$

$C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } =CH-C{ H }_{ 3 }$

Thus, possible structure of A are:

$\overset { \overset { OH }{ | } }{C}H_2-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } -CH_2-C{ H }_{ 3 }$ OR

$C{ H }_{ 3 }-\overset { \overset { OH }{ | } }{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } -CH_2-C{ H }_{ 3 }$

$C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } }H -\overset { \overset { OH }{ | } }{C}H-C{ H }_{ 3 }$

Thus possible B are:

$O={ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } -CH_2-C{ H }_{ 3 }$

$C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } =O$

and corresponding oxime are:

$HON={ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } -CH_2-C{ H }_{ 3 }$ =X

$C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } =NOH$ =Y

percent N in X is:(mass of N/molar mass of X)

$\times 100=\frac {14}{(12 \times 4+9+16+14)} \times 100=16.1$ %

percent N in Y is(:mass of N/molar mass of Y)

$\times 100=\frac {14}{(12 \times 3+7+16+14)} \times 100=19.18$ %

Therefore according to given data oxime obtained is

$C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } =NOH$

from ketone =B=

$C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } =O$

and possible alcohol are:

$C{ H }_{ 3 }-\overset { \overset { OH }{ | } }{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } -CH_2-C{ H }_{ 3 }$

$C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } }H -\overset { \overset { OH }{ | } }{C}H-C{ H }_{ 3 }$

Given that A gives ketone D on oxidation and with same number of carbons, thus it can only be a secondary alcohol as tertiary alcohol is resistant to oxidation. Thus A can only be:

$C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } }H -\overset { \overset { OH }{ | } }{C}H-C{ H }_{ 3 }$

overall reaction can be represented as:

$\underset {2.2g}{C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } }H -\overset { \overset { OH }{ | } }{C}H-C{ H }_{ 3 }}+CH_3MgI \rightarrow \underset {560mL\ at\ STP}{CH_4}$

$C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } }H -\overset { \overset { OH }{ | } }{C}H-C{ H }_{ 3 } \overset {(i)H^+(ii)Ozonolysis}{\rightarrow} C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } =O+O=CHCH_3$

$C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } =O+NH_2OH \rightarrow \underset {19.17N}{C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } =NOH}$

$C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } }H -\overset { \overset { OH }{ | } }{C}H-C{ H }_{ 3 } \overset {[O]}{\rightarrow}C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } }H -\overset { \overset { O }{ || } }{C}-C{ H }_{ 3 }$

Decreasing order of acidic strength of different (-OH) group is:

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$w > x > y > z$
0%
$w > z > x > y$
0%
$z > w > x > y$
0%
$z > x > w > y$

Explanation

$w$ will be the most acidic, as the anion (formed after the proton is removed) is stabilized by resonance in benzene ring.

$z$ will be the least acidic as the anion formed will experience a significant

$+I$ effect from the

$-CH_2$ group adjacent to it, thus destabilizing it.

$x$ will be comparatively more acidic than

$y$ as it experiences less

$+I$ effect to a less extent. Hence the order of acidic strength is:

$w>x>y>z$

Which of the following compounds will give tribromo derivative on treatment with bromine water?

Based on above isomers answer the following.

Which isomers on dehydration give most stable alkene?

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

What happens when tertiary butyl alcohol is passed over heated copper at

$300^0$ C ?

0%
Secondary butyl alcohol is formed.
0%
2-Methylpropene is formed .
0%
1-Butene is formed .
0%
Butanal is formed.

Explanation

Tertiary alcohol when passed over

$Cu$ at

$300^{\circ}C$ is dehydrated to alkene.

$CH_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ \overset { \overset { C{ H }_{ 3 } }{ | } }{ C } } }-OH\ \overset { Cu,\ 300^{ 0 }C }{ \longrightarrow } CH_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ \overset { \overset { C{ H }_{ 2 } }{ || } }{ C } } }$

$t-butyl\ alcohol \overset { Cu,\ 300^{ 0 }C }{ \longrightarrow } 2-methyl\ propene$

A compound

$X(C_8H_{10}O)$ upon treatment with an alkaline solution of iodine gives a yellow precipitate. The filtrate on acidification gives a white solid

$Y(C_7H_6O_2)$ . Write the structures of X and Y.

Explanation

Given that:

$X(C_8H_{10}O+NaOH+I_2 \rightarrow yellow\ ppt(Z)$

$X+H^+ \rightarrow Y(C_7H_6O_2)white\ solid$

From the molecular formula it is clear that X and Y have high degree of unsaturation. Thus they must have benzene ring.

Reaction of X with alkaline iodine is called iodoform test and the test is positive for molecules containing either

$CH_3CO$ or

$-CH(OH)CH_3$ group i.e methyl ketones.

Thus X should have the following moieties:

$C_6H_5\ and\ -COCH_3\ or\ -CH(OH)CH_3$

For molecular formula

$C_8H_{10}O$ its structure should be

$C_6H_{5}-CH(OH)CH_3$

The structure of X and Y are:

$C_6H_{5}-CH(OH)CH_3\ and\ C_6H_{5}-COOH$ , respectively.

Which of the following compounds is aromatic alcohol?

0%
I, II, III, IV
0%
I, IV
0%
II, III
0%
I

What is the principal product after product after the slurry obtained on testing benzenamine (auline) with potassium nitrite and

$HCl$ at

$0^{0}C$ has been added to

$4-$ ethyl phenol?

Explanation

When Aniline reacts with

$KNO_2+HCl$ at

$0°C$ , Benzene diazonium chloride is formed. This when reacted with

$4-ethyl phenol$ , we obtain an azo dye. As the para-position is occupied, coupling occurs at the ortho position of

$4-ethyl phenol$ .
Hence, Option A is the correct answer.

1055026_1073643_ans_3f94026add754147b641bc8062d3e820.png

Product (A) of the given reaction is?

A solution of

$Ph_{3}CCO_{2}H$ in conc.

$H_{2}SO_{4}$ gives

$(X)$ when poured into methanol

$X$ is :

0%
$Ph_3C-\overset{\overset{O}{||}}{C}-O-CH_3$
0%
$Ph_2CH-\overset{\overset{O}{||}}{C}-O=CH_3$
0%
$Ph_{3}C - OCH_{3}$
0%
$Ph_{3}C - CH_{3}$

Explanation

$Ph_3C-C-OOH+CH_3OH\xrightarrow []{conc.H_2SO_4}Ph_3C-OCH_3$

Mechanism:-

$Ph_3C^{\oplus}$ is a stable carbocation whicch breaks replaces

$H^+$ in

$CH_3OH$ to give

$Ph_3C-OCH_3$ ;

(Refer to Image)

1083184_993139_ans_1ceb370c872a4da2bdd3c0efb181df99.png

Dehydration of following alcohols will be in order :

0%
$1<2<3<4$
0%
$4>3>1>2$
0%
$4<2<1<3$
0%
$1>3>4>2$

$(CH_3)_2CHCH_2C \equiv N\xrightarrow[Heat]{HCl,H_2O} A \xrightarrow[II.H_2O]{I.LiAlH_4} B \xrightarrow [CH_2Cl_2]{PCC} C$ . The compound 'C' is:

0%
$(CH_3)_2 CH-{\overset{O}{\overset{||}C}} CH_3$
0%
$(CH_3)_2 CHCH_2{\overset{O}{\overset{||}C}}H$
0%
$(CH_3)_2 CHCH_2{\overset{O}{\overset{||}C}}OH$
0%
$(CH_3)_2 CH{\overset{O}{\overset{||}C}} - CH_2OH$

Consider the following sequence of reactions:

${ C }_{ 2 }{ H }_{ 5 }-C\equiv CH\xrightarrow { { CH }_{ 3 }MgBr } (A)\xrightarrow [ HCHO ]{ { H }_{ 2 }{ O } } (B)$
The product (B) is:

0%
${ CH }_{ 3 }{ CH }_{ 2 }OH$
0%
$C_2H_5 - \underset{\!\!CH_3}{\underset{|}{C}=} \,CH - CH_2OH$
0%
$C_2H_5 - C \equiv C - \underset{\!\!OH}{\underset{|}{C}H} - CH_3$
0%
${ C }_{ 2 }{ H }_{ 5 }-C\equiv C-{ CH }_{ 2 }OH$

Adipic acid

$\xrightarrow {P_2O_2 \triangle}{-CO_2-H-2O}A\xrightarrow {dil OH}{\triangle} B$
The compound B can be:

In the reaction

$CH_3CH_2ONa + CH_3CH_2OSO_2CH_3 \xrightarrow[heat]{THF}$ the product formed is

0%
$CH_3CH_2OCH_3$
0%
$CH_3CH_2OCH_2CH_3$
0%
$CH_3CH_2OSO_2OCH_2CH_3$
0%
$CH_3CH_2OSO_2OCH_3$

Which of the following does not gives alkane with

$R-Mg-X$

0%
$Ph-OH$
0%
$Cl-NH_{2}$
0%
$CH_{3}COOH$
0%
$HCl$

How many carbon atom are present in third homologue of methyl ether?

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

Methyl ether is

$CH_{3}OCH_{3}$

Next homologue is obtained by adding

$-CH_{2}$ group i.e.

$CH_{3}OCH_{2}CH_{3}$ .

$\therefore C$ atoms

$=3$

Again, adding 1 more

$-CH_{2}$ group,

$CH_{3}OCH_{2}CH_{2}CH_{3}$

$\therefore$ Number of carbon atoms in 3rd homologue

$=4$

Product

$'B'$ and phenol can be differentiated by:
a)

$NaOH$
b) Litmus paper
c)

$Br_2 / H_2O$
d)

$NahCO_3$

0%
$b$ and $d$
0%
$a, b$ and $d$
0%
$a , c$ and $d$
0%
$c$ and $d$

Explanation

R.E.F image

Solution -

Product (B) is benzoic acid

which can be differentiated from

phenol by

$NaHCO_{3}$

$\rightarrow$ Both by litmus paper test changes

blue litmus to red.

$\rightarrow$ Both (phenol & benzoic acid) reacts with

$NaOH$ &

$Br_{2}/H_{2}O$ These reagents are

not used for differentiation.

1121634_1167797_ans_00f59d21c5f4463f900a8e2043ff022f.jpg

What is order of decreasing reactivities of following alcohols towards nucleophilic substitution with HBr ?

Explanation

Higher the stability of carbocation, higher the reactivity towards Nucleophillic substitution with

$HBr$ .

If electron- donating group, is present higher the stability of carbocation.

Here, the above image is highly stable.

1100281_1225068_ans_77e2a40452e847efbef0b676c23e8686.PNG

In order to obtain diethyl ether from ethanol, the latter is taken in:

0%
In equal amount of sulphuric acid
0%
In slightly lesser amount of sulphuric acid
0%
In excess amount of sulphuric acid
0%
In far lesser amount of sulphuric acid

The major product is:

Among the following compounds, the most acidic is:

0%
p-nitrophenol
0%
o-hydroxybenzoic acid
0%
p-toluic acid
0%
p-hydroxybenzoic acid

Identify 'Z' in the following reaction series,

$CH_3CH_2CH_2Br \xrightarrow {aq. NaOH} [X] \xrightarrow[Heat]{Al_2O_3} [Y] \xrightarrow{HOCl} [Z]$

Which one of the following will most readily dehydrated in acidic medium?

0%
$1^0-alcohol$
0%
$2^0-alcohol$
0%
$3^0-alcohol$
0%
All are equal

Explanation

When alcohols are heated with conc.

$H_2SO_4$ , they undergo dehydration to form alkenes.

formation of alkenes depends upon two factors :-

1) Dehydration of alcohols occurs through internmeatiate carbocation formation and carbocation stability follows the order

$3^o > 2^o > 1^o$ . So,

$3^o$ carbonation intermediate will be most stable than

$2^o$ and

$1^o$ .

2) Saytzeff rule : - More highly subsituted alkene is the major product.

1107468_1178246_ans_49ee04ee0a104ae5b659f38ccdae6c2e.png

Phenol reacts with

$Cl_2$ to give:

0%
p-chlorophenol only
0%
Benzene
0%
o- and p-chlorophenol
0%
Triphenylphosphate

Phenol is:

0%
a neutral compound
0%
a base weaker than ammonia
0%
an acid stronger than carbonic acid
0%
an acid weaker than carbonic acid

Reaction - I :

Phenol

$\ \xrightarrow{NaOH}\ X\ \xrightarrow[{140^\circ}C]{C{O_2}}\ Y\ \xrightarrow{HCl}\ Z\ \xrightarrow[\Delta]{soda \ lime }\ P$

Reaction - II :

Phenol

$\ \xrightarrow[NaOH]{CC{l_4}}\ X\ \xrightarrow{HCl}\ Y\ \xrightarrow[\Delta]{soda \ lime }\ Q$

The wrong statement about

$P$ and

$Q$ is:

0%
Both the compounds give violet colour with FeCl $_3$
0%
Both the compounds on deoxygenation give benezene
0%
Both the compounds contain a phenolic $(-OH)$ group
0%
None of the above

The IUPAC name of the following compound is:

0%
$2$ -ethylbutan- $2$ -ol
0%
$1$ -ethy- $1$ -methylpropan- $1$ -ol
0%
$3$ -methyl pentan- $3$ -ol
0%
diethylethanol

Phenol (A)

$\xrightarrow[E{t_2}S{O_4}]{NaOH}$ (B)

$\xrightarrow[AlC{l_3}]{HCl+HCN}$ in the above sequence are-

Arrange the above phenols in increasing order of

$K_{a}$ value

0%
$i>iii>i$
0%
$i<ii<iii$
0%
$ii>i>iii$
0%
$iii>i<ii$

In the following reaction X is :-

$_{20}Ca^{40}+X \rightarrow _{21}Sc^{43} + _1H^1$

0%
Electron
0%
Positrom
0%
alpha particle
0%
proton

What is the total number of carbon atoms in

$P_1$ to

$P_7$ products ?.

0%
$91$
0%
$92$
0%
$93$
0%
$94$

Product C is:

0%
0%
0%
0%
Mixture of 1 and 2

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

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Practice Class 12 Medical Chemistry Quiz Questions and Answers

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