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:

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

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

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

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phenol
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benzene
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benzaldehyde
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acetophenone
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toluene

Write the IUPAC name of $$\begin{matrix} & OH & \\ & | & \\ C{ H }_{ 3 }-C{ H }_{ 2 }- & C & -C{ H }_{ 2 }-C{ H }_{ 3 } \\ & | & \\ & C{ H }_{ 3 } & \end{matrix}$$

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

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
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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 :

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Method 1 and 2 give same product.
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Method 1 gives resolvable racemic mixture while 2 will give non resolvable achiral product.
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Method 1 gives non resolvable optivally inactive compound while method 2 gives resolvable racemic mixture.
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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.

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?

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

The relationship between the reactant and the final product is:

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position isomers
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identical compounds
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enantiomers
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none of the above

Which of the following is not a characteristic of alcohol ?

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They are lighter than water.
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Their boiling points rise fairly uniformly with rising molecular weight.
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Lower members are insoluble in water and organic solvents but the solubility regularly increases with molecular mass.
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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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$$CH_3CH = CH_2, CH_3CH(OH)CH_3$$
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$$ CH_3CH = CHCH_3, CH_3CH_2CH_2OH$$
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$$CH_3CH_2CH = CH_2,CH_3CH_2CH_2OH$$
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$$CH_3CH_2CH_2CH_3, CH_3CH(OH)CH_2CH_3$$

Which of the following is not possible ?

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0%
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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:

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$$I > III > IV > II$$
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$$II > I > III > IV$$
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$$IV > I > III > II$$
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$$III > I > II > IV$$

Major product obtained in the reaction is :

Compare rate of dehydration of (i), (ii), and (iii) by conc. $$H_2SO_4$$ .

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(i) > (iii) > (ii)
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(i) > (ii) > (iii)
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(ii) > (i) > (iii)
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(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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$$CH_3-\underset{CH_3}{\underset{|}C}H-CH_2-\underset{OH}{\underset{|}C}H-CH_3$$
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$$CH_3-\underset{CH_3}{\underset{|}C}H-CH-\underset{OH}{\underset{|}C}H-CH_3$$
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$$CH_3-CH_2-\underset{OH}{\underset{|}C}H-CH_2-CH_3$$
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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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$$w > x > y > z$$
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$$w > z > x > y$$
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$$z > w > x > y$$
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$$z > x > w > y$$

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,
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d, e, f, g, h
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c,d, b, g
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None of these

What happens when tertiary butyl alcohol is passed over heated copper at $$300^0$$C ?

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

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.

Which of the following compounds is aromatic alcohol?

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

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 :

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

Dehydration of following alcohols will be in order :

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$$1<2<3<4$$
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$$4>3>1>2$$
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$$4<2<1<3$$
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$$1>3>4>2$$

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

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

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:

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

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

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

Which of the following does not gives alkane with $$R-Mg-X$$

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

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

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

Product $$'B'$$ and phenol can be differentiated by:
a) $$NaOH$$
b) Litmus paper
c) $$Br_2 / H_2O$$
d) $$NahCO_3$$

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

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

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

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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:

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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?

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

Phenol reacts with $$Cl_2$$ to give:

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

Phenol is:

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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:

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

The IUPAC name of the following compound is:

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$$2$$-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

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$$i>iii>i$$
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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