Alcohols Phenols and Ethers Test

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Alcohols Phenols and Ethers Test - 66

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Weekly Quiz Competition

Question 1
1 / -0

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

A
dipolar character of ether

B
alcohols having resonance structure

C
intermolecular hydrogen bonding in ethers

D
intermolecular hydrogen bonding in alcohols

Solution

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.
Question 2
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$$\underset{(major)}{Y}\overset{CH_3CH_2ONa}{\underset{(CH_3CH_2OH)}{\leftarrow}}H_3C-\overset{\overset{\displaystyle CH_3}{\displaystyle |}}{\underset{\displaystyle \underset{\displaystyle CH_3}{|}}{C}}-Br\overset{CH_3CH_2OH}{\rightarrow}\underset{(Major)}{X}$$
X and Y are.

A
$$\displaystyle H_3C-\overset{\overset{\displaystyle CH_3}{\displaystyle |}}{\underset{\displaystyle\underset{\displaystyle CH_3}{|}}{C}}-OCH_2CH_3$$ in both cases

B
$$\displaystyle CH_3\overset{\overset{\displaystyle CH_3}{|}}{C}=CH_2$$ in both cases

C
$$\displaystyle H_3C-\overset{\overset{\displaystyle CH_3}{|}}{\underset{\underset{\displaystyle CH_3}{|}}{C}}-OCH_2CH_3$$ and $$H_3C-\overset{\overset{\displaystyle CH_3}{\displaystyle |}}{\underset{\displaystyle \underset{CH_3}{\displaystyle ||}}{C}}$$

D
$$\displaystyle H_3C-\overset{\overset{\displaystyle CH_3}{|}}{\underset{\underset{\displaystyle CH_2}{||}}{C}}$$ and $$H_3C-\overset{\overset{\displaystyle CH_3}{|}}{\underset{\underset{\displaystyle CH_3}{||}}{C}}-OCH_2CH_3$$
Question 3
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Which of the following is not possible ?

A

B

C
Both A and B

D
None of these
Question 4
1 / -0

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:

A
phenol

B
benzene

C
benzaldehyde

D
acetophenone

E
toluene

Solution

This method is a commercial method for the manufacture of phenol.
Question 5
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In the preparation of alkene from alcohol using $$Al_{2}O_{3}$$ which is effective factor?

A
Porosity of $$Al_{2}O_{3}$$

B
Temperature

C
Concentration

D
Surface area of $$Al_{2}O_{3}$$

Solution

Temperature is the effective factor for dehydration of alcohols by $${ Al }_{ 2 }{ O }_{ 3 }$$.
$$R-C{ H }_{ 2 }-C{ H }_{ 2 }-OH\xrightarrow [ { 350 }^{ o }-{ 380 }^{ o }C ]{ { Al }_{ 2 }{ O }_{ 3 } }$$
$$ R-CH=C{ H }_{ 2 }+{ H }_{ 2 }O$$
At $${ 220 }^{ o }-{ 250 }^{ o }$$ it forms ether.
Question 6
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Write the IUPAC name of $$\begin{matrix} & OH & \\ & | & \\ C{ H }_{ 3 }-C{ H }_{ 2 }- & C & -C{ H }_{ 2 }-C{ H }_{ 3 } \\ & | & \\ & C{ H }_{ 3 } & \end{matrix}$$

A
3-Methylpentan-3-ol

B
3-Hydroxyhexane

C
3-Hydroxy-3-methyl pentane

D
All of the above

Solution

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.
Question 7
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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?

A
Propan-2-ol

B
Propan-1-ol

C
Ethane 1, 2 diol

D
Propane 1, 2-diol
Question 8
1 / -0

The relationship between the reactant and the final product is:

A
position isomers

B
identical compounds

C
enantiomers

D
none of the above

Solution
Question 9
1 / -0

Which of the following is not a characteristic of alcohol ?

A
They are lighter than water.

B
Their boiling points rise fairly uniformly with rising molecular weight.

C
Lower members are insoluble in water and organic solvents but the solubility regularly increases with molecular mass.

D
Lower members have a pleasant smell and burning taste, higher members are colourless and tasteless.

E
Higher members have a pleasant smell and burning taste, lower members are colourless and tasteless.

Solution

A. Alcohols are lighter than water because water molecules are closely packed together due to extensive H-bonding, which means that it has more mass in the same volume than either alcohol or oil. In addition, alcohol is made up of carbon and hydrogen atoms while water is made up of oxygen and hydrogen atoms.
B. Their boiling points rise fairly uniformly with rising molecular weight this is because boiling point depends on the intermolecular interactions such as H-bonding which decreases as the chain length decreases.
C. Lower members are soluble in water and organic solvents but the solubility regularly decreases with molecular mass because of ease of H-bonding in smaller alcohol molecules than those having long hydrocarbon chain.
D. Lower members have a pleasant smell and burning taste, higher members are colourless and tasteless.
Question 10
1 / -0

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 ?

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

B
$$CH_3-\underset{CH_3}{\underset{|}C}H-CH-\underset{OH}{\underset{|}C}H-CH_3$$

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

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

Solution

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:
$$\underset {2.2g}{C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } }H -\overset { \overset { OH }{ | } }{C}H-C{ H }_{ 3 }}+CH_3MgI \rightarrow \underset {560mL\ at\ STP}{CH_4}$$
$$C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } }H -\overset { \overset { OH }{ | } }{C}H-C{ H }_{ 3 } \overset {(i)H^+(ii)Ozonolysis}{\rightarrow} C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } =O+O=CHCH_3$$
$$C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } =O+NH_2OH \rightarrow \underset {19.17N}{C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } =NOH}$$
$$C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } }H -\overset { \overset { OH }{ | } }{C}H-C{ H }_{ 3 } \overset {[O]}{\rightarrow}C{ H }_{ 3 }-{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } }H -\overset { \overset { O }{ || } }{C}-C{ H }_{ 3 } $$