Alcohols Phenols and Ethers Test

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Alcohols Phenols and Ethers Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

Alcohols of low molecular weight are:

A
soluble in water

B
soluble in water on heating

C
insoluble in water

D
insoluble in all solvents.

Solution

Alcohols of low molecular weight are soluble in water while that of high molecular weight are insoluble in water .because of high hydrogen bonding in low molecular weight Alcinous.
Question 2
1 / -0

Which of the following synthesis given $$3$$-methyl-$$1$$-hexanol?

A
$$2-Bromohexane \xrightarrow[dry \, ether]{Mg} \,\, \xrightarrow[(ii) H_3 O^+]{(i) HCHO}$$

B

C
$$3-Bromopentane \xrightarrow[dry \, ether]{Mg} \,\, \xrightarrow[(ii) H_3O^+]{(i) CH_3 CHO}$$

D
$$1 - Bromobutane \xrightarrow[dry \, ether]{Mg} \,\, \xrightarrow[(ii) H_3 O^+]{(i) CH_3 COCH_3}$$

Solution
Question 3
1 / -0

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

A
3 -propylbutan-1-ol

B
2-Ethylpentan - 1 - ol

C
3- methyl hydroxyhexane

D
2- ethyl -2 - propyl ethanol

Solution

Steps to name an organic molecule:

1. Identify the functional group: OH in the given molecule. Alcohol i.e suffix "ol" will come.

2. Identify parent chain: The longest carbon chain that contains the functional group. Here its 5 C chain and not 4 C chain i.e pentane.

3. Assign locant number: lowest number to the main functional group and side chains i.e. 1-ol and 2-ethyl

4. Identify side chains/groups: its ethyl, in the given molecule.

IUPAC name is: 2-Ethylpentan-1-ol
Question 4
1 / -0

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

A
$$CH_3CH_2OH > CH_3CH_2CH_2OH > \underset {(Solubility)}{CH_3CH_2CH_2CH_2OH}$$

B
$$CH_3CH_2OH > CH_3CH_2CH_2OH > \underset {(Boiling point)}{CH_3CH_2CH_2CH_2OH}$$

C
$$CH_3CH_2CH_2OH > CH_3{\underset{CH_3}{\underset{|}{CH}}}-CH_2OH >\underset{Boiling \, point}{CH_3-\underset{CH_3}{\underset{|}{\overset{CH_3}{\overset{|}C}}}-OH}$$

D
$$CH_3-{\underset{CH_3}{\underset{|}{\overset{CH_3}{\overset{|}{C}}}}}-OH$$<$$CH_3{\underset{CH_3}{\underset{|}{CH}}}-CH_2OH$$ <$$CH_3CH_2CH_CH_2OH$$

Solution

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:
$$CH_3CH_2OH>CH_3CH_2CH_2OH>CH_3CH_2CH_2CH_2OH$$
Question 5
1 / -0

Give IUPAC name of the compound given below :
$$CH_3-\underset{Cl}{\underset{|}C}H-CH_2-CH_2-\underset{OH}{\underset{|}C}H-CH_3$$

A
2- Chloro - 5 - hydroxyhexane

B
2 - Hydroxy - 5 - chlorohexane

C
5 - Chlorohexan - 2 - ol

D
2 - Chlorohexan - 5 - ol

Solution

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
Question 6
1 / -0

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

A
primary alcohol

B
secondary alcohol

C
tertiary alcohol

D
dihydric alcohol

Solution

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
Question 7
1 / -0

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 :

A
$$C_2H_5OC_2H_5\, and \,CH_3OC_2H_5$$

B
$$CH_3OCH_3 \,and \,C_2H_5OCH_3$$

C
$$C_2H_5OH \,and \, CH_3OCH_3$$

D
$$CH_3OH \,and \, CH_3OCH_3$$

Solution

$$\underset{(X)}{C_2H_5OH} + HI \rightarrow C_2H_5I +H_2O$$
$$CH_3O\underset{(Y)}CH_3+HI \rightarrow CH_3I+CH_3OH$$
Question 8
1 / -0

Ether is obtained from ethyl alcohol in presence of $$H_2SO_4$$ at?

A
113 K

B
443 K

C
413 K

D
213 K

Solution

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.
Question 9
1 / -0

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?

A
ethyl iodide, sodium alkoxide , alkyl halide, Williamson's synthesis

B
ethanol, alcohol , alkyl halide , substitution

C
methyl iodide, Grignard's reagent , alkyl halide , Williamson's synthesis

D
ethyl iodide, phenol, ethyl iodide, esterification

Solution

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.
Question 10
1 / -0

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

A
$$(CH_3)_2C = CHCH_3$$

B
$$(CH_3)_{2}C = CH_{2}$$

C
$$(CH_3)_2CHCH_2CH_3$$

D
$$CH_3-CH_2-\underset{CH_3}{\underset{|}C}=CH_2$$

Solution

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

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Alcohols Phenols and Ethers Test - 42

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Question 1

soluble in water

soluble in water on heating

insoluble in water

insoluble in all solvents.

Solution

Question 2

$$2-Bromohexane \xrightarrow[dry \, ether]{Mg} \,\, \xrightarrow[(ii) H_3 O^+]{(i) HCHO}$$

$$3-Bromopentane \xrightarrow[dry \, ether]{Mg} \,\, \xrightarrow[(ii) H_3O^+]{(i) CH_3 CHO}$$

$$1 - Bromobutane \xrightarrow[dry \, ether]{Mg} \,\, \xrightarrow[(ii) H_3 O^+]{(i) CH_3 COCH_3}$$

Solution

Question 3

3 -propylbutan-1-ol

2-Ethylpentan - 1 - ol

3- methyl hydroxyhexane

2- ethyl -2 - propyl ethanol

Solution

Question 4

$$CH_3CH_2OH > CH_3CH_2CH_2OH > \underset {(Solubility)}{CH_3CH_2CH_2CH_2OH}$$

$$CH_3CH_2OH > CH_3CH_2CH_2OH > \underset {(Boiling point)}{CH_3CH_2CH_2CH_2OH}$$

$$CH_3CH_2CH_2OH > CH_3{\underset{CH_3}{\underset{|}{CH}}}-CH_2OH >\underset{Boiling \, point}{CH_3-\underset{CH_3}{\underset{|}{\overset{CH_3}{\overset{|}C}}}-OH}$$

$$CH_3-{\underset{CH_3}{\underset{|}{\overset{CH_3}{\overset{|}{C}}}}}-OH$$<$$CH_3{\underset{CH_3}{\underset{|}{CH}}}-CH_2OH$$ <$$CH_3CH_2CH_CH_2OH$$

Solution

Question 5

2- Chloro - 5 - hydroxyhexane

2 - Hydroxy - 5 - chlorohexane

5 - Chlorohexan - 2 - ol

2 - Chlorohexan - 5 - ol

Solution

Question 6

primary alcohol

secondary alcohol

tertiary alcohol

dihydric alcohol

Solution

Question 7

$$C_2H_5OC_2H_5\, and \,CH_3OC_2H_5$$

$$CH_3OCH_3 \,and \,C_2H_5OCH_3$$

$$C_2H_5OH \,and \, CH_3OCH_3$$

$$CH_3OH \,and \, CH_3OCH_3$$

Solution

Question 8

113 K

443 K

413 K

213 K

Solution

Question 9

ethyl iodide, sodium alkoxide , alkyl halide, Williamson's synthesis

ethanol, alcohol , alkyl halide , substitution

methyl iodide, Grignard's reagent , alkyl halide , Williamson's synthesis

ethyl iodide, phenol, ethyl iodide, esterification

Solution

Question 10

$$(CH_3)_2C = CHCH_3$$

$$(CH_3)_{2}C = CH_{2}$$

$$(CH_3)_2CHCH_2CH_3$$

$$CH_3-CH_2-\underset{CH_3}{\underset{|}C}=CH_2$$

Solution

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