Amines Test

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Amines Test - 53

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Question 1
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Directions For Questions

A hydrocarbon $$(A)$$ has the molecular formula $$({C}_{8}{H}_{10})$$. (A) is oxidized by a strong oxidizing agent to (B). (B) on dehydration and subsequent reaction with ammonia forms an imide, (C). (C) is then reacted with a strong inorganic base to form a compound that undergoes Hoffman bromamide reaction to give (D). (D) is treated with sodium nitrite in an ice cold acidic solution to form the product (E). (E) is a steam volatile compound and on nitration gives two mononitro derivatives. (E) is treated with sodium hydroxide to form the salt (F). On heating a solution of (F), bubble are formed due to release of gas. This gas does not burn. On analysis it was found that the gas has two components, one lighter than air and one heavier than air.

...view full instructions

Compound $$(A)$$ is:

A
Ethylbenzene

B
$$p-Xylene$$

C
$$m-Xylene$$

D
$$o-Xylene$$

Solution

D.U in $$(A)=\cfrac { \left( 2{ n }_{ C }+2 \right) -{ n }_{ H } }{ 2 } =\cfrac { \left( 8\times 2+2 \right) -10 }{ 2 } ={ 4 }^{ o }$$
4 D.U and $$C:H=1:1$$ suggest that $$(A)$$ contains benzene ring with two extra C atoms (i.e., two $$(Me)$$ groups]. Since compound $$(A)$$ is steam volatile and on nittration gives two nitro-derivatives, so $$(A)$$ is ortho-xylene
Question 2
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Identify the most basic amine among the above compounds :

A
i

B
ii

C
iii

D
iv

Solution
Question 3
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The following reaction is:

A
benzidine rearrangement

B
pinacol-pinacolone rearrangement

C
fries rearrangement

D
benzil-benzilic acid arrangement

Solution

Given reaction is Benzidine rearrangement, in which diphenylhydrazine reacts with mineral acids and induces a rearrangement reaction to form 4,4'-benzidine.
Question 4
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The decreasing order of the rate of bromination of the following compounds is:
$$I. Ph\overset { \oplus }{ N{ Me }_{ 3 } } $$
$$II. Ph{CH}_{3}\overset { \oplus }{ N{ Me }_{ 3 } } $$
$$III. PhMe$$
$$IV. PhN{Me}_{2}$$

A
$$(I)>(II)>(III)>(IV)$$

B
$$(IV)>(III)>(II)>(I)$$

C
$$(III)>(IV)>(I)>(II)$$

D
$$(III)>(IV)>(II)>(I)$$

Solution

Stronger the activating substituent (i.e EDG) faster is the rate of (bromination) reaction.

Order of $$\overline { e } $$ donating substituent
$$-\overset { \cdot \cdot }{ N } {Me}_{2}>-Me>-{CH}_{3}\overset { \oplus }{ N{ Me }_{ 3 } } >-\overset { \oplus }{ N{ Me }_{ 3 } } $$

Hence, the rate of bromination is as given in (B)
Question 5
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Directions For Questions

A hydrocarbon $$(A)$$ has the molecular formula $$({C}_{8}{H}_{10})$$. (A) is oxidized by a strong oxidizing agent to (B). (B) on dehydration and subsequent reaction with ammonia forms an imide, (C). (C) is then reacted with a strong inorganic base to form a compound that undergoes Hoffman bromamide reaction to give (D). (D) is treated with sodium nitrite in an ice cold acidic solution to form the product (E). (E) is a steam volatile compound and on nitration gives two mononitro derivatives. (E) is treated with sodium hydroxide to form the salt (F). On heating a solution of (F), bubble are formed due to release of gas. This gas does not burn. On analysis it was found that the gas has two components, one lighter than air and one heavier than air.

...view full instructions

Compound $$(E)$$ is:

A

B

C

D

Solution

D.U in $$(A)=\cfrac { \left( 2{ n }_{ C }+2 \right) -{ n }_{ H } }{ 2 } =\cfrac { \left( 8\times 2+2 \right) -10 }{ 2 } ={ 4 }^{ o }$$
4 D.U and $$C:H=1:1$$ suggest that $$(A)$$ contains benzene ring with two extra C atoms (i.e., two $$(Me)$$ groups]. Since compound $$(A)$$ is steam volatile and on nittration gives two nitro-derivatives, so $$(A)$$ is ortho-xylene
Question 6
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Gabriel synthesis is used for the preparation of:

A
$$\displaystyle { 1 }^{ o }$$ amine

B
$$\displaystyle { 2 }^{ o }$$ amine

C
$$\displaystyle { 3 }^{ o }$$ amine

D
all of the above
Question 7
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Directions For Questions

A hydrocarbon $$(A)$$ has the molecular formula $$({C}_{8}{H}_{10})$$. (A) is oxidized by a strong oxidizing agent to (B). (B) on dehydration and subsequent reaction with ammonia forms an imide, (C). (C) is then reacted with a strong inorganic base to form a compound that undergoes Hoffman bromamide reaction to give (D). (D) is treated with sodium nitrite in an ice cold acidic solution to form the product (E). (E) is a steam volatile compound and on nitration gives two mononitro derivatives. (E) is treated with sodium hydroxide to form the salt (F). On heating a solution of (F), bubble are formed due to release of gas. This gas does not burn. On analysis it was found that the gas has two components, one lighter than air and one heavier than air.

...view full instructions

Compound $$(D)$$ is:

A

B

C

D

Solution

D.U in $$(A)=\cfrac { \left( 2{ n }_{ C }+2 \right) -{ n }_{ H } }{ 2 } =\cfrac { \left( 8\times 2+2 \right) -10 }{ 2 } ={ 4 }^{ o }$$
4 D.U and $$C:H=1:1$$ suggest that $$(A)$$ contains benzene ring with two extra C atoms (i.e., two $$(Me)$$ groups]. Since compound $$(A)$$ is steam volatile and on nittration gives two nitro-derivatives, so $$(A)$$ is ortho-xylene
Question 8
1 / -0

Directions For Questions

A hydrocarbon $$(A)$$ has the molecular formula $$({C}_{8}{H}_{10})$$. (A) is oxidized by a strong oxidizing agent to (B). (B) on dehydration and subsequent reaction with ammonia forms an imide, (C). (C) is then reacted with a strong inorganic base to form a compound that undergoes Hoffman bromamide reaction to give (D). (D) is treated with sodium nitrite in an ice cold acidic solution to form the product (E). (E) is a steam volatile compound and on nitration gives two mononitro derivatives. (E) is treated with sodium hydroxide to form the salt (F). On heating a solution of (F), bubble are formed due to release of gas. This gas does not burn. On analysis it was found that the gas has two components, one lighter than air and one heavier than air.

...view full instructions

Compound $$(F)$$ is:

A

B

C

D

Solution

D.U in $$(A)=\cfrac { \left( 2{ n }_{ C }+2 \right) -{ n }_{ H } }{ 2 } =\cfrac { \left( 8\times 2+2 \right) -10 }{ 2 } ={ 4 }^{ o }$$
4 D.U and $$C:H=1:1$$ suggest that $$(A)$$ contains benzene ring with two extra C atoms (i.e., two $$(Me)$$ groups]. Since compound $$(A)$$ is steam volatile and on nittration gives two nitro-derivatives, so $$(A)$$ is ortho-xylene
Question 9
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Most basic compound in gaseous phase is :

A
$${NH}_{3}$$

B
$${CH}_{3}{CH}_{2}{NH}_{2}$$

C
$$({CH}_{3}{CH}_{2})_2NH$$

D
$${({CH}_{3}{CH}_{2})}_{3}N$$

Solution

$$\text { In gaseous phase (in case of amines) }$$

Stability of $$2^{\circ}$$ carbocation $$>3^{\circ}>1^{\circ}>\mathrm{NH}_{3}$$
Hence $$\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{2} \mathrm{NH}$$ is the most stable.
Question 10
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Among the following, the strongest base is :

A
$$C_6H_5NH_2$$

B
$$p-NO_2C_6H_4NH_2$$

C
$$m-NO_2C_6H_4NH_2$$

D
$$C_6H_5CH_2NH_2$$

Solution