Amines Test

Result

Amines Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

Which (isomeric) amine has lowest boiling point?

A
$$1^o$$ amine

B
$$2^$$ amine

C
$$3^o$$ amine

D
cannot predict

Solution

In $$3^o$$ amine, there is no hydrogen directly attached to nitrogen, due to which hydrogen bonding is not possible in them. Due to this absence of hydrogen bonding, less energy is required for boiling leading to lower boiling point.
Question 2
1 / -0

In sets a - d, only one of the set is incorrect regarding basic strength. Select it:

A

B

C

D

Solution

The basic character of amines is because of the lone pair of nitrogen. More the availability of lone pair, more will be the basic strength.
In compound as shown in IMAGE $$01$$, the lone pair of nitrogen is involved in making the ring aromatic and hence is less available due to which this compound will be least basic.
The compound as shown in IMAGE $$02$$ will be the most basic due to presence of $$2$$ nitrogen i.e., $$2$$ lone pairs instead of $$1$$ increase the basic strength.
The compound as shown in IMAGE $$03$$ will be intermediate between theses compounds.
This order is not followed in option $$C$$
Question 3
1 / -0

The amino ketone shown given undergoes a spontaneous cyclization on standing. What is the major product of this intramolecular reaction?

A

B

C

D

Solution
Question 4
1 / -0

Its basic strength is $$10^{10}$$ more than 1-dimethyl amino naphthalene due to:

A
resonance

B
steric inhibitation of resonance

C
ortho effect

D
hyperconjugation

Solution

For participation in resonance, the $$p-orbital$$ of $$N$$ should be perpendicular to the plane of the ring.
Presence of two bulky methyl groups does not allow the $$p-orbital$$ to get perpendicular to the plane and hence the lone pair of $$N$$ does not involve in resonance.
This increases the lone pair availability, thus increasing the basic strength.
Hence, $$1,8-bis(dimethylamino)$$ napthalene is more basic than $$1-dimethyl amino$$ napthalene due to steric inhibition of resonance.
Question 5
1 / -0

Which of the following arylamines will not form a diazonium salt on reaction with sodium nitrite in hydrochloric acid?

A
m-Ethylaniline

B
p-Aminoacetophenone

C
$$4$$-Chloro-$$2$$-nitroaniline

D
N-Ethyl-$$2$$-methyaniline

Solution

$$N-Ethyl-2methyl$$ aniline is a $${ 2 }^{ \circ }$$-secondary aniline (aryl).
But the remaining are primary-aryl-amines which would result in dizonium salt formation.
Question 6
1 / -0

Which of following is an example of Pinacol-Diazotization?

A

B

C

D

Solution

Amine groups are converted to $$N_2^+$$ (diazo) groups in presence of $$NaNO_2/HCl$$. When alcohols are adjacent to $$NH_2$$ group Ketones/Aldehydes are formed.
Question 7
1 / -0

In the given pair of compounds, in which pair second compound has higher boiling point than first compound?

A

B
$$HO - CH_2 - CH_2 - OH$$ and $$CH_3 - CH_2 - OH$$

C

D

Solution

The extent of hydrogen bonding is more in alcohol than in ether, as alcohol has a hydrogen directly attached to oxygen whereas ether has none. More the alcoholic groups more are the hydrogen bonding.
Now, in the case of amines, Secondary amines have more intermolecular hydrogen bonding than the tertiary amine, due to less steric crowding. More the hydrogen bonding, more the boiling point.
This is because there will be more association between the molecules and more energy will be required to break this association.
Considering the above points last option satisfied the condition.
Thus, it is the correct answer.
Question 8
1 / -0

In which of the following compound the methylenic hydrogens are the most acidic?

A
$$CH_3COCH_2CH_3$$

B
$$CH_3CH_2COOC_2H_5$$

C
$$CH_3CH_2CH(COOC_2H_5)_2$$

D
$$CH_3COCH_2CN$$

Solution

Here cyanide group pulled the bond pair electron from the chain by the inductive $$(-I$$ effect$$)$$. So, hydrogen of methyl group are because more acidic in nature. $$C-H$$ bond became labile due to pulling of electron by $$CN$$ group.
Question 9
1 / -0

Correct order of basic strength of given amines is:

A
$$\underset{2^o}{Me_2NM} > \underset{1^o}{MeNH_2} > \underset{3^o}{Me_3N > NH_3}$$ (Protic solvent)

B
$$\underset{2^o}{Et_2NH} > \underset{3^o}{Et_3H} > \underset{1^o}{EtNH_2} > NH_3$$ (Protic solvent)

C
$$Me_3N > Me_2NH > Me - NH_2 > NH_3$$ (Gas phase)

D
All are correct

Solution

Refer to Image
Basic strength (in gas phase)
Because Me is an electron donating group it increases the $$e^-$$ density on N and hence increase its tendency to donate lone pair.
Basicity of amine is directly related to the stability of amine after gaining $$H^+$$.
In the aqueous phase, the substituted ammonium cations gets stabilized not only by e^- releasing effect of the alkyl group (+I) but also by selvation with water molecules. The greater the size of ions, lesser will be the selvation and he less stabilised is the ion. Secondly, when the alkyl group is small like $$-CH_3$$ group, there is no steric hindrance to $$-H$$ bounding.
$$\therefore\quad \begin{matrix} Me_{ 2 }NH \\ 2^o \end{matrix}> \begin{matrix} Me_{ 2 }NH_2 \\1^o \end{matrix}> \begin{matrix} Me_{ 3 }N \\ 3^o \end{matrix}>NH_3$$
$$\therefore \quad \begin{matrix} Et_2NH \\ 2^o \end{matrix}>\begin{matrix} Et_3N \\ 3^o \end{matrix}>\begin{matrix} EtNH_2 \\ 1^o \end{matrix}>NH_3$$
Question 10
1 / -0

Which amine yield $$N-$$nitroso amine after treatment with nitrous acid $$(NaNO_2, HCl)?$$

A

B

C

D

Solution

$$\bullet$$ N-Nitroso amine is compound is which group is present.
$$\bullet$$ $$R,{ R }^{ 1 }$$ may be same or different alkyl or aryl group.