Amines Test

Result

Amines Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

The strongest base among the following is:

A
$${ C }_{ 6 }H_{ 5 }NH_{ 2 }$$

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

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

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

Solution

Due to the +I effect of alkyl groups, the electron density on nitrogen increases and thus the availability of the lone pair of electrons to proton increases and hence the basicity of amines also increases.

$$(CH_3)_2NH$$ this is answer
Question 2
1 / -0

Which of the following is the strong base?

A

B

C

D

Solution

Among the given options, compound "A" has the highest tendency to donate electrons. So, it is the strongest base.

Hence, Option "A" is the correct answer.
Question 3
1 / -0

Which nitrogen is prtonated redily in the guanidine?

A
1

B
2

C
3

D
None of these

Solution

Nitrogen which is attached with double bond i.e. 1 nitrogen is readily protonated because double-bonded nitrogen is $$sp^2$$ hybridized and on protonating double-bonded nitrogen enables effective resonance stabilization. The positive charge can be delocalized across the molecule

$$\mathbf{Hence\ the\ correct\ answer\ is\ option (A)}$$
Question 4
1 / -0

Strongest base is:

A

B

C

D

Solution

Order of basicity of given compounds:

A (lesser electronegativity of Nitrogen) > D (greater electronegativity of oxygen) > B (sp-2 hybridisation of nitrogen) > C (delocalisation of electrons of nitrogen)

Hence, Option "A"' is the correct answer.
Question 5
1 / -0

In the following compounds, piperidine (I) pyridine(II) morpholine (III) and pyrrole (IV) the order of basicity is:

A
IV > I > III > II

B
III > I > IV > II

C
II > I > III > IV

D
I > III > II > IV

Solution

Pyrrole is least basic because the lone pair of $$N$$ is involved in the resonance and thus they are not available for formation of a new bond with proton.

Pyridine is more basic than pyrrole as in pyridine the lone pair of $$N$$ is not involved resonance.

Piperidine is most basic because the lone pair is present in $$sp^3$$ hybrid orbital of $$N$$ while In pyridine it is present in $$sp^2$$ hybrid orbital. Greater the $$s$$ character of orbital more strongly the electrons will be bonded with the atom and less will be the basicity.

morpholine is less basic than piperidine but more basic than pyridine. In morphiline the electron withdrawing tendency of ring $$O$$ makes the lone pair of $$N$$ to be less available for the bonding with proton.

$$ Piperidine>\ morpholine>\ pyridine>\ pyrrole$$

$$\mathbf{I > III > II > IV}$$

$$\mathbf{Hence\ the\ correct\ answer\ is\ option\ (D)}$$
Question 6
1 / -0

The basicity of aniline is less than that of cyclohexylamine. This is due to:

A
-R effect of $$ - NH_2 \ group $$

B
-I effect of $$ - NH_2 \ group$$

C
+ R effect of $$ NH_2 \ group $$

D
Hyperconjugation effect

Solution

The basicity of aniline is less than that of cyclohexylamine because of + R effect of $$ NH_2 \ group $$ in aniline.
Hence, Option "C" is the correct answer.
Question 7
1 / -0

Gabriel phthalimide synthesis is used in the preparation of:

A
$$1^o$$ amine

B
$$2^o$$ amine

C
$$3^o$$ amine

D
$$4^o$$ amine

Solution

primary amine
Question 8
1 / -0

The compound

A
Alkyne, $$3^{\circ} amine$$

B
Alkene, $$2^{\circ} amine$$

C
Alkene, $$3^{\circ} amine$$

D
Alkyne, $$2^{\circ} amine$$

Solution
Question 9
1 / -0

Amines are more basic than:

A
alcohols

B
ethers

C
esters

D
all of these

Solution

-OH, -COOR,-COC group
all are correct
Question 10
1 / -0

Which of the following is the most basic amine ?

A
$$CH_3-NH_2$$

B
$$ClCH_2NH_2$$

C
$$Cl_2CH-NH_2$$

D
$$CCl_3-NH_2$$

Solution

$$CCl_3−NH_2$$ this is most basic amine