Amines Test

Result

Amines Test - 52

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Weekly Quiz Competition

Question 1
1 / -0

Arrange the following in the increasing order of their basicities.
I. p - Toluidine
II. N, N - Dimethyl - p - toluidine
III. p - Nitroaniline
IV. Aniline

A
III < IV < I < II

B
II < IV < I < III

C
III < II < I < IV

D
I < IV < III < II

Solution

N,N-Dimethyl p-toluidine is most basic whereas p-nitroaniline is least basic.
II +I effect of two Me groups of N atom and Me group at p position (+I and hyperconjugative. effects) are responsible for highest basicity.
I +I and H.C. effects of (Me) group at p position makes p-toluidine more basic than aniline.
IV (Standard) (Aniline)
III -I and -R effect of (-$$N{O}_{2}$$) group at p position are responsible for least basicity.
Question 2
1 / -0

N $$\displaystyle -$$ Ethyl pthalimide on hydrolysis gives:

A
methyl alcohol

B
ethyl amine

C
dimethyl amine

D
diethyl amine

Solution

The alkaline hydrolysis of N-ethylpthalimide gives ethyl amine.
Question 3
1 / -0

Which compound will liberate $$\displaystyle CO_{2}$$ from $$\displaystyle NaHCO_{3}$$ solution :

A
$$\displaystyle CH_{3}CO\:NH_{2}$$

B
$$\displaystyle CH_{3}NH_{2}$$

C
$$\displaystyle \left ( CH_{3} \right )_{4}N^{+}OH^{-}$$

D
$$\displaystyle CH_{3}N^{+}H_{3}Cl^{-}$$

Solution

$$\displaystyle CH_{3}N^{+}H_{3}Cl^{-}$$ will react with $$\displaystyle NaHCO_{3}$$ to liberate $$\displaystyle CO_{2}$$.
$$\displaystyle CH_{3}N^{+}H_{3}Cl^{-} + \displaystyle NaHCO_{3} \rightarrow CH_{3}NH_2 + NaCl + H_2O + \displaystyle CO_{2}$$
Question 4
1 / -0

Which of the following is the product of the reaction shown above?

A
$$(CH_3)_2CHCH_2CH_2NHNH_2$$

B
$$[(CH_3)_2CHCH_2CH_2]_2NH$$

C
$$(CH_3)_2CHCH_2CH_2NH_2$$

D
$$(CH_3)_2CHCH_2CH_2CONH_2$$

Solution
Question 5
1 / -0

Which of the following diazonium salt is relatively stable of $$0-5^\circ C$$- :

A
$$CH_{3}CH_2N\equiv N\left. \right \} ^\oplus Cl^{-} $$

B
$$CH_{3}-C(CH_{3})_{2}-N\equiv N\left. \right \} ^\oplus Cl^{-} $$

C
$$CH_{3}-N\equiv N\left. \right \} ^\oplus Cl^{-} $$

D
$$(CH_{3})_{3}C-N\equiv N\left. \right \} ^\oplus Cl^{-} $$

Solution

The diazonium salt $$CH_{3}-N\equiv N\left. \right \} ^\oplus Cl^{-}$$ obtained by treatment of methylamine with nitrous acid is relatively stable at $$0-5^\circ C$$.
Question 6
1 / -0

Which of the following diazonium salt is relatively stable at $$\displaystyle 0-5^{\circ}$$C?

A
$$\displaystyle CH_{3}-N\equiv N^{\oplus }Cl^{-}$$

B
$$\displaystyle CH_{3}CH\left ( CH_{3} \right )-N\equiv N ^{\oplus }Cl^{-}$$

C
$$\displaystyle C_{6}H_{5}-N\equiv N ^{\oplus }Cl^{-}$$

D
$$\displaystyle \left ( CH_{3} \right )_{3}C-N\equiv N ^{\oplus }Cl^{-}$$

Solution

Diazonium salts obtained from aliphatic primary amines are unstable. Those obtained from aromatic primary amines are relatively stable.
Benzene diazonium chloride $$(\displaystyle C_{6}H_{5}-N\equiv N ^{\oplus }Cl^{-})$$ is obtained from aromatic primary amine. Hence, it is relatively stable.
Question 7
1 / -0

Of the following statements :
$$(P)\:C_{6}H_{5}N=CH-C_{6}H_{5}$$ is a Schiff's base.
$$(Q)\:$$A dye is obtained by the reaction of aniline and $$C_{6}H_{5}N^+ \equiv NCl$$.
$$(R)\:C_{6}H_{5}CH_{2}NH_{2}$$ on treatment with $$[NaNO_{2}+HCl]$$ gives diazonium salt.
$$(S)\:p-$$Toluidine on treatment with $$[HNO_{2}+HCl]$$ gives diazonium salt.

A
only $$(P)\:and\:(Q)$$ are correct

B
only $$(P)\:and\:(R)$$ are correct

C
only $$(R)\:and\:(S)$$ are correct

D
$$(P),\:(Q)\:$$and$$\:(S)$$ are correct

Solution

A Schiff base is a substituted imine$$(R_2C=N-R')$$. It is used as antioxidant. Thus $$(P)\:C_{6}H_{5}N=CH-C_{6}H_{5}$$ is a Schiff's base. Thus statment $$(P)$$ is correct.

Aniline reacts with $$C_{6}H_{5}N^+\equiv NCl$$ to form higly coloured azo dye. This reaction is known as azo coupling. Thus the statement $$(Q)$$ is correct.

$$\:C_{6}H_{5}CH_{2}NH_{2}$$ is a primary aliphatic amine. When it is treated with $$[NaNO_{2}+HCl]$$, it gives unstable diazonium salt which decomposes to form benzyl alcohol. Hence, statement $$(R)$$ is incorrect.

p-Toluidne is an aromatic primary amine. Hence, it on treatment with $$[NaNO_{2}+HCl]$$ gives diazonium salt. Hence, statement $$(S)$$ is correct.

Option D is correct.
Question 8
1 / -0

Alkyl isocyanides ($$R N^{\bigoplus} \equiv C^{\ominus}$$) are reduced to 2$$^{\circ}$$ amines($$R-NH-CH_3$$) with:

A
$$LAH$$

B
$$NaBH_4$$

C
$$HI + P$$

D
$$H_2 / Pt$$

Solution

Alkyl isocyanides $$(RN≡C)$$ are reduced to $$ 2^o$$ amines with sodium and alcohol. Alkyl isocyanides (RN≡C) are reduced to $$2 ^o$$ amines with $$H_2 /Pt$$.
Question 9
1 / -0

Arrange the following in their decreasing order of acidity.

A
$$III > IV > I > II$$

B
$$I > II > III > IV$$

C
$$IV > III > II > I$$

D
$$II > III > I > IV$$

Solution

The decreasing order of acidity is $$III > IV > I > II$$.
In (III) $$+I$$ effect of two $$Me$$ groups makes it most basic
In (I) electron-withdrawing effect of the carbonyl group makes it less basic than ammonia.
In (II) electron-withdrawing effect of two ($$C = O$$) groups makes it least basic.
Question 10
1 / -0

The decreasing order of basic characters of the following is :

A
III > IV > I > II

B
II > I > IV > III

C
IV > III > II > I

D
I > II > III > IV