Amines Test

Result

Amines Test - 51

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Which of the following forms unstable diazonium ion when treated with $$NaNO_{2}$$ in aqueous $$HCl$$?

A
p-nitrotoluene

B
ethylamine

C
N,N-dimethyl aniline

D
All of the above

Solution

Aromatic diazonium salts are stable if kept cold $$(0-5^oC)$$ and in solution, but they often decompose violently when isolated. Their stability (compared to alkane diazonium salts) is due in part to resonance stabilization of the diazonium ion and in part to the very high energy of the aryl cation that would result from the loss of $$N_2$$(g).
Therefore except ethylamine all other given compounds forms stable diazonium ion when treated with $$NaNO_2$$ in aqueous $$HCl$$.

Hence, the correct option is B.
Question 2
1 / -0

In the reaction series: $${ C }_{ 6 }{ H }_{ 5 }Cl\xrightarrow [Cu _{ 2 }O,\: 200^oC ]{NH_3} X; X\xrightarrow [ { 0-5 }^{ 0 }C ]{ { HNO }_{ 2 } } Z; X+Z\rightarrow A$$, the number of $$\sigma $$ and $$\pi $$ bonds in $$A$$ are:

A
$$25$$$$\sigma $$ and $$6$$$$\pi $$

B
$$25 \sigma $$ and $$7 \pi $$

C
$$27\sigma $$ and $$7\pi $$

D
$$27\sigma $$ and $$6\pi $$

Solution

Nucleophilic substitution of chloro-benzene with ammonia gives aniline.

$${ C }_{ 6 }{ H }_{ 5 }Cl\xrightarrow [ { Cu }_{ 2 }O,\: { 200 }^{ 0 }C ]{ { NH }_{ 3 } } \underset {X}
{C_6H_5NH_2} +HCl$$

Diazotization of aniline gives benzene diazonium chloride.

$$ \underset {X} {C_6H_5NH_2} \xrightarrow [ { 0-5 }^{ 0 }C ]{ { HNO }_{ 2 } } \underset {Z} {C_6H_5N_2Cl}$$

Azo coupling of aniline with benzenediazonium chloride gives p-aminoazobenzene.

$$ \underset {X} {C_6H_5NH_2}+ \underset {Z} {C_6H_5N_2Cl}\rightarrow \underset {A}
{C_6H_5-N=N-C_6H_4-NH_2}$$

The number of $$\sigma $$ and $$\pi $$ bonds in $$C_6H_5-N=N-C_6H_4-NH_2$$ are $$27$$$$\sigma $$ and $$7$$$$\pi $$.

Hence, the correct option is C.
Question 3
1 / -0

The increasing order of basic strength for the following compounds is:
(I) $$CH_{3}-CH_{2}-NH_{2}$$
(II) $$CH_{3}-NH-CH_{3}$$
(III) $$CH_{3}-NH-Ph$$
(IV) $$CH_{3}-NH-CHO$$

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

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

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

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

Solution

Aliphatic amines are stronger bases than ammonia and aromatic amines are weaker bases than $$N{ H }_{ 3 }$$ and $$3°>2°>1°\rightarrow $$ basic strength in a gaseous phase.

Thus, the increasing order of basic strength from the options can be
$$IV<III<II>I$$

$$II,III,IV$$ are secondary amines. $$III$$ is aromatic, hence is less basic than $$II, IV$$. The lone pair of amine resonates with the $$-CHO$$ bond, hence it cannot be freely available, thus it is less basic.

Option C is correct.
Question 4
1 / -0

Predict the product (T) for the following reaction sequence.
$$C_6H_5-CH=O \xrightarrow{KCN / OH^-} R \xrightarrow{CrO_2 / H^{\oplus}}S \xrightarrow[(ii)\, Acidification]{(i)\, conc\, KOH}T$$ :

A
Cinnamic acid

B
Mandelic acid

C
Benzilic acid

D
Benzoic acid & Benzyl alcohol

Solution

This is an example of Benzil-benzilic acid rearrangement. Benzaldehyde reacts with $$KCN$$ in an alkaline medium to form benzaldehyde cyanohydrin.
Oxidation with acidified chromium trioxide gives benzil. Treatment with conc KOH followed by acidification gives rearranged product benzillic acid.
Question 5
1 / -0

The product $$U$$ is:

A

B

C

D

Solution

Benzoic acid reacts with thionyl chloride to form benzoyl chloride. Reaction with dinitrogen pentoxide introduces nitro group. Reaction with lithium dimethyl cuprate converts acid chloride to methyl ketone. The carbonyl group is protected with ethylene glycol, nitro group is hydrogenated and the protecting group is removed to form $$1-(4-aminophenyl)$$ $$ethanone$$. This is followed by diazotization and coupling with phenol to obtain the end product $$U$$.
Question 6
1 / -0

The correct basicity order of various atoms is :

A
1 < 2 < 3 < 4

B
2 < 3 < 4 < 1

C
3 < 2 < 1 < 4

D
3 < 2 < 4 < 1

Solution

Basicity order is: N $$sp^3$$ (localised lone pair) > N $$sp^2$$ (localised lone pair) > N $$sp^2$$ (Delocalised lone pair).
Question 7
1 / -0

The correct order of basicity of the above compounds is given.
Identify the correct statement(s).

A
In III, lone pair of electrons is involved in delocalisation but not in II.

B
In IV, lone pair of electrons is involved in delocalisation but not in III.

C
I is more basic than II because six membered ring is more stable than five membered ring.

D
In III, lone pair of electrons is present in sp orbital but in I, II and IV lone pair of electrons are present in $$sp^{2}$$ orbital.

Solution

In Pyrrole (IV), the lone pair electrons of the nitrogen atom is actively involved with the two carbon-carbon double bonds in the 5-member ring to form a conjugated system of pi electrons, leading to greater stability of the molecule.
Pyridine (III), on the other hand, already has a stable conjugated system of 3 double bonds in the aromatic hexagonal ring, like benzene. Hence the lone pair electrons on the N atom in Pyridine can be easily donated to a Hydrogen ion or a Lewis acid.
Therefore, Pyridine is a stronger base than Pyrrole.
Question 8
1 / -0

Which compound does not give positive test in Lassaigne's test for nitrogen?

A
Urea

B
Hydrazine

C
Azobenzene

D
Phenyl hydrazine

Solution

In Lassaigne's test, the organic compound is fused with sodium.
Carbon and nitrogen present in the organic compound gives sodium cyanide ($$NaCN$$).
This is further used in the test for nitrogen.
Hydrazine $$ (H_2N-NH_2)$$ do not contain carbon. It cannot form $$NaCN$$ on fusion with sodium.
Hence, hydrazine cannot give Lassaigne's test for nitrogen.
Question 9
1 / -0

Which of the following is weakest base?

A

B

C
$$H_2N-CH_2-NO_2$$

D
$$CH_3NH-CH=O$$

Solution

In amide lone pair electron of 'N' are in conjugation with carbonyl oxygen.

So amides are not basic they are neutral.

Option D is correct.
Question 10
1 / -0

The following product(s) will be obtained from the above reaction:

A

B

C

D

Solution

Both amino groups react with nitrous acid to form a diazonium salt.
The aliphatic diazonium salt decomposes to form alcohol whereas the aromatic diazonium salt is stable.