Chemistry Test

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Chemistry Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following gives benzoic acid on oxidation?

A

Chlorophenol

B

Chlorotoluene

C

Chlorobenzene

D

Benzyl chloride

Solution

Benzyl chloride gives benzoic acid on oxidation.
The reaction takes place as follows:
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{Cl}+2 \mathrm{KOH}+2[\mathrm{O}] \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOK}+\mathrm{KCl}+\mathrm{H}_{2} \mathrm{O}\)
Question 2
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Among the following, the aromatic compounds are

Choose the correct answer from the following options.

A
Only (A) and (B)

B
Only (B) and (C)

C
(B),(C) and (D)

D
(A),(B) and (C)

Solution

(b) For aromatic compounds following conditions should be satisfied

(i) planarity

(ii) complete delocalisation of \(\pi\)-electrons in the ring.

(iii) presence of \((4 n+2) \pi\)-electrons in the ring, where \(n=0,1,2, \ldots\).

This is called Huckel rule.

\(\text { e.g. } \mathrm{n}=0,2 \pi \mathrm{e}^{-} \)

\(\mathrm{n}=1,6 \pi \mathrm{e}^{-} \)

\(\mathrm{n}=2,10 \pi \mathrm{e}^{-}\)
Question 3
1 / -0

In which reaction acetamide is converted to methanamine?

A
Gabriel phthalimide synthesis

B
Carbylamine reaction

C
Stephen’s reaction

D
Hoffmann bromamide reaction

Solution

In Hoffmann bromamide reaction, acetamide is converted to methanamine.
When an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide, degradation of amide takes place leading to the formation of primary amine. This reaction involving degradation of amide and is popularly known as Hoffmann bromamide degradation reaction.
\(\underset{\text{Acetamide}}{\mathrm{CH}_{3} \mathrm{CONH}_{2}}+\mathrm{Br}_2+4 \mathrm{NaOH} \stackrel{\Delta}{\rightarrow}\underset{\text{Methanamine}} {\mathrm{CH}_{3} \mathrm{NH}_{2}}+2 \mathrm{NaBr}_{2}+\mathrm{Na}_{2} \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}\)
Question 4
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Pick out the option that is not a functional group from the following.

A
Hydroxyl group

B
Benzene group

C
Aldehyde group

D
Carboxylic acid group

Solution

Benzene group is not a functional group. Functional groups can be defined as an atom or group of atoms joined in a specified manner which is responsible for the characteristic chemical properties of the organic compound. Hydroxyl group (-OH), aldehyde group (-CHO), and carboxylic acid group (-COOH), all cause changes in chemical properties when attached to the hydrocarbon chains.
Question 5
1 / -0

Which one of the following is the functional group in Propanone?

A
Ketone

B
Carboxylic acid

C
Alcohol

D
Aldehyde

Solution

In propanone((CH₃)₂CO), the functional group present is a ketone (- C=O) group.
Acetone also is known as Propanone is an organic compound with the formula (CH₃)₂CO.
In propanol (C₃H₇OH), the functional group present is an alcohol (- OH) group.
In propanoic acid (C₂H₅COOH), the functional group present is a carboxylic acid (- COOH) group.
In ethanal (CH₃CHO), the functional group present is an aldehyde(- CHO) group.
In propanal (C₂H₅CHO), the functional group present is an aldehyde(- CHO) group.
Question 6
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______ is the only non-metal placed with alkali metals.

A
Francium

B
Hydrogen

C
Rubidum

D
Caesium

Solution

Hydrogen is the only non-metal placed with alkali metals.
Question 7
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Which of the following reaction does not involve Hoffmann bromamide degradation?

A

B

C

D

Solution

Methyl ketone gives haloform reaction to form carboxylic acid salt which on heating with \(\mathrm{NH}_3\) forms amide. Amide is reduced to amine by \(\mathrm{LiAl}_4\).

(a) Benzamide undergoes Hofmann bromamide degradation when treated with bromine and aqueous sodium hydroxide, to produces benzylamine.

(b) Benzene nitrile undergoes alkaline hydrolysis to form acidamide which further give Hofmann bromamide degradation.

(d) Acid chloride undergoes nucleophilic substitution with NH3 to form acid amide. Acid amide undergoes Hoffmann bromamide degradation.
Question 8
1 / -0

If a \(Zn ^{2+}/ Zn\) electrode is diluted \(100\) times, then the emf will show:

A
An increase of \(59 ~mV\)

B
A decrease of \(59 ~mV\)

C
An increase of \(29.5~ mV\)

D
A decrease of \(29.5 ~mV\)

Solution

We know,
\(E _{ cell }=\frac{0.059}{2} \log \frac{1}{ C }\)
where, \(C =100\) times
\(E _{\text {cell }}=-\frac{0.059}{2} \log \frac{1}{100}\) \((\because\log \frac{1}{100}=-2)\)
\(=-\frac{0.059}{2} \times -2\)
\(=0.059 V \)
\(=59~ mV\)
Thus, if the \(Zn ^{2+} / Zn\) electrode is diluted to \(100\) times than the emf is decreaseof \(59~ mV\).
Question 9
1 / -0

Find out the major products from the following reaction sequence.

A

B

C

D

Solution
Question 10
1 / -0

The value of \(\mathrm{K}_{c}\) for the reaction: \(\mathrm{A}+3 \mathrm{~B} \rightleftharpoons 2 \mathrm{C}\) at \(400^{\circ} \mathrm{C}\) is \(0.5\) then calculate the value of \(\mathrm{K} p\).

A
\(1.64 \times 10^{-4} \mathrm{~atm}\)

B
\(1.64 \times 10^{-6} \mathrm{~atm}\)

C
\(1.64 \times 10^{-5} \mathrm{~atm}\)

D
\(1.64 \times 10^{-3} \mathrm{~atm}\)

Solution

\(\mathrm{T}=673 \mathrm{~K}, \mathrm{~K}_{\mathrm{c}}=0.5\)
We know that:
\(\mathrm{R}=0.082 \text { litre bar } \mathrm{K}^{-1} \mathrm{~mol}^{-1}\)
\(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}[\mathrm{RT}]^{\mathrm{n}}\)
Where,
\(\mathrm{n}=\) Concentration of products - concentration of reactants
\(\mathrm{n}=2-4\)
\(\mathrm{n}=-2\)
Then,
\(\mathrm{K}_{\mathrm{p}}=0.5 \times(0.082 \times 673)^{-2}\)
\(\mathrm{~K}_{\mathrm{p}}=1.64 \times 10^{-4} \mathrm{~atm}\)