Chemistry Test 225

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Chemistry Test 225

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Weekly Quiz Competition

Question 1
4 / -1

Hydrogenation of alkenes can be carried out in the presence of

A
copper

B
zinc

C
aluminium

D
nickel
Solution
- Unsaturated hydrocarbons such as alkenes add Hydrogen in the presence of catalysts such as Ni to form saturated hydrocarbons. This is called an addition reaction.
- The reaction is commonly used in the hydrogenation of vegetable oils using a Nickel catalyst.
Question 2
4 / -1

Arrange the following compounds in increasing order of their basic strength.

A
III > IV > I > II

B
IV < III < II < I

C
III < II < I < IV

D
IV < II < III < I

Solution
Question 3
4 / -1

Benzene diazonium chloride when reacts with hypophosphorus acid produces

A
Benzene

B
Phenyl phosphate

C
Phenol

D
Phenyl isocyanide

Solution
Question 4
4 / -1

On hydrolysis maltose gives:

A
One molecule of glucose and one molecule of galactose

B
Two molecules of fructose

C
One molecule of glucose and one molecule of fructose

D
Two molecules of glucose

Solution

Maltose can be produced from starch by hydrolysis in the presence of the enzyme diastase. It can be broken down into two glucose molecules by hydrolysis. In living organisms, the enzyme maltase can achieve this very rapidly. In the laboratory, heating with a strong acid for several minutes will produce the same result.
Question 5
4 / -1

Which of the following reacts with alcohols to form esters besides carboxylic acids?

A
Aldehydes

B
Amines

C
Ethers

D
Acyl chloride
Solution
- The correct answer is option D
- Acyl chloride
- Esters are derived from carboxylic acids. A carboxylic acid contains the -COOH group, and in an ester, the hydrogen in this group is replaced by a hydrocarbon group of some kind.
- We shall just be looking at cases where it is replaced by an alkyl group, but it could equally well be an aryl group (one based on a benzene ring).
- A common ester - ethyl ethanoate
- The most commonly discussed ester is ethyl ethanoate. In this case, the hydrogen in the -COOH group has been replaced by an ethyl group.
Question 6
4 / -1

lUPAC names of the given structures are

A
(i) hexane, (ii) 3-methylbutane

B
(i) Isopentane, (ii) 2,2-dimethylbutane

C
(i) 3-ethylbutane, (ii) Isopentane

D
(i) 3-methylpentane, (ii) 2-methylbutane

Solution

Option D is correct.

For structure (i), identify the longest continuous carbon chain: it contains 5 carbon atoms, so the parent hydrocarbon is pentane.

Number the carbon chain from the end nearer the substituent so the substituent receives the lowest possible locant. A methyl substituent lies on carbon 3, giving the correct IUPAC name 3-methylpentane.

For structure (ii), the longest continuous carbon chain contains 4 carbon atoms, so the parent hydrocarbon is butane.

Number the chain to assign the lowest locant to the substituent. The methyl group is on carbon 2, giving the correct IUPAC name 2-methylbutane.

Therefore the pair of IUPAC names is 3-methylpentane and 2-methylbutane, matching option D.
Question 7
4 / -1

Write the IUPAC name of (CH₃)₂CHCHO?

A
2,2-Dimethylpropanal

B
3-Hydroxypropanal

C
But-3-en-2-one

D
None of these

Solution

The IUPAC name of (CH₃)₂CHCHO is 2-methylpropanal and its chemical name is Isobutyraldehyde.
Question 8
4 / -1

Which statement is true about the SN² mechanism?

A
The rate of reaction increases on increasing strength of the nucleophile.

B
The reaction is faster in polar aprotic solvents.

C
The rate of reaction increases as the leaving group ability increases.

D
All of the above.

Solution

Answer: D - All of the above.

Rate law: An S_N2 reaction is bimolecular in the rate-determining step and follows rate = k [nucleophile][substrate]. Therefore, increasing the strength or concentration of the nucleophile increases the rate.

Solvent effect: Polar protic solvents hydrogen-bond and solvate anionic nucleophiles, lowering their reactivity. Polar aprotic solvents do not solvate anions strongly, so nucleophiles remain more available and the S_N2 reaction is faster in polar aprotic solvents.

Leaving-group ability: A better leaving group stabilises the negative charge after bond breaking and lowers the activation energy. Thus, substrates with better leaving groups react faster by the S_N2 pathway; for example, I⁻ > Br⁻ > Cl⁻ as leaving groups.

All three statements are therefore correct, so the correct option is D.
Question 9
4 / -1

Carbonyl compounds when oxidized easily to form carboxylic group involves

A
aldehydes

B
ketones

C
haloalkanes

D
esters

Solution
Question 10
4 / -1

Propanamide on treatment with bromine in an aqueous solution of sodium hydroxide gives:

A
Propanamine

B
Butanamide

C
Ethanamine

D
Butanamie

Solution