General Organic Chemistry (GOC) - 1 - Free MCQ Test with solutions

If a molecule contains both carbon-carbon double or triple bonds, the two are treated as per in seeking the lowest number combination. However, if the sum of numbers turns out to be the same starting from either of the carbon chains, then the lowest number is given to the C=C double bond.

Thus, the correct name is pent-3-en-1-yne.

There are four double bonds. Hence, the number of π-electrons is 2 × 4 = 8.

- The correct order of priority for functional groups in the IUPAC system is determined by their reactivity and presence in naming conventions.
- Carboxylic acids -COOH have the highest priority due to their strong acidic nature.
- Sulfonic acids -SO₃H  come next because of similar acidity.
- Amides -CONH₂ follow due to their derivation from carboxylic acids.
- Aldehydes -CHO have lower priority than the above groups.
- Therefore, the correct order is: -COOH, -SO₃H, -CONH₂, -CHO, which corresponds to option D.

1º, 2º & 3º H’s are that which is attach at 1º , 2º & 3º carbon respectively.

The correct IUPAC name for the compound is:

2-Chloro-4,4-dimethylhexane

Here's how we arrive at this name:

• The main chain consists of six carbon atoms, making it a hexane derivative.
• There are two methyl groups attached to the fourth carbon in the chain.
• A chlorine atom is attached to the second carbon.

This systematic approach ensures that we accurately represent the structure of the compound through its IUPAC name.

To solve the question in the image regarding the number of correct names for given substituents, we need to analyze each structure and compare it to the IUPAC naming rules for organic substituents:

1. (a) Ethylmethyl: Correct. The substituent has an ethyl group attached to a central carbon, which is also connected to a methyl group.

2. (b) 1-Methylpropyl: Correct. The main chain has three carbons (propyl), with a methyl group attached to the first carbon.

3. (c) 2,3-Dimethylpropyl: Incorrect. The correct name should be isobutyl or 2-methylpropyl, as the main chain is a propyl with two methyls attached to the second carbon.

4. (d) 2,3-Dimethylbutyl: Correct. The main chain has four carbons (butyl), with methyl groups attached to the second and third carbons.

5. (e) Ethylidene: Incorrect. This group is actually an ethyl group with a double bond at the end carbon, so it should be called vinyl or ethenyl.

6. (f) 2-Methylethenyl: Correct. The main chain has two carbons with a double bond (ethenyl), and a methyl group attached to the second carbon.

7. (g) Ethynyl: Correct. This structure has a triple bond with a carbon at the end, which is characteristic of an ethynyl group.

8. (h) 2-Propenyl: Correct. The structure represents a three-carbon chain with a double bond starting from the first carbon; however, the correct name is allyl.

9. (i) Prop-1-ynyl: Correct. This name represents a three-carbon chain with a triple bond at the first carbon, which is accurately described.

Counting the correct names:

• Correct names: (a), (b), (d), (f), (g), (i)

• Incorrect names: (c), (e), (h)

There are 6 correct names among the options given. Thus, the answer to the question is Option B: 6.

A is correct: the molecule contains 5 secondary (2°) carbons.

A secondary carbon is bonded to exactly two other carbon atoms.

In the cyclohexane ring, the two carbons bearing the alkyl substituents are each bonded to three carbon atoms (two ring neighbours plus the substituent) and so are tertiary; the remaining four ring carbons are each bonded to two carbons and are therefore secondary.

The isopropyl substituent has a central carbon bonded to three carbons (tertiary) and two methyl carbons which are primary; it contributes 0 secondary carbons.

The 2-methylpropyl substituent (CH₂-CH(CH₃)-CH₃) has the CH₂ that attaches to the ring bonded to two carbons and is therefore secondary; the internal CH is tertiary and the methyls are primary, so this substituent contributes 1 secondary carbon.

Adding contributions gives 4 (ring) + 1 (side chain) = 5 secondary carbons; hence option A (5) is correct.

The IUPAC name of the molecule is:

2-Bromobenzene-1,4-dicarboxylic acid.

The structure can be described as follows:

• It features a bromine atom at the second position of the benzene ring.
• Two carboxylic acid groups are attached at the first and fourth positions of the ring.

This specific arrangement gives the molecule its distinct chemical properties and influences its reactivity. The use of "1,4" indicates the positions of the carboxylic acid groups relative to the bromine substituent.

The IUPAC name of the compound is:

2, 2, 5-Trimethyl-4-(1-methylpropyl) nonane.

• This name indicates the structure of the compound.
• It features a nonane backbone with specific methyl and propyl substituents.
• Understanding the nomenclature helps in identifying the compound's molecular arrangement.

Sigma bond is stronger than π-bond because of better overlap. All single bonds are σ bonds and all multiple bonds contain one σ and other π bonds.

Understanding Hyperconjugation: Hyperconjugation involves the interaction of sigma (σ) bonds (typically C-H bonds) of alkyl groups directly attached to a positively charged carbon (C⁺) or an unsaturated system (like a double bond).

Structure Analysis:

• In the given structure, the positively charged carbon (C⁺) is directly attached to two methyl groups (−CH₃) and a hydrogen atom.
• Each −CH₃ group contributes 3 hydrogens for hyperconjugation.
• The single hydrogen atom attached to the positively charged carbon also participates in hyperconjugation.

Total Hydrogens Involved:

• Contribution from the two methyl groups: 3 + 3 = 6
• Contribution from the single hydrogen atom on the positively charged carbon: 1
• Total: 6 + 1 = 7

Resonance and Benzene

Resonance in aromatic compounds like benzene involves the delocalisation of electrons. Certain groups can participate in this process, while others cannot.

• The key factor determining participation in resonance is the presence of a vacant orbital.
• Groups that lack a vacant orbital are unable to engage in resonance with benzene.
• Among the groups provided, identify which one does not meet this criterion.

In phenols, the presence of electron releasing groups decreases the acidity, whereas the presence of electron withdrawing groups increases the acidity, compared to phenol. Among the meta and para-nitrophenols, the latter is more acidic as the presence of -NO₂ group at the para position stabilizes the phenoxide ion to a greater extent than when it is present at the meta position. Thus, the correct order of acidity is:

Para-nitrophenol > meta-nitrophenol > phenol > methyl phenol

Geometrical isomerism requires restricted rotation around a double bond and different groups on each carbon of the double bond.

• CH₃CH₂CH=CH_2​ does not show geometrical isomerism because one carbon of the double bond has two identical groups (H and H).

• The other compounds have different groups on each carbon of the double bond, allowing geometrical isomerism.

Answer: Option A.

Primary (1°) carbon is bonded to one other carbon.

Secondary (2°) carbon is bonded to two other carbons.

Tertiary (3°) carbon is bonded to three other carbons.

There are three methyl carbons (two terminal methyl groups and one methyl substituent). Each methyl carbon is bonded to one carbon, so there are 3 primary carbons.

There are two methylene carbons (each bonded to two carbons), so there are 2 secondary carbons.

There is one methine carbon bonded to three carbons, so there is 1 tertiary carbon.

Final count: 3 primary, 2 secondary, 1 tertiary → Option A.