Some Basic Concepts of Organic Chemistry — NEET UG practice

101 questions

Practice NEET UG Some Basic Concepts of Organic Chemistry questions free — each with a detailed solution, graded instantly. Nothing is saved; log in to track your accuracy and build a streak.

Start free practice

Sample questions with solutions

Q1 · 2026

The correct IUPAC name of the following compound is:

A.
3-ethyl-5-methylheptane
B.
2,4-diethylhexane
C.
3-methyl-5-ethylheptane
D.
3,5-diethylhexane

Answer: A

In IUPAC naming, the longest continuous carbon chain is chosen as the parent chain. Here, the longest chain contains 7 carbons, so the parent name is heptane, not hexane.
Numbering of the parent chain must be done in the direction that gives the lowest set of locants to the substituents.

Given the positions of the ethyl and methyl groups on the chosen numbering,

\text{substituents are at position 3 (ethyl) and position 5 (methyl)}

When writing the name, IUPAC rules require substituent prefixes to be listed in alphabetical order (ethyl before methyl), not in order of their locants.

Therefore,

3\text{-ethyl-5-methylheptane}

Hence, the correct answer is option A.

Q2 · 2026

During Lassaigne's test, the elements present in an organic compound are converted from :

A.
Ionic form to ionic form
B.
Covalent form to ionic form
C.
Covalent form to covalent form
D.
Ionic form to covalent form

Answer: B

In an organic compound, elements like nitrogen, sulphur, and halogens are usually bonded through covalent bonds, so they cannot be detected by ordinary ionic tests directly.
Lassaigne's test converts these covalently bonded elements into ionic sodium salts by fusing the compound with sodium metal.

This fusion reaction can be represented as:

\text{Organic compound (covalent N, S, X)} + \text{Na} \longrightarrow \text{NaCN, Na}_2\text{S, NaX (ionic)}

Since these sodium salts are ionic, they readily dissociate in water and give the usual ionic reactions used for identification.

Hence, the elements are converted from covalent form to ionic form, making option B correct.

Q3 · 2026

The following carbocation is stabilized by the interaction of the empty $p$ orbital with:

A.
empty $\sigma^*$ and empty $\pi^*$ orbitals
B.
filled $\sigma$ and filled $\pi$ orbitals
C.
empty $\sigma$ and empty $\pi^*$ orbitals
D.
empty $\sigma^*$ and filled $\pi$ orbitals

Answer: B

A carbocation has an empty $p$ orbital, and it becomes more stable whenever a neighbouring filled orbital can donate electron density into this empty orbital.
If the carbocation is next to a double bond (or lone pair), electron density from the adjacent filled $\pi$ orbital delocalizes into the empty $p$ orbital. This is called resonance, and it spreads out the positive charge, adding stability.
If there are $\sigma$ C–H or C–C bonds adjacent to the empty $p$ orbital, their electrons can also overlap with it. This donation from a filled $\sigma$ orbital into the empty $p$ orbital is called hyperconjugation, and it too stabilizes the carbocation.
So the carbocation is stabilized by two filled orbitals donating into the empty $p$ orbital — a filled $\pi$ orbital (resonance) and filled $\sigma$ orbitals (hyperconjugation).

Hence, the correct answer is option B: filled $\sigma$ and filled $\pi$ orbitals.

Do more with a free account

Take it as a timed mock test
Build a daily streak
Track accuracy & progress
Save questions to your library