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Which among the following defines Meso forms of isomers?
  • a)
    Meso form is optically inactive due to external compensation.
  • b)
    The molecules of the meso isomers are chiral.
  • c)
    It can be separated into optically active enantiomeric pairs.
  • d)
    It is a single compound.
Correct answer is option 'D'. Can you explain this answer?

Rohan Singh answered
Meso forms of isomers are single compounds and their molecules are achiral hence they cannot be separated into pairs.
Additional Information: A meso compound or meso isomer is a non-optically active member of a set of stereoisomers, at least two of which are optically active. This means that despite containing two or more stereogenic centers, the molecule is not chiral. 

  • a)
    3
  • b)
    4
  • c)
    5
  • d)
    6
Correct answer is option 'D'. Can you explain this answer?

Krishna Iyer answered
General formula of ketone = CnH2nO
12x+2x+16 = 100
x = 6
So, the formula of compound is C6H12O. We have to make ketone only. These are as follow

  • a)
    3
  • b)
    4
  • c)
    5
  • d)
    7
Correct answer is option 'A'. Can you explain this answer?

Riya Banerjee answered
The compounds with each doubly bonded carbon attached to two different groups (like Cab=Cab, Cab=Ccd) exhibit geometrical isomerism i.e., cis and trans forms. The geometrical isomerism arises due to restricted rotation of double bond.
However, even though there is restricted rotation for triple bond, alkynes do not exhibit geometrical isomerism, since the triply bonded carbons are attached to one group each only.

 The observed order of the stability of the cabocation is:
  • a)
    (CH3)2CH+ < (CH3)3C+ 3+ < CH3CH2+
  • b)
    CH3+ < CH3CH2+ <(CH3)2CH+ < (CH3)3C+
  • c)
    CH3CH2+ <(CH3)2CH+ < (CH3)3C+ < CH3+
  • d)
    CH3+ < CH3CH2+ < (CH3)3C+ <(CH3)2CH+
Correct answer is option 'B'. Can you explain this answer?

Rajesh Gupta answered
Alkyl groups directly attached to the +vely charged carbon stabilize the carbocations due to inductive and hyperconjugation effects.
Inductive effect:
→ Stability of carbocation 
→ More number of +I group more stable carbocation.

Hyperconjugation:
Stability 
∝Number of canonical structures
              ∝Number of H (alpha hydrogen)

  • a)
    3
  • b)
    5
  • c)
    7
  • d)
    8
Correct answer is option 'C'. Can you explain this answer?

Neha Joshi answered
The correct answer is Option B.
 
5 isomers are possible. C5H10(CnH2n). Molecules having the CnH2n formulas are most likely to be cyclic alkanes. The five isomers are:

(1)   Cyclopentane 
(2) 1-Methylcyclobutane  
(3) 1-Ethylcyclopropane 
(4) 1,1-Dimethylcyclopropane 
(5) 1,2-Dimethylcyclopropane
 

[JEE Main 2014 Online Exam]
  • a)
    a
  • b)
    b
  • c)
    c
  • d)
    d
Correct answer is option 'B'. Can you explain this answer?

The correct answer is option 'B' - sp2 and sp.

Explanation:
In allene (C3H4), there are two carbon atoms, each with three sigma bonds. To determine the hybridization of the carbon atoms, we need to look at their bonding and lone pairs of electrons.

1. Hybridization of the first carbon atom:
- The first carbon atom is bonded to two other carbon atoms and one hydrogen atom.
- It has three sigma bonds, which means it will have three hybrid orbitals.
- The carbon atom also has one lone pair of electrons.
- The presence of three sigma bonds and one lone pair indicates sp2 hybridization.
- Therefore, the first carbon atom is sp2 hybridized.

2. Hybridization of the second carbon atom:
- The second carbon atom is also bonded to two other carbon atoms and one hydrogen atom.
- It also has three sigma bonds, which means it will have three hybrid orbitals.
- The carbon atom does not have any lone pairs of electrons.
- The presence of only three sigma bonds indicates sp hybridization.
- Therefore, the second carbon atom is sp hybridized.

In summary, in allene (C3H4), the type of hybridization of the carbon atoms is sp2 and sp.

Consider all possible isomeric ketones, including stereoisomers of Molecular weight 100. All these isomers are independently reacted with NaBH4
(Note stereoisomers are also reacted separately). The total number of ketones that give a racemic product(s) is/are
 
[JEE Advanced 2014] 
  • a)
    3
  • b)
    4
  • c)
    5
  • d)
    6
Correct answer is option 'C'. Can you explain this answer?

Pooja Shah answered
The general formula of isomeric ketone having molecular mass 100 is C6​H12​O [6×12+12×1+16].
The possible structure will be :
 
The number of ketones that gives racemic mixture with NaBH4​ is 5 as the ketone with chiral center will give diastereoisomer with NaBH4​

The isomer of ethanol is:
  • a)
    Dimethyl ether 
  • b)
    Diethyl ether  
  • c)
    Methanol
  • d)
    Acetone
Correct answer is option 'A'. Can you explain this answer?

The isomer of ethanol(CH3-CH2-OH) is dimethyl ether(CH3-O-CH3). This is a functional isomer of ethanol.
Explanation:
  • Functional isomers are the ones with the same molecular formula but different functional groups.
  • The molecular formula for dimethyl ether is C2H6O similar to ethanol.
  • Dimethyl ether has the structural formula CH3OCH3 and ethanol has the structural formula CH3CH2OH.
  • Therefore, dimethyl ether is the functional isomer of ethanol as both have the same molecular formula but one is ether and the other is alcohol. 

 Some meta-directing substituents in aromatic substitution are given. Which one is most deactivating? [NEET 2013]
  • a)
    –SO3H
  • b)
    –COOH
  • c)
    –NO2
  • d)
    –C ≡ N
Correct answer is option 'C'. Can you explain this answer?

Anu Bajaj answered
Decreasing order of deactivating effect of the given  m-directing group is > NO2 > – CN > – SO3H > – COOH
—NO2 group is most deactivating group due to strong – E, – I and – M effects.

How many structural isomers are possible with molecular formula C4H10O ?
  • a)
    3
  • b)
    5
  • c)
    7
  • d)
    8
Correct answer is option 'C'. Can you explain this answer?

Preeti Iyer answered
The formula that these isomers contain no rings or double bonds. The isomers must be alcohols and ethers.

Let's start by drawing the alcohols.
Start with four carbons in a row and put an  OH  group in every possible position.

This gives us

1. CH3CH2CH2CH2OH, butan 1-ol

and

2. CH3CH2CH(OH)CH3, butan-2-ol

Now use a 3-carbon chain with a CH3 on the middle carbon.

This gives
(CH3)2CHCH2OH, 2-methylpropan-1-ol

and

(CH3)3COH, 2-methylpropan-2-ol.

Now for the ethers.
Let's start with five atoms in a row.
This gives

CH3CH2CH2OCH3, 1-methoxypropane

and

CH3CH2OCH2CH3 , ethoxyethane

Finally, the only choice for a branched-chain ether is

(CH3)2CHOCH3 , 2-methoxpropane.

And there are your seven isomers of C4H10O.

A pair of enantiomers is possible for _______ isomer of 2,2-dibromobicyclo [2.2.1] heptane.
  • a)
    cis
  • b)
    trans
  • c)
    both cis and trans
  • d)
    none of the above
Correct answer is option 'A'. Can you explain this answer?

Tejas Verma answered
Only one pair of enantiomers is possible for cis-2,2-dibromobicyclo [2.2.1] heptane. The trans arrangement of one carbon bridge is structurally impossible. Such a molecule would have too much strain.

C3H6O represents an aldehyde and a ketone, is a type of:
  • a)
    Chain isomerism
  • b)
    Functional group isomerism
  • c)
    Metamerism
  • d)
    Position isomerism
Correct answer is option 'B'. Can you explain this answer?

Preeti Iyer answered
The aldehyde and ketone that can be represented by molecular formula C3​H6​are propanal and propanone respectively.
Aldehyde - Propanal - CH3​CH2​CHO
Ketone - Propanone - CH3​COCH3​
These two carbon compounds are called functional isomers. They do not have the same physical properties since their functional group is different.
When hydrogen is present on one side of C=O group and other side is R, (RCHO), then it is an aldehyde group. When R is present on both sides of C=O group, then it is a ketone.

 Heterolytic cleavage is a way to cleave the:
  • a)
    Non-ionic bonds
  • b)
    Ionic bonds
  • c)
    Covalent bonds
  • d)
    Polar bonds
Correct answer is option 'C'. Can you explain this answer?

Rahul Bansal answered
In heterolytic cleavage, a covalent bond breaks in such a way that one fragment gets both of the shared electrons. In homolytic cleavage, a covalent bond breaks in such a way that each fragment gets one of the shared electrons. The word heterolytic comes from the Greek heteros, "different", and lysis, "loosening".

Which of the following statements are correct?
I. A pair of positional isomers differs in the position of the same functional group.
II. A pair of. structural isomers have the same relative molar mass.
Ill. A pair of functional group isomers belongs to different homologous series. 
  • a)
    I, II and Ill 
  • b)
    I and Ill
  • c)
    II and III
  • d)
    I and II 
Correct answer is option 'A'. Can you explain this answer?

Preeti Iyer answered
According to me all statements are correct. For statement I, positional isomers have the same functional group. They just differ in the position of that group.
Statenmen II- Structural isomers have the same relative molar mass. Actually relative mass means the mass of one isomer is samee relative to another. Structural isomers differ in the structure of molecules. However they have the same molecular formula and hence; same molar mass.
Statement III- According to me, this statement is correct. As homologous series must contain the same functional group with the same physical & chemical properties. Also functional group isomers differ in the functional group attached with them. So, functional group isomers belong to different homologous series. ​

Which of the following are not functional isomers of each other?
  • a)
    CH3​CH2​NO2​ and CH3​CH2​ON=O
  • b)
    C2​H5​CHO and CH3​COCH3​
  • c)
    CH3​CH2​NH2​ and CH3​NHCH3​
  • d)
    C3​H7​NH2​ and (CH3​)2​CHNH2​
Correct answer is option 'D'. Can you explain this answer?

Anjali Sharma answered
Functional isomers are structural isomers that have the same molecular formula (that is, the same number of atoms of the same elements), but the atoms are connected in different ways so that the groupings are dissimilar or different functional groups.
10 amine to different 10 amines are not functional isomers. 
Hence, option ′d′ is the answer.
 

The number of isomers of C6H14 is:
  • a)
    4
  • b)
    5
  • c)
    6
  • d)
    7
Correct answer is option 'B'. Can you explain this answer?

Anu Mukherjee answered
**Explanation:**

To determine the number of isomers of C6H14, we need to consider the different ways in which carbon atoms can be arranged and the different ways in which hydrogen atoms can be distributed among these carbon atoms.

**1. Straight-chain isomers:**
The simplest isomer is the straight chain, where the carbon atoms are arranged in a linear sequence. For C6H14, there is only one straight-chain isomer.

**2. Branched-chain isomers:**
In branched-chain isomers, the carbon atoms are arranged in a branched manner. To determine the number of possible branched-chain isomers, we need to consider the different ways in which the branches can be arranged and the different positions at which the branches can be attached to the main chain.

In this case, one of the carbons in the straight chain can have three branches attached to it. We can have three different arrangements of these branches, which are:

- One branch attached to the first carbon and two branches attached to the second carbon.
- Two branches attached to the first carbon and one branch attached to the second carbon.
- Three branches attached to the first carbon and no branches attached to the second carbon.

For each of these arrangements, there are different possibilities for the positions of the branches along the main chain. Therefore, for the branched-chain isomers, there are a total of three possibilities.

**3. Cycloalkane isomers:**
Cycloalkanes are ring structures where the carbon atoms are arranged in a closed loop. For C6H14, we can have cycloalkane isomers with different ring sizes.

- Cyclohexane: In this case, all six carbon atoms form a single ring. There is only one possibility.

- Cyclopentane: In this case, five carbon atoms form a ring and the remaining carbon atom is attached to one of the carbons in the ring. There is only one possibility.

Therefore, the total number of isomers of C6H14 is the sum of the straight-chain isomer, branched-chain isomers, and cycloalkane isomers, which is 1 + 3 + 2 = 6.

Therefore, the correct answer is option **c) 6**.

Which among the following is not an aromatic compound(in specific)
  • a)
    Naphthalene
  • b)
    Aniline
  • c)
    Pyridine
  • d)
    Tropolone
Correct answer is option 'C'. Can you explain this answer?

Sai Mishra answered
Pyridine is heterocyclic aromatic compound. Whereas naphthalene and aniline are benzenoid aromatic compounds and tropolone is a non-benzenoid aromatic compound.

Organic compounds are broadly classified as
  • a)
    alicyclic compounds and acyclic compounds
  • b)
    Open chain compounds and linear chain compounds
  • c)
    Cyclic compounds and alicyclic compounds
  • d)
    Open chain compounds and closed compounds
Correct answer is option 'D'. Can you explain this answer?

Arpita Nambiar answered
The correct answer is option D
Organic compounds are broadly classified into open chain and closed chain compounds. Explanation: open chain compounds or acyclic compounds are otherwise called as aliphatic compounds.

Direction (Q. Nos. 1-18) This section contains 18 multiple choice questions. Each question has four
 choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.
Q.

  • a)
    3
  • b)
    5
  • c)
    6
  • d)
    7
Correct answer is option 'B'. Can you explain this answer?

Naina Sharma answered
CH10

 Two times double bond can’t appear as it has asked alkene and not alkadiene. So, these are the only types of alkenes that can be formed which are structural isomers.

[JEE Main 2013 Online Exam]
  • a)
    a
  • b)
    b
  • c)
    c
  • d)
    d
Correct answer is option 'C'. Can you explain this answer?

Om Desai answered
 In ethers, alcohols and amines, there is no double bond. However, esters have one double bond as same as monocarboxylic acid. So monocarboxylic  acid is a functional isomer of ester (only for no. of carbon greater than 2. As we need to have at least 3 carbon atoms to form ester.)

  • a)
    a
  • b)
    b
  • c)
    c
  • d)
    d
Correct answer is option 'B'. Can you explain this answer?

Om Desai answered
Double bond equivalent of C5H10O is 1. So, it will have either a double bond or a ring.So, it may be an aldehyde, ketone, alcohol( carbon chain having double bond) or epoxide.
Propyl ethanoate has 2 oxygen atoms. So, it will never be its isomer. HOwever option a and c will be its isomer as they are aldehyde.

Among the given compounds , the most susceptible to nucleophilic attack at the carbonyl group is: [2010]
  • a)
    CH3COOCH3
  • b)
    CH3CONH2
  • c)
    CH3COOCOCH3
  • d)
    CH3COCI
Correct answer is option 'D'. Can you explain this answer?

Rajat Kapoor answered
CH3COCl
Lesser the electron density of acyl carbon atom, more will be the susceptibility of a nucleophile to attack it. 
The Cl atom has strong - I effect and weakest +R effect because of the weak π-bond between the small sized C- atom and large sized Cl atom. Thus, in CH3COCl, acyl carbon has least electron density and hence, more susceptible for nucleophilic attack.

The organic reaction which proceeds through heterolytic bond cleavage are known as:
  • a)
    Covalent reactions
  • b)
    Ionic reactions
  • c)
    Free radical reaction
  • d)
    Polar reactions
Correct answer is option 'B'. Can you explain this answer?

Hansa Sharma answered
In heterolytic bond cleavage the bond breaks unevenly and the shared pair of electrons is accommodated by one of the products, which produces one or more ions.As heterolytic bond cleavage gives ions. So the reaction which proceeds through heterolytic bond cleavage is an ionic reaction.
Hence B is the correct answer.

Which of the following correctly ranks the cycloalkanes in order of increasing ring strain per methylene group? 
  • a)
    Cyclopropane < Cyclobutane < Cyclopentane < Cyclohexane 
  • b)
    Cyclohexane < Cyclopentane < Cyclobutane < Cyclopropane 
  • c)
    Cyclohexane < Cyclobutane < Cyclopentane < Cyclopropane 
  • d)
    Cyclopropane < Cyclopentane < Cyclobutane < Cyclohexane  
Correct answer is option 'B'. Can you explain this answer?

Rajesh Gupta answered
The correct answer is Option B.

The C-C-C bond angles in cyclopropane (60o) and cyclobutane (90o) are much different than the ideal bond angle of 109.5o.This bond angle causes cyclopropane and cyclobutane to have a high ring strain. However, molecules, such as cyclohexane and cyclopentane, would have a much lower ring strain because the bond angle between the carbons is much closer to 109.5o.

Which nomenclature is not according to IUPAC system? [2012]
  • a)
  • b)
    ​​​​
  • c)
     2-Methyl-3-phenylpentane
  • d)
     5-oxohexanoic acid
Correct answer is option 'A'. Can you explain this answer?

Rajesh Datta answered
In IUPAC system of nomenclature, preference is given to multiple bond than halogen substituent, so the correct name of
 is 3-Bromoprop-1-ene.

 Which among the following does not exhibit geometric isomerism
  • a)
    1-hexene
  • b)
    2-hexene
  • c)
    3-hexene
  • d)
    4-hexene
Correct answer is option 'A'. Can you explain this answer?

Alkenes like 1-hexene when flipped from top to bottom they have identical structures and also they have C=CH2 unit which does not exist as cis- trans isomers.

 Which among the following is a very unstable and reactive species:
  • a)
    Carbaanion
  • b)
    Polar ions
  • c)
    Carbocation
  • d)
    Free radical
Correct answer is option 'C'. Can you explain this answer?

Geethika Reddy answered
Actually carbanions are filled with octet, so they are stable and less reactive, in case of polar ions already they are stable, and free radicle is heptet i.e, near to octet whereas carbocation is sextet in nature, so in order to gain to octet it is more reactive and less stable

If there is no rotation of plane polarised light by a compound in a specific solvent, though to be chiral, it may mean that [2007]
  • a)
    the compound is certainly meso
  • b)
    there is no compound in the solvent
  • c)
    the compound may be a racemic mixture
  • d)
    the compound is certainly a chiral.
Correct answer is option 'A'. Can you explain this answer?

Rocky Handsome answered
Meso compound does not rotate plane polarised light. Compound which contains tetrahedral atoms with four different groups but the whole molecule is a chiral, is known as meso compound. It possesses a plane of symmetry and is optically inactive.

•One of the asymmetric carbon atoms turns the plane of polarised light to the right and other to the left and to the same extent so that the rotation due to upper half is compensated by the lower half, i.e., internally compensated, and finally there is no rotation of plane polarised light.

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