Structural Isomerism - 2 - Free MCQ Practice Test with solutions, JEE Chemistry

Let us consider organic acid as ‘A’.
Moles of A = mass of A/Molar mass = 6.6/132 = 0.05
Molarity of NaOH = 1.25M
And volume of NaOH  = 80 mL
Therefore, moles of NaOH = molarity × Volume
= 1.25×80×10^-3 = 0.1 moles.
SO, 0.05 moles of A requires 0.1 moles of NaOH
1 mole of acid requires x moles of NaOH
x = 0.1/0.05×1 = 2
So for neutralisation of 1 mole of A, we need 2 moles of NaOH. Therefore given acid is dibasic and contains 2 -COOH groups.

General formula
C_nH_2nO = 100
12×n+2×2n+16 = 100
n = 6
So, the formula of ketone is C₆H₁₂O

So, according to me, the correct answer is 6 and not 8.

The correct answer is Option A.
2-butene can show geometrical isomerism, due to different atoms/group of atoms attached on both sides of double bonded carbons as H and CH3 − groups.

We know that for ether, we need C-O-C connectivity. Now, we can see that this arrangement won't give any isomer.So, we will increase one carbon atom. 
We will get C-C-O-C. This also doesn’t give any isomer. SO we will add another carbon atom. This time, we will have C-C-C-O-C. With this arrangement, we can have 3 different isomers. They are :-
So, we havegeneral fomula as C₄H₉OH. The different constitutional isomers of this lcohol are as follow

We can see that there are 4 isomers of ths compound So option b is correct.

First, we will check the Double bond equivalent. In C₇H₈O, DBE = 4,this means that we will have 3 double bonds and a ring and no additional double bond outside the ring.

These are the structural isomers that exists for C₇H₈O

These are 8 isomers of C₄H₉ON.

The organic compound with molecular formula C₂​H₄​O₂ can have the following functional isomers:

• a) A carboxylic acid (e.g., acetic acid)
• b) An ester (e.g., methyl formate)
• d) A hydroxyaldehyde (e.g., hydroxyacetaldehyde)

c) A dialdehyde is not possible with this formula.

Thus, the correct answers are (a), (b), and (d).

a) It has eight isomers capable of forming intermolecular hydrogen bonds amongst themselves:

Hydrogen bonding occurs between a hydrogen atom bonded to an electronegative atom (such as oxygen) and a lone pair of electrons on another electronegative atom. In compounds with the formula C5H12O, there are multiple possibilities for hydrogen bonding.

The eight isomers of C₅H₁₂O are:

1. Pentanol (n-pentanol)

2. 2-methyl-1-butanol

3. 3-methyl-1-butanol

4. 2-methyl-2-butanol (tert-pentanol)

5. 3-methyl-2-butanol

6. 2,2-dimethyl-1-propanol

7. 2,3-dimethyl-1-butanol

8. 3,3-dimethyl-1-butanol

All of these isomers have the potential to form intermolecular hydrogen bonds with each other because they contain hydrogen bond donor (hydroxyl group -OH) and hydrogen bond acceptor (oxygen atom) functional groups.

b) Six of its isomers are only hydrogen bond acceptor but not hydrogen bond donor: Out of the eight isomers listed above, six of them act as hydrogen bond acceptors but not as hydrogen bond donors. This is because they have a lone pair of electrons on the oxygen atom, which can form a hydrogen bond with a hydrogen bond donor, but do not have a hydrogen atom bonded to an electronegative atom.

The six isomers that are only hydrogen bond acceptors are:

1. 2-methyl-1-butanol

2. 3-methyl-1-butanol

3. 2-methyl-2-butanol (tert-pentanol)

4. 3-methyl-2-butanol

5. 2,2-dimethyl-1-propanol

6. 2,3-dimethyl-1-butanol

c) One of its isomers has four methyl groups: Among the eight isomers of C₅H₁₂O, one of them, 2,2-dimethyl-1-propanol, has four methyl groups. This is evident from the name itself, where "dimethyl" indicates the presence of two methyl groups, and "propanol" indicates a three-carbon chain with an alcohol functional group.

d) None of its isomer pairs is metamers: Metamers are a type of structural isomers where the carbon skeleton remains the same, but the functional groups are attached to different carbon atoms. In the case of C5H12O, none of the isomer pairs have the same carbon skeleton with different functional group arrangements. Therefore, none of the isomers can be classified as metamers. Conclusion: In summary, the correct statements about the structural isomers of compounds with the chemical composition C5H12O are: - There are eight isomers capable of forming intermolecular hydrogen bonds amongst themselves. - Six of the isomers are only hydrogen bond accept.

The correct answer is option A,D

Ketones have the general formula R(C=O)R', where R and R' can be alkyl or aryl groups. In the given molecular formula, C3H6O, we can construct a ketone by placing a carbonyl group (C=O) in the middle of the carbon chain: CH3-C(=O)-CH3. This satisfies the given molecular formula, so option a is correct. An unsaturated ether would imply the presence of a pi (π) bond. In the case of the given molecular formula, C3H6O, an unsaturated ether could be represented as CH2=CHOCH3. This structure has a pi bond, so option d is correct.

So, option a, b and d are correct.

The correct answer is option a

C3H6O represents an aldehyde and a ketone

C3H4O represents an aldehyde and a ketone

C5H8O2 represents Carboxylic Acid, Carboxylic Acid Ester, Aldehyde and Ketone

C3H6O2 represents Carboxylic Acid, Carboxylic Acid Ester, Aldehyde and Ketone

The correct answer is option b

C4H6 represents a Cycloalkane,Bicyloalkane and Diene

C5H8 represents a Cycloalkane,Bicyloalkane,Spiro Compound and Diene

C6H10 represents a Cycloalkane,Bicyloalkane,Spiro Compound and Diene

C4H4 represents Diene

Double bond equivalent = 20-12/4 = 4
For phenyl ring, Double bond equivalent = 4(3 double bond and 1 ring)
So, we won’t have any double bond outside the ring.

Explanation:

1,1,1-trichloropropane;

1,2,3-trichloropropane;

1,2,2-trichloropropane;

1,1,2-trichloropropane; this has 2 stereoisomers, with the chiral centre at C2;

1,1,3-trichloropropane;

To determine the number of enols for heptane-2,4-dione, we need to consider the possible positions for the enolization (removal of a proton from a carbon adjacent to the carbonyl group) on the molecule.Enolization can occur at the α-carbon (carbon adjacent to the carbonyl group), leading to enols. Also, consider the possibility of keto-enol tautomerism, where the double bond in the enol form shifts to form a carbonyl group, and vice versa.