Test: Structural Isomerism - 2 - JEE MCQ

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).

All the 8 isomers of alcohol are intermolecular H-bonding. 
There are 6 ethers which have H-bond acceptor(i.e O atm) but not H-bond donor(i.e H atom)  
Also, the isomer with 4 methyl groups is mentioned above.
According to me, CH₃CH₂COCH₂CH₃ is metamer of CH₃COCH₂CH₂CH₃ OR CH₃COCH(CH₃)₂
So all options are correct.

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.