Test: Structural Isomerism - 2

We can see that for alkane, we atleast need alkane of 7 carbon atoms or C₇H₁₆.

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 correct answer is option A,B
Functional groups having two oxygen atoms are " Easter " and " Carboxylic Acids " .
Compounds having molecular formula C₂H₄O₂ have possible functional isomers. One is ethanoic acids or acetic acids ( Carboxylic Acids ) and other is METHYL Methanoate ( Ester ) ...
Acetic acids are colourless liquid organic compounds with the chemical formula CH₃COOH ( also written as CH₃CO₂H or C₂H₄O₂ ) .

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.

So, option a, b and d are correct.

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;