Answer
The following is the structure or name for each compound:
Work Step by Step
Question a:
Step 1: main prefix is ‘oct’. So, the no. of C in straight chain is 8. Draw 8 C in straight chain and join them with single bond and numbering 1 to 8 from right to left.
Step 2: ending name is ‘anoic acid’. So, Carboxylic acid group present here and it is present at no. 1 Carbon. Add the functional group at right end side
Step 3: The remaining valency is filled with H.
Question b:
Step 1: See the structure and identify the functional group. Ester group is present here. So, first we have to see what alkyl group is join with the O atom of the functional group at right side. Here, it is ‘methyl’ group that join with the O atom of the ester group. So, the 1st prefix name is ‘methyl’.
Step 2: Now see the remaining portion of the compound and count the total no. of C in remaining part. Here, the total no. of C is 2 in the remaining part. So, ‘eth’ is added as second prefix and a space is given between first prefix and second prefix.
Step 3: Now, for ester group the ending name is ‘anoate’. There should be no gap present between second prefix and ending name while naming ester.
So, the complete name is : methyl ethanoate
Question c:
Step 1: look at the ending name. It is ‘anoate’. That means this is ‘ester’. Draw the ester functional group.
Step 2: The first name is ‘ethyl’. So, add ethyl group on the right side of the functional group and the second prefix name is ‘but’. That means the total no. of C (excluding ethyl group on the right side of the functional group) is 4. One C is already present in the functional group. So, we have to add three C on the left side and the remaining valency of each C is then filled with H.
Question d:
Step 1: See the structure and count the no. of C in straight chain. Here, it is 7. So, the main prefix of the name is ‘hept’.
Step 2: Identify the functional group. Here, it is Carboxylic Acid. So, the ending name will be ‘anoic acid’.
So, the complete name is ‘heptanoic acid’.
