What should be the name of the given compound?
Answer
581.1k+ views
Hint: The nomenclature of bicyclio compounds are done by using bicyclo as a prefix and then the various bridge length measured between two carbons at ring junction is put in a square bracket in a decreasing manner followed by the parent hydrocarbon name.
Format – Bicyclo [x, y, z] parent hydrocarbon name
Complete step by step solution: The compound given in the question is a bicyclo compound with a total number of 7 carbon atoms. Thus, the parent hydrocarbon open-chain is heptane.
Now, we need to find the bridge lengths between two carbons present at the junctions of the ring. These two junctional carbon atoms are known as bridgeheads.
The bridge length is the number of carbon atoms that come between two bridgeheads. The different pathways will give different bridge lengths.
So, let us draw and find out the different pathways and the bridge length values for corresponding pathways.
1) – When we follow the pathway from B to E in the anti-clockwise direction, then we come across the carbon atoms at points G and F. Thus, the bridge length for this path is 2.
2) – If we go through the path BAE, then we end up with only 1 carbon atom between the two bridgeheads. So, in this case, the bridge length is 1.
3) – Lastly, if we go through the pathway from B to E in the clockwise direction, then again, we get carbon atoms at two points – C and D. so, again the bridge length is 2.
Now, we must write these bridge lengths in decreasing order in a square bracket followed by the parent carbon chain name and use bicyclo as a prefix. Putting all these things together we get the name – Bicyclo[2.2.1]heptane.
Hence, the name of the given compound is Bicyclo[2.2.1]heptane.
Note: The bridgeheads are the carbon atoms that join the three-ring system together and are present at the point of attachment of rings. These bridgehead carbons are not counted while calculating the bridge lengths.
Format – Bicyclo [x, y, z] parent hydrocarbon name
Complete step by step solution: The compound given in the question is a bicyclo compound with a total number of 7 carbon atoms. Thus, the parent hydrocarbon open-chain is heptane.
Now, we need to find the bridge lengths between two carbons present at the junctions of the ring. These two junctional carbon atoms are known as bridgeheads.
The bridge length is the number of carbon atoms that come between two bridgeheads. The different pathways will give different bridge lengths.
So, let us draw and find out the different pathways and the bridge length values for corresponding pathways.
1) – When we follow the pathway from B to E in the anti-clockwise direction, then we come across the carbon atoms at points G and F. Thus, the bridge length for this path is 2.
2) – If we go through the path BAE, then we end up with only 1 carbon atom between the two bridgeheads. So, in this case, the bridge length is 1.
3) – Lastly, if we go through the pathway from B to E in the clockwise direction, then again, we get carbon atoms at two points – C and D. so, again the bridge length is 2.
Now, we must write these bridge lengths in decreasing order in a square bracket followed by the parent carbon chain name and use bicyclo as a prefix. Putting all these things together we get the name – Bicyclo[2.2.1]heptane.
Hence, the name of the given compound is Bicyclo[2.2.1]heptane.
Note: The bridgeheads are the carbon atoms that join the three-ring system together and are present at the point of attachment of rings. These bridgehead carbons are not counted while calculating the bridge lengths.
Recently Updated Pages
The given figure shows two endocrine glands marked class 11 biology NEET_UG

Match columnI with columnII and select the correct class 11 biology NEET

Match column I with column II and select the correct class 11 biology NEET_UG

Which floral family has left 9 right + 1 arrangement class 11 biology NEET_UG

Which is not a variety of sheep A Lohi B Beetal C Nellore class 11 biology NEET_UG

Match column I with column II and select the correct class 11 biology NEET_UG

Trending doubts
Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Two of the body parts which do not appear in MRI are class 11 biology CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

10 examples of law on inertia in our daily life

