Q.Write structures and IUPAC names of different structural isomers of alkenes corresponding to C 5H10.

The key idea is Chain Isomerism — alkenes with the same molecular formula but different carbon skeletons.

Reasoning:

1. For $\ce{C5H10}$, the degree of unsaturation is 1 (one double bond or a ring). We only consider acyclic alkenes here.
2. Start with the longest possible carbon chain (5 carbons) and place the double bond at different positions.
3. Then shorten the main chain to 4 carbons, adding a methyl branch, and again vary the double bond position. Avoid creating the same skeleton twice.

Structures and IUPAC names:

The question asks for structural isomers of alkenes with the formula $\text{C}_5\text{H}_{10}$. Remember, an alkene must contain at least one carbon-carbon double bond. The general formula for an alkene is $\text{C}_n\text{H}_{2n}$, and here $n=5$, so we are looking at all possible arrangements of five carbon atoms that include one double bond. This is a classic problem in chain isomerism (different carbon skeletons) and position isomerism (the double bond located at different places along the same skeleton).

Let’s work through the possibilities systematically.

1. Start with the longest possible carbon chain. The maximum continuous chain of 5 carbons is a pentene. The double bond can be placed in two distinct positions:

   • Between carbon 1 and carbon 2: This gives pent-1-ene.
   • Between carbon 2 and carbon 3: This gives pent-2-ene.
   Watch out

   A common mistake is to think that placing the double bond between carbon 4 and 5 is a new isomer. It is identical to pent-1-ene because the chain is numbered from the end nearest the double bond. Always number to give the double bond the lowest possible locant.

2. Now, shorten the main chain to 4 carbons. This means we have a branched alkene. The main chain is a butene, and the fifth carbon becomes a methyl ($\text{CH}_3$) substituent. We must place the methyl group and the double bond such that the main chain remains the longest chain containing the double bond.

   • Consider the skeleton: a 4-carbon chain with a methyl branch. The double bond can be at the 1-position or the 2-position of the butene chain.
   • Case: Double bond at position 1 (but-1-ene). Where can the methyl group go?
     • Methyl on carbon 2: This gives 2-methylbut-1-ene.
     • Methyl on carbon 3: This gives 3-methylbut-1-ene.
     • Methyl on carbon 1? No, that would extend the main chain to 5 carbons (making it pent-1-ene again), which we already counted.
   • Case: Double bond at position 2 (but-2-ene). Where can the methyl group go?
     • Methyl on carbon 2: This gives 2-methylbut-2-ene.
     • Methyl on carbon 3: placing a methyl on carbon 3 makes the longest chain containing the double bond actually 5 carbons (pent-2-ene), not 4, so this is not a new structural isomer — it is the same as pent-2-ene.
     • Methyl on carbon 1? Again, this would extend the chain.

3. Check for a 3-carbon main chain. Can we have a main chain of 3 carbons (propene) with two methyl substituents? The formula would be $\text{C}_3\text{H}_5 + 2(\text{CH}_3) = \text{C}_5\text{H}_{10}$, so it’s possible.

   • The double bond must be in the 3-carbon chain (propene). The only position is between C1 and C2 (prop-1-ene; prop-2-ene is the same).
   • We need to place two methyl groups on this skeleton. The only way to keep the main chain as 3 carbons is to put both methyls on the central carbon (C2). This gives 2-methylbut-1-ene? No, that’s a 4-carbon chain. Let’s check: if we put two methyls on C2 of propene, the longest chain becomes 4 carbons (2-methylbut-1-ene or 2-methylbut-2-ene), which we already have. So no new isomer here.

4. List all unique structural isomers. From the steps above, we have five distinct structural isomers:

   1. Pent-1-ene
   2. Pent-2-ene
   3. 2-Methylbut-1-ene
   4. 3-Methylbut-1-ene
   5. 2-Methylbut-2-ene …