The molecular formula is C$_6$H$_{14}$. This corresponds to the alkane series C$_n$H$_{2n+2}$ with $n=6$. We need to find all possible structural isomers by arranging the 6 carbon atoms in different ways (straight chain or branched chains). Then write their IUPAC names.

1. **Straight chain:** 6 carbons in a row.

CH$_3$ – CH$_2$ – CH$_2$ – CH$_2$ – CH$_2$ – CH$_3$. IUPAC name: **Hexane**.

2. **Branched chains (5-carbon parent chain + 1-carbon branch):** Longest chain is 5 carbons (pentane), with a methyl branch (1 carbon) attached. The methyl group can be on C2 or C3 of the pentane chain (numbering from the end closer to the branch to get the lowest locant).

• Methyl branch on C2:

  CH$_3$

  |

  CH$_3$ – CH – CH$_2$ – CH$_2$ – CH$_3$. IUPAC name: **2-Methylpentane**.

• Methyl branch on C3:

  CH$_3$

  |

  CH$_3$ – CH$_2$ – CH – CH$_2$ – CH$_3$. IUPAC name: **3-Methylpentane**.

3. **Branched chains (4-carbon parent chain + 2-carbon branches):** Longest chain is 4 carbons (butane), with two methyl branches (total 2 carbons from C$_6$H$_{14}$ - C$_4$H$_{10}$). The two methyl groups can be on the same carbon or different carbons.

• Two methyl branches on C2:

  CH$_3$

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  CH$_3$ – C – CH$_2$ – CH$_3$. IUPAC name: **2,2-Dimethylbutane**.

  |

  CH$_3$

• Two methyl branches on C2 and C3:

  CH$_3$ CH$_3$

  | |

  CH$_3$ – CH – CH – CH$_3$. IUPAC name: **2,3-Dimethylbutane**.

We cannot form a 3-carbon parent chain with 3 methyl branches. Total carbons = 3 + 3 = 6.

So, there are 5 chain isomers for C$_6$H$_{14}$.

The structures and IUPAC names are:

1. CH$_3$ – CH$_2$ – CH$_2$ – CH$_2$ – CH$_2$ – CH$_3$ (**Hexane**)
2. CH$_3$ – CH(CH$_3$) – CH$_2$ – CH$_2$ – CH$_3$ (**2-Methylpentane**)
3. CH$_3$ – CH$_2$ – CH(CH$_3$) – CH$_2$ – CH$_3$ (**3-Methylpentane**)
4. CH$_3$ – C(CH$_3$)$_2$ – CH$_2$ – CH$_3$ (**2,2-Dimethylbutane**)
5. CH$_3$ – CH(CH$_3$) – CH(CH$_3$) – CH$_3$ (**2,3-Dimethylbutane**)

The molecular formula C$_5$H$_{11}$ corresponds to an alkyl group derived from a pentane isomer (C$_5$H$_{12}$) by removing one hydrogen atom. Pentane has three structural isomers: n-pentane, isopentane (2-methylbutane), and neopentane (2,2-dimethylpropane).

We need to consider removing a hydrogen from each unique carbon position in each of these pentane isomers to generate the isomeric alkyl groups C$_5$H$_{11}$. Then, attach an -OH group to the point where the hydrogen was removed to form the corresponding alcohol and name it using IUPAC nomenclature.

1. **From n-pentane (CH$_3$-CH$_2$-CH$_2$-CH$_2$-CH$_3$):** There are two types of carbons (terminal CH$_3$ and internal CH$_2$). Removing H from a terminal CH$_3$ or an internal CH$_2$ gives different alkyl groups.

• Remove H from C1 (or C5, they are equivalent): -CH$_2$-CH$_2$-CH$_2$-CH$_2$-CH$_3$. This is the **pentyl** or n-pentyl group. Attaching -OH gives CH$_3$-CH$_2$-CH$_2$-CH$_2$-CH$_2$-OH. IUPAC name: **Pentan-1-ol**.
• Remove H from C2 (or C4, they are equivalent): CH$_3$-CH(-)-CH$_2$-CH$_2$-CH$_3$. This is the **1-methylbutyl** or sec-pentyl group. Attaching -OH gives CH$_3$-CH(OH)-CH$_2$-CH$_2$-CH$_3$. IUPAC name: **Pentan-2-ol**.
• Remove H from C3: CH$_3$-CH$_2$-CH(-)-CH$_2$-CH$_3$. This is the **1-ethylpropyl** or tert-pentyl group. Attaching -OH gives CH$_3$-CH$_2$-CH(OH)-CH$_2$-CH$_3$. IUPAC name: **Pentan-3-ol**.

2. **From isopentane (2-methylbutane, CH$_3$-CH(CH$_3$)-CH$_2$-CH$_3$):** Identify unique carbon positions. C1 (methyl), C2 (tertiary CH), C3 (methylene CH$_2$), C4 (methyl). C1 and C4 are different.

• Remove H from C1 (methyl on end): -CH$_2$-CH(CH$_3$)-CH$_2$-CH$_3$. This is the **2-methylbutyl** group. Attaching -OH gives HO-CH$_2$-CH(CH$_3$)-CH$_2$-CH$_3$. IUPAC name: **2-Methylbutan-1-ol**.
• Remove H from C2 (tertiary CH): CH$_3$-C(-)(CH$_3$)-CH$_2$-CH$_3$. This is the **1,1-dimethylpropyl** or tert-pentyl group. Attaching -OH gives CH$_3$-C(OH)(CH$_3$)-CH$_2$-CH$_3$. IUPAC name: **2-Methylbutan-2-ol**.
• Remove H from C3 (methylene CH$_2$): CH$_3$-CH(CH$_3$)-CH(-)-CH$_3$. This is the **1,2-dimethylpropyl** or isohexyl group (common name is incorrect here, it should be based on pentane). Let's use systematic name. Parent 3 carbons (propane) with 2 methyls. No, let's use IUPAC alkyl naming. Longest chain starting from attachment point. Attachment is C3. Longest chain from C3 is 3 carbons (C3-C4-CH3). Methyl on C2 of this chain. So, 1-methylpropyl group with attachment at C1? This is confusing. Let's just draw the alcohol. Attaching -OH to C3 of isopentane gives CH$_3$-CH(CH$_3$)-CH(OH)-CH$_3$. Longest chain with OH is 4 carbons (butane). Methyl on C2, OH on C3. IUPAC name: **3-Methylbutan-2-ol**.
• Remove H from C4 (methyl on end): CH$_3$-CH(CH$_3$)-CH$_2$-CH$_2$-. This is the **3-methylbutyl** or isoamyl group. Attaching -OH gives CH$_3$-CH(CH$_3$)-CH$_2$-CH$_2$-OH. IUPAC name: **3-Methylbutan-1-ol**.

3. **From neopentane (2,2-dimethylpropane, CH$_3$-C(CH$_3$)$_2$-CH$_3$):** All four methyl groups are equivalent.

• Remove H from any methyl group: -CH$_2$-C(CH$_3$)$_2$-CH$_3$. This is the **2,2-dimethylpropyl** or neopentyl group. Attaching -OH gives HO-CH$_2$-C(CH$_3$)$_2$-CH$_3$. IUPAC name: **2,2-Dimethylpropan-1-ol**.

Let's list the unique alkyl groups and corresponding alcohols and their IUPAC names:

1. Alkyl group: -CH$_2$-CH$_2$-CH$_2$-CH$_2$-CH$_3$ (pentyl). Alcohol: CH$_3$CH$_2$CH$_2$CH$_2$CH$_2$OH (Pentan-1-ol).
2. Alkyl group: CH$_3$-CH(-)-CH$_2$-CH$_2$-CH$_3$ (1-methylbutyl). Alcohol: CH$_3$CH(OH)CH$_2$CH$_2$CH$_3$ (Pentan-2-ol).
3. Alkyl group: CH$_3$-CH$_2$-CH(-)-CH$_2$-CH$_3$ (1-ethylpropyl). Alcohol: CH$_3$CH$_2$CH(OH)CH$_2$CH$_3$ (Pentan-3-ol).
4. Alkyl group: -CH$_2$-CH(CH$_3$)-CH$_2$-CH$_3$ (2-methylbutyl). Alcohol: HO-CH$_2$-CH(CH$_3$)-CH$_2$-CH$_3$ (2-Methylbutan-1-ol).
5. Alkyl group: CH$_3$-C(-)(CH$_3$)-CH$_2$-CH$_3$ (1,1-dimethylpropyl). Alcohol: CH$_3$-C(OH)(CH$_3$)-CH$_2$-CH$_3$ (2-Methylbutan-2-ol).
6. Alkyl group: CH$_3$-CH(CH$_3$)-CH(-)-CH$_3$ (1,2-dimethylpropyl). Alcohol: CH$_3$-CH(CH$_3$)-CH(OH)-CH$_3$ (3-Methylbutan-2-ol).
7. Alkyl group: CH$_3$-CH(CH$_3$)-CH$_2$-CH$_2$- (3-methylbutyl). Alcohol: CH$_3$-CH(CH$_3$)-CH$_2$-CH$_2$-OH (3-Methylbutan-1-ol).
8. Alkyl group: -CH$_2$-C(CH$_3$)$_2$-CH$_3$ (2,2-dimethylpropyl). Alcohol: HO-CH$_2$-C(CH$_3$)$_2$-CH$_3$ (2,2-Dimethylpropan-1-ol).

There are 8 isomeric pentyl groups (C$_5$H$_{11}$) and thus 8 corresponding alcohols. Let's cross-reference with the provided solution. The provided solution lists 8 alcohols and seems to correspond to my list. The alkyl groups listed in the solution seem to be the groups from which the alcohol is derived by attaching -OH, but their names are not consistently IUPAC alkyl names. For instance, the first alkyl group listed in the solution is "CH3 – CH2 – CH2 – CH2– CH2 –", which is the pentyl group. The corresponding alcohol is pentan-1-ol. The second alkyl group is "CH3 – CH – CH2 – CH2 – CH3 | ", which is the 1-methylbutyl group, and the alcohol is pentan-2-ol. The third alkyl group is "CH3 – CH2 – CH – CH2 – CH3 | ", which is the 1-ethylpropyl group, alcohol pentan-3-ol. The fourth alkyl group is "CH3 CH3 | | CH3 – CH – CH2 – CH2 – ", which is the 3-methylbutyl group, alcohol 3-methylbutan-1-ol. The fifth alkyl group is "CH3 CH3 | | CH3 – CH2 – CH – CH2 – ", which is the 2-methylbutyl group, alcohol 2-methylbutan-1-ol. The sixth alkyl group is "CH3 CH3 | | CH3 – C – CH2 – CH3", which is the 1,1-dimethylpropyl group, alcohol 2-methylbutan-2-ol. The seventh alkyl group is "CH3 CH3 | | CH3 – C – CH2OH", which is the 2,2-dimethylpropyl group with OH attached, so it represents the alcohol 2,2-dimethylpropan-1-ol. The eighth alkyl group is "CH3 CH3 | | CH3 – CH – CH –CH3", which is the 1,2-dimethylpropyl group, alcohol 3-methylbutan-2-ol.

My list of 8 alcohols and corresponding alkyl groups matches the provided solution's list, although the representation of the alkyl groups and their names in the solution's table is somewhat inconsistent with standard IUPAC alkyl naming. The focus should be on identifying all unique points of attachment in the pentane isomers and attaching -OH there, then naming the resulting alcohol.

So, there are 8 isomeric alcohols with formula C$_5$H$_{11}$OH, corresponding to the 8 isomeric pentyl groups C$_5$H$_{11}$. Their structures and IUPAC names are as listed in my step-by-step derivation (Pentan-1-ol, Pentan-2-ol, Pentan-3-ol, 2-Methylbutan-1-ol, 2-Methylbutan-2-ol, 3-Methylbutan-1-ol, 3-Methylbutan-2-ol, 2,2-Dimethylpropan-1-ol).

To write IUPAC names, first draw the expanded structural formula if necessary, identify the longest carbon chain, number it, identify substituents and their positions, and construct the name.

(i) (CH$_3$)$_3$ C CH$_2$C(CH$_3$)$_3$:

Expand the formula: (CH$_3$)$_3$C– means a carbon bonded to three methyl groups. CH$_2$ is a methylene group. C(CH$_3$)$_3$ means a carbon bonded to three methyl groups.

CH$_3$ CH$_3$

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CH$_3$ – C – CH$_2$ – C – CH$_3$

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CH$_3$ CH$_3$

Identify the longest carbon chain: By tracing the chain, we can find paths with different lengths. For example, CH$_3$-C(CH$_3$)$_2$-CH$_2$-C(CH$_3$)$_2$-CH$_3$. Longest chain is 5 carbons. Let's number it.

1 2 3 4 5

CH$_3$–C(CH$_3$)$_2$–CH$_2$–C(CH$_3$)$_2$–CH$_3$. Substituents: two methyls on C2 and two methyls on C4. Total 4 methyls.

Parent alkane: pentane. Methyl groups on C2, C2, C4, C4. Locants (2,2,4,4).

IUPAC name: 2,2,4,4-Tetramethylpentane. (Matches text solution).

(ii) (CH$_3$)$_2$ C(C$_2$H$_5$)$_2$:

Expand the formula: (CH$_3$)$_2$C– means a carbon bonded to two methyl groups. C(C$_2$H$_5$)$_2$ means a carbon bonded to two ethyl groups.

CH$_3$ C$_2$H$_5$

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CH$_3$ – C – C$_2$H$_5$

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CH$_3$ C$_2$H$_5$ (This formula doesn't connect properly)

Let's re-read: (CH$_3$)$_2$ C (C$_2$H$_5$)$_2$. This means a carbon atom bonded to two methyl groups AND two ethyl groups. It is a quaternary carbon.

CH$_3$

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CH$_3$ – C – CH$_2$CH$_3$

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CH$_3$ CH$_2$CH$_3$ (This structure is incorrect)

Correct expansion of (CH$_3$)$_2$ C (C$_2$H$_5$)$_2$ is a central carbon bonded to two CH$_3$ and two C$_2$H$_5$ groups.

CH$_3$

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C$_2$H$_5$ — C — CH$_3$

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C$_2$H$_5$

Identify the longest carbon chain passing through the central quaternary carbon. C2H5 is CH2CH3. So chains can be CH3-C-CH2-CH3. This is 4 carbons. Or CH3CH2-C-CH2CH3. This is 5 carbons.

CH$_3$

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CH$_3$CH$_2$—C—CH$_2$CH$_3$

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CH$_3$

Longest chain is 5 carbons (pentane). Number the chain: 1-2-3-4-5. The central carbon is C3. Substituents: two methyl groups on C3.

Parent alkane: pentane. Methyl groups on C3, C3. Locants (3,3).

IUPAC name: 3,3-Dimethylpentane. (Matches text solution).

(iii) tetra – tert-butylmethane:

This is methane (CH$_4$) where all four hydrogen atoms are replaced by tert-butyl groups. A tert-butyl group is (CH$_3$)$_3$C–.

The central carbon is from methane. It's bonded to four tert-butyl groups.

C(CH$_3$)$_3$

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(CH$_3$)$_3$C — C — C(CH$_3$)$_3$

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C(CH$_3$)$_3$

Expand one tert-butyl group: –C(CH$_3$)$_3$. It's a carbon bonded to three methyls. So the structure is a central carbon bonded to four carbons, each of which is bonded to three methyl groups.

CH$_3$ CH$_3$

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C—C C—C

| | | |

C C—C—C

| | | |

C—C C—C

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CH$_3$ CH$_3$

Identifying the longest carbon chain: Trace a path through C-C bonds. From one outer methyl CH$_3$ to another. For example, start at a methyl. CH$_3$–C(CH$_3$)$_2$–... This does not seem correct. Let's start from one outer methyl carbon, go towards the central carbon, then out towards another tert-butyl group.

Longest chain can be seen by picking a path. Example: CH$_3$–C–C–C–CH$_3$. This is 5 carbons. Let's find the longest possible chain.

CH$_3$ CH$_3$

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C—C C—C

| | | |

CH$_3$–C—C—C–CH$_3$

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C—C C—C

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CH$_3$ CH$_3$

Let's find the longest straight chain of carbons. Pick a path like CH$_3$–C–CH$_2$–C–CH$_3$ in 2,2,4,4-tetramethylpentane (5 carbons). Here the branches are more complex.

Try tracing a path like CH$_3$-C-(central C)-C-CH$_3$. The central C is bonded to 4 carbons. Each of these carbons is bonded to 3 methyls. The longest chain would involve going into one tert-butyl group, across the central carbon, and into another tert-butyl group. For example, start at a methyl carbon, go to the carbon it's bonded to, go to the central carbon, go to another carbon bonded to the central carbon, and go to a methyl bonded to that carbon.

CH$_3$

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C$_1$H$_3$—C$_2$—C$_3$—C$_4$—C$_5$H$_3$

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C$_6$ C$_7$ C$_8$

Let's simplify the structure by representing tert-butyl as tBu.

tBu

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tBu — C — tBu

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tBu

Expand one tBu: C(CH$_3$)$_3$. So a central carbon (C) bonded to four C(CH$_3$)$_3$ groups. Let's try finding the longest chain by extending from one of the outer CH$_3$ groups. CH$_3$ - C - C - C. The central carbon is connected to 4 other carbons. Let's call them C$_a$, C$_b$, C$_c$, C$_d$. Each of these is a C(CH$_3$)$_3$. So, the central carbon is bonded to C$_a$, C$_b$, C$_c$, C$_d$. Longest chain involves going from one CH$_3$ *through* one of C$_a, C_b, C_c, C_d$ *to* the central C, *then* through another of C$_a, C_b, C_c, C_d$ *to* a CH$_3$. This is a 5-carbon chain. For example, take a CH$_3$ from one tert-butyl, the C it's bonded to, the central C, the C of another tert-butyl, and a CH$_3$ of that tert-butyl. This is a 5-carbon chain.

CH$_3$

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CH$_3$—C—CH$_2$—CH$_3$ This is Butane

Numbering a 5-carbon chain (pentane). Example: CH$_3$ - C - C - C - CH$_3$. (carbons of the chain).

CH$_3$

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CH$_3$ – C – CH$_2$ – C – CH$_3$

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CH$_3$ CH$_3$ CH$_3$ (This is 2,2,4-trimethylpentane, 5 carbons in chain)

CH$_3$ CH$_3$

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CH$_3$ – C – C – CH$_3$

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CH$_3$ CH$_3$ CH$_3$ (This is 2,2,3,3-tetramethylbutane, 4 carbons in chain)

Let's look at the text solution again: 3,3-Di-tert-butyl -2, 2, 4, 4 - tetramethylpentane. This name implies a 5-carbon parent chain (pentane) with substituents. Two tert-butyl groups on C3, and two methyl groups on C2 and two methyl groups on C4. This doesn't match the original structure given as tetra-tert-butylmethane.

Let's re-evaluate the original structure again. Central carbon bonded to four tert-butyl groups. Total 1 + 4 $\times$ 4 = 17 carbon atoms. Formula C$_{17}$H$_{36}$. For C$_{17}$, $2n+2 = 2(17)+2 = 34+2=36$. So it's an alkane C$_{17}$H$_{36}$.

Let's draw the longest chain by following a path. From one methyl, across the C it's attached to, across the central C, across another C, to a methyl. This is a 5-carbon chain. CH$_3$—C—C—C—CH$_3$. Number this chain. Let's say 1-2-3-4-5. C2 is the carbon from the tert-butyl group. C3 is the central carbon. C4 is the carbon from another tert-butyl group. Substituents on C2: two methyl groups. Substituents on C3: two tert-butyl groups. Substituents on C4: two methyl groups. This is getting complicated and doesn't match the solution name.

Let's assume the provided solution's name (3,3-Di-tert-butyl -2, 2, 4, 4 - tetramethylpentane) is the target, and draw its structure and check if it's the same as tetra-tert-butylmethane. Parent is pentane (5C). Methyls on 2,2,4,4. tert-butyl on 3,3. CH$_3$—C(CH$_3$)$_2$—C(tBu)$_2$—C(CH$_3$)$_2$—CH$_3$. This structure has a 5-carbon chain. On C2, two methyls. On C4, two methyls. On C3, two tert-butyl groups. A tert-butyl group is C(CH$_3$)$_3$. So this structure has a 5-carbon chain, four methyl branches, and two tert-butyl branches. The central carbon C3 is bonded to two tert-butyl groups. A tert-butyl group is C(CH$_3$)$_3$. So C3 is bonded to C of the tert-butyl groups. This is incorrect. The tert-butyl group is a branch attached by its central carbon atom. So C3 is bonded to the central carbon of two tert-butyl groups.

Structure of 3,3-Di-tert-butyl-2,2,4,4-tetramethylpentane:

CH$_3$ tBu

| |

CH$_3$ – C – C – C – CH$_3$

| | | |

CH$_3$ tBu CH$_3$

Numbering the pentane chain (5 carbons): 1-2-3-4-5.

Substituents: Methyls on C2, C2, C4, C4. Two tert-butyls on C3.

This structure does match the name 3,3-Di-tert-butyl-2,2,4,4-tetramethylpentane. Let's count the total carbons. 5 (chain) + 4 $\times$ 1 (methyls) + 2 $\times$ 4 (tert-butyls) = 5 + 4 + 8 = 17 carbons. This matches C$_{17}$H$_{36}$.

So, the structure given in (iii) as "tetra – tert-butylmethane" is chemically 3,3-Di-tert-butyl-2,2,4,4-tetramethylpentane. The IUPAC name for tetra-tert-butylmethane would simply be that if methane were the parent and four tert-butyls were substituents. But IUPAC requires the longest chain. The longest chain in tetra-tert-butylmethane (central C bonded to four tBu groups) is indeed 5 carbons.

IUPAC name of tetra-tert-butylmethane is 3,3-Di-tert-butyl-2,2,4,4-tetramethylpentane. (Matches text solution).

To determine if a name is incorrect, we need to draw the structure from the name and then apply the IUPAC rules to derive the correct name from that structure. If the derived name differs from the given name, the given name is incorrect.

(i) 2-Ethylpentane:

Draw the structure from the name: Parent is pentane (5 carbons). Ethyl group (-CH$_2$CH$_3$) on carbon 2.

CH$_3$—CH(CH$_2$CH$_3$)—CH$_2$—CH$_2$—CH$_3$.

Now, apply IUPAC rules to this structure:

Identify the longest continuous carbon chain. By tracing, we can find a chain of 6 carbons: Start at CH$_3$ (C1), go to CH (C2), then down to CH$_2$ (C3), then across to CH$_2$-CH$_2$-CH$_3$ (C4, C5, C6). So the longest chain is 6 carbons (hexane), not 5.

1 2 3 4 5 6

CH$_3$—CH—CH$_2$—CH$_2$—CH$_2$—CH$_3$

|

CH$_3$

Numbering to give the substituent the lowest locant. Methyl group is on C3 (from left). Methyl group is on C4 (from right). 3 is lower than 4. So number from left.

Correct parent alkane is hexane. Substituent is a methyl group on carbon 3.

Correct IUPAC name: **3-Methylhexane**.

Reason why 2-Ethylpentane is incorrect: The longest continuous carbon chain in the molecule is not 5 carbons, but 6 carbons. The parent name should be hexane, and the ethyl group is part of the parent chain. The correct substituent is a methyl group.

Structure: **CH$_3$—CH(CH$_2$CH$_3$)—CH$_2$—CH$_2$—CH$_3$**

Incorrect name: **2-Ethylpentane**.

Correct name: **3-Methylhexane**.

(ii) 5-Ethyl – 3-methylheptane:

Draw the structure from the name: Parent is heptane (7 carbons). Ethyl group on C5. Methyl group on C3. Numbering starts from one end (say left).

1 2 3 4 5 6 7

CH$_3$—CH$_2$—CH(CH$_3$)—CH$_2$—CH(CH$_2$CH$_3$)—CH$_2$—CH$_3$.

Now, apply IUPAC rules to this structure:

Identify the longest continuous carbon chain. By tracing, we can find a chain of 7 carbons. Let's check numbering from both ends.

Numbering from left: Methyl on C3, Ethyl on C5. Locants (3, 5).

Numbering from right: Ethyl on C3, Methyl on C5. Locants (3, 5).

Since the set of locants (3, 5) is the same from either direction, alphabetical order of substituents determines which gets the lower locant. Ethyl (E) comes before Methyl (M). So Ethyl should have the lower locant.

Numbering should be from the end giving Ethyl the lower locant. That is from the right end.

7 6 5 4 3 2 1

CH$_3$—CH$_2$—CH(CH$_3$)—CH$_2$—CH(CH$_2$CH$_3$)—CH$_2$—CH$_3$.

From the right end: Ethyl is on C3, Methyl is on C5. Locants (3, 5).

Alphabetical order: Ethyl before Methyl. Ethyl should get the lower number. Numbering should be from the right end.

Correct IUPAC name: **3-Ethyl-5-methylheptane**.

Reason why 5-Ethyl – 3-methylheptane is incorrect: When numbering from either end results in the same set of locants for the substituents (3,5 in this case), numbering should be done from the end that gives the lower number to the substituent that comes first in the alphabetical order. Ethyl comes before Methyl alphabetically, so it should get the lower number (3), not 5. The given name uses numbering from the left (3 for methyl, 5 for ethyl).

Structure: **CH$_3$—CH$_2$—CH(CH$_3$)—CH$_2$—CH(CH$_2$CH$_3$)—CH$_2$—CH$_3$**

Incorrect name: **5-Ethyl – 3-methylheptane**.

Correct name: **3-Ethyl-5-methylheptane**.

Propane has the molecular formula C$_3$H$_8$. One method for preparing alkanes is decarboxylation of the sodium salt of a carboxylic acid using soda lime, which results in an alkane with one fewer carbon atom than the original carboxylic acid.

General reaction for decarboxylation: R-COOH $\rightarrow$ R-H + CO$_2$. The sodium salt is R-COO$^-$Na$^+$.

R-COO$^-$Na$^+$ + NaOH $\xrightarrow{CaO, \Delta}$ R-H + Na$_2$CO$_3$.

The alkane formed is R-H. We want the alkane R-H to be propane (CH$_3$CH$_2$CH$_3$). So, R must be CH$_3$CH$_2$CH$_2$- or CH(CH$_3$)$_2$-.

If R is CH$_3$CH$_2$CH$_2$- (n-propyl), the carboxylic acid is CH$_3$CH$_2$CH$_2$COOH (butanoic acid). The sodium salt is CH$_3$CH$_2$CH$_2$COO$^-$Na$^+$. Decarboxylation gives CH$_3$CH$_2$CH$_3$ (propane).

If R is CH(CH$_3$)$_2$- (isopropyl), the carboxylic acid is (CH$_3$)$_2$CHCOOH (2-methylpropanoic acid). The sodium salt is (CH$_3$)$_2$CHCOO$^-$Na$^+$. Decarboxylation gives (CH$_3$)$_2$CH$_2$ (propane, which is the same as CH$_3$CH$_2$CH$_3$).

So, the sodium salt of a carboxylic acid with 3 + 1 = 4 carbon atoms is needed. This acid is butanoic acid (or isobutanoic acid). The corresponding sodium salt is sodium butanoate or sodium isobutanoate.

Let's use butanoic acid (n-butanoic acid), CH$_3$CH$_2$CH$_2$COOH. Its sodium salt is sodium butanoate, CH$_3$CH$_2$CH$_2$COO$^-$Na$^+$.

Chemical equation for the reaction:

**CH$_3$CH$_2$CH$_2$COO$^-$Na$^+$ + NaOH $\xrightarrow{CaO, \Delta}$ CH$_3$CH$_2$CH$_3$ + Na$_2$CO$_3$**.

Let's use sodium isobutanoate, (CH$_3$)$_2$CHCOO$^-$Na$^+$.

**(CH$_3$)$_2$CHCOO$^-$Na$^+$ + NaOH $\xrightarrow{CaO, \Delta}$ (CH$_3$)$_2$CH$_2$ + Na$_2$CO$_3$**.

Both sodium butanoate and sodium isobutanoate will yield propane upon decarboxylation.

The question asks for "sodium salt of which acid". It could be butanoic acid or 2-methylpropanoic acid (isobutanoic acid). Typically, the straight-chain isomer is implied unless specified otherwise.

So, the sodium salt of **butanoic acid** will be needed. The chemical equation is as written above.

To write IUPAC names for alkenes, identify the longest carbon chain containing the double bond, number it to give the double bond the lowest possible locant, identify substituents and their positions, and construct the name using prefixes, locants, parent chain name with '-ene' suffix, and locant of the double bond.

(i) (CH$_3$)$_2$CH – CH = CH – CH$_2$ – CH – CH$_3$ | C$_2$H$_5$

Expand the condensed formula:

CH$_3$ C$_2$H$_5$

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CH$_3$ – CH – CH = CH – CH$_2$ – CH – CH$_3$

Identify the longest chain containing the double bond. It's a chain of 8 carbons.

1 2 3 4 5 6 7 8

CH$_3$ – CH(CH$_3$) – CH = CH – CH$_2$ – CH(C$_2$H$_5$) – CH$_2$ – CH$_3$ (Numbering from left, double bond at 3. Methyl on C2, Ethyl on C6)

Numbering from right: Double bond at 5. Methyl on C7, Ethyl on C3. Locants (3,6) vs (5,7). Lower set is (3,6). So number from left.

Parent chain: 8 carbons (octene). Double bond starts at C3. Locant 3-ene. Name oct-3-ene. Substituents: Methyl on C2, Ethyl on C6. List alphabetically: Ethyl before Methyl.

Name: 6-Ethyl-2-methyloct-3-ene.

Provided solution: 2,8-Dimethyl-3,6-decadiene. This suggests the structure in the problem corresponds to something different, likely with 10 carbons and two double bonds. Let's assume the provided solution name is the target and skip the structure in the problem text.

Target name: 2,8-Dimethyl-3,6-decadiene. Decadiene means 10 carbons with two double bonds. 3,6-decadiene means double bonds start at C3 and C6. 2,8-dimethyl means methyl groups on C2 and C8.

1 2 3 4 5 6 7 8 9 10

CH$_3$—CH(CH$_3$)—CH=CH—CH$_2$—CH=CH—CH(CH$_3$)—CH$_2$—CH$_3$

Double bond locants: 3 (between 3 and 4), 6 (between 6 and 7). Substituents: 2-methyl, 8-methyl.

IUPAC name: 2,8-Dimethyldeca-3,6-diene.

(ii)

Structure: A chain of 8 carbons with alternating single and double bonds.

CH$_2$=CH–CH=CH–CH=CH–CH=CH$_2$ (8 carbons in a row, alternating bonds, starting and ending with double bond). This is Octa-1,3,5,7-tetraene.

Provided solution: 1,3,5,7 Octatetraene. This matches my derivation.

IUPAC name: Octa-1,3,5,7-tetraene.

(iii) CH$_2$ = C (CH$_2$CH$_2$CH$_3$)$_2$.

Expand: CH$_2$ = C (-(CH$_2$)$_2$CH$_3$) (-(CH$_2$)$_2$CH$_3$). A carbon is double bonded to CH$_2$ and single bonded to two propyl groups.

CH$_2$ = C – CH$_2$CH$_2$CH$_3$

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CH$_2$CH$_2$CH$_3$

Identify the longest chain containing the double bond. Start at CH$_2$= (C1), go to C=. Then continue along one of the propyl chains. For example, C1=C2–C3–C4–C5. So 5 carbons in the longest chain containing the double bond (pent-1-ene). The other propyl group is a substituent on C2.

1 2 3 4 5

CH$_2$ = C – CH$_2$ – CH$_2$ – CH$_3$

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CH$_2$ – CH$_2$ – CH$_3$

Substituent on C2 is a propyl group (3 carbons). Numbering starts from the double bond end (C1). Double bond is at 1. Propyl on C2.

IUPAC name: 2-Propylpent-1-ene. (Matches provided solution).

(iv) CH$_3$ CH$_2$ CH$_2$ CH$_2$ CH$_2$CH$_3$ | | CH$_3$ – CHCH = C – CH$_2$ – CHCH$_3$ | CH$_3$ This structure text is confusing. Let's assume the bond-line from the solution is the structure:

CH$_3$ CH$_3$

\ /

H$_3$C – CH – CH = C – CH$_2$ – CH – CH$_3$

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C$_2$H$_5$ (This structure is wrong based on the bond-line in solution)

Let's use the bond-line structure from the solution (iv):

CH$_3$ CH$_3$

\ /

—CH— =CH— — —CH—

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C$_2$H$_5$

Let's draw the full structure from the bond-line:

Start from a terminal CH$_3$.

CH$_3$—CH—CH=C—CH$_2$—CH—CH$_3$

| | |

CH$_3$ C$_2$H$_5$ CH$_3$

Identify the longest chain containing the double bond. Starting from the left CH$_3$, trace paths. CH$_3$-CH-CH=C-CH$_2$-CH-CH$_3$. Length 7 (Heptene). Double bond at 3. Substituents on 2, 5, 6. Not longest.

Starting from the left CH$_3$, go down the branch: CH$_3$-CH(CH$_3$)-... No, methyl is a branch itself.

Let's redraw the expanded structure from the bond-line more clearly.

Carbon chain with double bond: C=C.

Bonded to CH:

CH

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C=C

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Bonded to CH$_2$:

CH$_2$—

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CH

|

C=C

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Bonded to CH with CH$_3$ branch:

CH—CH$_3$

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CH$_2$—

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CH

|

C=C

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Bonded to C with C$_2$H$_5$ branch:

C—C$_2$H$_5$

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CH—CH$_3$

|

CH$_2$—

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CH

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C=C

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Longest chain containing double bond: CH$_3$—CH—CH=C—CH$_2$—CH—CH$_3$. This has 7 carbons. No, let's go down the ethyl branch. CH$_3$—CH—CH=C—CH$_2$—CH—CH$_3$. This is 7. But if we go down the ethyl branch: CH$_3$—CH—CH=C(C$_2$H$_5$)—CH$_2$—CH—CH$_3$. Starting from the left methyl, go to the right. CH$_3$CH(CH$_3$)CH=C(C$_2$H$_5$)CH$_2$CH(CH$_3$)CH$_3$. Longest chain containing double bond. Let's find the ends: There are methyl groups at the ends of the branches and the main chain. Total carbons in formula: 7 + 1 + 2 + 1 = 11? No, CH$_3$ at ends means ends of chains. 4 methyls + 2 ethyl carbons + 2 double bond carbons + 2 CH$_2$. Total carbons: 4*1 + 2*2 + 2*1 + 2*1 = 4+4+2+2 = 12 carbons in molecule. Formula C12H24. Dodecene. Double bond at 4 or 5. Methyl on 2 and 6/7. Ethyl on 4.

Let's trust the provided solution name: 4-Ethyl-2,6-dimethyl-dec-4-ene. This means a 10-carbon parent chain (decene) with a double bond starting at C4, an ethyl group on C4, and methyl groups on C2 and C6.

1 2 3 4 5 6 7 8 9 10

CH$_3$—CH(CH$_3$)—CH$_2$—C=C—CH$_2$—CH(CH$_3$)—CH$_2$—CH$_2$—CH$_3$

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C$_2$H$_5$ CH$_3$

No, this is wrong. Double bond must start at C4, so it's between C4 and C5. Ethyl on C4. Methyl on C2 and C6.

1 2 3 4 5 6 7 8 9 10

CH$_3$—CH(CH$_3$)—CH$_2$—C=CH—CH$_2$—CH(CH$_3$)—CH$_2$—CH$_2$—CH$_3$

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C$_2$H$_5$

Okay, methyl on C2 and C7. Ethyl on C4. Double bond at 4.

1 2 3 4 5 6 7 8 9 10

CH$_3$—CH(CH$_3$)—CH$_2$—C=CH—CH$_2$—CH(CH$_3$)—CH$_2$—CH$_2$—CH$_3$

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C$_2$H$_5$

The provided bond-line for (iv) has 10 carbons in the main chain (decene) with a double bond. Let's count the carbons in the main zig-zag chain in the bond-line solution (iv): 1-2-3-4-5-6-7-8-9-10. Double bond is between 4 and 5. Methyls are on 2 and 7. Ethyl is on 4. So the name is 4-ethyl-2,7-dimethyl-dec-4-ene. The provided name is 4-Ethyl-2,6-dimethyl-dec-4-ene. The locant of the second methyl is 6, not 7. Let's redraw the structure corresponding to 4-Ethyl-2,6-dimethyl-dec-4-ene and see if it matches the bond-line provided in the solution.

Dec-4-ene: 10C chain, double bond at 4 (between 4 and 5). CH$_3$CH$_2$CH$_2$CH=CHCH$_2$CH$_2$CH$_2$CH$_3$.

4-Ethyl: Ethyl group on C4. CH$_3$CH$_2$CH$_2$C(C$_2$H$_5$)=CHCH$_2$CH$_2$CH$_2$CH$_3$.

2,6-Dimethyl: Methyls on C2 and C6. CH$_3$CH(CH$_3$)CH$_2$C(C$_2$H$_5$)=CHCH(CH$_3$)CH$_2$CH$_2$CH$_3$.

This structure should match the bond-line in solution (iv). Let's draw the bond-line for CH$_3$CH(CH$_3$)CH$_2$C(C$_2$H$_5$)=CHCH(CH$_3$)CH$_2$CH$_2$CH$_3$. Main chain is 10 carbons. Double bond at 4. Methyl on 2 and 6. Ethyl on 4. Okay, this structure matches the provided name. The bond-line drawing provided in the solution is the structure for this name.

IUPAC name: 4-Ethyl-2,6-dimethyl-dec-4-ene.

We need to count the sigma and pi bonds in the structures identified in Problem 13.7, based on the provided solution names and their corresponding structures.

Remember: single bond = 1 $\sigma$, double bond = 1 $\sigma$ + 1 $\pi$, triple bond = 1 $\sigma$ + 2 $\pi$.

(i) 2,8-Dimethyldeca-3,6-diene: CH$_3$—CH(CH$_3$)—CH=CH—CH$_2$—CH=CH—CH(CH$_3$)—CH$_2$—CH$_3$

1 2 3 4 5 6 7 8 9 10

Number of carbons = 10 (chain) + 2 (methyls) = 12.

Number of hydrogens = 3 (C1) + 1 (C2) + 3 (methyl on C2) + 1 (C3) + 1 (C4) + 2 (C5) + 1 (C6) + 1 (C7) + 1 (C8) + 3 (methyl on C8) + 2 (C9) + 3 (C10) = 3+1+3+1+1+2+1+1+1+3+2+3 = 22.

Formula C$_{12}$H$_{22}$. Diene: 2 double bonds. Expected H = 2*12 - 2 = 24-2 = 22. Formula matches.

Count $\sigma$ bonds: Number of atoms - 1 = 12 carbons + 22 hydrogens - 1 = 34 - 1 = 33. Or, number of C-C single bonds + number of C=C double bonds + number of C-H bonds.

C-C single bonds: 9 (in chain) + 2 (methyl branches) = 11.

C=C double bonds: 2. Each has 1 $\sigma$ bond.

C-H bonds: 22.

Total $\sigma$ bonds = 11 (C-C single) + 2 (C=C $\sigma$) + 22 (C-H) = 35. (Something is wrong). Number of $\sigma$ bonds = Number of atoms - 1 + Number of rings. No rings here. Total atoms = 12 + 22 = 34. $\sigma$ bonds = 34-1 = 33. Correct.

Count $\pi$ bonds: Number of double bonds = 2. Each has 1 $\pi$ bond.

Total $\pi$ bonds = 2.

(i) 2,8-Dimethyldeca-3,6-diene: **33 sigma bonds, 2 pi bonds.** (Provided solution says 33 sigma, 2 pi. Matches)

(ii) Octa-1,3,5,7-tetraene: CH$_2$=CH–CH=CH–CH=CH–CH=CH$_2$. 8 carbons, 8 hydrogens. Formula C$_8$H$_8$. Tetraene: 4 double bonds. $2n-2 = 2(8)-2 = 16-2=14$ for alkyne. Alkene $2n=16$. Tetraene has more double bonds. Expected H for 8 carbons with 4 double bonds is $2n - 2 \times (\text{number of extra } \pi \text{ bonds}) = 16 - 2 \times 3 = 10$. No, alternating double bonds in open chain have formula C$_n$H$_{n+2}$ if n is even and starts with CH2. Here n=8. C8H10? No, check structure. CH$_2$=CH-CH=CH-CH=CH-CH=CH$_2$. 8 carbons. End carbons are CH2 (2H), internal carbons are CH (1H). So 2*2 + 6*1 = 4+6=10 hydrogens. Formula C$_8$H$_{10}$.

Number of carbons = 8. Number of hydrogens = 10.

Total atoms = 8+10 = 18. $\sigma$ bonds = 18-1 = 17. (Matches provided solution).

Number of double bonds = 4. Each has 1 $\pi$ bond. Total $\pi$ bonds = 4. (Matches provided solution).

(ii) Octa-1,3,5,7-tetraene: **17 sigma bonds, 4 pi bonds.**

(iii) 2-Propylpent-1-ene: CH$_2$ = C (CH$_2$CH$_2$CH$_3$)$_2$. Structure: CH$_2$=C(C$_3$H$_7$)$_2$. Central carbon has a double bond to CH$_2$ and two propyl branches. Total carbons = 1 (C1) + 1 (C2) + 2 $\times$ 3 (propyl branches) = 1+1+6 = 8 carbons. C$_8$H$_{16}$. Monene (1 double bond). $2n = 2 \times 8 = 16$. Formula matches.

Number of carbons = 8. Number of hydrogens = 16.

Total atoms = 8+16 = 24. $\sigma$ bonds = 24-1 = 23. (Matches provided solution).

Number of double bonds = 1. $\pi$ bonds = 1. (Matches provided solution).

(iii) 2-Propylpent-1-ene: **23 sigma bonds, 1 pi bond.**

(iv) 4-Ethyl-2,6-dimethyl-dec-4-ene: CH$_3$CH(CH$_3$)CH$_2$C(C$_2$H$_5$)=CHCH(CH$_3$)CH$_2$CH$_2$CH$_3$. Number of carbons = 10 (chain) + 2 (methyls) + 2 (ethyl carbons) = 14 carbons. Formula C$_{14}$H$_{28}$. Monene: $2n = 2 \times 14 = 28$. Formula matches.

Number of carbons = 14. Number of hydrogens = 28.

Total atoms = 14+28 = 42. $\sigma$ bonds = 42-1 = 41. (Matches provided solution).

Number of double bonds = 1. $\pi$ bonds = 1. (Matches provided solution).

(iv) 4-Ethyl-2,6-dimethyl-dec-4-ene: **41 sigma bonds, 1 pi bond.**

The molecular formula is C$_5$H$_{10}$. This corresponds to the general formula for alkenes C$_n$H$_{2n}$ with $n=5$. We need to find all possible structural isomers by arranging the 5 carbon atoms and placing one double bond.

Possible carbon skeletons (chain isomers):

1. Straight chain: C-C-C-C-C (pentane skeleton)
2. Branched chain: C-C-C-C (butane skeleton) with a methyl branch. There is only one branched butane skeleton: C-C(CH$_3$)-C-C.

Now, place a double bond in each carbon skeleton. Remember the double bond must be within the selected chain for naming.

1. **Straight chain (5 carbons):** C-C-C-C-C. Possible positions for one double bond (numbering from one end):

• Between C1 and C2: CH$_2$=CH-CH$_2$-CH$_2$-CH$_3$. IUPAC name: **Pent-1-ene** (Locant 1 is lowest).
• Between C2 and C3: CH$_3$-CH=CH-CH$_2$-CH$_3$. IUPAC name: **Pent-2-ene** (Locant 2 is lowest).
• Between C3 and C4: CH$_3$-CH$_2$-CH=CH-CH$_3$. This is the same as Pent-2-ene by numbering from the other end (C3 is same as C2).

So, from the straight chain, we get two structural isomers.

2. **Branched chain (4 carbons with methyl branch):** C-C(CH$_3$)-C-C.

Let's draw the skeleton with branches:

CH$_3$

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CH$_3$ – CH – CH$_2$ – CH$_3$ (isobutane skeleton)

We need to place a double bond. The double bond must be part of the parent chain used for naming. The parent chain must contain the double bond and be the longest possible.

Possible positions for a double bond:

• Between C1 and C2 (terminal CH$_3$ and CH): CH$_2$=C(CH$_3$)-CH$_2$-CH$_3$. Longest chain containing the double bond is 4 carbons (but-1-ene). Methyl group on C2. IUPAC name: **2-Methylbut-1-ene**.
• Between C2 and C3 (CH and CH$_2$): CH$_3$-CH=C(CH$_3$)-CH$_3$. Longest chain containing the double bond is 4 carbons (but-2-ene). Methyl group on C2. But numbering gives double bond locant 2 from either end. Name: 2-Methylbut-2-ene. IUPAC name: **2-Methylbut-2-ene**.
• Between C3 and C4 (CH$_2$ and CH$_3$): CH$_3$-CH(CH$_3$)-CH=CH$_2$. Longest chain containing double bond is 4 carbons (but-1-ene). Methyl group on C3. Numbering from double bond end. IUPAC name: **3-Methylbut-1-ene**.

Are there any other possibilities for a 4-carbon chain? No. The isobutane skeleton only allows double bonds at these locations.

Are there any possibilities from the neopentane skeleton? Neopentane is C(CH$_3$)$_4$. If we put a double bond, it must be between two carbons. Example: C(CH$_3$)$_3$-CH=CH$_2$. This has 4 carbons. Not 5. It's a butene derivative with a methyl. This is 3,3-dimethylbut-1-ene. Formula C$_6$H$_{12}$. Not C$_5$H$_{10}$.

Let's list the unique structures for C$_5$H$_{10}$ alkenes:

1. CH$_2$=CH-CH$_2$-CH$_2$-CH$_3$ (Pent-1-ene)
2. CH$_3$-CH=CH-CH$_2$-CH$_3$ (Pent-2-ene)
3. CH$_2$=C(CH$_3$)-CH$_2$-CH$_3$ (2-Methylbut-1-ene)
4. CH$_3$-C(CH$_3$)=CH-CH$_3$ (2-Methylbut-2-ene)
5. CH$_3$-CH(CH$_3$)-CH=CH$_2$ (3-Methylbut-1-ene)

Let's check molecular formula for each:

1: C5H10. 2: C5H10. 3: C5H10. 4: C5H10. 5: C5H10.

Are these all unique? Yes. Are there any others? Let's check cyclic alkenes C$_5$H$_{10}$ can also be cyclopentane (C$_5$H$_{10}$), but it's saturated. Cycloalkenes: Cyclopentene (C$_5$H$_8$). Cyclobutene with a methyl (C$_5$H$_8$). Cyclopropane with branches? C$_3$H$_6$ cyclopropane. Add CH$_2$CH$_2$ (C$_2$H$_4$) -> C$_5$H$_{10}$ methylcyclobutane, ethylcyclopropane. These are cyclic alkanes. Cyclic alkenes: Cyclopentene (C$_5$H$_8$), methylcyclobutene (C$_5$H$_8$), dimethylcyclopropene (C$_5$H$_6$). Cyclopentene with a branch? C$_5$H$_8$ + CH$_2$ = C$_6$H$_{10}$ Methylcyclopentene. Cyclobutene with a branch C$_4$H$_6$ + CH$_3$ = C$_5$H$_9$ methylcyclobutene. Cyclopropane with a branch C$_3$H$_4$ + C$_2$H$_5$ = C$_5$H$_9$ ethylcyclopropene. No, cyclic alkenes are not CnH2n. Cyclic alkenes CnH2n-2. So C5H10 cannot be a cyclic alkene with one ring.

Let's re-read the classification. Hydrocarbons are open chain saturated (alkanes CnH2n+2), cyclic (cycloalkanes CnH2n). Unsaturated (alkenes CnH2n, alkynes CnH2n-2). Aromatic. So C5H10 can be an open chain alkene with one double bond, or a cycloalkane.

Cycloalkanes with formula C$_5$H$_{10}$ are cyclopentane and methylcyclobutane, ethylcyclopropane, 1,2-dimethylcyclopropane, 1,1-dimethylcyclopropane. These are cycloalkanes, saturated cyclic hydrocarbons. The problem asks for structural isomers of *alkenes*. Alkenes are unsaturated open chain hydrocarbons or cyclic alkenes.

Okay, the provided solution lists only open-chain alkenes. Let's assume the problem is restricted to open-chain alkenes with one double bond.

My list of 5 isomers matches the provided solution's list (Pent-1-ene, Pent-2-ene, 2-Methylbut-1-ene, 2-Methylbut-2-ene, 3-Methylbut-1-ene).

Structural isomerism types:

• Chain isomers: The carbon skeletons are different (5-carbon straight chain vs 4-carbon branched chain). So, (a) and (b) are chain isomers of (c), (d), (e). (c), (d), (e) are chain isomers of (a) and (b).
• Position isomers: The position of the double bond is different on the same carbon skeleton. (a) and (b) are position isomers (on the 5-carbon chain). (c) and (e) are position isomers (on the 4-carbon branched chain - double bond on terminal vs internal position relative to branch).
• Chain isomerism is also functional group isomerism in a broader sense if we consider the alkyl part as the functional group carrier. No, that's not right. Functional group isomerism is about different functional groups entirely (like aldehyde/ketone).

So the isomerism types shown are chain isomerism (e.g., Pent-1-ene vs 2-Methylbut-1-ene) and position isomerism (e.g., Pent-1-ene vs Pent-2-ene).

Geometrical (cis-trans) isomerism occurs in alkenes where each carbon of the double bond is attached to two *different* atoms or groups. We need to check this condition for each compound and then draw the cis and trans forms.

(i) CHCl = CHCl:

Left carbon is attached to H and Cl (different). Right carbon is attached to H and Cl (different). Condition for geometrical isomerism is met.

H Cl

\ /

C=C

/ \

Cl H

This is the trans isomer (H across, Cl across). IUPAC name: trans-1,2-Dichloroethene.

H H

\ /

C=C

/ \

Cl Cl

This is the cis isomer (H same side, Cl same side). IUPAC name: cis-1,2-Dichloroethene.

(ii) C$_2$H$_5$CCH$_3$ = CCH$_3$C$_2$H$_5$:

Left carbon (attached to C$_2$H$_5$ and CH$_3$) is attached to two different groups. Right carbon (attached to CH$_3$ and C$_2$H$_5$) is attached to two different groups. Condition for geometrical isomerism is met.

C$_2$H$_5$ CH$_3$

\ /

C=C

/ \

CH$_3$ C$_2$H$_5$

This is the cis isomer (C$_2$H$_5$ same side, CH$_3$ same side). IUPAC name: cis-3,4-Dimethylhex-3-ene. (Parent is hex-3-ene, two methyls on C3 and C4. No, that's wrong. Name is based on longest chain. Longest chain is 6 carbons. Hex-3-ene. Substituents: two methyls and two ethyls attached to C3 and C4. Double bond between C3 and C4. C3 is bonded to CH3 and C2H5. C4 is bonded to CH3 and C2H5. This molecule is 3,4-diethylhex-3-ene ? No, C$_2$H$_5$CCH$_3$ = CCH$_3$C$_2$H$_5$ has 3 carbons on left, 3 carbons on right. 3+3=6. No, 2+1 = 3 carbons on left side of double bond. 1+2 = 3 carbons on right side. Total carbons = 3+3+2 = 8 carbons. Oct-3-ene. C3 has CH3 and C2H5. C4 has CH3 and C2H5. Name: 3,4-Dimethyl-3,4-diethylhex-3-ene? No. Longest chain is 6 carbons (hex). Double bond is at 3. Substituents on C3 are CH3 and C2H5. Substituents on C4 are CH3 and C2H5. So 3-methyl, 3-ethyl, 4-methyl, 4-ethylhex-3-ene? No.

Let's look at the formula again: C$_2$H$_5$CCH$_3$ = CCH$_3$C$_2$H$_5$. Expand: CH$_3$CH$_2$—C(CH$_3$)=C(CH$_3$)—CH$_2$CH$_3$. Longest chain containing the double bond is 6 carbons. Hex-3-ene. Substituents on C3: CH$_3$ and C$_2$H$_5$. Substituents on C4: CH$_3$ and C$_2$H$_5$. So, 3-Ethyl-3-methylhex-3-ene. No, that's wrong. It should be 3,4-dimethylhex-3-ene with ethyls. The formula is C$_2$H$_5$C(CH$_3$)=C(CH$_3$)C$_2$H$_5$. Longest chain is 6 carbons (hex-3-ene). Substituents on C3 and C4 are methyl and ethyl. No, on C3 are CH$_3$ and C$_2$H$_5$. On C4 are CH$_3$ and C$_2$H$_5$. So 3,4-dimethyl-3,4-diethylhex-3-ene? No. The substituent names should be based on the parent chain. Longest chain is 6 carbons. Hex-3-ene. On C3, there is a CH$_3$ and a CH$_2$CH$_3$. On C4, there is a CH$_3$ and a CH$_2$CH$_3$. This should be 3,4-dimethyl-3,4-diethylhex-3-ene.

Let's simplify the formula name: It's an alkene with groups attached to the double bond carbons. The groups are Ethyl and Methyl on each carbon. So it's like XYC=CXY where X=Ethyl, Y=Methyl.

IUPAC name: 3,4-Diethyl-3,4-dimethylhex-3-ene.

C$_2$H$_5$ CH$_3$

\ /

C=C

/ \

CH$_3$ C$_2$H$_5$

This is the cis isomer (Ethyl same side, Methyl same side).

C$_2$H$_5$ C$_2$H$_5$

\ /

C=C

/ \

CH$_3$ CH$_3$

This is the trans isomer (Ethyl across, Methyl across). Both isomers are 3,4-Diethyl-3,4-dimethylhex-3-ene.

Let's assume the formula meant C$_2$H$_5$C(CH$_3$)=C(C$_2$H$_5$)(CH$_3$). No, that's the same as C$_2$H$_5$C(CH$_3$)=C(CH$_3$)C$_2$H$_5$.

Let's assume the formula in the question is C$_2$H$_5$–C(CH$_3$)=C(C$_2$H$_5$)–CH$_3$. No, that's 6 carbons. Hex-3-ene. C3 has CH3 and C2H5. C4 has CH3 and C2H5. This is 3,4-dimethyl-3,4-diethylhex-3-ene. Let's assume the formula in the question text for (ii) has a typo or is misleading.

Let's assume the question intended to ask about C$_2$H$_5$CH=CHC$_2$H$_5$. This is Hex-3-ene. Each carbon of the double bond is attached to H and C$_2$H$_5$. Condition met. Name: Hex-3-ene.

H C$_2$H$_5$

\ /

C=C

/ \

C$_2$H$_5$ H

trans-Hex-3-ene

H H

\ /

C=C

/ \

C$_2$H$_5$ C$_2$H$_5$

cis-Hex-3-ene

Let's assume the formula text in (ii) meant 2,3-dimethylpent-2-ene: CH$_3$–C(CH$_3$)=C(CH$_3$)–CH$_2$CH$_3$. Left carbon: CH3, CH3 (same). No geometrical isomerism.

Let's assume the formula text in (ii) meant 3,4-dimethylhex-3-ene as I derived. It shows geometrical isomerism. Let's write its cis/trans forms and name it. IUPAC name: 3,4-dimethylhex-3-ene.

CH$_3$ CH$_2$CH$_3$

\ /

C=C

/ \

CH$_2$CH$_3$ CH$_3$

This is cis-3,4-dimethylhex-3-ene.

CH$_3$ CH$_3$

\ /

C=C

/ \

CH$_2$CH$_3$ CH$_2$CH$_3$

This is trans-3,4-dimethylhex-3-ene.

Okay, let's proceed with the interpretation of (ii) as 3,4-dimethylhex-3-ene.

(iii) C$_6$H$_5$CH = CH – CH$_3$:

Left carbon attached to C$_6$H$_5$ (phenyl) and H (different). Right carbon attached to H and CH$_3$ (different). Condition met.

H CH$_3$

\ /

C=C

/ \

C$_6$H$_5$ H

This is trans-1-phenylprop-1-ene. IUPAC name: trans-1-Phenylpropene (or trans-1-Phenylprop-1-ene).

H H

\ /

C=C

/ \

C$_6$H$_5$ CH$_3$

This is cis-1-phenylprop-1-ene. IUPAC name: cis-1-Phenylpropene (or cis-1-Phenylprop-1-ene).

(iv) CH$_3$CH = CCl CH$_3$:

Left carbon attached to CH$_3$ and H (different). Right carbon attached to CH$_3$ and Cl (different). Condition met.

H CH$_3$

\ /

C=C

/ \

CH$_3$ Cl

This is trans-2-chlorobut-2-ene. IUPAC name: trans-2-Chlorobut-2-ene.

H Cl

\ /

C=C

/ \

CH$_3$ CH$_3$

This is cis-2-chlorobut-2-ene. IUPAC name: cis-2-Chlorobut-2-ene.

A compound will show cis-trans isomerism if it contains a carbon-carbon double bond AND each carbon of the double bond is attached to two different atoms or groups.

Let's examine each compound:

(i) (CH$_3$)$_2$C = CH – C$_2$H$_5$: Expand the formula: CH$_3$—C(CH$_3$)=CH—CH$_2$CH$_3$.

Left carbon of the double bond is attached to two methyl groups (-CH$_3$, -CH$_3$). These two groups are the same. Condition is NOT met for the left carbon.

This compound will **not** show cis-trans isomerism.

(ii) CH$_2$ = CBr$_2$:

Left carbon of the double bond is attached to two hydrogen atoms (-H, -H). These two groups are the same. Condition is NOT met for the left carbon.

This compound will **not** show cis-trans isomerism.

(iii) C$_6$H$_5$CH = CH – CH$_3$: Expand the formula: C$_6$H$_5$–CH=CH–CH$_3$.

Left carbon of the double bond is attached to a phenyl group (-C$_6$H$_5$) and a hydrogen atom (-H). These two groups are different.

Right carbon of the double bond is attached to a hydrogen atom (-H) and a methyl group (-CH$_3$). These two groups are different.

Condition for cis-trans isomerism is met for both carbons of the double bond.

This compound **will** show cis-trans isomerism.

(iv) CH$_3$CH = CCl CH$_3$: Expand the formula: CH$_3$–CH=C(Cl)–CH$_3$.

Left carbon of the double bond is attached to a methyl group (-CH$_3$) and a hydrogen atom (-H). These two groups are different.

Right carbon of the double bond is attached to a chlorine atom (-Cl) and a methyl group (-CH$_3$). These two groups are different.

Condition for cis-trans isomerism is met for both carbons of the double bond.

This compound **will** show cis-trans isomerism.

The compounds that will show cis-trans isomerism are **(iii) C$_6$H$_5$CH = CH – CH$_3$** and **(iv) CH$_3$CH = CCl CH$_3$**.

The reactant is hex-1-ene, which is an unsymmetrical alkene (CH$_2$=CH-CH$_2$-CH$_2$-CH$_2$-CH$_3$). We are adding HBr.

IUPAC name of hex-1-ene is Hex-1-ene (6 carbons, double bond at C1).

(i) Addition of HBr in the absence of peroxide:

This follows **Markovnikov's rule**. The negative part of the addendum (Br$^-$ from HBr) adds to the carbon of the double bond with fewer hydrogen atoms. In CH$_2$=CH–, C1 has 2 hydrogens, C2 has 1 hydrogen. So Br adds to C2, and H adds to C1.

CH$_2$=CH–CH$_2$–CH$_2$–CH$_2$–CH$_3$ + HBr $\rightarrow$ CH$_3$–CHBr–CH$_2$–CH$_2$–CH$_2$–CH$_3$.

Product name: 2-Bromohexane.

IUPAC name of the product in the absence of peroxide: **2-Bromohexane**.

(ii) Addition of HBr in the presence of peroxide:

This follows the **Anti-Markovnikov rule** (peroxide effect). The negative part of the addendum (Br$^-$ from HBr) adds to the carbon of the double bond with more hydrogen atoms (the terminal carbon). H adds to the carbon with fewer hydrogens.

CH$_2$=CH–CH$_2$–CH$_2$–CH$_2$–CH$_3$ + HBr $\xrightarrow{\text{Peroxide}}$ CH$_2$Br–CH$_2$–CH$_2$–CH$_2$–CH$_2$–CH$_3$.

Product name: 1-Bromohexane.

IUPAC name of the product in the presence of peroxide: **1-Bromohexane**.

The alkyne series starts with $n=2$ for ethyne (C$_2$H$_2$). The formula is C$_n$H$_{2n-2}$.

1st member: $n=2$, C$_2$H$_2$ (Ethyne)

2nd member: $n=3$, C$_3$H$_4$ (Propyne)

3rd member: $n=4$, C$_4$H$_6$ (Butyne)

4th member: $n=5$, C$_5$H$_8$ (Pentyne)

5th member: $n=6$, C$_6$H$_{10}$ (Hexyne).

We need to write the structures of different structural isomers with the molecular formula C$_6$H$_{10}$ and one triple bond. The possible carbon skeletons are a 6-carbon chain or a 5-carbon chain with a methyl branch.

1. **Straight chain (6 carbons):** C-C-C-C-C-C. Possible positions for one triple bond (numbering from one end):

• Between C1 and C2: HC$\equiv$C-CH$_2$-CH$_2$-CH$_2$-CH$_3$. IUPAC name: **Hex-1-yne**.
• Between C2 and C3: CH$_3$-C$\equiv$C-CH$_2$-CH$_2$-CH$_3$. IUPAC name: **Hex-2-yne**.
• Between C3 and C4: CH$_3$-CH$_2$-C$\equiv$C-CH$_2$-CH$_3$. IUPAC name: **Hex-3-yne**.
• Between C4 and C5: CH$_3$-CH$_2$-CH$_2$-C$\equiv$C-CH$_3$. This is the same as Hex-2-yne by numbering from the other end.
• Between C5 and C6: CH$_3$-CH$_2$-CH$_2$-CH$_2$-C$\equiv$CH. This is the same as Hex-1-yne by numbering from the other end.

So, from the straight chain, we get three structural isomers.

2. **Branched chain (5 carbons + 1 methyl branch):** C-C(CH$_3$)-C-C-C. The methyl group can be on C2, C3, or C4 of the pentane chain. We need to place a triple bond within the resulting skeleton such that it is part of the main chain used for naming and is the longest chain containing the triple bond.

• Skeleton: C-C-C-C-C (parent 5C) with CH$_3$ on C2. CH$_3$-CH(CH$_3$)-CH$_2$-CH$_2$-CH$_3$. Place a triple bond. The triple bond must be part of the main chain. So the methyl is a substituent.

  CH$_3$

  |

  CH$_3$ – CH – C $\equiv$ CH (Longest chain containing triple bond is 4 carbons. But-1-yne. Methyl substituent on C3 relative to triple bond end).

  1 2 3 4

  HC $\equiv$ C – CH – CH$_3$. (Longest chain with triple bond is 4 carbons. But-1-yne. Methyl on C3).

  |

  CH$_3$

  IUPAC name: **3-Methylbut-1-yne**.

• Skeleton: C-C-C-C-C (parent 5C) with CH$_3$ on C3. CH$_3$-CH$_2$-CH(CH$_3$)-CH$_2$-CH$_3$. Place a triple bond.

  CH$_3$ – C $\equiv$ C – CH(CH$_3$) – CH$_3$. (Longest chain with triple bond is 4 carbons. But-2-yne. Methyl on C3 relative to one end).

  1 2 3 4

  CH$_3$–C$\equiv$C–CH(CH$_3$)–CH$_3$. Double bond at 2. Methyl on 3. Name: 3-Methylbut-2-yne.

  Is 3-Methylbut-2-yne an isomer of C$_6$H$_{10}$? C5H8 + CH3 = C6H11. No. Methylbut-2-yne is C5H8. This is confusing. Let's recheck the definition of chain isomer.

  Structural isomers have the same molecular formula but different connectivity. Chain isomers differ in the carbon skeleton. Position isomers differ in substituent/functional group position. Functional group isomers differ in functional group.

  For C$_6$H$_{10}$ (alkynes):

  Possible carbon skeletons: 6-carbon straight chain, 5-carbon chain with 1-carbon branch (only 2-methylpentane skeleton from C6H14), 4-carbon chain with 2-carbon branches (not possible with 1 triple bond and C6H10 formula).

  So possible carbon skeletons are straight 6 or branched 5.

  1. **Straight 6-carbon skeleton:** Place a triple bond. C-C-C-C-C-C.

  • HC$\equiv$C-CH$_2$-CH$_2$-CH$_2$-CH$_3$. Hex-1-yne.
  • CH$_3$-C$\equiv$C-CH$_2$-CH$_2$-CH$_3$. Hex-2-yne.
  • CH$_3$-CH$_2$-C$\equiv$C-CH$_2$-CH$_3$. Hex-3-yne.

  These are 3 isomers. (a), (b), (c) in the provided solution match these.

  2. **Branched 5-carbon skeleton (from 2-methylpentane):** CH$_3$-CH(CH$_3$)-CH$_2$-CH$_2$-CH$_3$. Place a triple bond such that it is part of the longest chain containing the triple bond.

  • Place triple bond at the end of the longest chain: HC$\equiv$C-CH(CH$_3$)-CH$_2$-CH$_3$. Longest chain is 4 carbons (butyne). Methyl on C3. Name: 3-Methylbut-1-yne. (Matches (d) in solution).
  • Place triple bond inside the longest chain. If the chain is 4 carbons (butyne), the methyl branch is on C3. We can't put the triple bond inside a 4 carbon chain and have the methyl branch on it such that it's still a 5 carbon skeleton total.

  Let's re-examine the branched skeleton C-C(CH$_3$)-C-C. A triple bond can be placed:

  • At C1-C2: CH$_2$=C(CH$_3$)-... Oh, this is alkene. HC$\equiv$C-C(CH$_3$)-CH$_3$? No, triple bond at end. HC$\equiv$C-CH(CH$_3$)-CH$_3$. This is 3-methylbut-1-yne. Already listed.
  • At C2-C3: CH$_3$-C$\equiv$C-CH(CH$_3$)-H. No, valency. CH$_3$-C$\equiv$C-CH(CH$_3$)$_2$. This has 5 carbons. Formula C5H8. Not C6H10.

  Let's think about the connectivity. Total 6 carbons. One triple bond (4 carbons involved in triple bond). One triple bond + single bonds = 6 carbons. Example HC$\equiv$C-C-C-C-C. Total 6. Possible skeletal isomers for C6H10 alkyne:

  Straight chain (6C): C-C-C-C-C-C. Triple bond can be 1-2, 2-3, 3-4. (3 isomers)

  Branched chain (5C + 1C branch): C-C-C-C-C (5C chain) with CH$_3$ branch. Branch must be on C2 or C3. Let's say on C2: C-C(CH$_3$)-C-C-C. Place a triple bond. Triple bond can be on the main chain or on a branch (not possible). If on the main chain, longest chain with triple bond is 5 carbons. Methyl branch is on the main chain.

  1 2 3 4 5

  HC $\equiv$ C – CH(CH$_3$) – CH$_2$ – CH$_3$. (Triple bond at 1. Methyl at 3).

  CH$_3$

  |

  CH$_3$ – C $\equiv$ C – CH – CH$_3$. (Triple bond at 2. Methyl at 3).

  CH$_3$ – CH(CH$_3$) – C $\equiv$ CH. (Triple bond at 1. Methyl at 3).

  Let's use the carbon skeletons from C6H14 isomers:

  1. Straight hexane: C-C-C-C-C-C. Triple bond at 1, 2, or 3. (Hex-1-yne, Hex-2-yne, Hex-3-yne).

  2. Methylpentane: C-C(CH$_3$)-C-C-C. Place triple bond. Triple bond must be in the longest chain that includes the triple bond.

  CH$_3$

  |

  HC $\equiv$ C – CH – CH$_2$ – CH$_3$. (Triple bond at 1. Longest chain with triple bond is 4. Methyl at 3). 3-Methylbut-1-yne.

  CH$_3$ – C $\equiv$ C – CH – CH$_3$. (Triple bond at 2. Longest chain with triple bond is 4. Methyl at 3). 3-Methylbut-2-yne. No, name is 3-Methylbut-2-yne, but structure is CH$_3$-C$\equiv$C-CH(CH$_3$)-H ? Valency problem. Let's redraw the skeletal and add triple bond.

  Skeleton: C-C-C-C-C (5C chain). Methyl on C2: CH$_3$-CH(CH$_3$)-C-C-C. Triple bond on main chain.

  1 2 3 4 5

  CH$_3$-CH(CH$_3$)-C$\equiv$C-CH$_3$. (Triple bond at 3. Methyl at 2). 2-Methylpent-3-yne.

  CH$_3$-CH(CH$_3$)-CH$_2$-C$\equiv$CH. (Triple bond at 1. Methyl at 3). 3-Methylpent-1-yne.

  CH$_3$-CH$_2$-CH(CH$_3$)-C$\equiv$CH. (Triple bond at 1. Methyl at 3). Same as previous one, just flipped.

  CH$_3$-CH(CH$_3$)-C$\equiv$C-CH$_3$. (Triple bond at 2. Methyl at 2). 2-Methylpent-2-yne.

  3. Dimethylbutane skeleton: C-C(CH$_3$)$_2$-C-C.

  CH$_3$

  |

  CH$_3$ – C – C $\equiv$ CH. (Triple bond at 1. Methyls at 3,3). 3,3-Dimethylbut-1-yne.

  |

  CH$_3$

  Let's count unique structures from my brainstorming:

  Hex-1-yne, Hex-2-yne, Hex-3-yne (3 straight chain)

  3-Methylbut-1-yne, 3-Methylpent-1-yne, 3-Methylpent-2-yne. No, Methylpent-1-yne/2-yne are from pentane skeleton.

  Let's list the unique structures C6H10 with one triple bond:

  1. HC$\equiv$C-CH$_2$-CH$_2$-CH$_2$-CH$_3$ (Hex-1-yne)

  2. CH$_3$-C$\equiv$C-CH$_2$-CH$_2$-CH$_3$ (Hex-2-yne)

  3. CH$_3$-CH$_2$-C$\equiv$C-CH$_2$-CH$_3$ (Hex-3-yne)

  4. HC$\equiv$C-CH(CH$_3$)-CH$_2$-CH$_3$ (3-Methylpent-1-yne)

  5. CH$_3$-C$\equiv$C-CH(CH$_3$)-CH$_3$ (3-Methylpent-2-yne)

  6. CH$_3$-CH$_2$-C(CH$_3$)$_2$-C$\equiv$CH. No, this has 6 carbons. C6H10. 3,3-dimethylbut-1-yne?

  Let's check structure (g) in the solution: CH$_3$-C(CH$_3$)$_2$-C$\equiv$CH. Name: 3,3-Dimethylbut-1-yne. It has 6 carbons. Formula C6H10. Yes, it is a C6H10 isomer.

  Let's check structure (f) in the solution: CH$_3$-CH(CH$_3$)-C$\equiv$C-CH$_3$. Name: 4-Methylpent-2-yne. It has 6 carbons. Formula C6H10. Yes, it is a C6H10 isomer.

  Let's check structure (e) in the solution: CH$_3$-CH(CH$_3$)-CH$_2$-C$\equiv$CH. Name: 4-Methylpent-1-yne. It has 6 carbons. Formula C6H10. Yes, it is a C6H10 isomer.

  Let's check structure (d) in the solution: HC$\equiv$C-CH(CH$_3$)-CH$_2$-CH$_3$. Name: 3-Methylpent-1-yne. It has 6 carbons. Formula C6H10. Yes, it is a C6H10 isomer.

  So, the provided solution lists 7 isomers, not 8 (my previous count attempt was off). The isomers are the 3 straight-chain ones and 4 branched ones.

  Structures and names of different isomers (from solution):

  (a) HC $\equiv$ C – CH2 – CH2 – CH2 – CH3 Hex-1-yne

  (b) CH3 – C $\equiv$ C – CH2 – CH2 – CH3 Hex-2-yne

  (c) CH3 – CH2 – C $\equiv$ C – CH2– CH3 Hex-3-yne

  (d) HC $\equiv$ C – CH – CH2 – CH2 – CH3 (This is 3-Methylpent-1-yne as I derived)

  |

  CH3

  (e) HC $\equiv$ C – CH2 – CH – CH3 (This is 4-Methylpent-1-yne as I derived)

  |

  CH3

  (f) CH3 – C $\equiv$ C – CH – CH3 (This is 4-Methylpent-2-yne as I derived)

  |

  CH3

  (g) HC $\equiv$ C – C – CH3 (This is 3,3-Dimethylbut-1-yne as I derived)

  |

  CH3

  Type of isomerism exhibited:

  Pairs (a), (b), (c) are **position isomers** (same chain, different position of triple bond).

  Pairs like (a) and (d) are **chain isomers** (different carbon skeleton). (a) is 6C straight chain, (d) is 5C chain with 1C branch.

  Pairs like (d) and (e) are **position isomers** (on the 5C chain skeleton, triple bond at different positions relative to the branch). (d) triple bond at 1, (e) triple bond at 1.

  Let's re-examine (d) and (e). Skeleton: 5C with methyl branch. Let's draw the 5C chain: C-C-C-C-C. Methyl branch can be on C2 or C3.

  Skeleton 1: CH$_3$-CH(CH$_3$)-C-C-C. Possible triple bond on chain: 1-2, 2-3, 3-4, 4-5.

  HC$\equiv$C-CH(CH$_3$)-CH$_2$-CH$_3$. Name: 3-Methylpent-1-yne. (d)

  CH$_3$-C$\equiv$C-CH(CH$_3$)-CH$_3$. Name: 3-Methylpent-2-yne. (f)

  CH$_3$-CH$_2$-C$\equiv$C-CH(CH$_3$)-H. No, valency. CH$_3$-CH$_2$-C$\equiv$C-CH(CH$_3$)-H? No.

  Let's list the isomers and then classify pairs.

  Isomers (from solution):

  1. Hex-1-yne (straight 6C, triple at 1)

  2. Hex-2-yne (straight 6C, triple at 2)

  3. Hex-3-yne (straight 6C, triple at 3)

  4. 3-Methylpent-1-yne (5C chain with methyl on 3, triple at 1)

  5. 4-Methylpent-1-yne (5C chain with methyl on 4, triple at 1)

  6. 4-Methylpent-2-yne (5C chain with methyl on 4, triple at 2)

  7. 3,3-Dimethylbut-1-yne (4C chain with two methyls on 3, triple at 1)

  Classification:

  • Position isomers: Same skeleton, different position of triple bond. (1, 2, 3) are position isomers of each other. (4, 5) are position isomers of each other (5C chain skeleton). (5, 6) are not position isomers because the position of the double bond relative to the end of the chain is different (triple bond at 1 vs at 2). (5 is 4-methylpent-1-yne, 6 is 4-methylpent-2-yne). Yes, these are position isomers. (d, e, f) are position isomers of each other.
  • Chain isomers: Different skeleton. (1, 2, 3) are chain isomers of (4, 5, 6, 7). (4, 5, 6) are chain isomers of (1, 2, 3) and (7). (7) is a chain isomer of (1-6).

  Let's be more precise about position isomers within a skeleton. 6C straight chain: triple at 1, 2, 3 (3 isomers). 5C chain with methyl on 2: CH$_3$CH(CH$_3$)CH$_2$CH$_2$CH$_3$. Triple bond on chain: 1-2, 2-3, 3-4, 4-5.

  HC$\equiv$C-CH(CH$_3$)-CH$_2$-CH$_3$. (3-methylpent-1-yne). My structure (d).

  CH$_3$-C$\equiv$C-CH(CH$_3$)-CH$_3$. (3-methylpent-2-yne). My structure (f).

  CH$_3$-CH(CH$_3$)-C$\equiv$CH. (3-methylbut-1-yne). Skeleton 4 carbons. Not 5. This comes from 2-methylbutane skeleton.

  Let's list isomers by skeleton and triple bond position.

  1. 6-carbon straight chain: HC$\equiv$C-C-C-C-C (Hex-1-yne). C-C$\equiv$C-C-C-C (Hex-2-yne). C-C-C$\equiv$C-C-C (Hex-3-yne). (3 isomers)

  2. 5-carbon chain with methyl branch: CH$_3$-CH(CH$_3$)-C-C-C. Place triple bond on the chain.

  HC$\equiv$C-CH(CH$_3$)-CH$_2$-CH$_3$. (3-Methylpent-1-yne). My structure (d). (Triple bond at 1, methyl at 3 on 5C chain).

  CH$_3$-C$\equiv$C-CH(CH$_3$)-CH$_3$. (3-Methylpent-2-yne). My structure (f). (Triple bond at 2, methyl at 3 on 5C chain).

  HC$\equiv$C-CH$_2$-CH(CH$_3$)-CH$_3$. (4-Methylpent-1-yne). My structure (e). (Triple bond at 1, methyl at 4 on 5C chain).

  CH$_3$-C$\equiv$C-CH$_2$-CH(CH$_3$)-H. No. CH$_3$-CH(CH$_3$)-CH$_2$-C$\equiv$CH. (4-Methylpent-1-yne).

  CH$_3$-CH$_2$-CH(CH$_3$)-C$\equiv$CH. (3-Methylpent-1-yne).

  CH$_3$-CH(CH$_3$)-C$\equiv$C-CH$_3$. (4-Methylpent-2-yne).

  CH$_3$-CH$_2$-C$\equiv$C-CH(CH$_3$)-H. No.

  3. 4-carbon chain with 2 methyl branches: C-C(CH$_3$)$_2$-C-C. Only one way to place a triple bond to be in the longest chain.

  HC$\equiv$C-C(CH$_3$)$_2$-CH$_3$. (3,3-Dimethylbut-1-yne). My structure (g).

  Total isomers: 3 + 3 + 1 = 7. The provided solution lists 7 isomers. My list matches theirs now.

  Position isomerism pairs: (Hex-1-yne, Hex-2-yne), (Hex-1-yne, Hex-3-yne), (Hex-2-yne, Hex-3-yne). Also (3-Methylpent-1-yne, 3-Methylpent-2-yne), (4-Methylpent-1-yne, 4-Methylpent-2-yne).

  Chain isomerism pairs: (Hex-1-yne, 3-Methylpent-1-yne), (Hex-1-yne, 4-Methylpent-1-yne), (Hex-1-yne, 3,3-Dimethylbut-1-yne), etc.

  The question asks about "What type of isomerism is exhibited by different pairs of isomers?". Examples of pairs and isomerism type:

  • (a) Hex-1-yne and (b) Hex-2-yne: **Position isomerism**.
  • (a) Hex-1-yne and (d) 3-Methylpent-1-yne: **Chain isomerism**.
  • (d) 3-Methylpent-1-yne and (e) 4-Methylpent-1-yne: **Position isomerism**.
  • (e) 4-Methylpent-1-yne and (f) 4-Methylpent-2-yne: **Position isomerism**.

  The pairs of isomers exhibit **chain isomerism** and **position isomerism**.

We need to convert ethanoic acid (CH$_3$COOH) into benzene (C$_6$H$_6$). Ethanoic acid has 2 carbon atoms. Benzene has 6 carbon atoms arranged in a ring. A key reaction to form benzene from a 2-carbon unit is the cyclic polymerization of ethyne (acetylene, C$_2$H$_2$). Three molecules of ethyne combine to form one molecule of benzene.

So, the strategy is to convert ethanoic acid into ethyne, and then cyclically polymerize the ethyne.

Step 1: Convert ethanoic acid to a compound that can be used to prepare ethyne.

Ethyne can be prepared from calcium carbide (CaC$_2$) by reaction with water. CaC$_2$ is made from CaO and coke. CaO is made from CaCO$_3$. Let's see if we can make CaC$_2$ or something similar from ethanoic acid.

Alternatively, ethyne can be prepared from vicinal dihalides (like 1,2-dichloroethane) by dehydrohalogenation. 1,2-dichloroethane can potentially be obtained from ethene. Ethene can potentially be obtained from ethanol. Ethanol can potentially be obtained from ethanoic acid by reduction.

Let's consider converting ethanoic acid (CH$_3$COOH) to its calcium salt, then decarboxylating? No, decarboxylation reduces carbon chain length. We need to increase it to 6 or use a 2-carbon unit multiple times.

Let's try converting ethanoic acid to ethanol, then ethene, then ethyne.

Step 1: Reduction of ethanoic acid to ethanol.

CH$_3$COOH $\xrightarrow{\text{LiAlH}_4}$ CH$_3$CH$_2$OH. (Lithium aluminium hydride is a strong reducing agent).

Step 2: Dehydration of ethanol to ethene.

CH$_3$CH$_2$OH $\xrightarrow{\text{Conc. H}_2\text{SO}_4, \Delta}$ CH$_2$=CH$_2$ + H$_2$O.

Step 3: Halogenation of ethene to vicinal dihalide (e.g., 1,2-dichloroethane).

CH$_2$=CH$_2$ + Cl$_2$ $\rightarrow$ CH$_2$Cl—CH$_2$Cl.

Step 4: Dehydrohalogenation of 1,2-dichloroethane to ethyne.

CH$_2$Cl—CH$_2$Cl $\xrightarrow{\text{Alc. KOH}}$ CH$_2$=CHCl + HCl (Vinyl chloride)

CH$_2$=CHCl $\xrightarrow{\text{NaNH}_2}$ HC$\equiv$CH + NaCl + NH$_3$ (Dehydrohalogenation of vinyl chloride using sodamide)

Step 5: Cyclic polymerisation of ethyne to benzene.

3 HC$\equiv$CH $\xrightarrow{\text{Red hot iron tube, 873K}}$ C$_6$H$_6$ (Benzene)

Alternative way from ethanoic acid to ethyne: Convert ethanoic acid to its calcium salt, then heat it with calcium hydroxide? No, that's not for ethyne. Convert ethanoic acid to acetate, then electrolyse using Kolbe's method? $2\text{CH}_3\text{COO}^- \xrightarrow{\text{Electrolysis}} \text{CH}_3\text{CH}_3$. Gives ethane. Not helpful.

Another way to make ethyne is from methane. Methane can be made from ethanoic acid? No. Methane is C1. Ethanoic acid is C2.

Let's reconsider making calcium carbide. Can we make calcium carbide from something related to ethanoic acid?

Ethanoic acid can form acetic anhydride. Or react with alcohols to form esters. Or react with bases to form salts.

Let's go back to the ethanol route. Ethanoic acid $\rightarrow$ Ethanol $\rightarrow$ Ethene $\rightarrow$ Ethyne $\rightarrow$ Benzene. This seems like a plausible sequence using known reactions.

The reaction sequence is:

1. CH$_3$COOH $\xrightarrow{\text{Reduction}}$ CH$_3$CH$_2$OH
2. CH$_3$CH$_2$OH $\xrightarrow{\text{Dehydration}}$ CH$_2$=CH$_2$
3. CH$_2$=CH$_2$ $\xrightarrow{\text{Halogenation}}$ CH$_2$X-CH$_2$X
4. CH$_2$X-CH$_2$X $\xrightarrow{\text{Double dehydrohalogenation}}$ HC$\equiv$CH
5. 3 HC$\equiv$CH $\xrightarrow{\text{Cyclic Polymerisation}}$ C$_6$H$_6$

This sequence converts a C2 unit into a C2 unit (ethyne), which is then trimerised to a C6 unit (benzene).