(a) HC≡CCH=CHCH₃

Expand the formula to show all bonds and atoms:

H—C≡C—CH=CH—CH₃

Count $\sigma$ bonds (all single bonds and the first bond in a multiple bond):

• C≡C triple bond has 1 $\sigma$ and 2 $\pi$.
• C=C double bond has 1 $\sigma$ and 1 $\pi$.
• All other C-C and C-H bonds are $\sigma$ bonds.

$\sigma$ bonds: (C-H) + (C-C in triple) + (C-C single) + (C-C in double) + (C-H) + (C-C single) + (C-H) + (C-H) + (C-H)

$\sigma$ bonds: 1 (C-H on C1) + 1 (C1-C2) + 1 (C2-C3) + 1 (C3-C4) + 1 (C4-H) + 1 (C4-C5) + 1 (C5-H) + 3 (C5-H) = 10 $\sigma$ bonds.

Alternatively, count atoms and use the formula: For a molecule with N atoms and C single bonds in a chain, number of $\sigma$ bonds = N-1 + (number of rings). Here no rings. Number of atoms = 1+1+1+1+1+3 = 8 (C₃H₅). No, 1+1+1+1+1+3 = 8, but it's C₆H₆. C: 1+1+1+1+1+1=6. H: 1+1+1+3=6. Total atoms = 6+6=12. Number of $\sigma$ bonds = 12-1 = 11 in a non-cyclic chain? No, that's for molecules with only single bonds. Let's just count the bonds carefully.

H—C $\overset{\sigma}{\equiv}$ C $\overset{\sigma}{-}$ CH $\overset{\sigma}{=}$ CH $\overset{\sigma}{-}$ CH₃

$\sigma$ bonds: C1-H, C1-C2, C2-C3, C3-C4, C3-H, C4-C5, C4-H, C5-H(3). Total $\sigma$: 1+1+1+1+1+1+1+3 = 10.

$\pi$ bonds: C≡C has 2 $\pi$. C=C has 1 $\pi$. Total $\pi$: 2+1 = 3.

Number of $\sigma$ bonds = 10, Number of $\pi$ bonds = 3.

(The provided solution is incorrect for part (a)). Re-counting: C1-H (1), C1-C2 ($\sigma$), C2-C3 ($\sigma$), C3-C4 ($\sigma$), C4-C5 ($\sigma$). Total C-C $\sigma$ = 4. C-H bonds: C1 (1), C3 (1), C4 (1), C5 (3). Total C-H $\sigma$ = 6. Total $\sigma = 4 + 6 = 10$. C≡C has 2 $\pi$. C=C has 1 $\pi$. Total $\pi = 2 + 1 = 3$. Still not matching the provided solution (σC–C: 4; σC–H : 6; πC=C :1; π C≡C:2). My interpretation of the provided solution is that it counts C-C sigma, C-H sigma, C=C pi, and C≡C pi separately. Let's break it down that way for part (a):

• C-C single $\sigma$ bonds: C2-C3, C4-C5. Total = 2. (Not 4 as in solution?) Oh, it might be counting all C-C sigma bonds in the structure. C1-C2 (in triple), C2-C3 (single), C3-C4 (in double), C4-C5 (single). Yes, 4 C-C sigma bonds.
• C-H sigma bonds: C1 (1), C3 (1), C4 (1), C5 (3). Total = 6. (Matches solution)
• C=C pi bonds: C3=C4. Total = 1. (Matches solution)
• C≡C pi bonds: C1≡C2. Total = 2. (Matches solution)

Total $\sigma = 4 + 6 = 10$. Total $\pi = 1 + 2 = 3$.

So, (a) HC≡CCH=CHCH₃ has 10 $\sigma$ bonds and 3 $\pi$ bonds.

(b) CH₂=C=CHCH₃

Expand the formula:

H₂C=C=CH—CH₃

$\sigma$ bonds:

• C-H sigma bonds: C1 (2), C3 (1), C4 (3). Total = 6. (Matches solution)
• C-C single $\sigma$ bonds: C2-C3, C3-C4. Total = 2. (Solution says 3 for C-C $\sigma$?) Oh, it's counting all C-C sigma bonds. C1=C2 ($\sigma$), C2=C3 ($\sigma$), C3-C4 ($\sigma$). Yes, 3 C-C sigma bonds.

Total $\sigma = 6 + 3 = 9$.

$\pi$ bonds:

• C=C pi bonds: C1=C2, C2=C3. Total = 2. (Matches solution)

Total $\sigma = 9$. Total $\pi = 2$.

The provided solution format breaks down $\sigma$ bonds into C-C and C-H types and lists $\pi$ bonds separately.

(a) HC≡CCH=CHCH₃: σC–C: 4; σC–H : 6; πC=C :1; π C≡C:2. (This is the breakdown, not the total count). Total $\sigma = 4+6=10$, Total $\pi = 1+2=3$.

(b) CH₂=C=CHCH₃: σC–C: 3; σC–H: 6; πC=C: 2. Total $\sigma = 3+6=9$, Total $\pi = 2$.

My bond counting aligns with the bond types provided in the solution, so I will format the answer in that style.

Answer based on provided solution format:

(a) HC≡CCH=CHCH₃:

• $\sigma$ C–C bonds: C1-C2 ($\sigma$ in triple), C2-C3 (single), C3-C4 ($\sigma$ in double), C4-C5 (single). Count = 4.
• $\sigma$ C–H bonds: C1-H, C3-H, C4-H, C5-H (3). Count = 1+1+1+3 = 6.
• $\pi$ C=C bond: C3=C4. Count = 1.
• $\pi$ C≡C bonds: C1≡C2. Count = 2.

So, (a) is σC–C: 4; σC–H : 6; πC=C :1; π C≡C:2.

(b) CH₂=C=CHCH₃:

• $\sigma$ C–C bonds: C1=C2 ($\sigma$ in double), C2=C3 ($\sigma$ in double), C3-C4 (single). Count = 3.
• $\sigma$ C–H bonds: C1-H (2), C3-H (1), C4-H (3). Count = 2+1+3 = 6.
• $\pi$ C=C bonds: C1=C2 ($\pi$), C2=C3 ($\pi$). Count = 2.
• $\pi$ C≡C bonds: None. Count = 0.

So, (b) is σC–C: 3; σC–H: 6; πC=C: 2.

Determine the hybridization based on the number of $\sigma$ bonds and lone pairs around each carbon atom (or simply the number of groups attached + lone pairs, excluding $\pi$ bonds for $\sigma$ framework count). Alternatively, count the number of regions of electron density (single bonds, multiple bonds, lone pairs) around the atom: 4 regions = sp³, 3 regions = sp², 2 regions = sp.

• (a) CH₃Cl: Carbon is bonded to 3 H atoms and 1 Cl atom. It forms 4 single ($\sigma$) bonds. Hybridization is $\text{sp}^3$.
• (b) (CH₃)₂CO: This is Propanone (Acetone). CH₃-C(=O)-CH₃.
  • The two CH₃ carbons are bonded to 3 H and 1 C. They form 4 single ($\sigma$) bonds. Hybridization is $\text{sp}^3$.
  • The central carbon is bonded to two CH₃ groups and one oxygen via a double bond. It forms 3 $\sigma$ bonds (C-C, C-C, C=O $\sigma$) and 1 $\pi$ bond (C=O $\pi$). It has 3 regions of electron density around it. Hybridization is $\text{sp}^2$.
  So, (b) carbons are $\text{sp}^3$, $\text{sp}^2$, $\text{sp}^3$. The solution lists them for each carbon, probably from left to right or by type: $\text{sp}^3$ (for both methyl carbons), $\text{sp}^2$ (for the carbonyl carbon). Order matters for specific positions. Assuming it lists the hybridization of the distinct types or sequentially: sp³, sp² (probably meaning the methyl carbons are sp³ and the carbonyl carbon is sp²). Let's check the provided answer's format interpretation. Provided answer (b) is sp³, sp². This likely means the methyl carbons are sp³ and the carbonyl carbon is sp².
• (c) CH₃CN: Acetonitrile. CH₃-C≡N.
  • The CH₃ carbon is bonded to 3 H and 1 C. It forms 4 single ($\sigma$) bonds. Hybridization is $\text{sp}^3$.
  • The carbon in the C≡N group is bonded to one CH₃ group and one N via a triple bond. It forms 2 $\sigma$ bonds (C-C single, C≡N $\sigma$) and 2 $\pi$ bonds (C≡N $\pi$). It has 2 regions of electron density around it. Hybridization is $\text{sp}$.
  So, (c) carbons are $\text{sp}^3$, $\text{sp}$. Provided answer: sp³, sp.
• (d) HCONH₂: Formamide. H-C(=O)-NH₂.
  • The carbon is bonded to one H, one O via double bond, and one N. It forms 3 $\sigma$ bonds (C-H, C=O $\sigma$, C-N single) and 1 $\pi$ bond (C=O $\pi$). It has 3 regions of electron density around it. Hybridization is $\text{sp}^2$.
  So, (d) carbon is $\text{sp}^2$. Provided answer: sp².
• (e) CH₃CH=CHCN: But-2-enenitrile. CH₃-CH=CH-C≡N.
  • C1 (CH₃): Bonded to 3 H and 1 C. 4 $\sigma$ bonds. $\text{sp}^3$.
  • C2 (CH=): Bonded to C1, C3, H. Forms 3 $\sigma$ bonds (C-C, C-C $\sigma$, C-H) and 1 $\pi$ bond (C=C $\pi$). 3 regions. $\text{sp}^2$.
  • C3 (=CH): Bonded to C2, C4, H. Forms 3 $\sigma$ bonds (C-C $\sigma$, C-C, C-H) and 1 $\pi$ bond (C=C $\pi$). 3 regions. $\text{sp}^2$.
  • C4 (CN): Bonded to C3 and N via triple bond. Forms 2 $\sigma$ bonds (C-C single, C≡N $\sigma$) and 2 $\pi$ bonds (C≡N $\pi$). 2 regions. $\text{sp}$.
  So, (e) carbons are $\text{sp}^3$, $\text{sp}^2$, $\text{sp}^2$, $\text{sp}$. Provided answer: sp³, sp², sp², sp.

The provided solution lists the hybridization of each carbon atom sequentially from left to right in the formula.

Answer:

(a) CH₃Cl: C is bonded to 4 atoms ($\text{sp}^3$). Hybridisation: sp³

(b) (CH₃)₂CO: CH₃-C(=O)-CH₃. C1(CH₃) - 4 bonds (sp³); C2(C=O) - 3 $\sigma$ regions (sp²); C3(CH₃) - 4 bonds (sp³). Hybridisation: sp³, sp², sp³

(c) CH₃CN: CH₃-C≡N. C1(CH₃) - 4 bonds (sp³); C2(C≡N) - 2 $\sigma$ regions (sp). Hybridisation: sp³, sp

(d) HCONH₂: H-C(=O)-NH₂. C1(H-C(=O)-N) - 3 $\sigma$ regions (sp²). Hybridisation: sp²

(e) CH₃CH=CHCN: CH₃-CH=CH-C≡N. C1(CH₃) - 4 bonds (sp³); C2(CH=) - 3 $\sigma$ regions (sp²); C3(=CH) - 3 $\sigma$ regions (sp²); C4(C≡N) - 2 $\sigma$ regions (sp). Hybridisation: sp³, sp², sp², sp

Determine hybridization based on $\sigma$ bonds and lone pairs (or regions of electron density) around carbon. Determine shape based on electron geometry (for ideal cases) or VSEPR model (considering lone pairs and different groups).

• (a) H₂C=O (Formaldehyde). Carbon is bonded to 2 H atoms and 1 O atom via a double bond. It forms 3 $\sigma$ bonds (C-H, C-H, C=O $\sigma$) and 1 $\pi$ bond (C=O $\pi$). It has 3 regions of electron density around it. Hybridization is $\text{sp}^2$. The electron geometry around the $\text{sp}^2$ carbon is trigonal planar. With no lone pairs on carbon, the molecular shape is also trigonal planar.
• (b) CH₃F (Fluoromethane). Carbon is bonded to 3 H atoms and 1 F atom via single bonds. It forms 4 $\sigma$ bonds. Hybridization is $\text{sp}^3$. The electron geometry around the $\text{sp}^3$ carbon is tetrahedral. With no lone pairs on carbon, the molecular shape is also tetrahedral.
• (c) HC≡N (Hydrogen Cyanide). Carbon is bonded to 1 H atom and 1 N atom via a triple bond. It forms 2 $\sigma$ bonds (C-H, C≡N $\sigma$) and 2 $\pi$ bonds (C≡N $\pi$). It has 2 regions of electron density around it. Hybridization is $\text{sp}$. The electron geometry around the $\text{sp}$ carbon is linear. With no lone pairs on carbon (ignoring lone pair on N for carbon's geometry), the molecular shape is linear.

Answer:

(a) H₂C=O: Carbon is sp² hybridised, shape is trigonal planar.

(b) CH₃F: Carbon is sp³ hybridised, shape is tetrahedral.

(c) HC≡N: Carbon is sp hybridised, shape is linear.

Expand the condensed formula by showing all atoms and bonds.

(a) CH₃CH₂COCH₂CH₃: This is a ketone, where CO represents a carbonyl group (>C=O). The formula is $\text{CH}_3-\text{CH}_2-\text{C}(=\text{O})-\text{CH}_2-\text{CH}_3$. Now, expand each CH₃ and CH₂ group:

$\begin{array}{ccccc} \text{H} & \text{H} & \text{O} & \text{H} & \text{H} \\ | & | & || & | & | \\ \text{H}—\text{C}—\text{C}—\text{C}—\text{C}—\text{C}—\text{H} \\ | & | & & | & | \\ \text{H} & \text{H} & & \text{H} & \text{H} \end{array}$

So, the complete structural formula for (a) is:

(b) CH₃CH=CH(CH₂)₃CH₃: This contains a double bond and a chain with three repeating CH₂ groups. The formula is $\text{CH}_3-\text{CH=CH}-(\text{CH}_2)_3-\text{CH}_3$. Expand the CH₃, CH=, and (CH₂)₃ groups:

$\text{CH}_3-\text{CH=CH}-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_3$. Now, expand each part showing all bonds:

$\begin{array}{ccccccc} \text{H} & \text{H} & \text{H} & \text{H} & \text{H} & \text{H} & \text{H} \\ | & | & | & | & | & | & | \\ \text{H}—\text{C}—\text{C}=\text{C}—\text{C}—\text{C}—\text{C}—\text{C}—\text{H} \\ | & | & | & | & | & | & | \\ \text{H} & & & \text{H} & \text{H} & \text{H} & \text{H} \end{array}$

So, the complete structural formula for (b) is:

(a) HOCH₂CH₂CH₂CH(CH₃)CH(CH₃)CH₃

Condensed formula: Group the atoms and indicate identical groups with subscripts. Start from the left: HO attached to $\text{CH}_2$, then $\text{CH}_2$, then $\text{CH}_2$, then $\text{CH}$ with $\text{CH}_3$ branch, then $\text{CH}$ with $\text{CH}_3$ branch, then $\text{CH}_3$.

$\text{HO}-(\text{CH}_2)_3-\text{CH}(\text{CH}_3)-\text{CH}(\text{CH}_3)_2$.

Alternatively, $\text{HOCH}_2\text{CH}_2\text{CH}_2\text{CH}(\text{CH}_3)\text{CH}(\text{CH}_3)_2$.

Bond-line formula: Draw the carbon chain as zig-zag lines. Show heteroatoms (O in HO) and hydrogens on heteroatoms. Methyl branches are shown as lines from the main chain. The main chain has 7 carbons (excluding the initial HO). The HO is at one end (implied C1). Branches are at C4 and C5 (counting from HO end). The last carbon is C7. The branching is $\text{CH}(\text{CH}_3)\text{CH}(\text{CH}_3)_2$. This is a CH with a methyl and then a CH with two methyls. Let's draw the 7-carbon chain first.

Start from HO-C1-C2-C3-C4-C5-C6-C7. Branches on C4 and C5.

Main chain C1-C7: HO-C-C-C-C-C-C-C. HO at C1 terminal. C4 has a methyl branch. C5 has two methyl branches.

Let's draw the zig-zag for 7 carbons: $\land\dots$. 7 carbons would have 6 segments and 7 vertices/ends. HO is at one end (vertex). C1. C2, C3, C4, C5, C6, C7. Let's number from the end with HO (implied C1).

Vertex 1 (HO-): C1. Vertex 2: C2. Vertex 3: C3. Vertex 4: C4. Vertex 5: C5. Vertex 6: C6. Vertex 7: C7 (terminal CH₃).

HO is at vertex 1. Branch CH₃ at vertex 4. Branch CH(CH₃)₂ at vertex 5. Branch CH(CH₃)₂ means CH with two methyls. This implies the main chain is not C1-C7. Let's find the longest carbon chain first. The chain is HO-C-C-C-C(CH₃)-C(CH₃)-CH₃. Longest chain is 6 carbons: C-C-C-C-C-C. The HO is attached to C1. CH₃ at C4, two CH₃ at C5. HO-C1-C2-C3-C4(CH₃)-C5(CH₃)-CH₃. Longest chain is 6 carbons.

Let's re-read the condensed formula: HOCH₂CH₂CH₂CH(CH₃)CH(CH₃)CH₃. Longest chain is from the carbon attached to HO. C1-C2-C3-C4-C5-C6. The groups CH(CH₃)CH(CH₃)CH₃ represent carbons 4, 5, 6, and branches. The chain is HO-C1-C2-C3. At C3 is CH₂. At C4 is CH with CH₃ branch. At C5 is CH with CH₃ branch. The chain continues CH₃. The longest chain is HO-C1-C2-C3-C4-C5-C6. No, the longest chain is HO-C1-C2-C3-C4(CH₃)-C5(CH₃)-C6(CH₃). Longest chain is 6 carbons. HO-C1-C2-C3-C4(CH₃)-C5(CH₃)2? No. Let's look at the provided structure (a). It has HO-(CH₂)₃-CH(CH₃)-CH(CH₃)₂. So the main chain is 3 CH₂ followed by CH, CH, CH₃. The total linear chain is 3+1+1+1=6 carbons. HO is at C1. CH₃ at C4. CH₃ at C5. The remaining CH₃ is C6. The branch at C5 is CH₃. The carbon attached to the branch at C4 is a CH. The carbon attached to the branch at C5 is a CH. It should be HO-C1-C2-C3-C4(CH₃)-C5(CH₃)-C6. No, the end is CH₃.

Let's write the full structure from the condensed formula: HO-CH₂-CH₂-CH₂-CH(CH₃)-CH(CH₃)-CH₃. Find the longest chain including the carbon attached to HO. C1(HO)-C2-C3-C4-C5-C6. Branches are at C4 (CH₃) and C5 (CH₃). It's a 6-carbon chain with HO at C1 and methyls at C4 and C5.

$\begin{array}{cccccc} & & & \text{CH}_3 \\ & & & | \\ \text{HO}—\text{CH}_2—\text{CH}_2—\text{CH}_2—\text{CH}—\text{CH}—\text{CH}_3 \\ & & & & | \\ & & & & \text{CH}_3 \end{array}$

Bond-line formula for a 6-carbon chain starting with HO at one end and methyls at C4 and C5 (counting from HO end):

Draw a 6-carbon zig-zag line starting with HO-:

HO-$\overset{1}{\curvearrowright}$C$\overset{2}{\curvearrowright}$C$\overset{3}{\curvearrowright}$C$\overset{4}{\curvearrowright}$C$\overset{5}{\curvearrowright}$C$\overset{6}{\curvearrowright}$C.

Vertex 1 has HO. Vertex 2 is C2. Vertex 3 is C3. Vertex 4 is C4. Vertex 5 is C5. Vertex 6 is C6 (terminal CH₃).

Branch CH₃ at C4 (vertex 4). Branch CH₃ at C5 (vertex 5).

Bond-line: HO-line-vertex-vertex-vertex-vertex-vertex. 6 carbons. Start with HO. Vertex 1 is C1. Vertex 6 is C6 (terminal CH₃). Vertex 4 gets a branch. Vertex 5 gets a branch.

Bond-line for 6 carbons is a zig-zag with 6 vertices (including ends). Example: $\overset{1}{\curvearrowright}\overset{2}{\curvearrowright}\overset{3}{\curvearrowright}\overset{4}{\curvearrowright}\overset{5}{\curvearrowright}\overset{6}{\curvearrowright}$. Ends are C1 and C6. Let's say HO is at C1.

HO attached to C1. C1 is a vertex. C2 is vertex 2. C3 is vertex 3. C4 is vertex 4. C5 is vertex 5. C6 is vertex 6.

Draw HO-. Then draw a zig-zag chain for 6 carbons. The first vertex is C1. The last vertex is C6. There are 4 intermediate vertices (C2, C3, C4, C5). Branches are on C4 and C5.

Correct bond-line should represent the skeleton. The longest chain is 6 carbons. HO- is attached to one end. Methyl branches are at positions 4 and 5 from that end. Draw a 6-carbon zig-zag. Put HO at one end vertex. Put lines (methyls) off the 4th and 5th vertices from that end.

Bond-line for (a):

(b) $\begin{array}{c} \text{CH}_3 \\ | \\ \text{H}—\text{C}—\text{OH} \\ | \\ \text{CN} \end{array}$

Condensed formula: Group atoms around each carbon. Methyl is CH₃. The carbon with OH and CN and CH₃ is CH(OH)(CN)(CH₃). No, it's CH with OH, CN, and CH₃ attached. So it's CH(CH₃)(OH)(CN). The H is attached to the carbon. The CH₃ is a branch. The OH is a branch. The CN is a branch. The central carbon is bonded to H, CH₃, OH, CN. It's a single carbon with 4 different groups attached.

$\text{CH}_3\text{CH(OH)CN}$. (This shows CH₃ attached to CH which has OH and CN). This matches the provided solution's interpretation.

Bond-line formula: The central carbon is bonded to CH₃, H, OH, CN. Only carbon and heteroatoms (O, N) are shown explicitly. H on C is implied. CH₃ is shown as a line terminal. OH is O-H. CN is C≡N. The carbon atom bonded to 4 groups is a vertex. One bond is to CH₃ (line terminal). One bond is to OH (explicit). One bond is to CN (explicit). One bond is to H (implied). It's a single carbon atom as the base, with branches. The base carbon can be shown as a vertex, but with multiple branches like this, it's sometimes shown as a single vertex with lines to terminals/heteroatoms.

Start with the central carbon as a point/vertex. Line to CH₃ terminal. Line to OH. Line to CN. Line to H (implied). Draw a point. Draw lines emanating from it. One line ends as a vertex (CH₃). One line goes to O with H. One line goes to C≡N.

Bond-line for (b):

Convert vertices and line ends to carbon atoms. Add hydrogen atoms to each carbon to satisfy carbon's tetravalency (4 bonds total). Remember: vertex with N bonds means it's bonded to 4-N hydrogens.

(a)

A 6-membered ring. Each vertex is a carbon. Each carbon in a cyclohexane ring forms 2 C-C bonds. So, each carbon is bonded to 2 hydrogens ($\text{CH}_2$).

Expanded structure:

(b)

A 5-membered ring with a double bond. Each vertex is a carbon. Carbons in the ring have varying numbers of bonds.

Let's number the carbons in the ring (arbitrarily, just to count bonds):

• C1: Bonded to C2 and C5 (2 C-C bonds). Needs 2 hydrogens ($\text{CH}_2$).
• C2: Bonded to C1 and C3 (2 C-C bonds) and a double bond to C3 (1 C=C bond $\sigma$, 1 C=C bond $\pi$). Total 3 bonds ($\sigma$ framework count). Needs 1 hydrogen ($\text{CH}$).
• C3: Bonded to C2 and C4 (2 C-C bonds) and a double bond to C2. Total 3 bonds ($\sigma$ framework). Needs 1 hydrogen ($\text{CH}$).
• C4: Bonded to C3 and C5 (2 C-C bonds). Needs 2 hydrogens ($\text{CH}_2$).
• C5: Bonded to C1 and C4 (2 C-C bonds). Needs 2 hydrogens ($\text{CH}_2$).

Expanded structure:

(c)

A chain with branches. Let's number the carbons starting from one end of the main chain (e.g., the left end of the longest chain):

• C1 (end): Bonded to C2 (1 C-C bond). Needs 3 hydrogens ($\text{CH}_3$).
• C2 (vertex): Bonded to C1, C3, and C branch (3 C-C bonds). Needs 1 hydrogen ($\text{CH}$).
• C branch (end): Bonded to C2 (1 C-C bond). Needs 3 hydrogens ($\text{CH}_3$).
• C3 (vertex): Bonded to C2 and C4 (2 C-C bonds). Needs 2 hydrogens ($\text{CH}_2$).
• C4 (vertex): Bonded to C3 and C5 (2 C-C bonds). Needs 2 hydrogens ($\text{CH}_2$).
• C5 (end): Bonded to C4 (1 C-C bond). Needs 3 hydrogens ($\text{CH}_3$).

Expanded structure:

(d)

A chain with a double bond and a branch. Numbering the main chain (containing the double bond) to give the double bond the lowest number:

• C1 (end): Bonded to C2 (1 C-C bond) and double bond to C2 (1 C=C). Total 2 bonds ($\sigma$ framework). Needs 2 hydrogens ($\text{CH}_2$).
• C2 (vertex): Bonded to C1, C3, and C branch (3 C-C bonds) and double bond to C1. Total 4 bonds ($\sigma$ framework count includes one bond from double bond). Needs 0 hydrogens ($\text{C}$). This carbon is bonded to C1 (double), C3 (single), C branch (single), and implicitly one hydrogen in the bond line? No, it's bonded to 4 carbons. Let's recount bonds from vertices carefully. Vertex = Carbon. Line = C-C single bond. Two lines = C=C double bond. Three lines = C≡C triple bond.
• C1 (end): One line to C2. This is $\text{CH}_3$.
• C2 (vertex): Bonded to C1, C3, and C4 branch. 3 lines. $\text{CH}$.
• C3 (vertex): Bonded to C2, C4, and C5 branch. 3 lines. $\text{CH}$.
• C4 (vertex): Bonded to C3 and C branch. This is a double bond =. So C4 is bonded to C3 (single), C branch (single), and C5 (double), C6 (double). Let's look at the structure again. The vertical line indicates a double bond.
• C1 (end): Attached to one line (to C2). $\text{CH}_3$.
• C2 (vertex): Attached to 3 lines (C1, C3, branch). $\text{CH}$.
• C3 (vertex): Attached to 3 lines (C2, C4, branch). $\text{CH}$.
• C4 (vertex): Attached to C3 and a vertical double line to C5. So C4 is bonded to C3 (single), C5 (double). This means C4 is bonded to 3 carbon atoms and a double bond. The vertical double bond means C4 is bonded to C5 and C6 via double bonds. C4 is bonded to C3 (single), C5 (double), C6 (double). 1 + 2 + 2 = 5 bonds? No, a carbon can only form 4 bonds. The structure is likely representing CH3-CH=C(CH3)-CH=CH2 or something similar. Let's break the bond-line structure differently.
• Vertex 1 (end): $\text{CH}_3$.
• Vertex 2: Bonded to vertex 1 and vertex 3. $\text{CH}_2$.
• Vertex 3: Bonded to vertex 2 and vertex 4. $\text{CH}_2$.
• Vertex 4: Bonded to vertex 3. Has a double bond to vertex 5. Has a branch to vertex 6. $\text{CH}$. (Bonded to C3 (single), C5 ($\sigma$ in double), C6 (single). Needs 1 H.) Wait, C4 is part of a double bond. So it should be CH= or C=. The vertical double bond means C4=C5. C4 is bonded to C3 (single), C5 (double), C6 (branch). $\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH=C}(\text{CH}_3)-\text{CH}=?$. No, the vertical double bond is between two carbons.

Let's count vertices as carbons and add hydrogens to satisfy valency 4. Horizontal line segment = C-C. Vertical line segment = C-C. Double line vertical = C=C. Branch = C. End of line = CH3 unless bonded to something explicit.

• C1 (end): 1 bond to C2. $\text{CH}_3$.
• C2 (vertex): 3 bonds (C1, C3, C4). $\text{CH}$.
• C3 (vertex): 3 bonds (C2, C4, branch). $\text{CH}$.
• C4 (vertex): 2 bonds (C3, C5). Plus a vertical double bond implies C4=C5. This interpretation makes C4 bonded to C3 (single), C5 (double), and implicitly something else? No.

Let's consider the double bond as a structural element connecting two vertices. The zig-zag forms a chain. The vertical double bond is a branch off a chain. The chain is C1-C2-C3-C4-C5. Length 5. The double bond is a branch off C4, connecting to a new carbon (C6) via a double bond.

• C1 (end): 1 bond to C2. $\text{CH}_3$.
• C2 (vertex): 2 bonds (C1, C3). $\text{CH}_2$.
• C3 (vertex): 2 bonds (C2, C4). $\text{CH}_2$.
• C4 (vertex): 2 bonds (C3, C5). Plus a branch. The branch is a vertical line ending in C6. This vertical line is a double bond. So C4 is bonded to C3 (single), C5 (single), and C6 (double). Wait, C4 has 1 single + 1 single + 1 double = 4 bonds. It is $\text{CH}$.
• C5 (end): 1 bond to C4. This implies $\text{CH}_3$.
• C6 (end): Bonded to C4 via a double bond. $\text{CH}_2$.

Let's redraw the skeleton based on this interpretation:

C1-C2-C3-C4-C5

$\quad\quad\quad\quad |=$

$\quad\quad\quad\quad$C6

This does not match the zig-zag pattern. The vertical double bond is connecting two carbons that are vertices in the main chain. The main chain is C1-C2-C3-C4-C5. The double bond is between C4 and C5.

• C1 (end): 1 bond to C2. $\text{CH}_3$.
• C2 (vertex): 2 bonds (C1, C3). $\text{CH}_2$.
• C3 (vertex): 2 bonds (C2, C4). Plus a branch to C branch. $\text{CH}$. (Bonded to C2, C4, and a branch).
• C branch (end): Bonded to C3. $\text{CH}_3$.
• C4 (vertex): 2 bonds (C3, C5). Plus a double bond connecting C4 to C5. This means C4 is bonded to C3 (single), C5 (double). $\text{CH}$.
• C5 (end): Bonded to C4 via double bond. $\text{CH}_2$.

Let's redraw based on this interpretation:

C1-C2-C3(branch CH₃)-C4=C5

$\text{CH}_3-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH=CH}_2$. This matches 3-methylpent-1-ene. Let's check the bond-line for this.

Yes, the provided bond-line formula matches 3-methylpent-1-ene.

So, let's write the expanded structure for 3-methylpent-1-ene:

$\text{CH}_3-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH=CH}_2$.

$\begin{array}{ccccccc} & & & \text{H} & \text{H} & \text{H} \\ & & & | & | & | \\ \text{H}—\text{C}—\text{C}—\text{C}—\text{C}═\text{C} \\ | & | & | & | & \\ \text{H} & \text{H} & \text{H} & \text{CH}_3 & \\ & & & | & \\ & & & \text{H} & \end{array}$. No, that's not right.

Let's expand $\text{CH}_3\text{CH}_2\text{CH}(\text{CH}_3)\text{CH=CH}_2$ properly.

$\begin{array}{cccccc} & & \text{H} & & \\ & & | & & \\ \text{H}—\text{C}—\text{C}—\text{C}—\text{C}═\text{C} \\ | & | & | & | & | \\ \text{H} & \text{H} & \text{CH}_3 & \text{H} & \text{H} \\ & & | & & \\ & & \text{H} & & \end{array}$. Still wrong. Let's expand group by group.

CH₃: $\begin{array}{c} \text{H} \\ | \\ \text{H}—\text{C}— \\ | \\ \text{H} \end{array}$

CH₂: $\begin{array}{c} \text{H} \\ | \\ —\text{C}— \\ | \\ \text{H} \end{array}$

CH with branch CH₃: $\begin{array}{c} \text{H} \\ | \\ —\text{C}— \\ | \\ \text{CH}_3 \end{array}$

CH= : $\begin{array}{c} \text{H} \\ | \\ —\text{C}= \end{array}$

=CH₂ : $=\text{CH}_2$

Putting it together: CH₃—CH₂—CH(CH₃)—CH=CH₂

$\begin{array}{ccccccc} \text{H} & \text{H} & & \text{H} & \\ | & | & & | & \\ \text{H}—\text{C}—\text{C}—\text{C}—\text{C}═\text{C} \\ | & | & | & | & | \\ \text{H} & \text{H} & \text{CH}_3 & \text{H} & \text{H} \\ & & | & & \\ & & \text{H} & & \end{array}$. This must be it. All valencies satisfied. Let's write it horizontally.

Expanded structure (d): $\text{CH}_3—\text{CH}_2—\begin{array}{c} \text{CH} \\ | \\ \text{CH}_3 \end{array}—\text{CH=CH}_2$

So, the expanded structural formula for (d) is:

Final Answer Format Check:

(a) $\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2$. Add H's. $\text{CH}_2$ in ring bonded to 2 C's, so 2 H's. Correct for cyclohexane.

(b) $\text{CH}_2\text{CH=CHCH}_2\text{CH}_2$. Add H's. C1-2H, C2-1H, C3-1H, C4-2H, C5-2H. Correct for cyclopentene.

(c) $\text{CH}_3\text{CH(CH}_3)\text{CH}_2\text{CH}_2\text{CH}_3$. Add H's. C1-3H, C2-1H (bonded to C1, C3, CH3), C3-2H, C4-2H, C5-3H, CH3-3H. Correct for 2-methylpentane.

(d) $\text{CH}_3\text{CH}_2\text{CH(CH}_3)\text{CH=CH}_2$. Add H's. C1-3H, C2-2H, C3-1H (bonded to C2, C4, CH3), C4-1H (bonded to C3, C5=), C5-2H (bonded to C4=), CH3-3H. Correct for 3-methylpent-1-ene.

The solution provided for part (d) matches my expanded structure, confirming the bond-line interpretation.

i)

This is an alcohol. The functional group is -OH. The longest carbon chain containing the -OH group has 8 carbons.

Let's number the chain to give the -OH group the lowest number:

The -OH group is on carbon 3. There is a methyl substituent (-CH₃) on carbon 6.

Parent alkane: Octane. Suffix: -ol (for alcohol). Position of -OH: 3.

Substituent: Methyl. Position: 6.

Name: 6-Methyloctan-3-ol.

IUPAC Name: 6-Methyloctan-3-ol.

ii)

This compound contains ketone (>C=O) functional groups. There are two of them, so the suffix will be -dione.

The longest chain containing both ketone groups has 6 carbons.

Number the chain to give the ketone carbons the lowest numbers. Starting from the left: 1, 2, 3, 4, 5, 6. Ketones at positions 2 and 4 (2,4). Starting from the right: 6, 5, 4, 3, 2, 1. Ketones at positions 3 and 5 (3,5). The lowest set of locants is 2,4.

Parent alkane: Hexane. Suffix: -dione. Positions: 2,4.

Name: Hexane-2,4-dione.

IUPAC Name: Hexane-2,4-dione.

iii)

This compound contains a carboxylic acid group (-COOH) and a ketone group (>C=O). Carboxylic acid has higher priority than ketone.

The parent chain contains the carboxylic acid group. Numbering starts from the carbon of the -COOH group (position 1).

The longest chain including C1 is 6 carbons.

Principal functional group: Carboxylic acid (-COOH), suffix -oic acid, position 1 (implied by parent chain name). Parent alkane: Hexane.

Subordinate functional group: Ketone (>C=O), prefix oxo-, position 5.

Name: 5-Oxohexanoic acid.

IUPAC Name: 5-Oxohexanoic acid.

iv)

This compound contains carbon-carbon double bonds (=) and a carbon-carbon triple bond (≡). The longest chain containing both multiple bonds has 6 carbons.

Number the chain to give the multiple bonds the lowest possible numbers. Double and triple bonds have equal priority; the lowest number is given to the multiple bond encountered first. Starting from the left, double bond is at 1. Starting from the right, triple bond is at 1. Left end gives double bond 1, triple bond 5 (1, 5). Right end gives triple bond 1, double bond 5 (1, 5). Let's use the lower number for the double bond.

Chain length: 6 carbons. Parent hydrocarbon: Hex-. Double bonds: diene, at positions 1, 3. Triple bond: -yne, at position 5. When both double and triple bonds are present, the name ends in -enyne. The double bond suffix comes before the triple bond suffix. The parent name is alkadienyne.

Parent name: Hexa-. Double bonds at 1, 3. Triple bond at 5.

Name: Hexa-1,3-dien-5-yne.

IUPAC Name: Hexa-1,3-dien-5-yne.

Construct the structure from the IUPAC name by identifying the parent chain and attaching functional groups/substituents at specified positions.

(i) 2-Chlorohexane:

• Parent name: hexane, indicates a 6-carbon chain (single bonds). C-C-C-C-C-C
• Substituent: Chloro-, at position 2.
• Place Cl on the second carbon. Then add hydrogens to satisfy carbon valency 4.

Structure: $\text{CH}_3-\text{CH(Cl)}-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_3$ or $\text{CH}_3\text{CHClCH}_2\text{CH}_2\text{CH}_2\text{CH}_3$.

(ii) Pent-4-en-2-ol:

• Parent name: pent-, indicates a 5-carbon chain. C-C-C-C-C
• Functional groups: -en- (double bond) at position 4, -ol (alcohol) at position 2.
• Number the chain 1 to 5 from one end. Place the double bond between C4 and C5. Place the -OH group on C2. Add hydrogens.

Structure: C1-C2(OH)-C3-C4=C5. $\text{CH}_3-\text{CH(OH)}-\text{CH}_2-\text{CH=CH}_2$.

(iii) 3-Nitrocyclohexene:

• Parent name: cyclohexene, indicates a 6-membered ring with one double bond.
• Number the ring such that the double bond carbons are 1 and 2 (lowest possible numbers for the multiple bond).
• Substituent: Nitro- (–$\text{NO}_2$) at position 3.
• Draw a 6-membered ring. Place a double bond. Number the carbons starting at one end of the double bond. Place the –$\text{NO}_2$ group on C3. Add hydrogens.

Structure:

(iv) Cyclohex-2-en-1-ol:

• Parent name: cyclohexene, 6-membered ring with one double bond.
• Functional group: -ol (alcohol) at position 1, -en- (double bond) starting at position 2.
• Number the ring to give the -OH the lowest number (1), then number the double bond carbons consecutively (2 and 3).
• Draw a 6-membered ring. Place the -OH group on C1. Place the double bond between C2 and C3. Add hydrogens.

Structure:

(v) 6-Hydroxyheptanal:

• Parent name: heptanal, indicates a 7-carbon chain with an aldehyde (-CHO) group. The carbon of the aldehyde group is C1.
• Substituent/Functional group: Hydroxy- (-OH) at position 6.
• Draw a 7-carbon chain with -CHO at C1. Place -OH on C6. Add hydrogens.

Structure: $\text{CH}_3-\text{CH(OH)}-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CHO}$.

(a) o-Ethylanisole:

• Base compound: Anisole (methoxybenzene, $\text{C}_6\text{H}_5\text{OCH}_3$). The $\text{OCH}_3$ group defines position 1.
• Substituent: Ethyl ($\text{CH}_2\text{CH}_3$) at the ortho position (position 2 relative to $\text{OCH}_3$).
• Draw a benzene ring. Attach an $\text{OCH}_3$ group. Attach a $\text{CH}_2\text{CH}_3$ group at the adjacent carbon.

Structure:

(b) p-Nitroaniline:

• Base compound: Aniline (aminobenzene, $\text{C}_6\text{H}_5\text{NH}_2$). The $\text{NH}_2$ group defines position 1.
• Substituent: Nitro (–$\text{NO}_2$) at the para position (position 4 relative to $\text{NH}_2$).
• Draw a benzene ring. Attach an $\text{NH}_2$ group. Attach a –$\text{NO}_2$ group at the carbon directly opposite.

Structure:

(c) 2,3 - Dibromo -1 - phenylpentane:

• Parent chain: pentane, a 5-carbon chain.
• Substituents: Dibromo- (two Br atoms) at positions 2 and 3, phenyl (–$\text{C}_6\text{H}_5$) at position 1.
• Draw a 5-carbon chain. Number it 1 to 5 from one end. Attach a phenyl group to C1. Attach Br atoms to C2 and C3. Add hydrogens.

Structure: $\text{C}_6\text{H}_5-\text{CH}_2-\text{CH(Br)}-\text{CH(Br)}-\text{CH}_2-\text{CH}_3$.

(d) 4-Ethyl-1-fluoro-2-nitrobenzene:

• Parent compound: benzene ring.
• Substituents: Ethyl (–$\text{CH}_2\text{CH}_3$), Fluoro- (–F), Nitro- (–$\text{NO}_2$).
• Number the ring. The lowest possible numbers should be used for the substituents. Alphabetical order: Ethyl, Fluoro, Nitro. F is at 1, $\text{NO}_2$ at 2, Ethyl at 4.
• Draw a benzene ring. Attach F at position 1. Attach –$\text{NO}_2$ at position 2. Attach –$\text{CH}_2\text{CH}_3$ at position 4.

Structure:

Heterolytic cleavage results in the electron pair going to one atom, forming charged ions. The curved arrow shows the movement of the electron pair. The tail of the arrow is at the bond, and the head points to the atom receiving the electron pair.

(a) CH₃–SCH₃: The bond is between carbon and sulfur. Sulfur is more electronegative than carbon. The electron pair goes to sulfur, forming a methyl carbocation and a methyl thiolate anion.

Reactive intermediates: CH₃⁺ (methyl carbocation) and CH₃S⁻ (methyl thiolate anion).

(b) CH₃–CN: The bond is between carbon and nitrogen. Nitrogen is more electronegative than carbon. The electron pair goes to nitrogen, forming a methyl carbocation and a cyanide anion.

Reactive intermediates: CH₃⁺ (methyl carbocation) and CN⁻ (cyanide anion).

(c) CH₃–Cu: The bond is between carbon and copper. Carbon is more electronegative than copper (Copper is a metal). The electron pair goes to carbon, forming a methyl carbanion and a copper(I) cation.

Reactive intermediates: CH₃⁻ (methyl carbanion) and Cu⁺ (copper(I) cation).

Categorize as nucleophile (electron-rich, electron pair donor) or electrophile (electron-deficient, electron pair acceptor).

• HS⁻: Has a negative charge and lone pairs on S. Electron-rich. Can donate an electron pair. Nucleophile.
• BF₃: Boron has only 6 valence electrons (incomplete octet). Electron-deficient. Can accept an electron pair. Electrophile.
• C₂H₅O⁻ (Ethoxide ion): Has a negative charge and lone pairs on O. Electron-rich. Can donate an electron pair. Nucleophile.
• (CH₃)₃N: (Trimethylamine): Neutral molecule with a lone pair on N. Electron-rich. Can donate an electron pair. Nucleophile.
• $^+{\text{CH}}_3$ (Methyl carbocation): Carbon has a positive charge and only 6 valence electrons (incomplete octet). Electron-deficient. Can accept an electron pair. Electrophile.
• H₂N:⁻ (Amide ion): Has a negative charge and lone pair on N. Electron-rich. Can donate an electron pair. Nucleophile.
• CH₃−$\overset{+}{\text{C}}$═O (Acylium ion): The carbon atom has a formal positive charge and is part of a double bond to oxygen. The carbon atom is electron-deficient. Can accept an electron pair. Electrophile.
• $^+{\text{NO}}_2$ (Nitronium ion): Nitrogen has a formal positive charge and is bonded to oxygen. The nitrogen atom is electron-deficient. Can accept an electron pair. Electrophile.

Nucleophiles: HS⁻, C₂H₅O⁻, (CH₃)₃N:, H₂N:⁻ (Justification: Possess negative charge or lone pairs, can donate electron pair).

Electrophiles: BF₃, $^+{\text{CH}}_3$, CH₃−$\overset{+}{\text{C}}$═O, $^+{\text{NO}}_2$ (Justification: Possess incomplete octet or formal positive charge, can accept electron pair).

An electrophilic center is an atom or site that is electron-deficient and susceptible to attack by a nucleophile. This often occurs when an atom is bonded to a more electronegative atom, creating a partial positive charge.

• CH₃CH=O (Acetaldehyde): The oxygen atom in the carbonyl group (C=O) is more electronegative than carbon, drawing electron density away from the carbon. This creates a partial positive charge ($\delta^+$) on the carbon atom. The carbonyl carbon is the electrophilic center.
• CH₃CN (Acetonitrile): Nitrogen is more electronegative than carbon, drawing electron density away from the carbon in the C≡N triple bond. This creates a partial positive charge ($\delta^+$) on the carbon atom triple bonded to nitrogen. This carbon triple bonded to nitrogen is the electrophilic center.
• CH₃I (Iodomethane): Iodine is more electronegative than carbon, drawing electron density away from the carbon in the C-I bond. This creates a partial positive charge ($\delta^+$) on the carbon atom. The carbon atom bonded to iodine is the electrophilic center.

Answer: The electrophilic centers are the carbon atoms marked with an asterisk (*): CH₃CH*═O (carbonyl carbon), CH₃C*≡N (carbon in CN group), CH₃*I (carbon bonded to I).

Bond polarity increases with the difference in electronegativity between the bonded atoms. We compare the electronegativity difference for the bonds in each pair.

Electronegativity values (approximate, Pauling scale): H (2.20), C (2.55), N (3.04), O (3.44), S (2.58), Br (2.96).

(a) H₃C-H vs H₃C-Br:

• C-H bond: $\Delta$EN = $|2.55 - 2.20| = 0.35$.
• C-Br bond: $\Delta$EN = $|2.55 - 2.96| = 0.41$.

Since $\Delta$EN is greater for the C-Br bond, the C–Br bond is more polar.

(b) H₃C-NH₂ vs H₃C-OH:

• C-N bond: $\Delta$EN = $|2.55 - 3.04| = 0.49$.
• C-O bond: $\Delta$EN = $|2.55 - 3.44| = 0.89$.

Since $\Delta$EN is greater for the C-O bond, the C–O bond is more polar.

(c) H₃C-OH vs H₃C-SH:

• C-O bond: $\Delta$EN = $|2.55 - 3.44| = 0.89$.
• C-S bond: $\Delta$EN = $|2.55 - 2.58| = 0.03$.

Since $\Delta$EN is greater for the C-O bond, the C–O bond is more polar.

The molecule is CH₃-CH₂-CH₂-Br. The polar C-Br bond (due to Br's electronegativity) exerts an electron-withdrawing (-I) inductive effect on the carbon chain.

Let's label the carbons starting from the one bonded to Br:

C³ — C² — C¹ — Br

The -I effect of Br is strongest on C¹ (directly attached). This polarity is transmitted through the $\sigma$ bonds to C², then to C³. The magnitude of the inductive effect decreases rapidly with distance (number of bonds) from the polar bond.

The C-C bonds are C²-C¹ and C³-C². The inductive effect originates at Br and influences the chain C¹ $\leftarrow$ C² $\leftarrow$ C³. Thus, the C¹-Br bond induces a partial positive charge on C¹, which induces a smaller partial positive charge on C², which induces an even smaller partial positive charge on C³.

The inductive effect is felt through the bonds. The C²-C¹ bond is closer to the origin of the effect (Br) than the C³-C² bond.

The inductive effect on the C²-C¹ bond is caused by the polarity of the C¹-Br bond. The inductive effect on the C³-C² bond is caused by the induced polarity of the C²-C¹ bond.

Therefore, the inductive effect (electron withdrawal) is strongest on the C²-C¹ bond and weaker on the C³-C² bond.

The C-C bond where the inductive effect originating from Br is least felt is the one furthest away through the chain. In this molecule, the carbons are C1 (attached to Br), C2, C3. The C-C bonds are C1-C2 and C2-C3.

The -I effect of Br polarizes C1. This polarity is transmitted to C2 through the C1-C2 bond. This polarity is further transmitted to C3 through the C2-C3 bond. The strength of the effect decreases with distance.

Therefore, the inductive effect is stronger on the C1-C2 bond than on the C2-C3 bond.

The C-C bond where the inductive effect is expected to be the least is the one furthest from the substituent, which is the C²–C³ bond in the chain starting from the carbon attached to Br (C¹).

The acetate ion ($\text{CH}_3\text{COO}^-$) has a carboxylate group with a negative charge delocalized over two oxygen atoms.

First, draw the Lewis structure. The carbon atom is bonded to three H atoms, one C atom, and two O atoms. The carboxylate group is C(=O)O⁻.

The negative charge is formally on one oxygen atom. This oxygen has three lone pairs. The carbon is double-bonded to the other oxygen, which has two lone pairs. The carbon is bonded to the methyl carbon.

Resonance involves the movement of $\pi$ electrons or lone pairs that are conjugated (alternating single and multiple bonds or lone pair adjacent to a $\pi$ system). Here, a lone pair on the negatively charged oxygen is adjacent to the C=O $\pi$ bond. The lone pair can move to form a $\pi$ bond with the carbon, and the $\pi$ bond in C=O can break, moving the electron pair onto the oxygen that was double-bonded.

Using curved arrows:

The resonance hybrid is an average of these two equivalent structures, where the negative charge is delocalized over both oxygen atoms, and the C-O bonds have partial double bond character.

The molecule is Propenal (Acrolein). It has a conjugated system: C=C-C=O. Resonance involves the delocalization of $\pi$ electrons across the conjugated system.

Start with the most stable structure (neutral structure with most bonds/octets). Then move $\pi$ electrons towards the more electronegative oxygen atom.

Structure I: Neutral, all carbons have 4 bonds (octet framework), Oxygen has 2 bonds and 2 lone pairs (octet). This is the main contributing structure.

Structure II: Formed by moving the C=O $\pi$ electron pair to oxygen. Oxygen gains a negative charge and a lone pair. Carbon gets a positive charge. A positive charge on carbon and charge separation make it less stable than I. However, all atoms (except H) have a complete octet in this structure (Carbon with + has 6 valence e⁻, but the definition is based on valence electrons including formal charge contributions and lone pairs). Let's count valence electrons around each atom excluding Hydrogens: C1 (4), C2 (4), C3 (4), O (6). Total = 18. In I: C1 bonded to 2H + C2. C2 bonded to C1, C3, H. C3 bonded to C2, O, H. O bonded to C3. Let's write full Lewis structure to be sure.

In structure II: C1(+): 2 bonds + 0 lone pairs = 4 valence electrons. C2: 3 bonds + 0 lone pairs = 6 valence electrons. C3: 3 bonds + 0 lone pairs = 6 valence electrons. O(-): 1 bond + 3 lone pairs = 7 valence electrons. Let's recount formal charges and octets carefully.

In Structure I: All atoms have formal charge 0. Octets are complete for C and O (counting shared pairs). C1 (2+2+4=8), C2 (4+2+2=8), C3 (2+4+2=8), O (4+4=8). Okay, Structure I is a valid Lewis structure with complete octets and no formal charges.

In Structure II (move C=O pi to O): C1(+): 1 bond + 0 lone pairs + formal charge +1 = 1 valence electron contributed + 3 shared electrons = 4 total valence e⁻. Not right. Formal charge calculation: Valence e⁻ - non-bonding e⁻ - ½ bonding e⁻. C1: 4 - 2 - ½(4) = 0. C2: 4 - 0 - ½(8) = 0. C3: 4 - 0 - ½(8) = 0. O: 6 - 4 - ½(4) = 0. Okay, formal charges are 0 in I.

Structure II: C1(+): formal charge +1. Carbon has 3 bonds, no lone pairs = 3 shared pairs = 6 valence electrons. Octet is incomplete (sextet). C2: neutral, 4 bonds. C3: neutral, 3 bonds. O(-): formal charge -1. Oxygen has 1 bond, 3 lone pairs = 1 shared pair + 6 non-bonding e⁻ = 7 valence e⁻. Octet complete.

Structure III: Formed by moving C=C $\pi$ electron pair to C1. C1 gets a negative charge and a lone pair. C2 gets a positive charge. Positive charge on carbon, charge separation, and positive charge on less electronegative atom make it less stable than I. C1(-): has 1 bond, 3 lone pairs = 1 shared pair + 6 non-bonding = 7 valence e⁻. Octet incomplete (sextet + lone pair, but formal charge makes it seem okay). Let's check formal charges. C1(-): 4 - 6 - ½(2) = -3 - 1 = -4. Something is wrong with the arrow movement/formal charge calculation in the provided image/solution. Let's recalculate using the moved electrons.

Starting from I: Move C=O $\pi$ to O. Move C=C $\pi$ to form C-C bond. This would not lead to II. Let's assume the arrows in the provided image are correct for generating II and III from I.

From I to II: Move C=O $\pi$ pair to O. Move C=C $\pi$ pair to form a C2-C3 $\pi$ bond. This makes C1 positive. C3 becomes negative. No, this gives C1(+)-C2=C3-O(-). Let's re-examine the first arrow in the image (I to II). It starts from the C=C double bond and ends *between* C2 and C3. This implies the double bond moves. The second arrow starts from the C=O double bond and ends *on* O. This implies the pi bond moves to form a lone pair on O. This is valid.

I to II: C=C $\pi$ moves to become C2-C3 $\pi$. C=O $\pi$ moves to become a lone pair on O. This makes C1 positive and O negative. Structure II is: CH₂⁺-CH=CH-O⁻. Check formal charges. C1(+): 4 - 2 - ½(2) = +1. C2: 4 - 0 - ½(8) = 0. C3: 4 - 0 - ½(8) = 0. O(-): 6 - 6 - ½(2) = -1. Charges: +1+0+0-1 = 0. Okay, formal charges are correct for CH₂⁺-CH=CH-O⁻.

I to III: C=C $\pi$ moves to become a lone pair on C1. C=O $\pi$ moves to become a lone pair on O. C2 gets a positive charge, C3 gets a positive charge. This cannot be right. Let's look at the arrows from I to III again. The first arrow starts from the C=C $\pi$ bond and ends *on* C1. This means C1 gets the $\pi$ pair as a lone pair and a negative charge. C2 gets a positive charge. The second arrow starts from the C=O $\pi$ bond and ends *on* O. This means O gets the $\pi$ pair as a lone pair and a negative charge. C3 gets a positive charge. Both C2 and C3 positive? No. Let's assume the arrows in the provided figure mean: move C=C $\pi$ to C1 (making C1 negative, C2 positive). Then move lone pair on O to form C=O $\pi$ bond (making C3 positive, O neutral). This would lead to C1(-)-CH=CH-C⁺=O. This structure has C1(-), C2(+), C3(+). Charge +1. Not valid unless the overall species is an ion. Let's assume the starting arrows in the provided solution figure are correct for generating structures II and III from I. The first arrow (I to II) moves C=C $\pi$ to form C2-C3 $\pi$, and C=O $\pi$ to O. This gives CH₂⁺-CH=CH-O⁻. The first arrow (I to III) moves C=C $\pi$ to C1, and C=O $\pi$ to O. This would give CH₂⁻-CH⁺-CH-C⁺-O⁻. This is too complicated.

Let's assume the intended structures II and III are correct as drawn in the solution figure, generated from I by valid electron movement arrows (even if the arrows in the figure are hard to follow). Let's check their relative stability based on the rules.

Structure I: Neutral. All atoms with octets (framework + pi + lone pairs). No formal charges.

Structure II: CH₂⁺-CH=CH-O⁻. Charge separation. Positive charge on C (less electronegative than O). Negative charge on O (more electronegative). Octet incomplete on C1 (+). Octet complete on O (-). All carbons have octets in their framework contribution + pi contribution. C1(+): forms 1 sigma bond + 0 lone pairs + formal charge +1 = 4 total valence electrons? No, it has 3 bonds (2 to H, 1 to C) = 3 shared pairs = 6 valence electrons. Incomplete octet. C2: 4 bonds = 8 valence e⁻. C3: 3 bonds = 6 valence e⁻. Oh, structure II in the figure seems to have a double bond between C2 and C3. Let's use the structures as drawn in the solution figure for I, II, III and evaluate them.

I: CH₂=CH-CH=O. Neutral. No formal charges. Complete octets. (C1, C2, C3, O all have 4 bonds or 2 bonds + 2 lone pairs). This is the most stable neutral structure.

II: CH₂⁺-CH=CH-O⁻. Charge separation. Positive charge on C, negative on O. Complete octets for C2, C3, O. Incomplete octet on C1 (+). Negative charge on more electronegative O, positive charge on less electronegative C is unfavorable compared to no charges. But positive on C1 is slightly better than positive on C2 or C3 (closer to delocalization). Incomplete octet is destabilizing.

III: CH₂⁻-CH=CH-O⁺. Charge separation. Negative charge on C, positive on O. Complete octets for C1, C2, C3, O. Positive charge on electronegative O and negative charge on electropositive C is very unfavorable compared to II. Also, carbon with a negative charge often has a complete octet (lone pair). Let's assume the formal charges are correct as implied by the arrows in generating III from I. The first arrow moves C=C pi to C1, making C1(-), C2(+). The second arrow moves O lone pair to form C=O pi, making C3(=O) and O(+). This would give CH₂⁻-CH⁺-CH=O⁺. This has two positive charges. This doesn't match structure III in the figure (CH₂⁻-CH=CH-O⁺). Let's assume structure III as drawn is CH₂⁻-CH=CH-O⁺. Formal charges: C1(-): 4-6-1 = -3? No. C1(-): bonded to 1C, 2H, has 1 lone pair. 1 bond + 1 lone pair = 1 shared pair + 2 non-bonding pairs = 3 regions = sp² hybridized with a lone pair in a p orbital? Octet complete. C2: 4 bonds = 8 valence e⁻. C3: 3 bonds = 6 valence e⁻. O(+): bonded to C3, has 3 lone pairs. 1 bond + 3 lone pairs = 1 shared pair + 6 non-bonding = 7 valence e⁻. Formal charge: 6 - 6 - 1 = -1? No. O is +1. 6 - 6 - ½(2) = +1. It has 3 lone pairs (6 non-bonding e⁻). Correct. C1 is -1. 4 - 2 - ½(2) = +1? No. C1(-): 4 valence e⁻, 2 H, 1 bond to C2. C1(-): 4 - 2 - ½(2) = +1? No. Formal charge: 4 (valence e⁻) - 2 (non-bonding, lone pair) - 1/2 * 2 (bonding, C1-C2 sigma) = 4-2-1 = +1? No. C1(-): 1 bond to C2, 2 H's, 1 lone pair. 4-2-1/2(2) = +1? No. C1(-): 4 valence e⁻. 2 H, 1 bond to C2 implies 3 bonds. CH₂ has 2 bonds. CH₂⁻ has 1 bond + 1 lone pair? Carbon with negative charge and 1 bond is CH₂⁻ bonded to something. CH₂⁻ forms 1 bond, has a lone pair. Formal charge: 4 - 2 - 1/2(2) = 1. Not -1. Let's assume structure III as drawn is correct CH₂⁻-CH=CH-O⁺. The negative charge is on C1, positive on O. Negative on electropositive C and positive on electronegative O is highly unfavorable.

Relative Stability: Structure I (neutral, no charge separation, complete octets) is the most stable and major contributor. Structures II and III involve charge separation and are less stable. Between II and III, II places negative charge on the more electronegative atom (O) and positive on the less electronegative atom (C), which is more favorable than the charge distribution in III. Also, the incomplete octet on C1(+) in II is a factor, but putting positive on O(+) in III is generally worse.

Relative stability based on standard rules: I > II > III. (I is neutral, no charge. II has charge separation, + on C, - on O. III has charge separation, - on C, + on O. Negative on O is better than negative on C. Positive on C is better than positive on O. But incomplete octet on C in II is unfavorable. Let's reconsider the rules. More covalent bonds is better. All are the same number of covalent bonds (single+double = 2 single + 1 double = 3, + 1 double = 4 total framework bonds). Octets: I is complete. II has C1 incomplete octet. III has O with incomplete octet. Placing + on electronegative atom is destabilizing. Placing - on electropositive atom is destabilizing. III is the least stable because it has + on O and - on C. II is more stable than III. I is most stable. So, I > II > III).

This matches the stability indicated in the provided solution figure's text (Stability: I > II > III).

The molecule is Methyl acetate ($\text{CH}_3\text{COOCH}_3$). Its structure is $\text{CH}_3-\text{C}(=\text{O})-\text{O}-\text{CH}_3$. The major contributing resonance structure is the neutral one with all atoms having complete octets and no formal charges.

Let's analyze structures I and II based on resonance rules and stability criteria.

Structure I:

(CH₃−$\overset{+}{\text{C}}$═O −O⁻−CH₃)

Formal charges: C1(CH₃) = 0. C2 = bonded to 3 atoms, double bond to O, single to O. C2(+): has 3 bonds, no lone pairs. Formal charge: 4 (valence) - 0 (non-bonding) - 1/2 * 6 (bonding) = +1. Carbon with + charge and 3 bonds has 6 valence electrons (incomplete octet). Oxygen (=O): 2 bonds, 2 lone pairs. Formal charge: 6 - 4 - 1/2 * 4 = 0. Oxygen (O⁻): 1 bond, 3 lone pairs, negative charge. Formal charge: 6 - 6 - 1/2 * 2 = -1. C4(CH₃) = 0.

Structure I has: Charge separation (+1 on C, -1 on O). Incomplete octet on the positively charged carbon. Positive charge on a carbon. Negative charge on an oxygen. This structure is less stable than a neutral structure with complete octets.

Structure II:

(CH₃−C(=O⁻)−$\overset{+}{\text{O}}$−CH₃)

Formal charges: C1(CH₃) = 0. C2 = bonded to 3 atoms, double bond to O⁻, single to O⁺. C2: 4 (valence) - 0 (non-bonding) - 1/2 * 6 (bonding) = +1. Wait, C2 should be neutral. Let's assume the formal charges are as shown in the provided image's generation of structure II from the neutral structure (which is not shown here). Assume the neutral structure is CH₃-C(=O)-O-CH₃. Resonance involves moving a lone pair from the oxygen adjacent to the carbonyl to form a pi bond, and moving the pi bond in the carbonyl to oxygen. Neutral: CH₃-C(=O)-O-CH₃. Move lone pair from O (single bond) to form C=O pi bond. Move C=O pi bond to O (double bond). This gives CH₃-C(O⁻)=O⁺-CH₃. Let's assume the provided structure II is CH₃-C(=O⁻)-O⁺-CH₃. Formal charges: C1(CH₃)=0. C2 (bonded to CH₃, C=O⁻, O⁺): C2: 4 - 0 - 1/2 * 6 = +1. Charge +1 on C2? No. Let's assume the diagram's arrows meant: move a lone pair from the oxygen with the single bond to form a double bond with the carbonyl carbon. Move the C=O pi bond to the oxygen of the carbonyl. This leads to CH₃-C(O⁻)=O⁺-CH₃. Okay, let's assume the structure II as drawn is CH₃-C(=O⁻)-O⁺-CH₃.

Structure II has: Charge separation (+1 on O, -1 on O). Complete octets for both oxygens and the carbon they are bonded to. Positive charge on an oxygen (electronegative atom) and negative charge on another oxygen (also electronegative atom). Having a positive charge on an electronegative atom is generally destabilizing, even if octets are complete.

Comparing I and II to a neutral structure (CH₃-C(=O)-O-CH₃): The neutral structure has no formal charges and complete octets on all atoms. This is the most stable contributing structure (major contributor).

Structures I and II cannot be major contributors because they involve charge separation and/or incomplete octets (Structure I has an incomplete octet on carbon). Structures with charge separation are always less stable than neutral structures (like the neutral ester structure) if the neutral structure satisfies the octet rule for all atoms. Therefore, I and II are only minor contributors (less stable resonance structures) to the real structure of methyl acetate.

Reason they are not major contributors: They involve charge separation and are therefore less stable than the neutral structure with complete octets. Specifically, Structure I also has an incomplete octet on the carbon atom, which is highly unfavorable. Structure II has a positive charge on an electronegative oxygen atom, which is also unfavorable.

The stability of carbocations is primarily explained by hyperconjugation and the inductive effect of alkyl groups.

• (CH₃)₃C⁺ (tert-butyl cation): The positively charged carbon is attached to three methyl groups. Each methyl group has three C-H bonds. These C-H $\sigma$ bonds are adjacent to the empty p orbital on the carbocation carbon and can undergo hyperconjugation. There are $3 \times 3 = 9$ such C-H bonds available for hyperconjugation. The $+I$ inductive effect of the three methyl groups also helps to stabilize the positive charge by donating electron density through $\sigma$ bonds.
• CH₃CH₂⁺ (ethyl cation): The positively charged carbon is attached to one methyl group. This methyl group has three C-H bonds adjacent to the empty p orbital, available for hyperconjugation. There are 3 such C-H bonds. The $+I$ inductive effect of one ethyl group also stabilizes the positive charge.
• C⁺H₃ (methyl cation): The positively charged carbon is only bonded to hydrogen atoms. Hydrogen atoms attached to the carbocation carbon itself are not involved in hyperconjugation. There are no adjacent alkyl groups with C-H bonds available for hyperconjugation. It also lacks the $+I$ inductive effect stabilization from alkyl groups. The positive charge is concentrated on the single carbon atom.

Comparing the hyperconjugation and inductive effects:

• (CH₃)₃C⁺ benefits from hyperconjugation from 9 C-H bonds and the +I effect of 3 alkyl groups.
• CH₃CH₂⁺ benefits from hyperconjugation from 3 C-H bonds and the +I effect of 1 alkyl group.
• C⁺H₃ has no hyperconjugation and no +I inductive effect from alkyl groups.

Therefore, the stability order is (CH₃)₃C⁺ > CH₃CH₂⁺ > C⁺H₃. (CH₃)₃C⁺ is the most stable due to maximum hyperconjugation and inductive effect, and C⁺H₃ is the least stable due to the absence of these stabilizing effects.