NCERT Solutions · Class 11 Chemistry Organic Chemistry – Some Basic Principles and Techniques

Hybridisation of a carbon atom is fixed by how many sigma bonds it makes, not by how many bonds are drawn in the structure. Count only the sigma (single-bond) connections around a carbon — every double bond contributes exactly one sigma and one extra pi bond, and a triple bond contributes one sigma and two extra pi bonds. Then read off the state:\[\text{4 sigma bonds, 0 pi} \;\to\; sp^3 \qquad \text{3 sigma bonds, 1 pi} \;\to\; sp^2 \qquad \text{2 sigma bonds, 2 pi} \;\to\; sp \]Here \(\displaystyle sp^3\), \(\displaystyle sp^2\), \(\displaystyle sp\) are the hybrid orbital sets formed by mixing one \(\displaystyle 2s\) orbital with three, two, or one \(\displaystyle 2p\) orbitals of carbon respectively. The mistake people make is counting bonds drawn instead of sigma bonds: a carbon with two double bonds to it (like the central carbon of an allene) still has only $\displaystyle 2$ sigma bonds, so it is \(\displaystyle sp\), even though it looks "more bonded" on paper.1. \(\displaystyle \mathrm{CH_2=C=O}\) (ketene)Label the terminal carbon \(\displaystyle \mathrm{C_{1}}\) (the \(\displaystyle \mathrm{CH_2}\)) and the central carbon C2. 2. \(\displaystyle \mathrm{CH_3-CH=CH_2}\) (propene)Label the carbons \(\displaystyle \mathrm{C_{1}}\) (\(\displaystyle \mathrm{CH_3}\)), \(\displaystyle \mathrm{C_{2}}\) (\(\displaystyle =\mathrm{CH}-\)), \(\displaystyle \mathrm{C_{3}}\) (\(\displaystyle =\mathrm{CH_2}\)). 3. \(\displaystyle \mathrm{(CH_3)_2CO}\) (propanone / acetone) 4. \(\displaystyle \mathrm{CH_2=CH-CN}\) (prop-$\displaystyle 2$-enenitrile / acrylonitrile)Label \(\displaystyle \mathrm{C_{1}}\) (\(\displaystyle \mathrm{CH_2}=\)), \(\displaystyle \mathrm{C_{2}}\) (\(\displaystyle =\mathrm{CH}-\)), \(\displaystyle \mathrm{C_{3}}\) (the nitrile carbon, \(\displaystyle -\mathrm{C \equiv N}\)). 5. \(\displaystyle \mathrm{C_6H_6}\) (benzene)Every ring carbon is joined by a sigma bond to one hydrogen and by sigma bonds to its two neighbouring ring carbons; one of those two ring C–C connections also carries a pi bond in any single Kekulé structure (in reality the six pi electrons are delocalised equally around the ring, but the sigma-bond count is unaffected either way). That gives each carbon $\displaystyle 3$ sigma bonds and $\displaystyle 1$ pi bond, so all six carbons of benzene are \(\displaystyle sp^2\).Answer: \(\displaystyle \mathrm{CH_2=C=O}\): \(\displaystyle =\mathrm{CH_2}\) carbon is \(\displaystyle sp^2\), central \(\displaystyle =\mathrm{C}=\) carbon is \(\displaystyle sp\). \(\displaystyle \mathrm{CH_3CH=CH_2}\): \(\displaystyle \mathrm{CH_3}\) is \(\displaystyle sp^3\); both \(\displaystyle =\mathrm{CH}-\) and \(\displaystyle =\mathrm{CH_2}\) are \(\displaystyle sp^2\). \(\displaystyle \mathrm{(CH_3)_2CO}\): both \(\displaystyle \mathrm{CH_3}\) carbons are \(\displaystyle sp^3\); the carbonyl carbon is \(\displaystyle sp^2\). \(\displaystyle \mathrm{CH_2=CHCN}\): \(\displaystyle \mathrm{CH_2}=\) and \(\displaystyle =\mathrm{CH}-\) are \(\displaystyle sp^2\); the nitrile carbon \(\displaystyle -\mathrm{C \equiv N}\) is \(\displaystyle sp\). \(\displaystyle \mathrm{C_6H_6}\): all six ring carbons are \(\displaystyle sp^2\).

A single bond between two atoms is always exactly one σ (sigma) bond, formed by head-on overlap of orbitals along the bond axis; a double bond is one σ bond plus one π (pi) bond, formed by sideways overlap of unhybridised p orbitals — never two π bonds or two σ bonds. So the whole problem is: draw the correct skeleton for each molecule, count every line as one σ bond, then add one π bond for every double bond drawn.\(\displaystyle C_6H_6\) — benzene. Six carbons form a ring, and in the Kekulé structure the ring bonds alternate single and double: \(\displaystyle C_1{=}C_2{-}C_3{=}C_4{-}C_5{=}C_6{-}C_1\), with one H on every carbon. \(\displaystyle C_6H_{12}\) — cyclohexane. Paired with benzene, this is the saturated six-membered ring: six \(\displaystyle CH_2\) units joined only by single bonds. \(\displaystyle CH_2Cl_2\) — dichloromethane. One tetrahedral (sp\(\displaystyle ^3\)) carbon at the centre, singly bonded to two hydrogens and two chlorines — no multiple bonds at all. \(\displaystyle CH_2{=}C{=}CH_2\) — allene (propa-$\displaystyle 1,2$-diene). Three carbons in a row, with the central carbon doubly bonded to both end carbons, and each end carbon carrying two hydrogens. \(\displaystyle CH_3NO_2\) — nitromethane. Written as \(\displaystyle CH_3{-}NO_2\): the methyl carbon is singly bonded to N; N in turn is doubly bonded to one oxygen and singly bonded to the other (the standard Lewis structure carries a formal \(\displaystyle +\) charge on N and \(\displaystyle -\) charge on the singly-bonded O). \(\displaystyle HCONHCH_3\) — N-methylformamide. Read the formula as \(\displaystyle H{-}C(=O){-}NH{-}CH_3\): a formyl carbon (bearing one H and a double-bonded O) is singly bonded to an \(\displaystyle NH\) nitrogen, which is singly bonded to a methyl carbon. Answer: \(\displaystyle C_6H_6\) (benzene): $\displaystyle 12$ σ, $\displaystyle 3$ π. \(\displaystyle C_6H_{12}\) (cyclohexane): $\displaystyle 18$ σ, $\displaystyle 0$ π. \(\displaystyle CH_2Cl_2\): $\displaystyle 4$ σ, $\displaystyle 0$ π. \(\displaystyle CH_2{=}C{=}CH_2\) (allene): $\displaystyle 6$ σ, $\displaystyle 2$ π. \(\displaystyle CH_3NO_2\) (nitromethane): $\displaystyle 6$ σ, $\displaystyle 1$ π. \(\displaystyle HCONHCH_3\) (N-methylformamide): $\displaystyle 8$ σ, $\displaystyle 1$ π.

A bond-line (skeletal) formula shows only the carbon skeleton as a zigzag of straight lines — every line-end and every kink stands for one carbon atom, hydrogens sitting on that carbon are left out and understood to fill up its valence of four, and only atoms other than C and H — here oxygen — are written in explicitly. So to "write" a bond-line formula you first have to fix, from the name, exactly how many carbons are in the chain and exactly which carbon carries the functional group or branch — that placement is the whole content of the drawing.1. Isopropyl alcoholThe IUPAC name is propan-$\displaystyle 2$-ol: "propan" fixes a $\displaystyle 3$-carbon chain, and "-$\displaystyle 2$-ol" puts the \(\displaystyle \mathrm{-OH}\) group on carbon $\displaystyle 2$, the middle carbon.Full structural formula, every bond shown: \[\mathrm{CH_3-CH(OH)-CH_3} \] Here \(\displaystyle \mathrm{CH_3}\) is a carbon bonded to three hydrogens (a methyl group), and \(\displaystyle \mathrm{CH(OH)}\) is a carbon bonded to one hydrogen, one \(\displaystyle \mathrm{-OH}\) group, and the two neighbouring carbons.A common slip here is to put the \(\displaystyle \mathrm{OH}\) on an end carbon — that would be propan-$\displaystyle 1$-ol (n-propyl alcohol), a different, primary alcohol. Isopropyl alcohol is specifically the secondary alcohol, with \(\displaystyle \mathrm{OH}\) on \(\displaystyle \mathrm{C_{2}}\), and since propane has only three carbons, \(\displaystyle \mathrm{C_{2}}\) is unavoidably the middle one no matter which end you start numbering from.As a bond-line drawing: two line segments meeting at a central vertex (the classic "V" for a $\displaystyle 3$-carbon chain) — the left line-end is \(\displaystyle \mathrm{C_{1}}\) (a \(\displaystyle \mathrm{CH_3}\), left unlabelled), the middle kink is \(\displaystyle \mathrm{C_{2}}\), and the right line-end is \(\displaystyle \mathrm{C_{3}}\) (also unlabelled \(\displaystyle \mathrm{CH_3}\)). Only the label \(\displaystyle \mathrm{OH}\) is written, coming off the middle vertex.2. $\displaystyle 2,3$-Dimethylbutanal"Butanal" is a $\displaystyle 4$-carbon chain with an aldehyde group, \(\displaystyle \mathrm{-CHO}\) (a carbon double-bonded to one O and single-bonded to one H). By the rule for naming aldehydes, the carbonyl carbon must sit at the very end of the chain and is always numbered \(\displaystyle \mathrm{C_{1}}\) — there is no choice here, unlike a ketone, because \(\displaystyle \mathrm{-CHO}\) can only exist at a chain terminus.So the butanal backbone is \[\underset{C1}{\mathrm{CHO}}-\underset{C2}{\mathrm{CH_2}}-\underset{C3}{\mathrm{CH_2}}-\underset{C4}{\mathrm{CH_3}} \] "$\displaystyle 2,3$-dimethyl" replaces one hydrogen on \(\displaystyle \mathrm{C_{2}}\) and one on \(\displaystyle \mathrm{C_{3}}\) with a methyl (\(\displaystyle \mathrm{CH_3}\)) branch each: \[\mathrm{OHC-CH(CH_3)-CH(CH_3)-CH_3} \] Counting atoms: $\displaystyle 4$ chain carbons + $\displaystyle 2$ branch carbons = $\displaystyle 6$ carbons, and the hydrogens work out to $\displaystyle 12$, so the molecular formula is \(\displaystyle \mathrm{C_6H_{12}O}\), consistent with the drawn structure.As a bond-line drawing: a $\displaystyle 4$-vertex zigzag for the main chain. The left end \(\displaystyle \mathrm{(C_{1})}\) is not a bare vertex — it carries an explicit short line up to a doubly-bonded \(\displaystyle \mathrm{O}\) (the aldehyde group), since the carbonyl oxygen is always shown even in skeletal formulas. From the second vertex \(\displaystyle \mathrm{(C_{2})}\) draw one extra line ending in nothing, representing the first methyl branch; from the third vertex \(\displaystyle \mathrm{(C_{3})}\) draw another such line for the second methyl branch. The right end \(\displaystyle \mathrm{(C_{4})}\) is a plain, unlabelled vertex (\(\displaystyle \mathrm{CH_3}\)).3. Heptan-$\displaystyle 4$-one"Heptane" fixes a straight $\displaystyle 7$-carbon chain, and "-$\displaystyle 4$-one" places the ketone carbonyl (\(\displaystyle \mathrm{C=O}\)) at carbon 4. With $\displaystyle 7$ carbons numbered $\displaystyle 1$ to $\displaystyle 7$, \(\displaystyle \mathrm{C_{4}}\) is exactly the middle carbon, so this chain is symmetric on both sides of the \(\displaystyle \mathrm{C=O}\): three carbons on the left, three on the right.Full structural formula: \[\mathrm{CH_3-CH_2-CH_2-\underset{C4}{C}(=O)-CH_2-CH_2-CH_3} \] often abbreviated with "CO" standing for that carbonyl carbon: \[\mathrm{CH_3CH_2CH_2-CO-CH_2CH_2CH_3} \] This carbon \(\displaystyle \mathrm{(C_{4})}\) has no hydrogen on it at all — its four bonds are used up by two single bonds to the neighbouring \(\displaystyle \mathrm{CH_2}\) carbons and one double bond to O. Because the two halves of the chain are identical (a propyl group on each side of the carbonyl), heptan-$\displaystyle 4$-one is also known as dipropyl ketone.A point worth flagging: this symmetry is exactly why numbering from either end of the chain still gives the carbonyl carbon the locant $\displaystyle 4$ — there's no lower-locant choice to make, unlike an unsymmetrical ketone (say hexan-$\displaystyle 2$-one), where you must number from whichever end puts the \(\displaystyle \mathrm{C=O}\) at the lower number.As a bond-line drawing: a $\displaystyle 7$-vertex zigzag. The 4th vertex (dead centre) has a straight line drawn off it (up or down) ending in the label \(\displaystyle \mathrm{O}\), with a double bond marked to that oxygen. All six remaining vertices — three on each side — are left as plain, unlabelled kinks/ends (\(\displaystyle \mathrm{CH_2}\) or, at the two chain ends, \(\displaystyle \mathrm{CH_3}\)).Answer: Isopropyl alcohol is \(\displaystyle \mathrm{CH_3-CH(OH)-CH_3}\) (OH on the middle carbon of a $\displaystyle 3$-carbon chain); $\displaystyle 2,3$-dimethylbutanal is \(\displaystyle \mathrm{OHC-CH(CH_3)-CH(CH_3)-CH_3}\) (CHO fixed at \(\displaystyle \mathrm{C_{1}}\), methyl branches at \(\displaystyle \mathrm{C_{2}}\) and \(\displaystyle \mathrm{C_{3}}\), \(\displaystyle \mathrm{C_6H_{12}O}\)); heptan-$\displaystyle 4$-one is \(\displaystyle \mathrm{CH_3CH_2CH_2-CO-CH_2CH_2CH_3}\), i.e. dipropyl ketone (C=O at the middle carbon, \(\displaystyle \mathrm{C_{4}}\), of a straight $\displaystyle 7$-carbon chain, \(\displaystyle \mathrm{C_7H_{14}O}\)).

The idea: find the longest chain that carries the principal group, number it from whichever end gives that group (or the substituents) the lower locants, then name the branches alphabetically.Two rules do all the work here. First, the principal characteristic group (–CN, –CHO, –OH) must sit in the parent chain and gets the lowest possible number — it outranks the branches. Second, when there is no such group (parts (a), (c), (d)), you number from the end that gives the set of substituent locants the lower value at the first point of difference.(a) \(\displaystyle \mathrm{C_6H_5CH_2CH_2CH_3}\)The ring is joined to the chain at a carbon that is at the end of the $\displaystyle 3$-carbon chain (counting from the free end, the ring sits on carbon $\displaystyle 3$ — the last one). So the chain hangs off the ring by a terminal carbon: the group is a straight propyl, \(\displaystyle \mathrm{-CH_2CH_2CH_3}\), not propan-$\displaystyle 2$-yl.Take the benzene ring as the parent, since the ring here outranks the short chain. Only one substituent is present, so no locant is needed (a monosubstituted benzene needs no "$\displaystyle 1$-").Name: propylbenzene. (Watch the trap: had the ring been attached at the middle carbon it would be isopropylbenzene, a different compound.)(b) \(\displaystyle \mathrm{CH_3CH_2CH(CH_3)CH_2CN}\)The –CN is the principal group, and in a nitrile the carbon of the CN counts as part of the parent chain — even though it is drawn only as a label, not as a vertex. So count it in:\[\mathrm{\overset{5}{C}H_3-\overset{4}{C}H_2-\overset{3}{C}H(CH_3)-\overset{2}{C}H_2-\overset{1}{C}\!\equiv\!N}\]That is $\displaystyle 5$ chain carbons \(\displaystyle \Rightarrow\) parent = pentanenitrile.Which end? No choice at all — the nitrile carbon must be C-$\displaystyle 1$ by definition. Numbering from the CN end puts the methyl branch on carbon 3.Name: $\displaystyle 3$-methylpentanenitrile.(c) \(\displaystyle \mathrm{CH_3CH(CH_3)CH_2CH_2CH(CH_3)CH_2CH_3}\)Longest chain = $\displaystyle 7$ carbons \(\displaystyle \Rightarrow\) heptane. Two methyl branches, and no principal group, so the numbering is decided purely by the branches. Compare at the first point of difference: \(\displaystyle 2 < 3\), so the left-hand end wins.Name: $\displaystyle 2,5$-dimethylheptane.(d) \(\displaystyle \mathrm{CH_3CH_2C(Cl)(Br)CH_2CH_2CH_2CH_3}\)Longest chain = $\displaystyle 7$ carbons \(\displaystyle \Rightarrow\) heptane. Both halogens sit on the same carbon, so there is a single locant to minimise. \(\displaystyle 3 < 5\), so number from the left. Now list the substituents alphabetically — bromo before chloro — remembering that alphabetical order decides the order of citation in the name, never the numbering (the numbering was already fixed above).Name: $\displaystyle 3$-bromo-$\displaystyle 3$-chloroheptane.(e) \(\displaystyle \mathrm{ClCH_2CH_2CHO}\)The –CHO is the principal group. In an aldehyde the carbonyl carbon is part of the chain and must be C-$\displaystyle 1$ — there is nothing to choose:\[\mathrm{\overset{3}{C}H_2Cl-\overset{2}{C}H_2-\overset{1}{C}HO}\]Three chain carbons \(\displaystyle \Rightarrow\) parent = propanal (the "-al" suffix already tells you C-$\displaystyle 1$ carries the CHO, so you never write "$\displaystyle 1$-al"). Numbering from the CHO end puts Cl on carbon 3.Name: $\displaystyle 3$-chloropropanal.(f) \(\displaystyle \mathrm{Cl_2CHCH_2OH}\)Here –OH is the principal group, so the carbon bearing it takes the lowest locant. The chain is $\displaystyle 2$ carbons \(\displaystyle \Rightarrow\) ethanol.\[\mathrm{\overset{2}{C}HCl_2-\overset{1}{C}H_2OH}\]Number from the \(\displaystyle \mathrm{CH_2OH}\) end so that OH is on carbon $\displaystyle 1$ (numbering from the other end would give $\displaystyle 2$-ol, a higher locant — not allowed). Both chlorines then sit on carbon $\displaystyle 2$, and because there are two of them on the same carbon the locant is repeated: $\displaystyle 2,2$-, with the multiplying prefix di-.Name: $\displaystyle 2,2$-dichloroethan-$\displaystyle 1$-ol (also written $\displaystyle 2,2$-dichloroethanol).Answer: (a) propylbenzene; (b) $\displaystyle 3$-methylpentanenitrile; (c) $\displaystyle 2,5$-dimethylheptane; (d) $\displaystyle 3$-bromo-$\displaystyle 3$-chloroheptane; (e) $\displaystyle 3$-chloropropanal; (f) $\displaystyle 2,2$-dichloroethan-$\displaystyle 1$-ol.

A locant is not decoration — IUPAC numbering is decided by a fixed order of priority: ($\displaystyle 1$) give the lowest number to the principal characteristic group if there is one, ($\displaystyle 2$) then to multiple bonds as a set, ($\displaystyle 3$) then to all the substituent prefixes as a set (compared at the first point where the two possible numberings differ), and only if that is still tied, ($\displaystyle 4$) give the lower number to whichever substituent is cited first alphabetically in the name. Each part below is the same skeleton named two ways — from the two ends of the chain, or with one locant left out — and the rule above is what picks the winner.(a) $\displaystyle 2,2$-Dimethylpentane or $\displaystyle 2$-DimethylpentaneThe skeleton is a five-carbon chain (pentane, \(\displaystyle \text{C}_1\text{-C}_2\text{-C}_3\text{-C}_4\text{-C}_5 \)) carrying two methyl (\(\displaystyle \text{CH}_3 \)) branches on the same carbon. Numbering from the end nearer that carbon puts it at position $\displaystyle 2$ (numbering from the far end would put it at position $\displaystyle 4$, since \(\displaystyle 6-2=4\)); $\displaystyle 2$ is lower, so that carbon is C2.The error in "$\displaystyle 2$-Dimethylpentane" is not the numbering — it is that only one locant was written for two substituents. The prefix "di" tells you how many methyl groups there are, but IUPAC still requires one locant per substituent cited, even when both substituents are identical and sit on the same carbon. Two methyls need two numbers: $\displaystyle 2$ and $\displaystyle 2$, joined as "$\displaystyle 2,2$-". Writing "$\displaystyle 2$-dimethyl" reads as if only one locant exists, which is not a legal IUPAC citation.So the compound \(\displaystyle \text{CH}_3\text{-C}(\text{CH}_3)_2\text{-CH}_2\text{-CH}_2\text{-CH}_3 \) (a pentane chain with both methyls on C2) is correctly named $\displaystyle 2,2$-Dimethylpentane.(b) $\displaystyle 2,4,7$-Trimethyloctane or $\displaystyle 2,5,7$-TrimethyloctaneOctane is an eight-carbon chain, so numbering it from the opposite end replaces every locant \(\displaystyle n\) by \(\displaystyle 9-n\). Checking the first set: \(\displaystyle 2\to 9-2=7\), \(\displaystyle 4\to 9-4=5\), \(\displaystyle 7\to 9-7=2\). That turns \(\displaystyle \{2,4,7\}\) into \(\displaystyle \{7,5,2\}=\{2,5,7\}\) — exactly the second option. So both names describe one and the same molecule, just numbered from its two ends, and the choice comes down to rule ($\displaystyle 3$): compare the two locant sets term by term at the first point where they differ.\[\{2,4,7\} \quad \text{vs} \quad \{2,5,7\} \]The first terms are equal (\(\displaystyle 2=2\)); at the second term, \(\displaystyle 4<5\). The set \(\displaystyle \{2,4,7\}\) is lower at that first point of difference, so it wins outright — the alphabetical tie-break in rule ($\displaystyle 4$) is never needed here, because there is no tie.$\displaystyle 2,4,7$-Trimethyloctane is the correct name.(c) $\displaystyle 2$-Chloro-$\displaystyle 4$-methylpentane or $\displaystyle 4$-Chloro-$\displaystyle 2$-methylpentanePentane has five carbons, so numbering it from the other end sends locant \(\displaystyle n\) to \(\displaystyle 6-n\): here \(\displaystyle 2\to4\) and \(\displaystyle 4\to2\). Both proposed names use the same pair of numbers, \(\displaystyle \{2,4\}\) — only which substituent (chloro or methyl) sits on which number changes. This is the case people get wrong: because the two locant sets are identical, rule ($\displaystyle 3$) (compare the sets) is a tie and gives no answer by itself — you cannot decide by "the set with the smaller first number," because both sets ARE \(\displaystyle \{2,4\}\). You have to drop down to rule ($\displaystyle 4$).Rule ($\displaystyle 4$) says: when locant sets tie, the substituent that comes first in alphabetical order gets the lower number. Comparing the names alphabetically, "chloro" (c) precedes "methyl" (m), so chloro must receive the lower locant, $\displaystyle 2$, and methyl takes 4.$\displaystyle 2$-Chloro-$\displaystyle 4$-methylpentane is the correct name.(d) But-$\displaystyle 3$-yn-$\displaystyle 1$-ol or But-$\displaystyle 4$-ol-$\displaystyle 1$-yneHere the chain has both a principal characteristic group — the alcohol, expressed as the suffix "-ol" — and a triple bond, expressed as "-yne/-yn-". Rule ($\displaystyle 1$) outranks rule ($\displaystyle 2$): the carbon bearing –OH gets the lowest possible locant before the triple bond's position is even considered. Numbering the four-carbon chain from the –OH end gives \(\displaystyle \text{HO-CH}_2\text{-CH}_2\text{-C}{\equiv}\text{CH} \) as C1(–OH)–C2–C3≡\(\displaystyle \mathrm{C_{4}}\), so the hydroxyl sits at position $\displaystyle 1$ and the triple bond starts at position $\displaystyle 3$: but-$\displaystyle 3$-yn-$\displaystyle 1$-ol. Numbering from the other end instead would put the triple bond at position $\displaystyle 1$ but push the hydroxyl out to position $\displaystyle 4$ — which breaks rule ($\displaystyle 1$), since the principal characteristic group no longer has the lowest locant available to it."But-$\displaystyle 4$-ol-$\displaystyle 1$-yne" fails on two counts, not one: it gets the numbering priority backwards (letting the triple bond claim position $\displaystyle 1$ at the expense of –OH), and it also puts the suffixes in the wrong order — in IUPAC format the locant for unsaturation is inserted right before "-yne/-yn-" and the locant for the principal characteristic group right before its own suffix at the very end of the name (…-yn-ol), never with "-ol" sitting in the middle ahead of "-yne".But-$\displaystyle 3$-yn-$\displaystyle 1$-ol is the correct name.Answer: (a) $\displaystyle 2,2$-Dimethylpentane (b) $\displaystyle 2,4,7$-Trimethyloctane (c) $\displaystyle 2$-Chloro-$\displaystyle 4$-methylpentane (d) But-$\displaystyle 3$-yn-$\displaystyle 1$-ol

A structural formula's whole job is to show which atom is bonded to which — a condensed formula does that by writing each carbon's attached groups in a row, and a bond-line (skeletal) formula does the same thing by drawing the carbon skeleton as a zig-zag of lines and writing in only the atoms that are not carbon or hydrogen. Since only text renders here, each bond-line formula below is described stroke by stroke — every vertex and branch it asks you to draw — right alongside the condensed formula, which carries exactly the same information in words.(a) $\displaystyle 2,2,4$-TrimethylpentaneName the parent chain first: "pentane" is a $\displaystyle 5$-carbon chain, numbered \(\displaystyle \mathrm{C_{1}}\) to C5. "$\displaystyle 2,2,4$-trimethyl" places three \(\displaystyle \mathrm{CH_3}\) (methyl) branches on it — two of them on \(\displaystyle \mathrm{C_{2}}\), one on C4.A trap worth naming: the "tri" in trimethyl adds three extra carbons on top of the five in "pentane," so the molecule has \(\displaystyle 5+3=8\) carbons total, not five. Missing this is the most common way to under-count this molecule's formula.Laying the chain out carbon by carbon: Condensed structural formula: \[\mathrm{CH_3-C(CH_3)_2-CH_2-CH(CH_3)-CH_3} \]Bond-line formula, described: draw a zig-zag of five vertices for C1–C5 (four line segments). From the second vertex, draw two short branch lines ending in free vertices — the two methyl groups on C2. From the fourth vertex, draw one short branch line ending in a free vertex — the methyl on C4. No letters are written anywhere; every vertex is carbon, and hydrogens are never drawn in a skeletal formula — you read off how many belong on each vertex from carbon's valence being 4.Counting the H's on the condensed structure above ($\displaystyle 3$+$\displaystyle 0$+$\displaystyle 3$+$\displaystyle 3$+$\displaystyle 2$+$\displaystyle 1$+$\displaystyle 3$+$\displaystyle 3$) gives $\displaystyle 18$ hydrogens on $\displaystyle 8$ carbons, i.e. \(\displaystyle \mathrm{C_8H_{18}}\). Checking with the index of hydrogen deficiency (which counts total rings + \(\displaystyle \pi\) bonds), \[\text{IHD} = \frac{2C+2-H}{2} = \frac{2(8)+2-18}{2} = 0, \] confirms there is no ring and no \(\displaystyle \pi\) bond anywhere in the molecule.Functional group: none. This is a plain, saturated alkane — every bond is a single C–C or C–H sigma bond, so no functional group is present. (It is also familiar by its common name, isooctane — the compound that defines $\displaystyle 100$ on the octane-rating scale — but "alkane" is its only functional classification.)(b) $\displaystyle 2$-Hydroxy-$\displaystyle 1,2,3$-propanetricarboxylic acidParent chain: "propane," a $\displaystyle 3$-carbon chain, C1–C2–C3. The suffix "$\displaystyle 1,2,3$-tricarboxylic acid" puts a \(\displaystyle \mathrm{-COOH}\) (carboxyl) group at each of the three propane carbons; "$\displaystyle 2$-hydroxy" puts an \(\displaystyle \mathrm{-OH}\) (hydroxyl) group additionally on C2.The trap here is the mirror image of part (a)'s: when "carboxylic acid" is used as a suffix on more positions than a chain has ends — three \(\displaystyle \mathrm{-COOH}\) groups can't all sit at the termini of a $\displaystyle 3$-carbon chain — each \(\displaystyle \mathrm{-COOH}\) carbon is cited as an extra carbon hung off its numbered propane carbon; it is not itself one of \(\displaystyle \mathrm{C_{1}}\), \(\displaystyle \mathrm{C_{2}}\), C3. So the carbon count is \(\displaystyle 3\ (\text{propane}) + 3\ (\text{three COOH carbons}) = 6\), not 3. (Contrast this with an ordinary acid like propanoic acid, \(\displaystyle \mathrm{CH_3CH_2COOH}\), where the COOH carbon is \(\displaystyle \mathrm{C_{1}}\) of the chain — the distinction only shows up once there are more \(\displaystyle \mathrm{-COOH}\) groups than a chain has ends to hold them.)Carbon by carbon: Condensed structural formula: \[\mathrm{HOOC-CH_2-C(OH)(COOH)-CH_2-COOH} \]Bond-line formula, described: a $\displaystyle 3$-vertex zig-zag for C1–C2–C3. From \(\displaystyle \mathrm{C_{1}}\), a line down to a carboxyl vertex, which itself carries a double line up to O and a single line to an \(\displaystyle \mathrm{OH}\) label. From \(\displaystyle \mathrm{C_{3}}\), the same carboxyl group. From the middle vertex, \(\displaystyle \mathrm{C_{2}}\), two branches: one plain line to an \(\displaystyle \mathrm{OH}\) label, and one line down to a third carboxyl group drawn exactly like the other two.This is citric acid — the acid in citrus fruit. Its molecular formula, read off the condensed structure, is \(\displaystyle \mathrm{C_6H_8O_7}\): $\displaystyle 8$ H made up of $\displaystyle 2$ on \(\displaystyle \mathrm{C_{1}}\), $\displaystyle 2$ on \(\displaystyle \mathrm{C_{3}}\), and $\displaystyle 1$ on each of the four \(\displaystyle \mathrm{-OH}\) oxygens (the C2–OH plus the three carboxyl –OH's). Checking, \[\text{IHD} = \frac{2(6)+2-8}{2} = 3, \] matches exactly the three C=O double bonds in the three carboxyl groups, with no ring — the account balances.Functional groups: carboxylic acid (\(\displaystyle \mathrm{-COOH}\)), present three times, and hydroxyl/alcohol (\(\displaystyle \mathrm{-OH}\)), present once. That one alcohol is a tertiary alcohol — the carbon carrying the \(\displaystyle \mathrm{-OH}\), \(\displaystyle \mathrm{C_{2}}\), is itself attached to three other carbon atoms (C1, \(\displaystyle \mathrm{C_{3}}\), and the third \(\displaystyle \mathrm{-COOH}\) carbon), which is what "tertiary" means for an alcohol. It is easy to misclassify this as secondary by counting only the two chain neighbours and forgetting the branch.(c) HexanedialParent chain: "hexane," a $\displaystyle 6$-carbon chain. The suffix "-dial" means two \(\displaystyle \mathrm{-CHO}\) (aldehyde) groups, and an aldehyde suffix — unlike the carboxylic-acid case above — is always carried by a carbon that IS one of the numbered chain carbons: an aldehyde carbon has only one bonding position left after its \(\displaystyle \mathrm{C=O}\) and its one H, so it can only ever sit at a chain terminus, never as a branch off the middle. With no locants given, "-dial" defaults to both chain ends, \(\displaystyle \mathrm{C_{1}}\) and C6.So all six carbons of "hexane" are still all six carbons of hexanedial — nothing extra is added on top, unlike part (b).Carbon by carbon: Condensed structural formula: \[\mathrm{OHC-CH_2-CH_2-CH_2-CH_2-CHO} \]Bond-line formula, described: a zig-zag of six vertices, \(\displaystyle \mathrm{C_{1}}\) through \(\displaystyle \mathrm{C_{6}}\) (five line segments). At each of the two end vertices, draw a line up to O with a double bond, and write the H explicitly next to that end vertex — the H on an aldehyde carbon is always written in, even in a skeletal formula, because leaving it off would make the carbonyl carbon look like it has only three bonds. The four middle vertices carry no labels at all.Molecular formula: \(\displaystyle \mathrm{C_6H_{10}O_2}\) — $\displaystyle 10$ H made up of one on each aldehyde carbon and two on each of the four \(\displaystyle \mathrm{CH_2}\) carbons. Checking, \[\text{IHD} = \frac{2(6)+2-10}{2} = 2, \] matches the two C=O bonds, one per aldehyde, with no ring.Functional group: aldehyde (\(\displaystyle \mathrm{-CHO}\)), present twice — hexanedial is a dialdehyde.Answer: (a) $\displaystyle 2,2,4$-Trimethylpentane, \(\displaystyle \mathrm{CH_3-C(CH_3)_2-CH_2-CH(CH_3)-CH_3}\), \(\displaystyle \mathrm{C_8H_{18}}\) — no functional group (a saturated alkane). (b) $\displaystyle 2$-Hydroxy-$\displaystyle 1,2,3$-propanetricarboxylic acid (citric acid), \(\displaystyle \mathrm{HOOC-CH_2-C(OH)(COOH)-CH_2-COOH}\), \(\displaystyle \mathrm{C_6H_8O_7}\) — three carboxylic acid groups and one tertiary hydroxyl group. (c) Hexanedial, \(\displaystyle \mathrm{OHC-CH_2-CH_2-CH_2-CH_2-CHO}\), \(\displaystyle \mathrm{C_6H_{10}O_2}\) — aldehyde group, present twice. Bond-line skeletons for all three are described vertex-by-vertex above.

The idea: a functional group is a specific atom or bond arrangement, so read the structure atom by atom and name every one you meet — including the benzene ring itself.A useful habit: walk around the ring first (naming what is attached at each substituted carbon), then walk out along any side chain from the ring to its far end. Nothing gets missed that way.(a) \(\displaystyle \mathrm{C_8H_8O_3}\) — the trisubstituted ring, \(\displaystyle \mathrm{CHO}\) at \(\displaystyle \mathrm{C_{1}}\), \(\displaystyle \mathrm{OMe}\) at \(\displaystyle \mathrm{C_{3}}\), \(\displaystyle \mathrm{OH}\) at \(\displaystyle \mathrm{C_{4}}\)Walk the ring clockwise from the top: Note the \(\displaystyle \mathrm{OMe}\) and \(\displaystyle \mathrm{OH}\) are on adjacent carbons (C3, \(\displaystyle \mathrm{C_{4}}\) — ortho to each other), and \(\displaystyle \mathrm{OMe}\) is meta to the \(\displaystyle \mathrm{CHO}\). That arrangement makes this $\displaystyle 4$-hydroxy-$\displaystyle 3$-methoxybenzaldehyde — vanillin.(b) \(\displaystyle \mathrm{C_{13}H_{20}N_2O_2}\) — para-disubstituted ring, \(\displaystyle \mathrm{NH_2}\) at \(\displaystyle \mathrm{C_{1}}\), ester side chain at \(\displaystyle \mathrm{C_{4}}\)Start at the top of the ring and work outward along the chain: So this molecule has two different nitrogen groups, and missing one of them is the usual slip here: a primary aromatic amine on the ring and a tertiary amine at the chain end. (This is procaine.)(c) \(\displaystyle \mathrm{C_8H_7NO_2}\) — \(\displaystyle \mathrm{C_6H_5-CH=CH-NO_2}\)Walk outward from the ring along the two-carbon chain: Answer: (a) aldehyde \(\displaystyle (\mathrm{-CHO}\) carbonyl at C1), ether (methoxy \(\displaystyle \mathrm{-O-}\) at C3), phenolic \(\displaystyle \mathrm{-OH}\) at \(\displaystyle \mathrm{C_{4}}\), and an aromatic (benzene) ring; (b) primary aromatic amine \(\displaystyle (\mathrm{-NH_2}\) N on ring C1), ester \(\displaystyle (\mathrm{-C(=O)-O-}\) linkage at ring C4), tertiary amine (the chain-end \(\displaystyle \mathrm{N}\) of \(\displaystyle \mathrm{-N(C_2H_5)_2}\)), and an aromatic (benzene) ring; (c) nitro group \(\displaystyle (\mathrm{-NO_2}\) N on the terminal chain carbon), alkene \(\displaystyle \mathrm{C=C}\) in the side chain, and an aromatic (benzene) ring.

A negative charge becomes more stable when it is spread out (dispersed) over several atoms, and less stable when it stays concentrated on one atom.Look at what is attached to the charged oxygen in each ion. The two substituents pull electrons in opposite directions along the chain of bonds. Now apply the "spread the charge" idea from the opening line.In \(\displaystyle \text{O}_2\text{NCH}_2\text{CH}_2\text{O}^- \), the \(\displaystyle -I \) pull of \(\displaystyle -\text{NO}_2 \) drags some of the electron density sitting on the charged oxygen back along the chain, toward itself. The negative charge that started out fully localized on one oxygen atom is now spread over the oxygen and the nitro group. Dispersing a charge over more atoms lowers the ion's energy, so this ion is stabilized.In \(\displaystyle \text{CH}_3\text{CH}_2\text{O}^- \), there is no electron-withdrawing group to do this job. If anything, the methyl group's weak \(\displaystyle +I \) effect pushes a little extra electron density onto the already-negative oxygen, concentrating (intensifying) the charge on that single atom instead of spreading it. Concentrating a charge raises the ion's energy, so this ion is comparatively less stable.The step people get wrong: it is tempting to recall "electron-donating groups are stabilizing" from carbocation chemistry and apply it here too — but that rule is for a positive charge, where donated electrons fill an electron deficiency. For a negative charge the direction reverses: an electron-withdrawing group helps (it removes some of the excess electron density), and an electron-donating group hurts (it adds to an already-excessive electron density).So the ion carrying the electron-withdrawing nitro group, which disperses its negative charge over more atoms, is the more stable one.Answer: \(\displaystyle \text{O}_2\text{NCH}_2\text{CH}_2\text{O}^- \) (the $\displaystyle 2$-nitroethoxide ion) is more stable than \(\displaystyle \text{CH}_3\text{CH}_2\text{O}^- \) (the ethoxide ion), because the electron-withdrawing nitro group ( \(\displaystyle -\text{NO}_2 \), a \(\displaystyle -I \) group) pulls electron density through the sigma-bond framework and disperses the negative charge over more atoms, whereas the ethyl group's weak \(\displaystyle +I \) effect in the second ion instead concentrates the charge on a single oxygen atom.

Alkyl groups feed electron density into a neighbouring π system in two ways: a small, short‑range inductive push through the σ‑bond framework, and a much larger σ–π donation called hyperconjugation, in which a C–H σ bond on the carbon touching the π system tilts its electron pair into that system's p orbitals.1. The inductive (+I) contribution. Compare ethene, \(\displaystyle \text{H}_2\text{C}=\text{CH}_2 \), with propene, \(\displaystyle \text{CH}_3-\text{CH}=\text{CH}_2 \), where one vinylic H has been replaced by a methyl group. Along a chain of σ bonds, a carbon atom holds its bonding electrons slightly more tightly than a hydrogen atom does in the same position, so the alkyl group relays a small amount of extra electron density onto the sp² carbon it is attached to. This inductive effect is real but weak, and it fades quickly as you move further away — it cannot, by itself, explain the size of the effect alkyl groups actually have.2. Hyperconjugation — the dominant contribution. This requires a C–H σ bond sitting on a carbon that is directly attached to the π system (this carbon's hydrogens are called α‑hydrogens). Take the ethyl cation, \(\displaystyle \text{CH}_3-\overset{+}{\text{C}}\text{H}_2 \): the positively charged carbon has an empty p orbital, and the methyl carbon next to it carries three C–H bonds. When one of those C–H bonds happens to be oriented parallel to the empty p orbital, its bonding electron pair can spread sideways into that orbital — exactly the geometry by which two p orbitals overlap to form an ordinary π bond. Because the electrons are now shared between the σ bond and the empty orbital, the same cation can be drawn as a second contributing (resonance) structure in which that particular C–H bond is shown broken ("no bond"), a new π bond has formed between the two carbons, and the positive charge has shifted onto the hydrogen itself: \[\text{CH}_3-\overset{+}{\text{C}}\text{H}_2 \;\longleftrightarrow\; \text{H}_2\text{C}=\text{CH}_2\;\cdots\;\text{H}^{+} \] Nothing has actually moved or been released here — both structures describe the same ion, same nuclei, same overall +$\displaystyle 1$ charge; only the drawn position of one electron pair (and hence of the formal charge) differs. That is why this effect is also called no‑bond resonance: the "H⁺" on the right is not a free proton, it is the same hydrogen, now pictured without its bond, carrying the charge that used to sit on carbon. Because the electron pair has partly moved from the alkyl group's C–H bond into the π system, the alkyl group has effectively donated electron density to it.Aside — the step that gets skipped. It is tempting to stop at "alkyl groups are +I groups" and leave it there, but induction alone cannot explain why more alkyl substitution helps so much. The tert‑butyl cation, \(\displaystyle (\text{CH}_3)_3\text{C}^{+} \), has nine α C–H bonds available to hyperconjugate (three on each of three methyl groups) and is dramatically more stable than the ethyl cation above, which has only three. Hyperconjugation counts α‑hydrogens directly — each one gives another no‑bond resonance structure, so more α C–H bonds mean more delocalization of charge into the π system. This is the reason chemists rank alkyl groups' donating power (methyl < ethyl < isopropyl < tert‑butyl, as attached groups) by the number of α‑hydrogens they carry, not by inductive reasoning alone, and it is also why more heavily alkyl‑substituted alkenes are the more stable ones (Zaitsev's rule) — the same σ–π overlap raises the electron density, and hence the energy, of the double bond in a neutral alkene too, not only in a carbocation.3. Why not resonance/mesomeric donation? A group like \(\displaystyle -\text{OH} \) or \(\displaystyle -\text{NH}_2 \) donates into a π system by resonance because it has its own lone pair or π bond to conjugate. An alkyl group has neither — only C–H and C–C σ bonds — so it cannot show a true mesomeric (+M) effect. Its electron donation to an attached π system is entirely inductive-plus-hyperconjugative, as described above.Answer: Alkyl groups act as electron donors toward an adjacent π system chiefly through hyperconjugation — a C–H σ bond on the carbon attached to the π system (an α C–H bond) aligns with the system's p orbitals and delocalizes its electron pair into them, which can be drawn as a "no‑bond" resonance structure, e.g. \(\displaystyle \text{CH}_3-\overset{+}{\text{C}}\text{H}_2 \leftrightarrow \text{H}_2\text{C}=\text{CH}_2\cdots\text{H}^{+} \) for the ethyl cation — reinforced by a smaller inductive (+I) push through the σ‑bond framework; since more α C–H bonds mean more such delocalization, alkyl groups with more α‑hydrogens (e.g. tert‑butyl over ethyl) are stronger electron donors.