301. Study the name-structure pairs below.
| Structure | Proposed name |
|---|
| P. \( \mathrm{C_6H_5NO_2} \) | \( \mathrm{nitrobenzene} \) |
| Q. \( \mathrm{C_6H_5NH_2} \) | \( \mathrm{aniline} \) |
| R. \( \mathrm{C_6H_5OH} \) | \( \mathrm{phenol} \) |
| S. \( \mathrm{C_6H_5CH_3} \) | \( \mathrm{chlorobenzene} \) |
Which row contains the unsuitable proposed name?
ⓐ. P
ⓑ. S
ⓒ. Q
ⓓ. R
Correct Answer: S
Explanation: \( \mathrm{C_6H_5CH_3} \) is benzene with a methyl group attached, so its common name is toluene and its systematic description is methylbenzene. Chlorobenzene would require \( \mathrm{Cl} \) attached to the benzene ring, giving \( \mathrm{C_6H_5Cl} \). Row P correctly identifies \( \mathrm{C_6H_5NO_2} \) as nitrobenzene. Row Q correctly identifies \( \mathrm{C_6H_5NH_2} \) as aniline, and Row R correctly identifies \( \mathrm{C_6H_5OH} \) as phenol. The attached group on the benzene ring must be read directly instead of assuming all monosubstituted benzene compounds have the same name pattern.
302. A benzene ring has two substituents, \( \mathrm{NO_2} \) and \( \mathrm{Cl} \), on adjacent ring carbons. The locant pattern and alphabetical prefix order give the name:
ⓐ. \( \mathrm{1\text{-}chloro\text{-}2\text{-}nitrobenzene} \)
ⓑ. \( \mathrm{2\text{-}chloro\text{-}1\text{-}nitrobenzene} \)
ⓒ. \( \mathrm{1\text{-}nitro\text{-}2\text{-}chlorobenzene} \)
ⓓ. \( \mathrm{chloronitrobenzene} \) with no locants
Correct Answer: \( \mathrm{1\text{-}chloro\text{-}2\text{-}nitrobenzene} \)
Explanation: Adjacent substituents on benzene receive the locant set \(1,2\). When two different substituents can be assigned carbon \(1\) with the same locant set, alphabetical order helps decide the numbering. The prefix \( \mathrm{chloro} \) begins with \( \mathrm{c} \), while \( \mathrm{nitro} \) begins with \( \mathrm{n} \), so \( \mathrm{chloro} \) is written first and receives locant \(1\). The name \( \mathrm{1\text{-}chloro\text{-}2\text{-}nitrobenzene} \) gives both positions and prefix order. Omitting locants would be unclear because several chloronitrobenzene arrangements are possible.
303. In a compound containing both \( \mathrm{-COOH} \) and \( \mathrm{-NO_2} \), the naming approach at this level should treat:
ⓐ. \( \mathrm{-COOH} \) as the principal group and \( \mathrm{-NO_2} \) as a nitro prefix
ⓑ. \( \mathrm{-NO_2} \) as the principal group and \( \mathrm{-COOH} \) as a carboxy prefix only
ⓒ. both groups as identical because both contain oxygen
ⓓ. the compound as an ether because two oxygen-containing groups are present
Correct Answer: \( \mathrm{-COOH} \) as the principal group and \( \mathrm{-NO_2} \) as a nitro prefix
Explanation: The group \( \mathrm{-COOH} \) is a high-priority principal functional group in basic IUPAC naming. It is named with the suffix \( \mathrm{-oic\ acid} \) when it is the main group. The nitro group \( \mathrm{-NO_2} \) is generally written as a prefix, \( \mathrm{nitro} \), when another principal group controls the name. The two groups are not identical just because both contain oxygen. The naming decision depends on functional-group priority and actual connectivity.
304. A structure is described as \( \mathrm{O_2NCH_2CH_2COOH} \), where the carboxyl carbon is counted as carbon \(1\). The best name is:
ⓐ. \( \mathrm{1\text{-}nitropropanoic\ acid} \)
ⓑ. \( \mathrm{3\text{-}nitropropanoic\ acid} \)
ⓒ. \( \mathrm{nitroethanoic\ acid} \)
ⓓ. \( \mathrm{3\text{-}aminopropanoic\ acid} \)
Correct Answer: \( \mathrm{3\text{-}nitropropanoic\ acid} \)
Explanation: The structure contains the carboxyl group \( \mathrm{-COOH} \), so the compound is named as a carboxylic acid. The carboxyl carbon is carbon \(1\), and the chain contains three carbon atoms in total, giving the parent name \( \mathrm{propanoic\ acid} \). The \( \mathrm{-NO_2} \) group is attached to carbon \(3\), so it is written as \( \mathrm{3\text{-}nitro} \). It is not an amino compound because \( \mathrm{-NO_2} \) is nitro, not \( \mathrm{-NH_2} \). Counting from the carboxyl carbon is necessary because the principal group fixes the numbering direction.
305. A compound named \( \mathrm{3\text{-}nitropropanoic\ acid} \) is analysed by formula. What is its molecular formula and molar mass? Use \( \mathrm{C}=12 \), \( \mathrm{H}=1 \), \( \mathrm{N}=14 \), and \( \mathrm{O}=16 \).
ⓐ. \( \mathrm{C_3H_7NO_4} \), \(121\,\text{g mol}^{-1}\)
ⓑ. \( \mathrm{C_3H_5NO_2} \), \(87\,\text{g mol}^{-1}\)
ⓒ. \( \mathrm{C_3H_5NO_4} \), \(119\,\text{g mol}^{-1}\)
ⓓ. \( \mathrm{C_4H_7NO_4} \), \(133\,\text{g mol}^{-1}\)
Correct Answer: \( \mathrm{C_3H_5NO_4} \), \(119\,\text{g mol}^{-1}\)
Explanation: \( \textbf{Parent acid:} \) \( \mathrm{propanoic\ acid} \) has formula \( \mathrm{C_3H_6O_2} \).
\( \textbf{Nitro substitution:} \) The prefix \( \mathrm{3\text{-}nitro} \) means one hydrogen on carbon \(3\) is replaced by \( \mathrm{-NO_2} \).
\( \textbf{Formula change:} \)
\[
\mathrm{C_3H_6O_2}-\mathrm{H}+\mathrm{NO_2}=\mathrm{C_3H_5NO_4}
\]
\( \textbf{Carbon mass:} \)
\[
3(12)=36
\]
\( \textbf{Hydrogen mass:} \)
\[
5(1)=5
\]
\( \textbf{Nitrogen mass:} \)
\[
1(14)=14
\]
\( \textbf{Oxygen mass:} \)
\[
4(16)=64
\]
\( \textbf{Total molar mass:} \)
\[
36+5+14+64=119\,\text{g mol}^{-1}
\]
\( \textbf{Final answer:} \) The formula is \( \mathrm{C_3H_5NO_4} \), and the molar mass is \(119\,\text{g mol}^{-1}\).
The acid group already supplies two oxygen atoms, and the nitro group adds two more after replacing one hydrogen.
306. Consider the following statements about substitutive nomenclature.
Statement I: A halogen substituent is usually named with a prefix such as \( \mathrm{chloro} \) or \( \mathrm{bromo} \).
Statement II: A nitro group is usually named with the prefix \( \mathrm{nitro} \).
Statement III: A carboxylic acid group is normally ignored when a nitro group is present.
Which statements are valid?
ⓐ. II and III only
ⓑ. I and III only
ⓒ. I, II, and III
ⓓ. I and II only
Correct Answer: I and II only
Explanation: Statement I is valid because halogens are commonly named as prefixes in organic nomenclature. Statement II is also valid because the group \( \mathrm{-NO_2} \) is written as the prefix \( \mathrm{nitro} \). Statement III is not valid because \( \mathrm{-COOH} \) is a principal functional group in basic naming and is not ignored. When \( \mathrm{-COOH} \) is present as the main group, the suffix \( \mathrm{-oic\ acid} \) is used. Prefix groups are still included, but they do not replace the acid suffix.
307. A compound has the name \( \mathrm{4\text{-}bromo\text{-}3\text{-}nitrobutanoic\ acid} \). Which structural reading is correct?
ⓐ. butanoic acid with 3-nitro and 4-bromo
ⓑ. A four-carbon alcohol with bromine and nitrogen at carbon \(1\)
ⓒ. A benzene ring with \( \mathrm{Br} \) and \( \mathrm{NO_2} \) attached
ⓓ. An ester with a bromoalkyl group and no carboxyl carbon
Correct Answer: butanoic acid with 3-nitro and 4-bromo
Explanation: The parent name \( \mathrm{butanoic\ acid} \) shows a four-carbon carboxylic acid chain. The carboxyl carbon is counted as carbon \(1\). The prefix \( \mathrm{3\text{-}nitro} \) places \( \mathrm{-NO_2} \) at carbon \(3\), and the prefix \( \mathrm{4\text{-}bromo} \) places \( \mathrm{Br} \) at carbon \(4\). The name does not describe an aromatic compound because the parent is not benzene. The suffix \( \mathrm{-oic\ acid} \) controls the principal family and the numbering direction.
308. In the name \( \mathrm{4\text{-}bromo\text{-}3\text{-}nitrobutanoic\ acid} \), why is \( \mathrm{bromo} \) written before \( \mathrm{nitro} \) even though its locant is larger?
ⓐ. The larger locant must always be written first
ⓑ. Nitro groups are never written in names
ⓒ. alphabetical prefix order: bromo before nitro
ⓓ. Carboxylic acids cannot have prefixes
Correct Answer: alphabetical prefix order: bromo before nitro
Explanation: In IUPAC names, substituent prefixes are generally arranged alphabetically. The word \( \mathrm{bromo} \) begins with \( \mathrm{b} \), while \( \mathrm{nitro} \) begins with \( \mathrm{n} \). Therefore, \( \mathrm{bromo} \) is written before \( \mathrm{nitro} \), even though its locant is \(4\) and the nitro locant is \(3\). The locants stay attached to their own substituents and do not decide alphabetical order. The parent suffix \( \mathrm{-oic\ acid} \) remains at the end because it names the principal group.
309. Read the case below and answer the question.
A student writes the name \( \mathrm{3\text{-}nitro\text{-}2\text{-}chloropentane} \) for a compound with a pentane chain, chlorine at carbon \(2\), and nitro at carbon \(3\). The structure is correct, but the written order of prefixes is questioned.
What is the best corrected name?
ⓐ. \( \mathrm{3\text{-}nitro\text{-}2\text{-}chloropentane} \)
ⓑ. \( \mathrm{2\text{-}chloro\text{-}3\text{-}nitropentane} \)
ⓒ. \( \mathrm{2\text{-}nitro\text{-}3\text{-}chloropentane} \)
ⓓ. \( \mathrm{chloronitropentane} \)
Correct Answer: \( \mathrm{2\text{-}chloro\text{-}3\text{-}nitropentane} \)
Explanation: The parent chain is \( \mathrm{pentane} \). The substituents are \( \mathrm{chloro} \) at carbon \(2\) and \( \mathrm{nitro} \) at carbon \(3\). Prefixes are arranged alphabetically, so \( \mathrm{chloro} \) comes before \( \mathrm{nitro} \). The corrected name is \( \mathrm{2\text{-}chloro\text{-}3\text{-}nitropentane} \). The locants must remain with the correct groups; alphabetical order changes the order of writing, not the positions in the molecule.
310. The bond \( \mathrm{C-Cl} \) in chloroethane is polar mainly because:
ⓐ. carbon and chlorine have identical electronegativity
ⓑ. chlorine transfers all its electrons completely to carbon
ⓒ. the bond contains no shared electron pair
ⓓ. chlorine pulls electron density toward itself
Correct Answer: chlorine pulls electron density toward itself
Explanation: A covalent bond between atoms of different electronegativity is polar. Chlorine is more electronegative than carbon, so the shared electron pair in the \( \mathrm{C-Cl} \) bond is pulled more toward chlorine. This gives chlorine a partial negative character and carbon a partial positive character. The bond is still covalent, so it should not be treated as complete electron transfer. This polarity becomes important when discussing electronic effects and reaction centres in organic molecules.
311. In the fragment \( \mathrm{CH_3-CH_2-Cl} \), the inductive effect of chlorine is best described as:
ⓐ. \( \mathrm{+I} \) effect, electron donation through \( \pi \)-bonds
ⓑ. \( \mathrm{-I} \), electron withdrawal through \( \sigma \)-bonds
ⓒ. resonance effect through a benzene ring
ⓓ. hyperconjugation from a carbocation
Correct Answer: \( \mathrm{-I} \), electron withdrawal through \( \sigma \)-bonds
Explanation: Chlorine is more electronegative than carbon, so it withdraws electron density through the sigma-bond framework. This is called a negative inductive effect, written as \( \mathrm{-I} \). The inductive effect operates through \( \sigma \)-bonds and decreases with distance along the chain. It is not a resonance effect because the given fragment has no conjugated \( \pi \)-system. The key feature here is polarisation of sigma bonds due to electronegativity difference.
312. A group \( \mathrm{-NO_2} \) attached to an organic skeleton strongly withdraws electron density. In electronic-effect language, it is mainly:
ⓐ. an electron-donating \( \mathrm{+I} \) alkyl group
ⓑ. a group with no effect on electron density
ⓒ. a group that always donates electrons through \( \sigma \)-bonds
ⓓ. an electron-withdrawing \( \mathrm{-I} \) group
Correct Answer: an electron-withdrawing \( \mathrm{-I} \) group
Explanation: The nitro group \( \mathrm{-NO_2} \) contains highly electronegative oxygen atoms attached through nitrogen. It withdraws electron density from the carbon framework. This electron-withdrawing tendency is often described as a \( \mathrm{-I} \) effect when discussed through sigma bonds. It should not be grouped with alkyl groups, which generally show electron-releasing \( \mathrm{+I} \) character. The sign of the inductive effect depends on whether the group withdraws or releases electron density relative to carbon.
313. A comparison of substituent effects is shown below.
| Group | Expected inductive behaviour |
|---|
| P. \( \mathrm{-CH_3} \) | electron-releasing \( \mathrm{+I} \) |
| Q. \( \mathrm{-Cl} \) | electron-withdrawing \( \mathrm{-I} \) |
| R. \( \mathrm{-NO_2} \) | electron-withdrawing \( \mathrm{-I} \) |
| S. \( \mathrm{-CH_3} \) | electron-withdrawing \( \mathrm{-I} \) |
Which row contains the unsuitable inductive-effect description?
ⓐ. S
ⓑ. P
ⓒ. Q
ⓓ. R
Correct Answer: S
Explanation: Alkyl groups such as \( \mathrm{-CH_3} \) are generally electron-releasing by the \( \mathrm{+I} \) effect. Halogens such as \( \mathrm{-Cl} \) withdraw electron density through the sigma framework because they are more electronegative than carbon. The nitro group \( \mathrm{-NO_2} \) is also strongly electron withdrawing. Therefore, Row S is unsuitable because it describes \( \mathrm{-CH_3} \) as \( \mathrm{-I} \). The same group should not be assigned opposite inductive signs in the same simple comparison.
314. The influence of an inductive effect becomes weaker as the distance from the substituent increases. For the chain \( \mathrm{Cl-CH_2-CH_2-CH_2-COOH} \), which carbon is most directly affected by the \( \mathrm{-I} \) effect of chlorine?
ⓐ. the carbonyl carbon of \( \mathrm{-COOH} \) only
ⓑ. only the farthest carbon from \( \mathrm{Cl} \)
ⓒ. the carbon directly bonded to \( \mathrm{Cl} \)
ⓓ. no carbon, because inductive effects cannot pass through \( \sigma \)-bonds
Correct Answer: the carbon directly bonded to \( \mathrm{Cl} \)
Explanation: The inductive effect is transmitted through \( \sigma \)-bonds. Its strength is greatest near the substituent causing the polarity and decreases with distance. In \( \mathrm{Cl-CH_2-CH_2-CH_2-COOH} \), the carbon directly attached to \( \mathrm{Cl} \) experiences the strongest electron-withdrawing influence. Carbons farther away are affected less strongly. This distance dependence is why the position of an electronegative substituent can influence properties differently in related compounds.
315. The acid strength order among the following carboxylic acids is best represented as:
ⓐ. \( \mathrm{CH_3COOH} \gt \mathrm{ClCH_2COOH} \gt \mathrm{Cl_2CHCOOH} \gt \mathrm{CCl_3COOH} \)
ⓑ. \( \mathrm{ClCH_2COOH} \gt \mathrm{CH_3COOH} \gt \mathrm{CCl_3COOH} \gt \mathrm{Cl_2CHCOOH} \)
ⓒ. \( \mathrm{CH_3COOH} \gt \mathrm{CCl_3COOH} \gt \mathrm{ClCH_2COOH} \gt \mathrm{Cl_2CHCOOH} \)
ⓓ. \( \mathrm{CCl_3COOH} \gt \mathrm{Cl_2CHCOOH} \gt \mathrm{ClCH_2COOH} \gt \mathrm{CH_3COOH} \)
Correct Answer: \( \mathrm{CCl_3COOH} \gt \mathrm{Cl_2CHCOOH} \gt \mathrm{ClCH_2COOH} \gt \mathrm{CH_3COOH} \)
Explanation: Chlorine atoms show a strong \( \mathrm{-I} \) effect and withdraw electron density through \( \sigma \)-bonds. In substituted acetic acids, this withdrawal stabilises the carboxylate ion formed after loss of \( \mathrm{H^+} \). A more stable conjugate base corresponds to a stronger acid. As the number of chlorine atoms on the alpha carbon increases, the electron-withdrawing effect becomes stronger. Thus \( \mathrm{CCl_3COOH} \) is stronger than \( \mathrm{Cl_2CHCOOH} \), which is stronger than \( \mathrm{ClCH_2COOH} \), and \( \mathrm{CH_3COOH} \) is the weakest in this set. The comparison is based on conjugate-base stabilisation, not on the number of hydrogen atoms in the molecular formula.
316. A graph is described with distance of a substituent from the \( \mathrm{-COOH} \) group on the horizontal axis and acid-strength increase due to \( \mathrm{-I} \) effect on the vertical axis. For \( \mathrm{ClCH_2CH_2CH_2COOH} \), \( \mathrm{CH_3CH(Cl)CH_2COOH} \), and \( \mathrm{CH_3CH_2CH(Cl)COOH} \), what trend should the graph show as chlorine moves farther from \( \mathrm{-COOH} \)?
ⓐ. the acid-strength increase becomes smaller
ⓑ. the acid-strength increase becomes larger
ⓒ. the acid-strength increase remains exactly constant
ⓓ. the compound stops being a carboxylic acid
Correct Answer: the acid-strength increase becomes smaller
Explanation: The \( \mathrm{-I} \) effect is transmitted through \( \sigma \)-bonds and becomes weaker with distance. A chlorine atom closer to the \( \mathrm{-COOH} \) group withdraws electron density more effectively from the carboxylate region. This better stabilises the conjugate base and increases acid strength more strongly. When chlorine is farther away, its effect still exists but is less influential. The graph should therefore show a decreasing acid-strength contribution with increasing distance from the carboxyl group. The structural position of chlorine matters even when the molecular formula remains similar.
317. Compare the following acids:
| Acid | Position of \( \mathrm{Cl} \) relative to \( \mathrm{-COOH} \) |
|---|
| P. \( \mathrm{CH_3CH_2CH(Cl)COOH} \) | nearest among the three |
| Q. \( \mathrm{CH_3CH(Cl)CH_2COOH} \) | middle distance |
| R. \( \mathrm{ClCH_2CH_2CH_2COOH} \) | farthest among the three |
What is the expected acid strength order?
ⓐ. \( \mathrm{R \gt Q \gt P} \)
ⓑ. \( \mathrm{Q \gt R \gt P} \)
ⓒ. \( \mathrm{P \gt Q \gt R} \)
ⓓ. \( \mathrm{P = Q = R} \)
Correct Answer: \( \mathrm{P \gt Q \gt R} \)
Explanation: The chlorine atom withdraws electron density by the \( \mathrm{-I} \) effect. This withdrawal stabilises the conjugate base of the carboxylic acid and increases acid strength. The inductive effect decreases as the number of \( \sigma \)-bonds between chlorine and the carboxyl group increases. In P, chlorine is closest to \( \mathrm{-COOH} \), so its effect is strongest. In R, chlorine is farthest, so its acid-strengthening effect is weakest. The order follows distance from the carboxyl group rather than only the total number of atoms in the molecule.
318. A substituted acid has molar mass data and acid-strength clues as follows: Acid P is \( \mathrm{CH_3COOH} \), Acid Q is \( \mathrm{ClCH_2COOH} \), and Acid R is \( \mathrm{CCl_3COOH} \). Using \( \mathrm{C}=12 \), \( \mathrm{H}=1 \), \( \mathrm{O}=16 \), and \( \mathrm{Cl}=35.5 \), which option correctly combines molar mass and expected acid strength?
ⓐ. \(M_P=60\,\text{g mol}^{-1}\), \(M_Q=95.5\,\text{g mol}^{-1}\), \(M_R=162.5\,\text{g mol}^{-1}\); acid strength \( \mathrm{P \gt Q \gt R} \)
ⓑ. \(M_P=61\,\text{g mol}^{-1}\), \(M_Q=94.5\,\text{g mol}^{-1}\), \(M_R=163.5\,\text{g mol}^{-1}\); acid strength \( \mathrm{Q \gt R \gt P} \)
ⓒ. \(M_P=60\,\text{g mol}^{-1}\), \(M_Q=94.5\,\text{g mol}^{-1}\), \(M_R=163.5\,\text{g mol}^{-1}\); acid strength \( \mathrm{P \gt Q \gt R} \)
ⓓ. \(M_P=60\,\text{g mol}^{-1}\), \(M_Q=94.5\,\text{g mol}^{-1}\), \(M_R=163.5\,\text{g mol}^{-1}\); acid strength \( \mathrm{R \gt Q \gt P} \)
Correct Answer: \(M_P=60\,\text{g mol}^{-1}\), \(M_Q=94.5\,\text{g mol}^{-1}\), \(M_R=163.5\,\text{g mol}^{-1}\); acid strength \( \mathrm{R \gt Q \gt P} \)
Explanation: \( \textbf{Formula of P:} \) \( \mathrm{CH_3COOH}=\mathrm{C_2H_4O_2} \).
\( \textbf{Molar mass of P:} \)
\[
2(12)+4(1)+2(16)=24+4+32=60\,\text{g mol}^{-1}
\]
\( \textbf{Formula of Q:} \) \( \mathrm{ClCH_2COOH}=\mathrm{C_2H_3ClO_2} \).
\( \textbf{Molar mass of Q:} \)
\[
2(12)+3(1)+35.5+2(16)=24+3+35.5+32=94.5\,\text{g mol}^{-1}
\]
\( \textbf{Formula of R:} \) \( \mathrm{CCl_3COOH}=\mathrm{C_2HCl_3O_2} \).
\( \textbf{Molar mass of R:} \)
\[
2(12)+1(1)+3(35.5)+2(16)=24+1+106.5+32=163.5\,\text{g mol}^{-1}
\]
\( \textbf{Acid-strength reasoning:} \) More chlorine atoms produce a stronger \( \mathrm{-I} \) effect near \( \mathrm{-COOH} \).
\( \textbf{Conjugate-base effect:} \) Stronger \( \mathrm{-I} \) stabilises the carboxylate ion more effectively.
\( \textbf{Final answer:} \) The molar masses are \(60\), \(94.5\), and \(163.5\,\text{g mol}^{-1}\), and the acid strength order is \( \mathrm{R \gt Q \gt P} \).
The mass calculation and acidity comparison use different ideas, so the heavier compound is not chosen only because it has greater mass.
319. The stability order of the simple alkyl carbocations \( \mathrm{(CH_3)_3C^+} \), \( \mathrm{(CH_3)_2CH^+} \), \( \mathrm{CH_3CH_2^+} \), and \( \mathrm{CH_3^+} \) is:
ⓐ. \( \mathrm{CH_3^+ \gt CH_3CH_2^+ \gt (CH_3)_2CH^+ \gt (CH_3)_3C^+} \)
ⓑ. \( \mathrm{(CH_3)_2CH^+ \gt (CH_3)_3C^+ \gt CH_3^+ \gt CH_3CH_2^+} \)
ⓒ. \( \mathrm{(CH_3)_3C^+ \gt (CH_3)_2CH^+ \gt CH_3CH_2^+ \gt CH_3^+} \)
ⓓ. \( \mathrm{CH_3CH_2^+ \gt CH_3^+ \gt (CH_3)_3C^+ \gt (CH_3)_2CH^+} \)
Correct Answer: \( \mathrm{(CH_3)_3C^+ \gt (CH_3)_2CH^+ \gt CH_3CH_2^+ \gt CH_3^+} \)
Explanation: Alkyl groups generally release electron density by the \( \mathrm{+I} \) effect. A carbocation is electron-deficient at the positively charged carbon. More alkyl groups attached to the positively charged carbon help disperse and reduce this electron deficiency. A tertiary carbocation has three alkyl groups, a secondary carbocation has two, a primary carbocation has one, and a methyl carbocation has none. Therefore, the stability order is \(3^\circ \gt 2^\circ \gt 1^\circ \gt \mathrm{methyl}\). This order is for simple alkyl carbocations where resonance effects are not being introduced.
320. A carbon species is negatively charged, and the attached alkyl groups release electron density by \( \mathrm{+I} \) effect. For the simple alkyl carbanions \( \mathrm{CH_3^-} \), \( \mathrm{CH_3CH_2^-} \), \( \mathrm{(CH_3)_2CH^-} \), and \( \mathrm{(CH_3)_3C^-} \), the expected stability order is:
ⓐ. \( \mathrm{(CH_3)_3C^- \gt (CH_3)_2CH^- \gt CH_3CH_2^- \gt CH_3^-} \)
ⓑ. \( \mathrm{CH_3^- \gt CH_3CH_2^- \gt (CH_3)_2CH^- \gt (CH_3)_3C^-} \)
ⓒ. \( \mathrm{CH_3CH_2^- \gt CH_3^- \gt (CH_3)_3C^- \gt (CH_3)_2CH^-} \)
ⓓ. \( \mathrm{(CH_3)_2CH^- \gt CH_3CH_2^- \gt CH_3^- \gt (CH_3)_3C^-} \)
Correct Answer: \( \mathrm{CH_3^- \gt CH_3CH_2^- \gt (CH_3)_2CH^- \gt (CH_3)_3C^-} \)
Explanation: A carbanion has excess electron density on carbon. Alkyl groups show \( \mathrm{+I} \) effect and push electron density toward the negatively charged carbon. This increases electron crowding and usually destabilises a simple alkyl carbanion. Therefore, adding more alkyl groups decreases stability in the simple alkyl series. The methyl carbanion has no alkyl group pushing extra electron density toward the negative centre, so it is the most stable among these. This trend is opposite to simple alkyl carbocations because the charge being stabilised is negative rather than positive.
