Grade 11 : Chemistry - Organic Chemistry Basics (Set 4)

Explanation:The longest continuous carbon chain containing the hydroxyl ($-OH$) functional group has 5 carbon atoms (pentan). Numbering starts from the end closest to the $-OH$ group to give it the lowest locant ($C_2$). The methyl substituent is located at $C_4$. Thus, the IUPAC name is 4-methylpentan-2-ol.

Explanation:In propadiene ($H_2C=C=CH_2$), the terminal carbons form two $\sigma$ bonds with hydrogen and one $\sigma$ bond with carbon (3 $\sigma$ domains), making them $sp^2$ hybridised. The central carbon forms two $\sigma$ bonds and two $\pi$ bonds (2 $\sigma$ domains), making it sp hybridised. Therefore, the order is $sp^2, sp, sp^2$.

Explanation:Moles ratio: $C = 40/12 = 3.33$, $H = 6.67/1 = 6.67$, $O = 53.33/16 = 3.33$. Dividing by smallest number (3.33) gives ratio $C:H:O = 1:2:1$. Empirical formula is $CH_2O$ (formula mass $= 12+2+16 = 30\text{ g/mol}$). $n = 60/30 = 2$. Thus, molecular formula is $(CH_2O)_2 = C_2H_4O_2$.

Explanation:Carbocation stability increases with hyperconjugation and $+I$ inductive effect from attached alkyl groups. Tertiary carbocations ($3^\circ$, 9 $\alpha$-H) are more stable than secondary ($2^\circ$, 6 $\alpha$-H), primary ($1^\circ$, 3 $\alpha$-H), and methyl carbocations ($0^\circ$). Hence, $(CH_3)_3C^+ > (CH_3)_2CH^+ > CH_3CH_2^+ > CH_3^+$.

Explanation:A nucleophile is an electron-rich species capable of donating an electron pair. Cyanide ion ($CN^-$) carries a negative charge and unshared electron pairs, making it a nucleophile. $NO_2^+$ is positively charged (electrophile), while $BF_3$ and $AlCl_3$ are electron-deficient Lewis acids (electrophiles).

Explanation:Single bonds contain $1\ \sigma$. Double bond contains $1\ \sigma + 1\ \pi$. Triple bond contains $1\ \sigma + 2\ \pi$. C-H bonds $= 6\ \sigma$. C-C bonds $= 4\ \sigma$. Total $\sigma$ bonds $= 6 + 4 = 10\ \sigma$. Total $\pi$ bonds $= 1\ \pi \text{ (double bond)} + 2\ \pi \text{ (triple bond)} = 3\ \pi$.

Explanation:Metamerism arises due to unequal distribution of alkyl chains on either side of a polyvalent functional group (such as $-CO-$, $-O-$, $-NH-$). Pentan-2-one has methyl and propyl groups attached to $-CO-$, whereas pentan-3-one has two ethyl groups. Thus, they are metamers.

Explanation:Normality of $1.0m H_2SO_4 = 2 \times 1.0 = 2.0\text{ N}$. Formula for nitrogen percentage: $%N = \frac{1.4 \times N \times V}{w} = \frac{1.4 \times 2.0 \times 10.0}{0.50} = \frac{28.0}{0.50} = 56%$.

Explanation:The electron-withdrawing inductive ($-I$) effect strength follows electronegativity and positive character on central atoms. The established $-I$ order is: $-NO_2 > -CN > -F > -Cl$. Nitro group has strong electron withdrawal due to formal positive charge on nitrogen and highly electronegative oxygen atoms.

Explanation:$\alpha$-hydrogens are attached to $sp^3$ carbons directly adjacent to double-bonded ($sp^2$) carbons. In 2-methylbut-2-ene, there are two methyl groups at $C_2$ ($3+3=6$) and one methyl group at $C_3$ (3), total $= 9\ \alpha$-H. In propene, there is one methyl group adjacent to the double bond ($3\ \alpha$-H).

Explanation:Both statements are correct. The electromeric effect ($E$ effect) is temporary and vanishes when the attacking reagent is removed. $+E$ effect occurs when $\pi$-electrons transfer toward the atom where the attacking electrophile attaches (e.g., addition of $H^+$ to alkenes).

Explanation:Carboxylic acid ($-COOH$) has higher principal priority than hydroxyl ($-OH$). The chain is numbered starting at $-COOH$ carbon as $C_1$: $C_1=COOH$, $C_2=CH(CH_3)$, $C_3=CH_2$, $C_4=CH_2OH$. Methyl is at $C_2$ and hydroxy is at $C_4$. The IUPAC name is 4-hydroxy-2-methylbutanoic acid.

Explanation:Mass of Cl in $\text{AgCl} = \frac{35.5}{143.5} \times 0.287g = 0.071g$. Mass percentage of Cl $= \frac{\text{Mass of Cl}}{\text{Mass of compound}} \times 100 = \frac{0.071}{0.30} \times 100 = 23.67%$.

Explanation:In homolytic cleavage (homolysis), the covalent bond breaks symmetrically such that each bonded atom retains one electron of the shared pair, generating neutral chemical species with an unpaired electron known as free radicals ($R^\bullet$).

Explanation:Methanol ($CH_3OH$) and ethanol ($C_2H_5OH$) both contain the primary alcohol group ($-OH$) and differ by a single $-CH_2-$ unit (14 u molecular mass difference). Thus, they belong to the same homologous series of alcohols.

Explanation:In a simple methyl free radical ($^\bullet CH_3$), carbon forms three $\sigma$ bonds with hydrogen atoms using $sp^2$ hybrid orbitals and holds the unshared odd electron in an unhybridised $p$-orbital. This gives it a planar, trigonal planar geometry with bond angles of $120^\circ$.

Explanation:Paper chromatography is a type of partition chromatography where water trapped in the cellulose fibers of the paper acts as the stationary liquid phase, and the mobile solvent carries components according to their differential partition coefficients between phases.

Explanation:Aldehyde ($-CHO$) has higher IUPAC priority than ketone ($-CO-$). The main chain carbon 1 is assigned to $-CHO$: $C_1=CHO$, $C_2=CH_2$, $C_3=CH_2$, $C_4=CO$, $C_5=CH_3$. The ketone oxygen at $C_4$ is designated by prefix 'oxo'. Thus, the IUPAC name is 4-oxopentanal.

Explanation:The hydroxyl group ($-OH$) possesses unshared lone pairs on the oxygen atom that can be donated into the $\pi$-system, exhibiting $+R$ (positive resonance) effect. Conversely, $-NO_2$, $-CHO$, and $-COOH$ withdraw electron density via $\pi$-bonds ($-R$ effect).

Explanation:$C_5H_{12}$ has exactly 3 structural isomers: 1) Pentane ($CH_3CH_2CH_2CH_2CH_3$), 2) Isopentane / 2-methylbutane ($CH_3CH(CH_3)CH_2CH_3$), and 3) Neopentane / 2,2-dimethylpropane ($(CH_3)_4C$).