JEE Main Practice — Chemistry Set 5

Explanation: Hofmann bromamide degradation: $\mathrm{RCONH_2+Br_2+4NaOH\to RNH_2+Na_2CO_3+2NaBr+2H_2O}$, removing one carbon. $\mathrm{CH_3CH_2CONH_2\to CH_3CH_2NH_2}$ (ethanamine).

Explanation: Direct bromination of aniline gives the 2,4,6-tribromo product because $-\mathrm{NH_2}$ is strongly activating. Acetylation gives acetanilide, whose $-\mathrm{NHCOCH_3}$ is only moderately activating (lone pair delocalised onto C=O) and bulky, so monobromination occurs mainly para. Acid hydrolysis then removes the acetyl group: 4-bromoaniline. (4-Bromoacetanilide is the intermediate after step ii.)

Explanation: Primary: propan-1-amine, propan-2-amine. Secondary: N-methylethanamine. Tertiary: N,N-dimethylmethanamine (trimethylamine). Total 4.

Explanation: $\mathrm{C_6H_5NO_2}$ (123 g mol$^{-1}$) $\to\mathrm{C_6H_5NH_2}$ (93 g mol$^{-1}$), 1:1. 123 g = 1 mol nitrobenzene gives 1 mol = 93 g aniline.

Explanation: The Lucas test goes through a carbocation ($\mathrm{S_N1}$). Tertiary alcohols form stable carbocations and give turbidity immediately; secondary alcohols take about 5 min; primary alcohols give no turbidity at room temperature.

Explanation: Aldehydes reduce Tollens' reagent to a silver mirror; ketones generally do not. Ethanal is an aldehyde, propanone a ketone.

Explanation: Carboxylate is stabilised by resonance over two equivalent O atoms ($\mathrm pK_a\approx4.8$); phenoxide by delocalisation into the ring ($\approx10$); water ($\approx15.7$); ethoxide is destabilised by the +I ethyl group (ethanol $\approx16$).

Explanation: The iodoform test needs a $\mathrm{CH_3CO{-}}$ group or a $\mathrm{CH_3CH(OH){-}}$ group that can be oxidised to it. Ethanol, propan-2-ol and acetophenone ($\mathrm{C_6H_5COCH_3}$) qualify; pentan-3-one ($\mathrm{C_2H_5COC_2H_5}$) has no methyl ketone unit.

Explanation: The Cannizzaro reaction (disproportionation into alcohol + carboxylate) occurs with aldehydes having no $\alpha$-hydrogen, e.g. benzaldehyde. Ethanal and propanal have $\alpha$-H and undergo aldol condensation instead; propanone is a ketone.

Explanation: After protonation of the ether oxygen, the C–O bond to the tertiary group breaks by an $\mathrm{S_N1}$ pathway because the tertiary carbocation is stable; $\mathrm{I^-}$ captures it, giving $\mathrm{(CH_3)_3CI}$ and $\mathrm{CH_3OH}$. (For a methyl ether whose other group is primary or secondary, $\mathrm{S_N2}$ attack at the methyl carbon would give $\mathrm{CH_3I}$ instead; option A describes that case, which does not apply to a tertiary group.)

Explanation: $\mathrm{CH_3{-}CO{-}CH_2{-}CH_3}$: the carbons adjacent to C=O are the $\mathrm{CH_3}$ (3 H) and the $\mathrm{CH_2}$ (2 H). Total 5.

Explanation: $\mathrm{CH_3COOH}$ (60) $+\mathrm{C_2H_5OH\to CH_3COOC_2H_5}$ (88) $+\mathrm{H_2O}$. 30 g = 0.5 mol acid gives 0.5 mol ester $=44$ g.

Explanation: NO (like $\mathrm{N_2O}$) is neutral. $\mathrm{N_2O_3}$ and $\mathrm{N_2O_5}$ are acidic anhydrides of $\mathrm{HNO_2}$ and $\mathrm{HNO_3}$; $\mathrm{NO_2}$ is acidic (mixed anhydride).

Explanation: Acid strength of HX increases down the group as the H–X bond dissociation enthalpy decreases: HF < HCl < HBr < HI.

Explanation: Normally bond enthalpy falls as atoms get larger, but $\mathrm{F_2}$ is anomalously weak because of strong repulsion between lone pairs on the very small F atoms. Approximate values (kJ/mol): $\mathrm{Cl_2}$ 242, $\mathrm{Br_2}$ 192, $\mathrm{F_2}$ 158, $\mathrm{I_2}$ 151.

Explanation: The E–H bond weakens as E gets larger, so thermal stability falls and reducing character rises. The lone pair becomes more diffuse (less available), so basicity falls ($\mathrm{NH_3}$ most basic). Bond angle decreases (107° in $\mathrm{NH_3}$ to about 90° lower down).

Explanation: Down group 14, the +2 state becomes more stable than +4 (reluctance of the $ns^2$ pair to take part in bonding). $\mathrm{Pb^{4+}}$ readily gains two electrons to become the stable $\mathrm{Pb^{2+}}$, so $\mathrm{PbO_2}$ is a strong oxidant; $\mathrm{SiO_2}$ is not oxidising at all.

Explanation: Boron has no d orbitals in its valence shell, so its maximum covalency is 4 ($\mathrm{[BF_4]^-}$); aluminium can use 3d orbitals and forms $\mathrm{[AlF_6]^{3-}}$. The other statements are correct: small N forms strong $p\pi$–$p\pi$ bonds; F is the most electronegative element; C–C (~348 kJ/mol) is much stronger than Si–Si (~297 kJ/mol).

Explanation: $\mathrm{H_3PO_2}$: one P–OH and two P–H (monobasic). $\mathrm{H_3PO_3}$: two P–OH and one P–H (dibasic). $\mathrm{H_3PO_4}$: three P–OH (tribasic). Total $1+2+3=6$.

Explanation: Xe: 8 valence electrons, 4 used in bonds, 2 lone pairs. Each F has 3 lone pairs: $4\times3=12$. Total $=2+12=14$.

Explanation: $2(+1)+2x+7(-2)=0\Rightarrow2x=12\Rightarrow x=+6$.

Explanation: $E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}=0.34-(-0.76)=1.10$ V.

Explanation: Mn: $+7\to+4$ (gains 3 e$^-$). I: $-1\to+5$ (loses 6 e$^-$). Electron balance: 2 $\mathrm{MnO_4^-}$ per $\mathrm{I^-}$. Balanced: $\mathrm{2MnO_4^-+I^-+H_2O\to2MnO_2+IO_3^-+2OH^-}$.

Explanation: Cell reaction $\mathrm{Zn+Cu^{2+}\to Zn^{2+}+Cu}$, $n=2$. $E=E^\circ-\frac{0.059}{2}\log\frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]}=1.10-0.0295\log(10^{-2})=1.10+0.059=1.159\approx1.16$ V.

Explanation: $\mathrm{Al^{3+}+3e^-\to Al}$. 2.7 g = 0.1 mol Al needs 0.3 mol e$^-$ $=0.3\times96500=28950$ C.