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Chemistry
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Showing 1–25 of 45 questions
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The reaction that does not give benzoic acid as the major product is:$K_2Cr_2O_7$ and $KMnO_4/H^+$ oxidise benzyl alcohol to benzoic acid, and acetophenone undergoes the haloform reaction with NaOCl to give benzoic acid (+ $CHCl_3$). PCC is a mild oxidising agent and stops at benzaldehyde ($C_6H_5CHO$).
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The amine that reacts with Hinsberg's reagent to give an alkali insoluble product is:Secondary amines react with benzenesulphonyl chloride to give N,N-dialkylbenzenesulphonamides with no acidic H on N, so they are insoluble in alkali: $[(CH_3)_2CH]_2NH+ClSO_2Ph\rightarrow[(CH_3)_2CH]_2NSO_2Ph+HCl$. Primary amines give alkali-soluble products and tertiary amines do not react.
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Which structure(s) of proteins remain(s) intact during denaturation process?During denaturation the $2^\circ$ and $3^\circ$ structures are destroyed but the $1^\circ$ structure remains intact.
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The polymer that is used as a substitute for wool in making commercial fibres is:Polyacrylonitrile is used as a substitute for wool in making commercial fibres such as Orlon or Acrilan.
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The artificial sweetener stable at cooking temperature and does not provide calories is:Sucralose is stable at cooking temperature and does not provide calories.
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The liquified gas that is used in dry cleaning along with a suitable detergent is:
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The hydrolysis reaction that takes place at the slowest rate, among the following is:Aryl halides do not undergo nucleophilic substitution easily at room temperature (partial double bond character of C-Cl due to resonance), so the chlorotoluene reacts slowest.
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When vapours of a secondary alcohol is passed over heated copper at 573 K, the product formed is:$CH_3-CH(OH)-CH_3\xrightarrow{Cu/573\ K}CH_3-CO-CH_3$. Secondary alcohols are dehydrogenated to ketones.
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The major products C and D formed in the following reactions respectively are: $H_3C-CH_2-CH_2-O-C(CH_3)_3\xrightarrow[\Delta]{excess\ HI}C+D$The ether cleaves via the stable tertiary carbocation to give $CH_3CH_2CH_2OH+(CH_3)_3C-I$. With excess HI the propanol is further converted to $CH_3CH_2CH_2-I$. So C and D are $CH_3CH_2CH_2I$ and $(CH_3)_3CI$.
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Match the oxide given in Column A with its property given in Column B. Which option has all correct pairs?
Pick the matching item from Column II for each row.
Column I
- I. $Na_2O$
- II. $Al_2O_3$
- III. $N_2O$
- IV. $Cl_2O_7$
Column II
- A. Neutral
- B. Basic
- C. Acidic
- D. Amphoteric
Correct answer: I → B, II → D, III → A, IV → C
$Na_2O\rightarrow$ basic; $Al_2O_3\rightarrow$ amphoteric; $N_2O\rightarrow$ neutral; $Cl_2O_7\rightarrow$ acidic. -
Match the catalyst with the process. Which of the following is the correct option?
Pick the matching item from Column II for each row.
Column I
- I. $V_2O_5$
- II. $TiCl_4+Al(CH_3)_3$
- III. $PdCl_2$
- IV. Nickel complexes
Column II
- A. The oxidation of ethyne to ethanal
- B. Polymerisation of alkynes
- C. Oxidation of $SO_2$ in the manufacture of $H_2SO_4$
- D. Polymerisation of ethylene
Correct answer: I → C, II → D, III → A, IV → B
$V_2O_5$ - oxidation of $SO_2$ in the contact process; $TiCl_4+Al(CH_3)_3$ (Ziegler-Natta) - polymerisation of ethylene; $PdCl_2$ - Wacker oxidation to ethanal; nickel complexes - polymerisation of alkynes. -
The most stable carbocation, among the following is:Stability of these carbocations depends on hyperconjugation (+H effect), i.e. the number of $\alpha$-hydrogens. Option (3) has the maximum number of $\alpha$-H (5).
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The alkane that gives only one mono-chloro product on chlorination with $Cl_2$ in presence of diffused sunlight is:
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In the following reaction, $H_3C-C\equiv CH\xrightarrow[873\ K]{red\ hot\ iron\ tube}A$ the number of sigma ($\sigma$) bonds present in the product A is:Propyne trimerises in a red hot iron tube at 873 K to give mesitylene (1,3,5-trimethylbenzene, $C_9H_{12}$). $\sigma$ bonds: 6 ring C-C + 3 C-$CH_3$ + 12 C-H = 21.
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Aluminium chloride in acidified aqueous solution forms a complex 'A', in which hybridisation state of Al is 'B'. What are 'A' and 'B', respectively?$AlCl_3$ in acidified aqueous solution forms $[Al(H_2O)_6]^{3+}$. $Al^{3+}$ = [Ne], with no inner d orbitals available, so it uses outer 3d orbitals: $sp^3d^2$.
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Which of the following compounds is used in cosmetic surgery?
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Identify the incorrect statement.The contaminated undesired material present in an ore is called gangue; gangue is not the ore itself.
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A compound 'X' upon reaction with $H_2O$ produces a colorless gas 'Y' with rotten fish smell. Gas 'Y' is absorbed in a solution of $CuSO_4$ to give $Cu_3P_2$ as one of the products. Predict the compound 'X'$Ca_3P_2+6H_2O\rightarrow3Ca(OH)_2+2PH_3$ (colourless gas Y). $3CuSO_4+2PH_3\rightarrow Cu_3P_2+3H_2SO_4$. Hence X is calcium phosphide, $Ca_3P_2$.
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Which of the following oxoacids of phosphorus has strongest reducing property?Reducing property $\propto$ number of P-H bonds. $H_3PO_2$ has two P-H bonds, the maximum among these.
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Identify the correct formula of oleum from the followingOleum is pyrosulphuric acid, $H_2S_2O_7$.
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When neutral or faintly alkaline $KMnO_4$ is treated with potassium iodide, iodide ion is converted into 'X'. 'X' isIn neutral or faintly alkaline medium: $2MnO_4^-+I^-+H_2O\rightarrow2MnO_2+IO_3^-+2OH^-$. So X is iodate, $IO_3^-$.
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The Crystal Field Stabilisation Energy (CFSE) for $[CoCl_6]^{4-}$ is 18000 cm$^{-1}$. The CFSE for $[CoCl_4]^{2-}$ will be$\Delta_t=\frac{4}{9}\Delta_o=\frac{4}{9}\times18000=8000$ cm$^{-1}$.
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Following limiting molar conductivities are given as $\lambda^0_{m(H_2SO_4)}=x$ S cm$^2$ mol$^{-1}$, $\lambda^0_{m(K_2SO_4)}=y$ S cm$^2$ mol$^{-1}$, $\lambda^0_{m(CH_3COOK)}=z$ S cm$^2$ mol$^{-1}$ $\lambda^0_m$ (in S cm$^2$ mol$^{-1}$) for $CH_3COOH$ will be:By Kohlrausch's law, $\lambda^0_{CH_3COOH}=\lambda^0_{CH_3COO^-}+\lambda^0_{H^+}=\lambda^0_{CH_3COOK}+\frac{1}{2}\lambda^0_{H_2SO_4}-\frac{1}{2}\lambda^0_{K_2SO_4}=z+\frac{x}{2}-\frac{y}{2}=\frac{(x-y)}{2}+z$.
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A first order reaction has a rate constant of $2.303\times10^{-3}$ s$^{-1}$. The time required for 40 g of this reactant to reduce to 10 g will be [Given that $\log_{10}2=0.3010$]$t_{1/2}=\frac{0.693}{k}=\frac{0.693}{2.303\times10^{-3}}=301$ s. 40 g to 10 g is two half-lives (75% completion): $t=2\times301=602$ s.
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For a reaction, activation energy $E_a=0$ and the rate constant at 200 K is $1.6\times10^6$ s$^{-1}$. The rate constant at 400 K will be [Given that gas constant R = 8.314 J K$^{-1}$ mol$^{-1}$]$\log\frac{K_2}{K_1}=\frac{E_a}{2.303R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)=0$ since $E_a=0$. So $K_2=K_1=1.6\times10^6$ s$^{-1}$.