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JEE Main Organic Chemistry Flashcards
51 question-and-answer cards covering Organic Chemistry as it is examined in JEE Main. 24 of them are printed below, taken from across the deck — no signup, no paywall on the preview.
24 sample cards from the Organic Chemistry deck
Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.
How is sulphur detected in the sodium fusion extract?
By two tests: (1) Adding sodium nitroprusside gives a violet/purple colouration: $$\ce{Na2S + Na2[Fe(CN)5NO] -> Na4[Fe(CN)5NOS]}$$ (2) Acidifying with acetic acid and adding lead acetate gives a black precipitate of $\ce{PbS}$: $\ce{S^{2-} + Pb^{2+} -> PbS v}$.
In a compound containing both nitrogen and sulphur, what colour is obtained in the Lassaigne's test and why?
A blood-red colouration is obtained (not Prussian blue) because sodium thiocyanate (NaSCN) forms during fusion. With $\ce{Fe^{3+}}$ it gives ferric thiocyanate: $$\ce{Fe^{3+} + SCN^- -> [Fe(SCN)]^{2+}} \;(\text{blood red})$$ If excess sodium is used, NaSCN decomposes to NaCN and $\ce{Na2S}$, giving the usual N and S tests separately.
How is phosphorus detected in an organic compound?
The compound is heated with an oxidising agent (sodium peroxide, $\ce{Na2O2}$), converting phosphorus to sodium phosphate. Boiling with conc. $\ce{HNO3}$ and adding ammonium molybdate gives a yellow precipitate (or colouration) of ammonium phosphomolybdate, confirming phosphorus.
Write the reaction for the detection of phosphorus using ammonium molybdate.
Phosphorus is first oxidised to phosphate: $\ce{P ->[\text{Na2O2}] Na3PO4}$. Then $$\ce{Na3PO4 + 3HNO3 -> H3PO4 + 3NaNO3}$$ $$\ce{H3PO4 + 12(NH4)2MoO4 + 21HNO3 -> (NH4)3PO4.12MoO3 v + 21NH4NO3 + 12H2O}$$ The yellow $\ce{(NH4)3PO4.12MoO3}$ confirms phosphorus.
How are halogens detected by the silver nitrate (Lassaigne's) test?
The sodium fusion extract is acidified with dilute $\ce{HNO3}$ (to decompose NaCN and $\ce{Na2S}$ which would interfere) and then $\ce{AgNO3}$ is added: white ppt (soluble in $\ce{NH4OH}$) = chlorine; pale yellow ppt (sparingly soluble in $\ce{NH4OH}$) = bromine; yellow ppt (insoluble in $\ce{NH4OH}$) = iodine.
Why must the sodium fusion extract be boiled with dilute $\ce{HNO3}$ before testing for halogens with $\ce{AgNO3}$?
If nitrogen and sulphur are present, NaCN and $\ce{Na2S}$ would react with $\ce{AgNO3}$ to give precipitates of AgCN and $\ce{Ag2S}$, interfering with the test. Boiling with dilute $\ce{HNO3}$ decomposes these as $\ce{HCN}$ and $\ce{H2S}$ gases, removing the interference.
What is quantitative analysis of an organic compound?
Quantitative analysis is the determination of the percentage composition (relative amounts) of each element present in an organic compound, used to deduce its empirical and molecular formula.
On what principle is the estimation of carbon and hydrogen (Liebig's method) based?
A known mass of the compound is burnt completely in excess oxygen over heated $\ce{CuO}$. Carbon is oxidised to $\ce{CO2}$ (absorbed in KOH) and hydrogen to $\ce{H2O}$ (absorbed in anhydrous $\ce{CaCl2}$). From the masses of $\ce{CO2}$ and $\ce{H2O}$, the percentages of C and H are calculated.
In the estimation of carbon and hydrogen, what absorbs $\ce{H2O}$ and what absorbs $\ce{CO2}$?
Water vapour is absorbed in a weighed U-tube containing anhydrous calcium chloride ($\ce{CaCl2}$). Carbon dioxide is absorbed in a weighed tube containing potassium hydroxide (KOH) solution. The increases in mass give the masses of $\ce{H2O}$ and $\ce{CO2}$ formed.
Give the formula for the percentage of carbon in the combustion (Liebig) method.
If $m$ g of compound gives $m_1$ g of $\ce{CO2}$: $$\%\,\text{C} = \frac{12}{44} \times \frac{m_1}{m} \times 100$$ (since $44$ g of $\ce{CO2}$ contains $12$ g of carbon).
Give the formula for the percentage of hydrogen in the combustion method.
If $m$ g of compound gives $m_2$ g of $\ce{H2O}$: $$\%\,\text{H} = \frac{2}{18} \times \frac{m_2}{m} \times 100$$ (since $18$ g of $\ce{H2O}$ contains $2$ g of hydrogen).
Name the two methods used for the estimation of nitrogen and the type of compounds each suits.
Dumas method (gives nitrogen as $\ce{N2}$ gas; applicable to all nitrogen-containing compounds) and Kjeldahl's method (converts nitrogen to $\ce{(NH4)2SO4}$ then $\ce{NH3}$; applicable to compounds where nitrogen converts to ammonium sulphate, but not to those with N in rings, $\ce{-NO2}$, or azo groups).
Describe the principle of the Dumas method for nitrogen estimation.
A known mass of compound is heated with $\ce{CuO}$ in an atmosphere of $\ce{CO2}$. Nitrogen is converted to $\ce{N2}$ gas (any oxides of nitrogen are reduced to $\ce{N2}$ by hot copper gauze). The $\ce{N2}$ is collected over a KOH solution and its volume measured: $$\ce{C_xH_yN_z + (2x + y/2)CuO -> xCO2 + y/2 H2O + z/2 N2 + (2x+y/2)Cu}$$
In the Dumas method, how is the percentage of nitrogen calculated from the volume of $\ce{N2}$?
The volume of $\ce{N2}$ is converted to STP, giving volume $V$ (mL). Since $22400$ mL of $\ce{N2}$ at STP weighs $28$ g: $$\%\,\text{N} = \frac{28}{22400} \times \frac{V}{m} \times 100$$ where $m$ is the mass (g) of the compound taken.
Describe the principle of Kjeldahl's method for nitrogen estimation.
The compound is heated with conc. $\ce{H2SO4}$ (with $\ce{K2SO4}$ and $\ce{CuSO4}$ catalyst); nitrogen is converted to ammonium sulphate. This is treated with excess NaOH to liberate $\ce{NH3}$, which is distilled into a known excess of standard acid. The acid not neutralised by $\ce{NH3}$ is back-titrated to find the $\ce{NH3}$, hence nitrogen.
Give the formula for percentage of nitrogen in Kjeldahl's method.
If $m$ g of compound liberates $\ce{NH3}$ neutralising acid equivalent to $V$ mL of $M$ molar $\ce{H2SO4}$ (a dibasic acid): $$\%\,\text{N} = \frac{1.4 \times M \times 2 \times V}{m}$$ (Each mole of $\ce{H2SO4}$ neutralises $2$ moles of $\ce{NH3}$; $1.4 = \frac{14}{1000}\times 100$.)
Why is Kjeldahl's method not applicable to certain nitrogen compounds?
It cannot be applied to compounds containing nitrogen in rings (e.g. pyridine), nitro ($\ce{-NO2}$) groups, or azo ($\ce{-N=N-}$) groups, because the nitrogen in these is not quantitatively converted to ammonium sulphate during digestion with $\ce{H2SO4}$.
Describe the principle of Carius method for the estimation of halogens.
A known mass of compound is heated with fuming $\ce{HNO3}$ in the presence of $\ce{AgNO3}$ in a sealed Carius tube. The halogen is converted to silver halide (AgX), which is filtered, washed, dried, and weighed. From its mass the percentage of halogen is found.
Give the formula for the percentage of chlorine in Carius method.
If $m$ g of compound gives $m_1$ g of AgCl (molar mass $143.5$, of which $35.5$ is Cl): $$\%\,\text{Cl} = \frac{35.5}{143.5} \times \frac{m_1}{m} \times 100$$ Analogous expressions use AgBr ($\frac{80}{188}$) for bromine and AgI ($\frac{127}{235}$) for iodine.
Describe the principle of estimation of sulphur by Carius method.
A known mass of compound is heated with fuming $\ce{HNO3}$ in a Carius tube; sulphur is oxidised to sulphuric acid, which is precipitated as barium sulphate ($\ce{BaSO4}$) by adding $\ce{BaCl2}$. The $\ce{BaSO4}$ is filtered, washed, dried, and weighed.
Give the formula for the percentage of sulphur in the Carius (barium sulphate) method.
If $m$ g of compound gives $m_1$ g of $\ce{BaSO4}$ (molar mass $233$, containing $32$ g of S): $$\%\,\text{S} = \frac{32}{233} \times \frac{m_1}{m} \times 100$$
How is phosphorus estimated quantitatively?
The compound is oxidised with fuming $\ce{HNO3}$ to convert phosphorus to phosphoric acid, which is precipitated either as ammonium phosphomolybdate $\ce{(NH4)3PO4.12MoO3}$ or as magnesium ammonium phosphate $\ce{MgNH4PO4}$. The latter is ignited to magnesium pyrophosphate $\ce{Mg2P2O7}$, which is weighed.
Give the formula for the percentage of phosphorus when estimated as $\ce{Mg2P2O7}$.
If $m$ g of compound gives $m_1$ g of magnesium pyrophosphate $\ce{Mg2P2O7}$ (molar mass $222$, containing $62$ g of P, i.e. $2\times 31$): $$\%\,\text{P} = \frac{62}{222} \times \frac{m_1}{m} \times 100$$
How is oxygen usually estimated in an organic compound?
Oxygen is usually estimated indirectly by difference: $$\%\,\text{O} = 100 - (\text{sum of percentages of all other elements present})$$ Direct estimation is also possible by passing the decomposition gas products over heated coke to form $\ce{CO}$, which is converted to $\ce{CO2}$ and weighed.
What this deck covers
The Organic Chemistry deck follows the JEE Main Organic Chemistry syllabus — 8 chapters and 39 topics — so questions land on material that is genuinely examinable rather than trivia around it. That works out to roughly 6.4 cards per chapter.
Answers are written to be recallable, not just readable — averaging about 263 characters, which is long enough to carry the reasoning and short enough to say out loud.
A deck like this earns its keep on the second and third pass. Read the syllabus first so you know the shape of the subject, then use the cards to find the specific facts that have not stuck.
Organic Chemistry flashcards FAQ
How many Organic Chemistry flashcards are in this JEE Main deck?
51 cards. This page previews 24 of them, sampled evenly across the deck so you can judge the difficulty before installing anything.
Are these JEE Main flashcards free?
Yes. The preview here is free to read with no signup, and the full 51-card deck is free inside the Examius app.
What do the Organic Chemistry cards cover?
They follow the JEE Main Organic Chemistry syllabus — 8 chapters and 39 topics — so the questions track what is actually examinable.
How should I use these flashcards?
Read the syllabus first so you know the shape of the subject, then drill the deck. Examius schedules each card with spaced repetition, so cards you keep missing come back sooner and ones you know drift further apart.