A to Z Chemistry (Finding unknowns)

Chemistry Level 4

Warning: This question is pretty long. Treat it as a puzzle rather than a problem. At the end problem is simple but you have to solve for all the unknowns to crack the answer. Please don't try to guess the answer. You will enjoy finding the unknowns. I have provided hints at the end.

Part-1

A metal ( A ) is reacted with ( B ) (a neutral oxide, liquid at room temperature), gives off ( C ) and ( D ), and ( D ) burns to give ( B ). ( C ) on heating at around 50 0 C 500^\circ C gives ( E ) and ( B ). ( E ) when treated with ( F ) in electric arc furnace gives ( G ) and ( H ). ( H ) is an extremely poisonous, tasteless, odorless, colorless gas that burns with a blue flame. ( G ) on treatment with ( B ) gives ( I ) and ( C ). ( I ) is a highly flammable gas and decolorizes bromine water. ( I ) on heating in red hot copper tube trimerises to form ( J ) as the major product.

Part-2

Another metal ( K ) reacts with ( F ) in an electric arc furnace at about 150 0 C 1500^\circ C to 200 0 C 2000^\circ C to form ( L ). ( L ) reacts with ( B ) to form ( M ) and ( N ). ( N ) is basic in nature and sparingly soluble in water. ( M ) reacts with one equivalent of greenish yellow gas ( O ) in the presence of UV light to give ( P ). ( O ) also reacts with ( D ) in the presence of UV light to give ( Q ), aqueous solution of which reacts with ( N ) to give ( S ) which is a lewis acid and fumes in moist air.

Part-3

( J ) is reacted with ( P ) with ( S ) as a catalyst to give ( T ) which when treated with conc. H 2 S O 4 \ce{H}_2 \ce{SO}_4 gives ( U ). ( U ) is again treated with ( P ) with ( S ) as a catalyst to give ( V ). ( V ) is then treated with dilute acid solution to give ( W ). ( W ) is first treated with hot alkaline K M n O 4 \ce{KMnO}_4 and then very strongly heated to give ( X ).

Final Part

( J ) is reacted with ( O ) in the presence of iron as a catalyst and then treating the product with N a O H \ce{NaOH} at very high temperature yields ( Y ).

Finally 2 equivalents of ( Y ) are reacted with one equivalent of ( X ) in the presence of acid catalyst to yield ( Z ).

Find the color of ( Z ) in dilute aqueous solution of ( C ).

Hints.

1) Metal ( A ) gives brick red flame. It's carbonate is insoluble in water.

2) Cation of metal ( K ) belongs to third group of basic radicals (the N H 4 O H \ce{NH}_4\ce{OH} group).

3) ( F ) and ( H ) are used in metallurgy.

4) ( X ) is an anhydride.

5) ( B ) is a neutral, polar solvent.

Blue Pink Red White Yellow Orange Green Colourless

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2 solutions

Ronak Agarwal
Apr 28, 2015

A = C a A=Ca

B = H 2 O B={H}_{2}O

C = C a ( O H ) 2 C=Ca{(OH)}_{2}

D = H 2 D={H}_{2}

E = C a O E=CaO

F = C o k e F=Coke

G = C a C 2 G=Ca{C}_{2}

H = C O H=CO

I = C 2 H 2 I={C}_{2}{H}_{2}

J = B e n z e n e J=Benzene

K = A l K=Al

L = A l 4 C 3 L={Al}_{4}{C}_{3}

M = C H 4 M=C{H}_{4}

N = A l ( O H ) 3 N=Al{(OH)}_{3}

O = C l 2 O={Cl}_{2}

P = C H 3 C l P={CH}_{3}Cl

Q = H C l Q=HCl

S = A l C l 3 S=Al{Cl}_{3}

T = T o l u e n e T=Toluene

U=Para-Methyl Benzene Sulphonic Acid \text{U=Para-Methyl Benzene Sulphonic Acid}

V=3-4 DiMethyl Benzene Sulphonic Acid \text{V=3-4 DiMethyl Benzene Sulphonic Acid}

W=Ortho-Xylene \text{W=Ortho-Xylene}

X=Pthalic Anhydride \text{X=Pthalic Anhydride}

Y=Phenol \text{Y=Phenol}

Z= Phenolphthalein \text{Z= Phenolphthalein}

Salute ! _/_

Krishna Ar - 6 years, 1 month ago

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Have you tried the problem. Krishna

Ronak Agarwal - 6 years, 1 month ago

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No, I couldn't find a few answers! ( Some of them are really advanced!)

Krishna Ar - 6 years, 1 month ago

had fun solving it

Oliver Robb - 5 years, 11 months ago

Great problem!

You missed out 'R' though, @Ronak Agarwal

great problem . just missed the last rxn (phenolphthalein) ........... plz post more chemistry.................

rajat kharbanda - 6 years, 1 month ago

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can you explain +fe / naoh reaction... reading it here for the first time...

rajat kharbanda - 6 years, 1 month ago

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plz ans...

rajat kharbanda - 6 years, 1 month ago

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@Rajat Kharbanda What's this Fe/NaOH , are you referring to reaction used in the preparation of phenol.

Ronak Agarwal - 6 years, 1 month ago

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@Ronak Agarwal yes. the O to Y REACTION. could u plz expain

rajat kharbanda - 6 years, 1 month ago

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@Rajat Kharbanda Firstly I convert Benzene to chlorobenzene by reacting it with chlorine in presence of Fe as a catalyst.

Then it is treated with base NaOH which replaces the Cl with OH via benzyne mechanism.

Ronak Agarwal - 6 years, 1 month ago

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@Ronak Agarwal thanks.. keep posting chemistry

rajat kharbanda - 6 years, 1 month ago

@Ronak Agarwal Actually you are slightly wrong here. NaOH Cant do Benzyne Mechanism . For Benzyne we take stronger base like Sodamide or phenyl lithium

Prakhar Bindal - 4 years, 9 months ago

Oh no how can I miss R , well is this problem two easy. Or are the hints too much.

Ronak Agarwal - 6 years, 1 month ago

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I saw a photo of someone in the newspaper,who resembled you and his name was same as yours.....................CONGRATS! AND ALL THE BEST!

Siddharth Singh - 6 years, 1 month ago

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Yes, it is the same @Ronak Agarwal . If you didn't already know, he is one of the most intelligent people here.

Krishna Ar - 6 years, 1 month ago

Salute to you too! _/_

Krishna Ar - 6 years, 1 month ago

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