Class 12th

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4 months ago

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A
alok kumar singh

Contributor-Level 10

(a) is true, (b) is true.

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A
alok kumar singh

Contributor-Level 10

Acidity: b (carboxylic acid)>c (phenol)>d (enol)>a (alkyne).

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R
Raj Pandey

Contributor-Level 9

A? A = I
⇒ a²+b²+c²=1 and ab+bc+ca=0
Now, (a+b+c)²=1 ⇒ a+b+c=±1
So, a³+b³+c³-3abc = (a+b+c) (a²+b²+c²-ab-bc-ca) = (±1) (1-0)=±1
⇒ 3abc = 2±1 = 3,1
⇒ abc = 1, 1/3

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alok kumar singh

Contributor-Level 10

This is electrophilic substitution reaction which is determine by electronic effect of

 

 

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alok kumar singh

Contributor-Level 10

From I&II, rate∝ [B]². From I&III, rate∝ [A]¹.
From IV: 7.2e-2 = k (X) (0.2)². From II: 2.4e-2=k (0.1) (0.2)². X=0.3.
From V: 2.88e-1=k (0.3) (Y)². k=2.4e-2/ (0.1*0.04)=6.
2.88e-1 = 6 (0.3)Y². Y²=0.16. Y=0.4.

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alok kumar singh

Contributor-Level 10

Cast iron is used to make wrought iron and steel.

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alok kumar singh

Contributor-Level 10

E2 elimination. Most acidic proton is removed. Fluorine is more electronegative.

 

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alok kumar singh

Contributor-Level 10

Ligand field strength: NH? > NCS? > F? Stronger ligand, higher Δ, lower λ_max.
So λ (NH? ) < (NCS? ) < (F? ). A= (F? ), B= (NCS? ), C= (NH? ). A-ii, B-i, C-iii.

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alok kumar singh

Contributor-Level 10

Seliwanoff's test distinguishes aldoses from ketoses. Sucrose hydrolyzes to glucose (aldose) and fructose (ketose). Fructose gives a red color.

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alok kumar singh

Contributor-Level 10

gives iodoform test and slow Lucas test, so it's a methyl secondary alcohol. D gives fast Lucas test, so it's a tertiary alcohol. The Grignard products must be tertiary and secondary alcohols. So A and B must be an aldehyde and a ketone.

 

 

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