The major products obtained from the reactions in List-II are the reactants for the named reactions mentioned in List-I. Match List-I with List-II and choose the correct option.
P →2; Q→4 ; R →1; S →3
P →1; Q→3 ; R →5; S →2
P →3; Q→2 ; R →1; S →4
P →3; Q→4 ; R →5; S →2
To solve the problem, we'll analyze the named reactions in List-I and identify the corresponding major products from List-II. Let's match each:
1. Aldol Condensation | A reaction involving the formation of β-hydroxy ketone (aldol) followed by dehydration resulting in α,β-unsaturated ketone. In List-II, this product is associated with compound 4. |
2. Finkelstein Reaction | A halogen exchange reaction used to convert alkyl chlorides/bromides to iodides using sodium iodide, corresponding to compound 5 in List-II. |
3. Sandmeyer Reaction | A substitution reaction that converts diazonium salts to aryl halides, often involving the replacement of a diazo group with a halide ion. The right match here is compound 2 from List-II. |
4. Williamson Ether Synthesis | A reaction for the synthesis of ethers from alkoxides and alkyl halides, matching with compound 3 in List-II. |
Based on these connections, we get:
Thus, the correct option is: P →3; Q→4 ; R →5; S →2
Correct option is (D) P → 3; Q → 4; R → 5; S → 2 (P) → (3)
Let $ S $ denote the locus of the point of intersection of the pair of lines $$ 4x - 3y = 12\alpha,\quad 4\alpha x + 3\alpha y = 12, $$ where $ \alpha $ varies over the set of non-zero real numbers. Let $ T $ be the tangent to $ S $ passing through the points $ (p, 0) $ and $ (0, q) $, $ q > 0 $, and parallel to the line $ 4x - \frac{3}{\sqrt{2}} y = 0 $.
Then the value of $ pq $ is
Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ is ________.
Organic Chemistry is a subset of chemistry dealing with compounds of carbon. Therefore, we can say that Organic chemistry is the chemistry of carbon compounds and is 200-225 years old. Carbon forms bond with itself to form long chains of hydrocarbons, e.g.CH4, methane and CH3-CH3 ethane. Carbon has the ability to form carbon-carbon bonds quite elaborately. Polymers like polyethylene is a linear chain where hundreds of CH2 are linked together.
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