Compound A formed in the following reaction reacts with B gives the product C. Find out A and B.\[ \text{CH}_3 - \text{C} \equiv \text{CH} + \text{Na} \xrightarrow{} \text{A}\xrightarrow{B} \text{CH}_3 - \text{C} \equiv \text{C} - \text{CH}_2 - \text{CH}_2 + \text{NaBr}(C)\]
The Correct Option is C
Approach Solution - 1
The problem presents a two-step reaction sequence and asks to identify the intermediate compound A and the reactant B. The starting material is propyne (\(CH_3-C \equiv CH\)), and the final products are a substituted alkyne (C) and sodium bromide (\(NaBr\)).
Concept Used:
The solution involves two key reactions in alkyne chemistry:
1. Formation of an Acetylide Anion: Terminal alkynes (alkynes with a hydrogen atom attached to a triply-bonded carbon) are weakly acidic. They can be deprotonated by a strong base, such as sodium metal (\(Na\)) or sodamide (\(NaNH_2\)), to form a sodium acetylide salt. In this reaction, the sodium acts as a reducing agent, producing the sodium salt and hydrogen gas.
\[ 2 R-C \equiv CH + 2Na \longrightarrow 2 R-C \equiv C^-Na^+ + H_2(g) \]
2. Nucleophilic Substitution by an Acetylide Anion: The acetylide anion (\(R-C \equiv C^-\)) is a strong nucleophile. It can react with a primary alkyl halide in a nucleophilic substitution reaction (specifically, an \(S_N2\) reaction) to form a new carbon-carbon bond, resulting in a longer, internal alkyne.
\[ R-C \equiv C^-Na^+ + R'-X \longrightarrow R-C \equiv C-R' + NaX \]
Step-by-Step Solution:
Step 1: Determine the structure of compound A.
The first reaction is the reaction of propyne (\(CH_3-C \equiv CH\)) with sodium metal (\(Na\)). Propyne is a terminal alkyne, and the hydrogen atom on the terminal carbon is acidic. Sodium metal will deprotonate the alkyne to form a sodium propynide salt.
\[ CH_3 - C \equiv CH + Na \longrightarrow CH_3 - C \equiv C^-Na^+ + \frac{1}{2}H_2(g) \]
Therefore, the intermediate compound A is sodium propynide, \(CH_3 - C \equiv C^-Na^+\).
Step 2: Determine the structure of compound B.
\(CH_3-C \equiv C-CH_2-CH_2-CH_3\) with a \(CH_3\) branch on the second carbon of the added chain. This is likely a typographical error in the problem's depiction of product C. Let's assume the product is the straight-chain alkyne \(CH_3-C \equiv C-CH_2-CH_2-CH_3\) (hex-2-yne), as would be formed from 1-bromopropane, which is a more common reactant in textbook examples and aligns with the options.
Assuming the product is hex-2-yne, the added group is \( -CH_2-CH_2-CH_3 \) (a propyl group). This would mean B is 1-bromopropane, \(CH_3-CH_2-CH_2-Br\). This matches the structure of B given in option (1).
Let's proceed by assuming the product shown has a typo and the intended product is \(CH_3-C \equiv C-CH_2CH_2CH_3\). Reaction: \[ CH_3 - C \equiv C^-Na^+ + CH_3CH_2CH_2-Br \longrightarrow CH_3-C \equiv C-CH_2CH_2CH_3 + NaBr \]
Final Computation & Result:
Step 3: Match the identified structures of A and B with the given options.
- Compound A is sodium propynide: \(A = CH_3 - C \equiv C^-Na^+\)
- Compound B is 1-bromopropane: \(B = CH_3 - CH_2 - CH_2 - Br\)
Let's evaluate the other options:
- Option (2): A is incorrect (propene), B is incorrect.
- Option (3): A is incorrect (propane), B is incorrect.
- Option (4): A is correct, but B is incorrect (propane).
The correct choice is (1) \(A = CH_3 - C \equiv C^-Na^+, B = CH_3 - CH_2 - CH_2 - Br\).
Approach Solution -2
The given reaction suggests that sodium acetylide reacts with an alkyl halide to yield the final product. Here, compound A is formed as sodium acetylide:
CH3 – C ≡ CH + Na → CH3 – C ≡ CNa
Then, it reacts with compound B (1-bromopropane):
CH3 – C ≡ CNa + CH3CH2CH2Br → CH3 – C ≡ C – CH2CH2CH3 + NaBr
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main Polarity of bonds Questions
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,